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Quants Geometry

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Quants Geometry

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Geometry

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What are CATKing Bible LOD 1 Books?

CATKing Bible LOD 1 Books are specially designed books which are useful
in getting students Boosted Up and Ready for All Management Entrance
Tests (CAT/ CET/ NMAT/ CMAT/ SNAP/ TISSNET/ MICAT/ IIFT). These
books cover all the topics and are distributed section wise: Quantitative
Ability, Data Interpretation, Logical Reasoning and Verbal Ability. They
are recommended for all students who wish to get their Basics clear in
any section for any Management Entrance Test.

How to make best use of CATKing Bible LOD 1 Books?

i. Attend the CATKing Concept Builder Classes to gain an idea of what all
are the basic pointers of the chapters.
ii. Go through that chapter in the CATKing Bible LOD 1 Books and read all
the Theory and Formulae provided in the Introduction of the chapter.
iii. Make a note of all the formulae and try to learn them by Heart, this
will be your ready reckoner to revise before exams.
iv. Go through all the Solved Examples and solve them simultaneously
while referring the solutions provided to understand the best way to
solve each type of questions.
v. Solve all the Practice Questions provided on your own and then refer
to the solutions at the end so as to verify if you have solved the questions
correctly or is there a better smarter approach for the same question.
vi. If you are able to solve majority questions correctly, then move to the
next step of preparation by taking the Topicwise Tests.
vii. Once you are done with good set of 4-5 Topics, give the Sectional and
Full length Mocks and see where you stand.
viii. While you analyze the mocks see to it that you highlight your weak
and strong areas. For all weak areas, you must get back to these CATKing
Bible LOD 1 Books, read up your formulae, check out the Solved Exams
and your concepts will be revised.
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Geometry
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Table of Contents

Introduction 1
Solved Examples 19
Practice Questions 45
Answer Key 60
Solutions 60

Let’s get started...

Introduction
Lines and Angles:
• Sum of all angles on a straight line is 180°.
• Vertically opposite angles are congruent (equal)
• If any point is equidistant from the end points of a segment, then it
must lie on a perpendicular bisector.
• When 2 parallel lines are intersected by a transversal, corresponding
angles are equal, alternate angles are equal and co-interior angles
supplementary. (All acute angles formed are equal to each other and
all obtuse angles are equal to each other)
Concept: The ratio of intercepts formed by a transversal intersecting 3
parallel lines is equal to the ratio of corresponding intercepts formed by
any other transversal.

a c e
= =
b d f

Triangles:
• Sum of interior angles of a triangle is 180° and sum of exterior angles is
360°.
• Exterior Angle= Sum of remote interior angles.

• Sum of two sides is always greater than the third side and the
difference between the two sides is always lesser than the third side.
• Side opposite to the biggest angle is longest and the side opposite to
the smallest angle is shortest.

Area of a triangle:

1
Area = x base x height
2

1
Area = x Product of sides x Sine of included ∠le
2

Area = :(s)(s − a)(s − b)(s − c)

a+b+c
where, s (semi − perimeter) =
2

Area = r x s [r is the radius of the incircle)

abc
Area = [R is radius of circumcircle]
4R

A Median of a triangle is a line segment joining a vertex to the midpoint

of the opposing side. The three medians intersect in a single point, called
the Centroid of the triangle. Centroid divides the median in the ratio of
2:1.
An Altitude of a triangle is a straight line through a vertex and
perpendicular to the opposite side or an extension of the opposite side.
The three altitudes intersect in a single point, called the Orthocenter of
the triangle.
A Perpendicular Bisector is a line that forms a right angle with one of the
triangle's sides and intersects that side at its midpoint. The three
perpendicular bisectors intersect in a single point, called the
Circumcenter of the triangle. It is the center of the circumcircle which
passes through all the vertices of the triangle.
An Angle Bisector is a line that divides the angle at one of the vertices in
two equal parts. The three angle bisectors intersect in a single point,
called the Incentre of the triangle. It is the center of the incircle which
touches all sides of a triangle.

Points to note:
Centroid and incentre will always lie inside the triangle.
• For an acute angled triangle, the Circumcenter and the Orthocenter
will lie inside the triangle.
• For an obtuse angled triangle, the Circumcenter and the Orthocenter
will lie outside the triangle.
• For a right-angled triangle, the Circumcenter will lie at the mid-point
of the hypotenuse and the orthocenter will lie at the vertex at which
the angle is 90°

Points to note:
The orthocenter, centroid, and circumcenter always lie on the same line
known as Euler line.
• The orthocenter is twice as far from the centroid as the circumcenter
is.
• If the triangle is isosceles, then the incentre lies on the same line
• If the triangle is equilateral, all four are the same point.

Theorems:

Mid-Point Theorem: The line joining the midpoint of any two sides is
parallel to the third side and is half the length of the third side.

Basic proportionality theorem:

If DE//BC. Then:
AD AE
=
DB EC

Apollonius’ Theorem:
AB2 + AC2= 2 (AD2 + BD2)

Interior Angle Bisector Theorem:

AE BA
=
ED BD

Special Triangles:
1) Right angled triangle:

ΔABC ∼ ΔADB ∼ ΔBDC

BD2 = AD x DC and AB x BC = BD x AC

2) Equilateral Triangle:

All angles are equal to 60°. All sides are also equal.
√"
a) Height = x side
#
√"
b) Area = x side2
$
%
c) Inradius = x Height
"
#
d) Circumradius = x Height
"

3) Isosceles Triangle:

Angles opposite to equal sides are equal.

c
Area = x √(4a# − c # )
4

4) 30°-60°-90° Triangle:

√3
Area = x (X)#
2

5) 45°-45°-90° Triangle

x#
Area =
2

6) 30°-30°-120° Triangle:

√3
Area = x (X)#
4

Similarity of triangles:
Two triangles are similar if their corresponding angles are congruent and
corresponding sides are in proportion.
Tests of similarity:
1. AA
2. SSS
3. SAS

For similar triangles, if the sides are in the ratio a:b

• Corresponding heights are in the ratio a:b
• Corresponding medians are in the ratio a:b
• Circumradii are in the ratio a:b
• Inradii are in the ratio a:b
• Perimeters are in the ratio a:b
• Areas are in the ratio a2:b2

Congruency of Triangles:
Two triangles are congruent if their corresponding sides and angles are
congruent.
Tests of congruency:
1. SSS
2. SAS
3. AAS
4. ASA
All ratios mentioned in similar triangles are now 1:1.

Polygons:
• Sum of interior angles= (n-2)x180°= (2n-4)x90°
• Number of diagonals= nC2-n= n(n-3)/2
• Number of triangles which can be formed by ‘n’ non-collinear
vertices=nC3

Regular Polygon:
• If all sides and all angles are equal, it is a regular polygon.
• All regular polygons can be inscribed in or circumscribed about a circle.
%
• Area = x Perimeter x inradius
#
(Note: inradius is the perpendicular from centre to any side)
('(#)*%+,°
• Each interior Angle =
'
".,°
• Each exterior angle =
'

Quadrilaterals:

• Sum of the interior angles = Sum of the exterior angles = 360°

• Area of a quadrilateral= (1/2) x d1 x d2 x Sin θ

Cyclic Quadrilateral:

• If all vertices of a quadrilateral lie on the circumference of a circle, it is

known as a cyclic quadrilateral.
• Opposite angles are supplementary.
/010203
• Area = :(s − a)(s − b)(s − c)(s − d) , where s =
#

Parallelogram

• Opposite sides are parallel and congruent.

• Opposite angles are congruent and consecutive angles are
supplementary.
• A parallelogram inscribed in a circle is always a Rectangle. A
parallelogram circumscribed about a circle is always a Rhombus.
• Diagonals of a parallelogram bisect each other.
• Perimeter = 2(Sum of adjacent sides)
• Area = Base x Height = AD x AE = BC x AE
• Each diagonal divides a parallelogram in two triangles of equal area.
• Sum of squares of diagonals = Sum of squares of four sides
AC2 + BD2= AB2 + BC2 +CD2 + DA2
• A Rectangle is formed by intersection of the four angle bisectors of a
parallelogram.

Rhombus:

• A parallelogram with all sides equal is a Rhombus. Its diagonals bisect

at 90°.
• Perimeter= 4a
• Area= (1/2) x d1 x d2
• Area = √4a# − 𝑑1# x √4a# − 𝑑2#

Rectangle
• A parallelogram with all angles equal (90°) is a Rectangle. Its diagonals
are congruent.
• Perimeter= 2(l + b)
• Area= l x b

Square
• A parallelogram with sides equal and all angles equal is a square. Its
diagonals are congruent and bisect at 90°.
• Perimeter= 4a
• Area= a2
• Diagonal= a √2
• Square has the least perimeter of all polygons. For any quadrilaterals
with a given perimeter, the square has the greatest area.

Kite:

• Two pairs of adjacent sides are congruent.

• The longer diagonal bisects the shorter diagonal at 90°
4563728 6: 3;/<6'/=>
• Area =
#

Trapezium/Trapezoid:

• A quadrilateral with exactly one pair of sides parallel is known as a

Trapezoid. The parallel sides are known as bases and the non-parallel
sides are known as lateral sides.
%
• Area = x (sum of parallel sides)x height
#
• Median, the line joining the midpoints of lateral sides, is half the sum
of parallel sides.
• Sum of the squares of the length of the diagonals = Sum of squares of
lateral sides + 2 Product of bases.
AC2 + BD2 = AD2 +BC2 + 2 x AB x CD
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Isosceles Trapezium:

• If a trapezium is inscribed in a circle, it has to be an isosceles

trapezium. If a circle can be inscribed in a trapezium, Sum of parallel
sides = Sum of lateral sides.
• The non-parallel sides (lateral sides) are equal in length. Angles made
by each parallel side with the lateral sides are equal.

Hexagon (Regular, with side ‘a’)

• Number of diagonals = 9 {3 big and 6 small}

• Length of big diagonals (3) = 2a
• Area of a Pentagon = 1.72a2
• Area of an Octagon = 2(√2 + 1)a2
• A regular hexagon can be considered as a combination of six
equilateral triangles. All regular polygons can be considered as a
combination of ‘n’ isosceles triangles.

Circles

• Diameter = 2r; Circumference = 2πr; Area =πr2

• Chords equidistant from the center of a circle are equal. A line from
the center, perpendicular to a chord, bisects the chord.
• Equal chords subtend equal angles at the center.
• The diameter is the longest chord of a circle.
• A chord /arc subtends equal angle at any point on the circumference
and double of that at the center.
• Chords / Arcs of equal lengths subtend equal angles.

Chord AB divides the circle into two parts: Minor Arc AXB and Major Arc
AYB
Measure of arc AXB = ∠AOB =θ
?
Length (arc AXB) = x 2πr
".,
?
Area (sector OAXB) = x πr2
".,
Area of Minor Segment = Shaded Area in above figure
=Area of Sector OAXB - Area of triangle OAB
@? A;'?
=r2 [ - ]
"., #

Properties of Tangents, Secants, and Chords

The radius and tangent are perpendicular to each other.

There can only be two tangents from an external point, which are equal
in length PA = PB

PA x PB = PC x PD
θ= ½ [m (Arc AC) – m (Arc BD)]

PA x PB = PC x PD
θ= ½ [m (Arc AC) + m (Arc BD)]

PA x PB = PC2
θ = ½ [m(Arc AC) - m(Arc BC)]

The angle made by the chord AB with the tangent at A (PQ) is equal to
the angle that it subtends on the opposite side of the circumference.
∠ BAQ = ∠ACB

Common Tangents
Two Circles No. of Common Distance between
Tangents Centers
One is completely inside
other 0 <r1-r2
Touch Internally 1 =r1-r2
Intersect 2 r1-r2<d<r1+r2
Touch Externally 3 =r1+r2
One is completely outside
other 4 >r1+r2

Length of the Direct Common Tangent (DCT)

AD=BC=√d2-(r1-r2)2

Length of the Transverse Common Tangent (TCT)

RT = SU =√d2-(r1+r2)2
The two centers (O and O’), point of intersection of DCTs (P)and point of
intersection of TCTs (Q) are collinear. Q divides OO’ in the ratio r1: r2
internally whereas P divides OO’ in the ratio r1: r2 externally.

Solved Examples
1) If ST // QR, then find the value of w and x

a) 52°, 58°
b) 48°,58°
c) 58°,69°
d) 65°,42°
e) None of these
Answer 1: a (52°, 58°)
Explanation:
y= 180° - ∠AQP= 180°-128°= 52° (sum of angles on a straight line=180°)
m= 180°-70°-y= 180°-70°- 52°= 58° (sum of angles of a triangle=180°)
∴ w=52° (corresponding angles)
∴x= 58° (corresponding angles)

2) The heights of two triangles are in the ratio 3:4. Whereas, the ratios
of their areas is 5:8. What is the ratio of their bases?
a) 5:6
b) 2:3
c) 4:5
d) 6:5
e) 5:4
Answer 2: a (5:6)
Explanation:
Let height of first triangle be 3x.
Let height of second triangle be 4x.
Let base of first triangle by a
Let base of second triangle be b.
a:b=?
1 1 5
∴ Q x 3X x aR ÷ Q x 4X x bR =
2 2 8
On solving, we get:
a:b= 5:6

3) ∠PSR= 2 ∠PQR, PQ= PR, ∠SPR= 30°. Find the value of ∠QPS?

a) 60°
b) 50°
c) 45°
d) 30°
e) 75°
Answer 3: b (50°)
Explanation:
∠PRQ= ∠PQR=x (Opposite angles of equal sides)
∠PRQ + ∠PSR +30= 180 (Sum of all angles of a triangle is 180° )
∴2x + x + 30= 180
∴3x=150
∴x=50
∠PSR= ∠PQR + ∠QPS
2x= x + ∠QPS
x =∠QPS
∠QPS=x=50°

4) In triangle ABC, ∠A=60° and the angle bisectors of ∠B and ∠C meet at

X, then ∠BXC (in degrees) is= ?

a) 60°
b) 90°
c) 120°
d) 45°
e) 30°

Answer 4: c (120°)
Explanation:

Let ∠ABC=2x
Let ∠AYB= 2y
∴∠ABX=XBY=x
∴∠AYX=∠XYB=y

∴ 2x + 2y + 60= 180 (Sum of all angles of a triangle=180)

2x + 2y= 120
x+y= 60

∴∠XBY + ∠XYB + ∠ BXY = 180

∴x+y + ∠BXY=180
∴60+ ∠BXY= 180
∴ ∠BXY= 120

5) In the figure, AB is parallel to CD and RD// SL// TM// AN and

BR:RS:ST:TA=3:5:2:7. If it is known that CN= 1.3333 BR. Find the ratio of
BF:FG:GH:HI:IC?

a) 3:7:2:5:4
b) 3:5:2:7:4
c) 4:7:2:5:3
d) 4:7:2:5:3
e) 4:5:2:7:3

Answer 5: b (3:5:2:7:4)
Explanation:
Let AT=7x, TS=2x, SR=5x. RB=3x
△BRF ~ △ CNI (By AAA Test)
BR BF
∴ = (AAA test of similarity)
CN CI

3X BF
=
1.333x3X CI

∴On solving, we get:

BF:CI=3:4

Let BF= 3y and CI= 4y

In △BSG,
RF// SG
BR BF
∴ = (Using basic proportionality theoram)
RS FG
3 3y
=
5 FG
∴FG= 5y

Similarly, GH=2y, HI= 7y.

∴ BF:FG:GH:HI: IC= 3y:5y:2y:7y:4y = 3:5:2:7:4

6) In the figure given below is the circle with the centre O. The radius of
the circle is 6.5 cm with PR being the diameter. In triangle PQR, PQ=5
cm. What is the area of the triangle PQR?

a) 30 cm2
b) 60 cm2
c) 90 cm2
d) 120 cm2
e) 150 cm2
Answer 6: a (30 cm2)
Explanation:
In △ PQR, ∠PQR= 90 (The angle subtended at arc is equal to 90 in a semi-
circle)
PR= diameter= 2 x radius= 2x 6.5 cm = 13 cm.
PR2= PQ2 + QR2 (Using Pythagoras Theorem)
132= 52 + QR2
169= 25 + QR2
144= QR2
∴12= QR
Area of △PQR= 1/2 x base x height= 1/2 x 5 x 12 = 30 cm2

7) Find the value of “x”?

a) 6 cm
b) 12 cm
c) 18 cm
d) 4 cm
e) 5 cm
Answer 7: d (4 cm)
Explanation:
PT2= PA x PB
62= x (x+5)
36= x (x+5)
36=x2 + 5x
On solving, we get:
x=4 or x=-9
Since x cannot be negative.
∴ x= 4 cm.

8) RPQ is a quarter circle and a circle is inscribed in it and if PQ=1 cm,

radius of smaller circle is

a) √2 – 1
b) (√2+1)/2
c) √2-1/2
d) 1-2√2
e) None of these

Answer 8: a (√2-1)
Explanation:

On constructing, perpendiculars from O to PR and PQ, we get a square

AODP
Let PE= R= 1 cm
∴ PO= R-r= 1 cm- r
Using Pythagoras theorem, we get:
(R-r)2= r2 + r2
(1-r)2= 2r2
1 + r2 – 2r= 2r2
∴0= r2+ 2r-1
∴r2 + 2r-1= 0
On solving, we get r= √2-1 or r= -√2- 1
Since r cannot be negative.
We get r=√2-1

9) Find the area of the circle excluding the triangle COB.

a) 91.25
b) 85.41
c) 43.11
d) 54.22
e) 67.74
Answer 9: e (67.74)
Explanation:
22 22x25
r#
Area of circle = π = Q R x 5x5 = = 78.57
7 7
1 1 √3
Area of triangle = Q R x r # xsin60 = x 5x 5 x = 10.83
2 2 2
∴ Area of circle (excluding triangle) = 78.57- 10.83= 67.74

10) In the given figure, AB is the diameter of the circle. ∠PAB= 25°. Find
the value of x.

a) 25°
b) 50°
c) 65°
d) 70°
e) 45°
Answer 10: c (65°)
Explanation:
∠APB=90 (Angle subtended at the arc of semi-circle is equal to 90)
∠ABP= 180 - ∠PAB- ∠APB
∠ABP= 180- 25-90= 180-115=65
∴ x = ∠TPA= 65° (the angle that a chord makes with the tangent is
subtended by the chord on the circumference in the alternate segment)

11) In quadrilateral PQRS, PQ=24 cm, QR=32 cm, RS=PS=25 cm. ∠PQR=
90. Find the area of the quadrilateral (in m2)?

a) 768
b) 534
c) 696.5
d) 684
e) 700

Answer 11: d (684)

Explanation:
Construction: Join P and R and construct ST⊥PR

PR# = 24# + 32# (Using Pythagoras theoram)

PR# = 576 + 1024 = 1600 m
∴ PR = √1600 = 40 m
PT= TR= PR/2 (perpendicular line between two equal sides bisects the
base)
40
∴ PT = TR = = 20 m
2
PT # + ST # = PS # (By pythagoras theoram)
∴ 20# + TS # = 25#
∴ TS # = 25# − 20# = 625 − 400 = 225
∴ TS = √225 = 15
Area of a PQRS= Area of △PQR + Area of △PSR
1 1
∴ Area of PQRS = x PQxQR + xPRxST
2 2
1 1
∴ Area of PQRS = x24x32 + x40x15
2 2
∴ Area of PQRS = (12x32) + (20x15)
∴ AREA OF PQRS = 384 + 300 = 684

12) ABCD is a Quadrilateral. E is the mid-point of AB. AB//DC and

DE//BC. Find the area of quadrilateral (in cm2) ABCD if area of triangle
ADE is 50cm2.

a) 100
b) 150
c) 200
d) 250
e) 300
Answer 12: b (150)
Explanation:
Let AE=EB=b
1
Area of △ ADE = x b x h = 50
2
∴ b x h = 100 … … … (1)
Area of parallelogram EBCD = base x height = EBxh = bxh
∴ Area of parallelogram EBCD = bxh = 100 (from 1)
∴Area of quadrilateral ABCD= Area of △ADE + Area of EBCD
∴Area of quadrilateral ABCD= 50 +100= 150 cm2

13) ABCD is a rhombus, such that AB=5 cm, AC=8 cm. Find the area of
ABCD?

a) 36 cm2
b) 18 cm2
c) 12 cm2
d) 24 cm2
e) 48 cm2
Answer 13: d (24 cm2)
Explanation:
OC= AC/2 = 8/2 = 4
∠DOC=90
OD# + OC # = CD#
OD# + 4# = 5#
OD# = 9
OD = √9
OD = 3cm
BD= 2 x OD= 2 x 3 = 6 cm.
1
Area of ABCD = x 6 x 8 = 24 cm#
2

14) The figure below shows a set of concentric squares. If the diagonal
of the innermost square is 2 units, and if the distance between the
corresponding corners of any two successive squares is 1 unit, find the
difference between the areas of the ninth and the tenth squares,
counting from the innermost square.

a) 34
b) 36
c) 38
d) 40
e) 42
Answer 14: c (38)
Explanation:
The distance between two successive corners is 1 unit.
∴ We can conclude that the diagonals of successive squares increase by 2
units in an AP (1 unit of one end and 1 unit of the other end)
So,
d1= 2 units
d2= 2 + 2 = 4 units
d3= 4 +2 = 6units
d4= 6 +2 = 8 units
d5= 8 +2= 10 units
d6=10+2=12 units

d7=12+2=14 units
d8=14+2=16 units
d9=16+2=18 units
d10=18+2=20 units
Note: “d” stands for diagonals and the numbers in subscript refer to the
diagonals of respective squares.
Square 9 Square 10
d9 =18 units d10=20 units
(diagonal)2= 2x (side)2 (diagonal)2= 2x (side)2
(18)2= 2 x (side)2 (20)2= 2x (side)2
324=2 x (side)2 400= 2 x (side)2
162= (side)2 200= (side)2
Area= (side)2=162……(1) Area= (side)2=200……(2)
∴Difference between the areas of squares 9 & 10 = (1)-(2)
=200-162=38 units2

15) In quadrilateral PQRS, PQ=QR and PS=RS. PR and QS are the

diagonals of the quadrilateral. QS= 10 cm and PR= 5cm. Find the area of
the quadrilateral.

a) 10 cm2
b) 15 cm2
c) 20 cm2
d) 25 cm2
e) 30 cm2
Answer 15: d (25 cm2)
Explanation:
PQRS has a kite like structure, so its diagonals intersect each other
perpendicularly.
1 1
∴ Area = (product of diagonals) = x 10x5 = 25 cm#
2 2

16) If the length of each side of a hexagon is 8 cm, then the area of the
hexagon is:
a) 96√3
b) 54√3
c) 48√3
d) 192√3
e) 64√3
Answer 16: a (96√3)
Explanation:
√3
Area of a hexagon = 6x x (side)#
4
√3
Area of hexagon = 6x x (8)# = 96√3
4

17) PQR is a right angled triangle with PQ=8 cm and QR= 6 cm. A circle
with center O and radius “x” has been inscribed in the triangle PQR.
What is the value of x?

a) 2
b) 3
c) 4
d) 5
e) 6

Answer 17 : a (2)
Explanation:
PR2= PQ2 + QR2 (Using Pythagoras theorem)
PR2= 62 + 82 = 36 + 64 = 100
PR= √100 =10
Semi-perimeter= (10+6+8)/2=12
Area= (1/2) x base x height= (1/2) x 6x8=24
∴in-radius=x= Area/semi-perimeter= 24/12=2

18) In triangle PQR, point S lies on PQ and point T lies on PR such that ST
is parallel to QR. ST:QR=3:5. STRQ is a trapezium. Find the ratio of area
of triangle PST to the area of the trapezium.
a) 9:16
b) 9:25
c) 3:5
d) 3:15
e) 4:25

Answer 18: a (9:16)

Explanation:

Given: ST// QR
In △PST and △PQR,
∠PST=∠PQR (corresponding angles)
∠PTS=∠PRQ (corresponding angles)
∠SPT=∠QPR (common angles)
∴△PST ∼ △PQR (By AAA test of similarity)
Area of △ PST (ST)# (3)# 9
∴ = = =
Area of △ PQR (QR)# (5)# 25
Let area of △PST=9x
Let area of △ PQR=25x
∴Area of trapezium STRQ= Area of △PQR – Area of △PST
=25x-9x=16x
∴Area of △PST: Area of trapezium STRQ= 9x:16x= 9:16

19) In the given figure, PQRS is a cyclic quadrilateral. ∠POR= 136°. Find
∠PSR.

a) 112°
b) 56°
c) 44°
d) 124°
e) 68°
Answer 19: a (112°)
Explanation:
∠PQR= (1/2) ∠POR
∠PQR= (1/2) x 136°
∴∠PQR= 68°
∠PQR + ∠ PSR= 180° (sum of opposite angles of a cyclic
quadrilateral=180°)
68°+ ∠PSR= 180°
∴∠PSR= 112°

20) Square STUV is inscribed in the triangle PQR. Area of square STUV is
64 cm2. ∠SQT= 45° and ∠QPR=90°. What is the area of triangle PQR?

a) 144 cm2
b) 121 cm2
c) 100 cm2
d) 169 cm2
e) 200 cm2
Answer 20: a (144 cm2)
Explanation:

∠PQR=45°
Area of square= 64 cm2
(side)2= (8 cm)2
∴side= 8 cm
∠QTS + ∠STU=180° (sum of angles on a straight line=180
∠QTS+90=180
∠QTS=180-90=90

Similarly,
∠RUV= 90
∠PQR + ∠QRP + ∠RPQ= 180 (sum of all angles of a triangle is 180)
90 +45 + ∠QRP=180
∴QRP=45
∴ △ QPR is an isosceles triangle with QP=PR (Opposite sides of ∠QRP and
∠PQR)
In △ QTS,
Tan 45= ST/QT
∴1= 8/QT
∴QT=8 cm
Similarly, UR=8 cm
∴QR= QT + TU + UR= 8cm + 8cm + 8cm = 24cm
Let PR=QP=x
(QR)2= (PR)2 + (QP)2 (By Pythagoras theorem)
(24)2= x2 + x2
576= 2x2
288=x2
1 1
∴ Area of △ PQR = x PRxPQ = x (X)#
2 2
1
∴ Area of △ PQR = x288 = 144 cm#
2

Practice Questions
1) △PQR and △PRS are right angled triangles with ∠PQR and ∠PRS
being right angles. PQ=x cm, QR=y cm and RS= z cm. x.y=z and x, y and z
have minimum integral values. What is the area of PQRS?
(Note: all sides have integer values)

a) 25 cm2
b) 36 cm2
c) 49 cm2
d) 64 cm2
e) 81 cm2

2) In △ABC, ∠ABC =120°. BD is the angle bisector of ∠ABC. BD=6 cm and

BC=9 cm. What is the length of AB?

a) 6 cm
b) 9 cm
c) 12 cm
d) 15 cm
e) 18 cm

3) In the figure given below AB= 2cm, DC=3cm and BC=5cm.

AB//EF//DC. Find the value of EF?

a) 1.0 cm
b) 1.2 cm
c) 1.4 cm
d) 1.6 cm
e) 1.8 cm

4) In △PQR, side PQ is a perfect cube. Side QR is 397 cm and side PR is a

power of 3. Side PR is thrice that of side PQ. What is the perimeter of
the triangle?
a) 3657 cm
b) 3400 cm
c) 3225 cm
d) 3122 cm
e) 3133 cm

5) If in the given figure, AB//CD//EF, then which of the given options is

true?

𝟏 𝟏 𝟏
a) − =
𝐚 𝐜 𝐛
𝟏 𝟏 𝟏
b) + =
𝐚 𝐛 𝐜
𝟏 𝟏 𝟏
c) − =
𝐛 𝐚 𝐜
𝟏 𝟏 𝟏
d) − =
𝐜 𝐛 𝐚
e) None of these

6) In the given figure, a semi-circle has been inscribed in triangle ABC

with BO= 2cm and OC= 4cm, where O is the centre of the semi-circle.
What is the diameter of the semi-circle?

a) √5/4
b) 8/√5
c) 18/√5
d) √5/6
e) 3/√5

7) In the diagram given below ED is a tangent to the circle and line AE

intersects the circle at point C. B is a point on AC such that DB is angle
bisector of ∠ADC. If ∠ADB=30° and ∠EDC: ∠ECD=2:5, then what is the
ratio of ∠ECD:∠CDE:∠DEC?

a) 4:4:10
b) 10:4:4
c) 2:2:8
d) 8:2:2
e) 8:4:4

8)In the given figure, O is the centre of the circle and ∠POR=140° Find
the value of ∠PQR.

a) 110°
b) 40°
c) 80°
d) 55°
e) 160°

9) In the given figure, ∠PRO is equal to 30° and “O” is the centre of the
circle. Find the value of ∠PQR.

a) 120°
b) 60°
c) 30°
d) 45°
e) 90°

10) A circle is inscribed in an equilateral triangle; the radius of the circle

is 2 cm. Find the area of triangle.
a) 12√3
b) 15√3
c) 12√2
d) 18√3
e) 12√2

11)ABCD is a square and △BCE is an equilateral triangle. What is the

value of ∠DEA?

a) 150°
b) 100°
c) 130°
d) 120°
e) 125°

12) Rectangle ABCD contains 4 congruent rectangles. If the smaller

dimension of one of the small rectangles is 4 units, what is the area of
rectangles ABCD in square units?

a) 192
b) 121
c) 100
d) 225
e) 256

13) In the given figure AD//IG//BC and AB//DC. EB intersects IG at H

and EC intersects IG at F. The ratio of area of △AHE to area of △BCH is
9:25. What is the ratio of area of quadrilateral EFGD to area of △FGC?
(Note: Points A, H and C are collinear)

a) 38:25
b) 39:25
c) 49:25
d) 25:49
e) 81:64

14) In rhombus PQRS, diagonal QS is equal to each side of the rhombus.

What is the ratio of the diagonals of the rhombus?
a) √3:1
b) 4: √3
c) √2:1
d) 4: √2
e) √3:2

15) In the diagram, PQRS is a rectangle with PT=TU=UQ. What is the

ratio of area of △RTU to area of rectangle PQRS?

a) 1:9
b) 1:8
c) 1:6
d) 1:4
e) 1:3

16) The number of diagonals in a polygon with 20 sides is?

a) 160
b) 170
c) 180
d) 190
e) 200

17) ABCD is a square of side 4 cm. E, F, G, and H are mid-points of AB,

BC, CD, and AD respectively leading to formation of square EFGH. I,J,K
and L are mid-points of HE,EF,FG and GH respectively leading to
formation of square IJKL. This process is continued till infinity. What is
the sum of areas of all the squares?

a) 8 cm2
b) 16 cm2
c) 32 cm2
d) 64 cm2
e) 128 cm2

18) In the given figure, PQ is extended up to R. ∠PRS=30° and ∠QSR

=50°. RS is a tangent to the circle. Find the value of ∠POQ?

a) 90°
b) 120°
c) 140°
d) 100°
e) 45°

19) The sum of all interior angles of a polygon is 1080. What is the
number of sides it has?
a) 5
b) 6
c) 7
d) 8
e) 9

20) In the given figure, ∠ABC is equal to 90° and diameter of the larger
circle is 4 cm. Find the diameter of the smaller circle?

a) 6- 3√2
b) 2-√2
c) 6-4√2
d) 8-2√2
e) 12-8√2

Answer Key
1(b) 2(e) 3(b) 4(e) 5(c)
6(b) 7(b) 8(a) 9(a) 10(a)
11(a) 12(a) 13(b) 14(a) 15(c)
16(b) 17(c) 18(d) 19(d) 20(e)

Solutions
Answer 1: b (36 cm2)
Explanation:
The values of the sides are integers that are whole numbers.
Let PQ, QR and PR be Pythagoras triplets with values PQ, QR and PR being
3 cm, 4 cm and 5 cm respectively. (Since the condition requires minimum
values)
z= x.y= 3x4= 12 cm.
PR, RS and PS forms another Pythagoras triplet of 5,12 and 13
respectively.
Hence, x= 3 cm
y= 4 cm
z= 12 cm
1 1
Area of △ PQR = x PQx QR = x 3x4 = 6 cm#
2 2
1 1
Area of △ PRS = x PR x RS = x 5x12 = 30 cm#
2 2
∴Area of PQRS= 30 + 6 = 36 cm 2

Answer 2: e (18 cm)

Explanation:
Area of △ABC= Area of △ ABD + Area of △ DBC
1 1 1
x ABxBC x sin120 = x ABxBDx sin60 + xBDxBCxsin60
2 2 2
1 1 1
x AB x 9 x sin60 = x ABx 6x sin60 + x 6 x 9 x sin 60
2 2 2
9AB sin60
sin60 = (6 AB + 54)
2 2
9 AB = 6AB + 54
3AB = 54
∴ AB = 18

Answer 3: b (1.2 cm)

Explanation:
Let BE= x
∴CE= BC-BE=5-x
In △ CEF and △CBA,
∠ECF= ∠BCA (common angles)
∠CEF= ∠CBA (corresponding angles)
∠CFE=∠CAB (Corresponding angles)
∴△CEF ∼△CBA (By AAA test)
CE EF
= (Ratios of sides of similar triangles)
BC AB
5 − x EF
=
5 2
10 − 2x
= EF … … … … … … (1)
5

In △BEF and △ BCD,

∠EBF= ∠CBD (Common Angles)
∠BEF= ∠BCD (Corresponding angles)
∠BFE= ∠BDC (Corresponding angles)
∴△BEF∼△BCD (By AAA test)
BE EF
= (Ratio of sides of similar triangles)
BC DC
x EF
=
5 3
3x
= EF … … … … … … . (2)
5
From (1) and (2), we get:
10 − 2x 3x
=
5 5
∴ 10 − 2x = 3x
∴ 10 = 5x
∴2=x
3x 3x2 6
∴ EF = = = = 1.2 cm
5 5 5

Answer 4: e (3133 cm)

Explanation:
Let PQ= x3
QR=397 cm…………..(1)
PR= 3 raised to a power= 3n (assuming power to be n)
∴PR= 3n……………………(2)
PR=3 PQ
3n= 3x3
3(n-1)=x3
∴PQ= x3= 3(n-1)…………………….(3)
PQ+ QR + PR= Perimeter
From (1), (2) and (3), we get:
3(n-1) + 397 cm + 3n= P
3(n-1) + 3n = P-397
3(n-1) x (1+3)= P-397
3(n-1)x 4= P-397
From the above equation, we can say that (P-397) is a multiple of 4 and 3.
So now, we substitute the values of options in the given equation: (P-397)
and determine the value by dividing by 12. Whichever value satisfies the
condition is the perimeter.
Options P (P-397)/12 Final answer
a 3657 (3657-397)/12 271.67
b 3400 (3400-397)/12 250.25
c 3225 (3225-397)/12 235.67
d 3122 (3122-397)/12 227.08
e 3133 (3133-397)/12 228
Since option (e) satisfies the condition. The perimeter is 3133 cm.

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Geometry
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𝟏 𝟏 𝟏
Answer 5: c h − i =
𝐛 𝐚 𝐜
Explanation:
Let BF=1
Let BD=x
∴DF=1-x
In △ DCF and △FAB,
∠FCD=∠FAB (Corresponding angles)
∠FDC= ∠FBA (Corresponding angles)
∠CFD= ∠AFB (Common Angles)
∴△DCF∼△BAF (By AAA is test)
DC DF
= (Ratio of sides of similar triangles)
AB BF
b 1−x
= … … … … … … … . (1)
a 1
In △ BDC and △BFE,
∠BDC=∠BFE (Corresponding angles)
∠BCD=∠BEF (Corresponding angles)
∠CBD=∠FBE (Common Angles)
∴△BDC ∼△BFE (AAA test of similarity)
CD BD
∴ = (ratio of sides of similar triangles)
FE BF
b x
= … … … … . . . (2)
c 1
Substituting (2) in (1), we get:
b b
=1−
a c
b b
+ =1
a c

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Geometry
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1 1
Q + R (b) = 1
a c
1 1 1
+ =
a c b
1 1 1
∴ = −
c b a
1 1 1
∴ − =
b a c

Answer 6: b(8/√5)
Explanation:
Construction:
Construct OD and OE (where D and E are the points of intersection of the
tangent and the circle)

OD⊥AB and OE⊥AC (The radius is ⊥ to the tangent)

∠ODA=∠AEO=90°
∠ADO+∠DAE + ∠AEO + ∠DOE= 360° (Sum of all angles of a
quadrilateral=360°)
90°+90°+90°+∠DOE=360°
270°+ ∠DOE=360°
∴∠DOE=360°-270°=90°
Let radius=r
∴ DO=OE=r
Let ∠DOB=z°
∠DOB+∠DOE+∠EOC=180° (Sum of angles on a straight line is 180°)
z+90°+∠EOC=180°
∠EOC= 180°-90°-z°
∴∠EOC=(90-z)°

DO r
cos z = =
BO 2
r
cos(90 − z) =
4
r
∴ Sin z =
4
r # r #
sin# z + cos # z = h i +h i
4 2
r# r#
1= +
16 4
r # + 4r #
1=
16
5r #
1=
16
16
= r#
5
4
∴ =r
√5

4 8
∴ Diameter = 2x radius = 2x r = 2x =
√5 √5

Answer 7: b (10:4:4)
Explanation:
∠ADB=∠BDC=30° (As DB is angle bisector of ∠ADC)
Let ∠EDC:∠ECD=2x:5x

∴∠DAB=∠CDE=2x (Alternate segment theorem)

∴∠BCD=180°-5x (Sum of angles on a straight line is 180°)
∴∠DAC+∠ACD+∠ADC=180° (Sum of angles of a triangle)
∴2x + 180°-5x + 60°=180°
On solving, we get x=20°
∠DCE +∠CDE +∠CED =180°
5x+2x + ∠CED=180°
7x+ ∠CED=180°

7x20° + ∠CED=180°
140°+∠CED=180°
∴∠CED=40°
∴∠ECD:∠CDE:∠DEC=100°:40°:40°=10:4:4

Answer 8: a (110°)
Explanation:
Construction: Construct PS and RS

∠PSR=(1/2)∠POR (Angle subtended by an arc on the circumference is half

of what it subtends at the centre)
∴∠PSR=(1/2) x 140°= 70°
∠PSR +∠PQR= 180° (Opposite angles of cyclic quadrilateral are
supplementary)
∴70°+ ∠PQR=180°
∴∠PQR=180°-70°=110°

Answer 9: a (120°)
Explanation:
In △POR,
∠OPR=∠ORP=30° (Opposite angles of equal sides)
∠OPR + ∠ORP + ∠POR=180° (sum of all angles of a triangle is 180°)
30°+30° +∠POR = 180°
∠POR=180°-60°=120°
Construction:
Construct PS and RS

∠PSR=∠POR/2
∴∠PSR=120°/2=60°(Angle subtended by an arc on the circumference is
half of what it subtends at the centre)
∠PSR + ∠PQR=180° (Opposite angles of cyclic quadrilateral are
supplementary)
∴60° +∠PQR=180°
∴∠PQR=180°-60°=120°

Answer 10: a(12√3)

Explanation:
Inradius= Area ÷ s
√3 # 3a
2= a ÷
4 2
a
2=
2√3
∴ 4√3 = a
√3 # √3 # √3
Area = a = x k4√3l = x 48 = 12√3cm#
4 4 4

Answer 11: a (150°)

Explanation:
∠EBC=∠BEC=∠ECB=60° (angles of equilateral △)
∠ABE=∠ABC-∠EBC=90°-60°=30°……………(1)
∠ECD=∠BCD-∠ECB=90°-60°=30°…………….(2)
EB=BC=EC (Sides of equilateral triangle)……(3)
AB=BC=CD=AD (sides of square)……………(4)
So from (3) and (4), we can conclude that:
EB=BC=EC=AB=CD=AD
Let ∠BAE= y
In △BEA and △CED,
AB=CD (sides of square)
∠ABE=∠ECD=30 [from (1) and (2)]
BE=EC (Sides of equilateral △BEC)
∴ △BEA ≅ △CED (By SAS test of congruency)
∴ ∠BAE= ∠EDC (Angles of congruent triangles)

∴∠BEA=∠BAE=y (Opposite angles of equal sides)

∴∠DEC=∠EDC=y (Opposite angles of equal sides)

∠BAC +∠BEA +∠ABE=180 (Sum of all angles of a △ is 180)

y+y+30=180
2y=150
y=75.
∴∠AED + ∠AEB + ∠BEC + ∠CED= 180
∴∠AED+ y+60+y=360 (sum of all angles around a center is 360)
∴∠AED+75+60+75=360
∴∠AED +210=360
∴∠AED=360-210=150

Answer 12: a (192)

Explanation:
Let AE= 4 units.
∴ AB= AE+EF+FB= 4 units + 4units + 4units=12 units.
∴AB=DC= 12 units.
GD=4 units (Since smaller side of every rectangle is 4 units)
∴ Area of rectangle GDCJ= GD x DC= 4 x 12= 48 unit2
Area of 4 rectangles=48 units2 x 4= 192 units2

Answer 13: b (39:25)

Explanation:
Given:
A(△AHE):A(△BHC)=9:25
AD//IG//BC
AB//DC
In △AHE and △CHB
∠HAE=∠HCB (Alternate angles)
∠HEA= ∠HBC (Alternate angles)
∠AHE=∠BHC (Vertically opposite angles)
∴ △HEA ∼ △ HBC (By AAA test of similarity)
AH HE AE A(△ AHE) 9 3
∴ = = =o = o = … … . . (1)
CH HB BC A(△ BHC) 25 5
HE EF 3
= = (By Basic proportionality theoram)
HB FC 5
EF 3
=
FC 5
FC 5
∴ =
EF 3

Let FC=5x and EF=3x

∴CE=8x
In △CGF and △CDE,
∠GCF=∠DCE (Common angles)
∠CGF=∠CDE (Corresponding angles)
∠CFG=∠CED (Corresponding angles)
∴△CGF∼△CDE (By AAA test)
A(△ CGF) FC # 5x # 25
∴ =Q R =Q R =
A(△ CDE) CE 8x 64
Let A(△CGF) =25y
Let A(△CDE) =64y
∴A (EFGD)= A(△CDE)-A(△CGF) =64y-25y=39y
A(EFGD) 39y 39
∴ = =
A(△ CGF) 25y 25

Answer 14: a(√3:1)

Explanation:
Let QS=PQ=QR=RS=PS= a
Let PR= b
Let the diagonals intersect at point O.
PO=OR=b/2 (diagonals of a rhombus bisect each other)
SO=OQ=a/2 (diagonals of a rhombus bisect each other)

∠POS=∠SOR=∠ROQ=∠QOP= 90° (Diagonals of a rhombus intersect at

90°)
∴ △ PSO is a right-angled triangle.
∴PS2= PO2 + OS2
#
a # b
∴ a# = h i + Q R
2 2
# #
#
a +b
∴a =
4
∴ 4a = a + b#
# #

∴ 3a# = b#
∴ √3a = b
a 1
∴ =
b √3
∴a:b=1:√3

Answer 15: c (1:6)

Explanation:
Let PT=TU=UQ=x
Let PS=RQ=y
1 1 xy
∴ Area of △ TUR = x base x height = x X x Y =
2 2 2
Area of rectangle = base x height = 3x. y = 3xy
Area of △ RTU xy
∴ = ÷ 3xy
Area of rectangle PQRS 2
Area of △ RTU xy 1
∴ = =
Area of rectangle PQRS 6xy 6
∴Area of △RTU: Area of rectangle PQRS=1:6

Answer 16: b (170)

Explanation:
Number of diagonals in a polygon with n sides=nC2-n
∴Number of diagonals in a polygon with 20 sides=20C2-20
20x19
∴ Number of diagonals in the polygon = Q R − 20
2x1
∴ Number of diagonals in the polygon = 190 − 20
∴ Number of diagonals in the polygon = 170.

Answer 17: c (32 cm2)

Explanation:
Area of ABCD= 4cm x 4cm= 16cm2…………(1)
AE=EB=BF=FC=2 cm
(EF)2= (EB)2 + (BF)2 (By Pythagoras theorem)
(EF)2=(2)2+(2)2=4+4=8
∴EF=√8= 2√2
Area of square EFGH= (side)2=(EF)2=(2√2)2= 8…….(2)
HI=IE=EJ=JF = √2
(IJ)2= (EI)2+(EJ)2 (By Pythagoras theorem)
(IJ)2= (√2)2 + (√2)2
(IJ)2= 2 + 2= 4
∴IJ= √4=2
Area of square IJKL= (side)2=(IJ)2=(2)2= 4………….(3)
From (1), (2) and (3), we can conclude that the areas of squares reduce in
a geometric progression.
Series= 16,8,4……………..∞
r=n2/n1= 8/16=1/2=0.5
In an infinite GP,
S∞= a/(1-r)= 16/(1-0.5)=16/0.5=32 cm2

Answer 18: d (100°)

Explanation:
∠RSQ + ∠SRQ+ ∠SQR=180 (Sum of all angles of a △is equal to 180)
50+30+∠SQR=180
80+∠SQR=180
∠SQR=100
∠QSR=∠SPQ=50 (Alternate segment theorem)
∠QPS + ∠PSQ= ∠SQR
50+∠PSQ=100
∠PSQ=100-50=50
∠PSQ= ∠POQ/2(Angle subtended by an arc on the circumference is half
of what it subtends at the center)
∴50=∠POQ/2
∴100=∠POQ

Answer 19: d (8)

Explanation:
Let the number of sides of the polygon be n.
∴Sum of all interior angles= (n-2)x180
∴1080= (n-2) x 180
∴6= (n-2)
∴8=n

Answer 20: e (12-8√2)

Explanation:
Radius of larger circle= d/2=4/2=2 cm

DE=EF= 2 cm
BE = :(2)# + (2)# = √4 + 4 = √8 = 2√2
Let the radius of the smaller circle be r
∴ BO = :r # + r # = :2r # = √2r
∴ √2r + r + 2 = 2√2
On solving, we get:
r=6-4√2
∴d=2 x r = 2 x (6-4√2)= 12-8√2

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How to make best use of CATKing Bible LOD 1 Books?

i. Attend the CATKing Concept Builder Classes to gain an idea of what all are the
basic pointers of the chapters.
ii. Go through that chapter in the CATKing Bible LOD 1 Books and read all the
Theory and Formulae provided in the Introduction of the chapter.
iii. Make a note of all the formulae and try to learn them by Heart, this will be your
ready reckoner to revise before exams.
iv. Go through all the Solved Examples and solve them simultaneously while
referring the solutions provided to understand the best way to solve each type of
questions.
v. Solve all the Practice Questions provided on your own and then refer to the
solutions at the end so as to verify if you have solved the questions correctly or is
there a better smarter approach for the same question.
vi. If you are able to solve majority questions correctly, then move to the next step
of preparation by taking the Topicwise Tests.
vii. Once you are done with good set of 4-5 Topics, give the Sectional and Full
length Mocks and see where you stand.
viii. While you analyze the mocks see to it that you highlight your weak and strong
areas. For all weak areas, you must get back to these CATKing Bible LOD 1 Books,
read up your formulae, check out the Solved Exams and your concepts will be
revised.

Introduction 2
Solved Examples 4
Practice Questions 27
Answer Key 37
Solutions 37

Let’s get started...

Introduction
Sometimes, we have to find the height of a tower, building, tree etc.
Though we cannot measure them easily, we can determine these by
using trigonometric ratios. In a right-angled triangle, side opposite to

given angle (θ) is called perpendicular (p), the longest side is called
hypotenuse (h) and the adjacent side is called base (b). then,
Sin θ = p/b
Cos θ = b/h
Tan θ = p/b

Trigonometric Laws
sin θ = Opposite Side/Hypotenuse
sec θ = Hypotenuse/Adjacent Side
cos θ = Adjacent Side/Hypotenuse
tan θ = Opposite Side/Adjacent Side
cosec θ = Hypotenuse/Opposite Side
cot θ = Adjacent Side/Opposite Side
Reciprocal Identities
cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ
sin θ = 1/cosec θ
cos θ = 1/sec θ
tan θ = 1/cot θ

Trigonometric Identities
sin2 θ + cos2 θ = 1
1 + tan2 θ = sec2 θ
1 + cot2 θ = cosec2 θ
Trigonometric Table
Its a table that you can refer to if you are not so sure about the values of
different angles. Below is the table for trigonometry formulas of different
angles which are commonly used for solving an ample number of
problems.
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Angles (In Degrees) 0° 30° 45° 60° 90° 180° 270° 360°
Angles (In Radians) 0° π/6 π/4 π/3 π/2 π 3π/2 2π
sin 0 1/2 1/√2 √3/2 1 0 -1 0
cos 1 √3/2 1/√2 1/2 0 -1 0 1
tan 0 1/√3 1 √3 ∞ 0 ∞ 0
cot ∞ √3 1 1/√3 0 ∞ 0 ∞
cosec ∞ 2 √2 2/√3 1 ∞ -1 ∞
Sec 1 2/√3 √2 2 ∞ -1 ∞ 1

Angle of Elevation

Suppose a man from a point O looks up at an object P, placed above the

level of his eye. Then, the angle which the line of sight makes with the
horizontal through O, is called the angle of elevation of P as seen from O.
Angle of elevation of P from O = AOP.

Angle of Depression

Suppose a man from a point O looks down at an object P, placed below

the level of his eye, then the angle which the line of sight makes with the
horizontal through O, is called the angle of depression of P as seen from
O.
Angle of depression of P from O = AOP.

Solved Examples
1) A antenna is on top of a building. Rajesh is standing on the ground at
a distance of 50 m from the building. The angles of elevation to the top
of the antenna and bottom of the antenna are 45° and 30° respectively.
What is the height of the antenna?
a) 50(3 - 1)/ 3 m
b) 503 m
c) 25/3 m
d) None of these
Answer: a
Explanation:
!
ð tan θ =
"
P

Q N
Where, P = perpendicular, B = Base
ð tan 30° = 1/√3 ; tan 45° = 1
ð tan 60° = √3 ; tan 90° = not defined
In ∆PNQ , tan 45° = PQ/NQ
ð PQ = NQ = 50m
In ∆MNQ , tan 30° = MQ/NQ
ð 1/√3 = MQ/NQ
ð MQ = 50/√3
#$ #$
ð h = PM = PQ − MQ = 50 − = =√3 − 1>
√& √&

2) Two men are standing on the two sides of a Building. The angle of
elevation of the top of the building is observed by the two man are 30°
and 45° respectively. If the building is 100 m high, the distance between
the two men is:
a) 173 m
b) 200 m
c) 273 m
d) 300 m
Answer:-c
Explanation:
Let AB be the building and C and D be the positions of the man.

Then, AB = 100 m, ACB = 30° and ADB = 45°.

ð AB/AC
ð tan 30°
ð 1/√3
ð AC= 100√3
ð AB/AD= tan 45° = 1
ð AD = AB = 100 m.
ð CD = (AC + AD) = (100√3 + 100) m
= 100(√3 + 1)
= (100 x 2.73) m
= 273 m.

3) A ship sailing at a point P is watching the lighthouse, which makes an

angle of elevation of 30° from ship. The ship sails some distance
towards the lighthouse and the new angle from its top and the angle of
the elevation becomes 60°. What is the distance between the base of
the lighthouse and the point P?
a) 43 units
b) 8 units
c) 12 units
d) Data inadequate
e) None of these
Answer: d
Explanation:
One of AB, AD and CD must have given.

So, the data is inadequate

4) The angle of elevation of a broken wire leaning against a pole is 60°

and the foot of the pole is 4.6 m away from the wire touching the
ground. The length of the wire is:
a) 2.3 m
b) 4.6 m
c) 7.8 m
d) 9.2 m

Answer: d
Explanation:
Let AB be the pole and BC be the wire.

Then, ACB = 60° and AC = 4.6 m.

'(
ð = cos 60°
"(
ð½
ð BC = 2 x AC
= (2 x 4.6) m
= 9.2 m.

5) A window 1.6 m above the ground is 20√3 m away from a tower. The
angle of elevation from the window to the top of the tower is 30°. The
height of the tower is:
a) 21.6 m
b) 23.2 m
c) 24.72 m
d) None of these

Answer: a
Explanation:
Let AB be the window and CD be the tower.

Draw BE
Then, CE = AB = 1.6 m,
BE = AC = 20√3 m.
ð DE/BE
ð tan 30°=1/√3
ð DE =(20√3)/√3m = 20 m.
ð CD = CE + DE = (1.6 + 20) m = 21.6 m.

6) From a point P on a level ground, the angle of elevation of the top

tree is 30°. If the tree is 100 m high, the distance of point P from the
foot of the tree is:
a) 149 m
b) 156 m
c) 173 m
d) 200 m

Answer: c
Explanation:
Let AB be the tree.

Then, APB = 30° and AB = 100 m.

AB/AP = tan 30°
ð AP = (AB x √3) m
ð 100√3 m
ð (100 x 1.73) m
ð 173 m.

7) A man standing at √3 times of building height from the foot of the

building. Find the angle of elevation that man will make with the top of
the building.
a) 30°
b) 45°
c) 60°
d) 90°

Answer: a
Explanation:
Let AB be the building height and AC be the distance of man from the
building.

Then AB/(AC )= √(3 )=cot θ

=> 30°

8) A man on the top of the tower observes a car moving with uniform
speed toward it. When the car covers a distance 70m the angle changes
from 30° to 60°. What is the height of the tower?
a) 60.6 m
b) 140 m
c) 35 m
d) 20.2 m

Answer: a
Explanation:
Let AD be the tower, BD be the initial position of car and CD be the final
position of car.
Given that BC = 70 m, ABD = 30°, ACD = 60°,

Let CD = x, AD = h
From the right CDA,
tan60° = AD/CD = √3 = h/x ⋯(eq:1)
From the right BDA,
tan30°= AD/BD = 1/√3 = h/(x+70) ⋯(eq:2)
From eq1 and eq2
⇒ √3/(1/√3) = (h/x)⁄(h/(x+70))
⇒ 3 = (x+70)/x
⇒ 2x=70
⇒ x=35
⇒ √3 = h/x
⇒ h= 35 x 1.73
⇒ 60.55 approx. 60.6

9) An Eagle when 900 m high passes vertically above another Aeroplane

at an instant when their angles of elevation at same observing point are
60° and 45° respectively. Approximately, how many meters higher is the
Eagle than the Aeroplane?
a) 381 m
b) 211 m
c) 169 m
d) 254 m
Answer: a
Explanation:
Let C be the position of the Eagle and D be the position of Aeroplane.
Given that CB = 900 m, CAB = 60°, DAB = 45°
From the right angled ∆ ABC,
ð tan60° = CB/AB
ð √3= 900/AB
ð AB = (900 √3)/3
ð From the right ABD,
ð tan45°= DB/AB
ð 1= DB/AB
ð DB = AB = 300√3
Required height
ð CD = (CB-DB)
ð (900−300√3)
ð (900−300×1.73)
ð 900- 519 =381

10) A building of is divided in the ratio 1:9 by a mark on it with lower

part shorter than the upper part. If the two parts subtend equal angles
at a place on the ground, 15 m away from the base of the pole, what is
the height of the building?
a) 15√5m
b) 60√3m
c) 60√5 m
d) 15√3m
Answer: c
Let CB be the building and point D divides it such that BD : DC = 1 : 9

Given that AB = 15 m
Let the two parts subtend equal angles at point A such that
CAD = BAD = θ
From "Angle Bisector Theorem", we have
ð BD⁄DC = AB⁄AC
=> 1⁄9 = 15⁄AC
=> AC = 15 × 9 m ...(eq: 1)
From the right angled ∆ABC,
ð CB=√(AC^2-AB^2 )
ð √([(15 x 9)]2-[15]2 )
ð √(([15]2 (9) − 1)))
ð 15 x 4 √5
ð 60 √5 m

11) From the foot and the top of a tree of height 230 m, a bird observes
the top of a tower with angles of elevation of b and a respectively.
What is the distance between the top of tree and building if tan a =
5/12 and tan b = 4/5
a) 400 m
b) 650 m
c) 600 m
d) 250 m
Answer: b
Explanation:
Let ED be the Tree and AC be the building.

Given that ED = 230 m, ADC = b, AEB = a

Also given that tan a = 5/12 and tan b = 4/5
Let AC = h
Required Distance = Distance between the top of the building and tree =
AE
From the right ABE,
=> tan (a)= AB/BE
=> 5/12= (h-230)/BE
=> BE =(12(h-230))/5⋯(eq:1)
From the right ACD,
=> tan(b )= AC/CD
=> 4/5 = h/CD

=> CD =5h/4⋯ (eq:2)

From the diagram, BE = CD
⇒ 48h - (4 x 12 x 230) =25h
⇒23h = (4×12×230)
⇒h = ((4×12×230))/23=480 m
AB = (AC - BC)
= (480 - 230) [∵ Since AC=h=480(from eq:3) and BC=ED=230 m(given)]
= 250 m
In the triangle ABE, tan(a) = 5/12. Let's figure out the value of sin(a) now.
Consider a triangle with opposite side = 5 and adjacent side = 12 such
that tan(a) = 5/12
hypotenuse = √(5^2+[12]^2 ) = 13
i.e., sin(a) = 5/13
We have seen that sin(a) = 5/13
⇒AB/AE = 5/13
⇒AE = AB×135
=250× 13/5=650 m
i.e., Distance between the top of the building and tree = 650 m

12) A poster is placed on top of a tower. The Poster and tower subtend
equal angles at a point on level ground which is 200 m away from the
foot of the tower. If the height of the poster is 50 m and the height of
the tower is h, which of the following is true?
a) None of these
b) h3- 50h2 - (200)2(h) + (200)2(50) = 0
c) h3+ 50h2 + (200)2(h)- (200)2(50) = 0
d) h3- 50h2+ (200)2(h) - (200)2(50) = 0

Answer: c
Explanation:
Let AD be the poster and CD be the tower.

Assume that the poster and tower subtend equal angles at point B.
Given that AD = 50 m, CD = h and BC = 200 m
Let angle ABD = θ, DBC = θ (∵ poster and tower subtend equal angles at a
point on level ground).
Then, ABC = 2θ
from the right angled ∆BCD,
ð tanθ = DC/BC = h/200⋯(eq:1)
From the right BCA,
'( '*+*( #$+,
ð tan2θ = = =
"( )$$ )$$
)-./0 #$+,
ð =
12-./! 0 )$$

ð 2h =(1-h2/[200] 2) (50 + h)
ð h3+50h2+[(200)] 2 h-50([200] 2)

13) A Ship sailing, is standing exactly midway between two icebergs,

observes the top of the two icebergs at angle of elevation of 22.5° and
67.5°. What is the ratio of the height of the taller iceberg to the height
of the shorter iceberg? (Given that tan 22.5° = √2−1)2−1)
a) 3 - 2√2 :1
b) 1+2√2 :1
c) 1−2√2 :1
d) 3 + 2√2 :1
Answer: d
Explanation:
Let ED be the taller iceberg and AB be the shorter iceberg.

Let C be the point of observation.

Given that ACB = 22.5° and DCE = 67.5°
Given that C is the midpoint of BD.
Hence, BC = CD
From the right ABC,
ð tan22.5° = AB/BC (eq:1)
From the right ∆ CDE,
ð tan67.5° = ED/CD (eq:2)
ð (tan67.5°)/(tan22.5°) =(ED/CD)/(AB/BC)
ð cot 22.5° / tan 22.5° = ED/AB …….. (CD=BC)
ð ED/AB = 1/(tan2(22.5)°)

ð 1/[(√2-1)]2
ð ((√2+1)/(2-1))2 ……….. (rationalize)
ð (2 + 2√2 +1)/1
ð (3 + 2√2)/1
ð ED: AB = ( 3 + 2√2)/1

14) The angle of elevation of the top tree from a point on the ground is
[sin] -1(3/5). If the point of observation is 20 meters away from the foot
of the tree, what is the height of the tree?
a) 15 m
b) 12m
c) 9 m
d) 18 m
Answer: a
Explanation:
Solution 1
Consider a right-/ triangle PQR as shown below.

Let QR = 3 and PR = 5 such that sinθ = 3/5 [i.e., θ=sin−1(35)]

PQ = √(PR2-QR2 )
i.e., when θ=[sin] -1(3/5)
PQ : QR = 4 : 3 ...(eq:1)
Now Let's solve the question. Let P be the point of observation and QR be
the tree.
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Height and Distance
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Given that θ=[sin]-1 (3/5) and PQ = 20 m

We know that PQ: QR = 4: 3 (from eq:1)
i.e., 20 : QR = 4 : 3
=> 20 × 3 = QR × 4
=> QR = 15 m
i.e., height of the tower = 15 m

15) Find the angle of elevation of the sun when the shadow of a pole of
18 m height is 6√3m long?
a) 30°
b) 60°
c) 45°
d) None of these
Answer: b
Explanation:
Let RQ be the pole and PQ be the shadow

Given that RQ = 18 m and PQ =6√3 m

Let the angle of elevation, RPQ = θ
From the right PQR,
ð tanθ = RQ/PQ = 18/(6√3) = 3/√3
ð tanθ = √3
=> θ = 60°

16) Two rats are on either sides of a tower of height 50 m. The rat
observers the top of the tower at an angle of elevation of 30° and 60°. If
a car crosses these two rat in 10 seconds, what is the speed of the car?
a) 20√33 km/hr
b) 24√3 km/hr
c) None of these
d) 24√3 km/hr
Answer: d
Explanation:
Let BD be the tower and A and C be the positions of the rats.

Given that BD = 50 m, BAD = 30°, BCD = 60°

From the right ∆ ABD,
ð tan 30° = BD/BA
ð 1/√3 = 50/BA
ð BA=50√3
From the right ∆ CBD,
ð tan 60° = BD/BC
ð √3=50/BC
ð BC=50/√3
Distance between the two Rats
ð AC = BA + BC
ð 50√3+ 50/√3
ð (200√3)/3

Speed of the car =Distance/Time

ð ((200√3)/3)/10 = (20√3)/3 m/s
ð (20√3)/3 x 18/5 km/hr
ð 24 √3 km/hr

17) The elevation of the summit of a mountain from its foot is 45°. After
ascending 2 km towards the mountain upon an incline of 30°,the
elevation changes to 60°. What is the approximate height of the
mountain?
a) 0.6 km
b) 2.7 km
c) 1.2 km
d) 1.4 km
Answer: b
Explanation:

Let A be the foot and C be the summit of a mountain.

Given that angle CAB = 45°
From the diagram, CB is the height of the mountain. Let CB = x
Let D be the point after ascending 2 km towards the mountain such that
AD = 2 km and given that angle DAY = 30°
It is also given that from the point D, the elevation is 60°
i.e., angle CDE = 60°
From the right ∆ ABC,
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=> tan45°=CB/AB
=> 1=x/AB [∵ CB = x(the height of the mountain)]
=> AB = x ... (eq:1)
From the right angled ∆ AYD,
sin30°=DY/AD
=> 1/2=DY/2 (∵ Given that AD = 2)
=> DY = 1 ... (eq:2)
cos30°=AY/AD
=> √3/2 =AY/2
=> AY = √3 ... (eq:3)
From the right angled ∆CED,
tan60°=CE/DE
⇒tan60°= (CB - EB)/YB
⇒tan60°= (CB - DY)/ (AB - AY) [∵ EB=DY and YB=(AB-AY)]
⇒tan60°=(x - 1)/(x−√3) [∵ CB=x, DY=1(eq:2), AB=x(eq:1) and AY =
√3(eq:3)]
⇒√3=(x - 1)/(x−√3)
=√3x -3 = x-1
= x(√3-1) =2
⇒ x= 2/0-73 =2.7
i.e., the height of the mountain = 2.7 km

18) To a man standing outside his house, the angles of elevation of the
top and bottom of a Poster above the building are 60° and 45°
respectively. If the height of the man is 180 cm and he is 5 m away from
the building, what is the length of the poster?
a) 3.65 m
b) 2.5 m
c) 8.65 m
d) 2 m

Answer: a
Explanation:
Let AB be the man and CD be the window

Given that the height of the man, AB = 180 cm, the distance between the
man and the building, BE = 5 m,
Angle DAF = 45°, angle CAF = 60°
From the diagram, AF = BE = 5 m
From the right ∆ AFD,
tan45°=DF/AF = DF/5
ð DF = 5⋯(1)
From the right ∆ AFC,
ð tan60°=CF/AF
ð √3=CF/5
ð CF=5√3 ⋯ (2)
Length of the poster
ð CD = (CF - DF)
ð 5√3−5 [∵ Substituted the value of CF and DF from
(1) and (2)]
ð 5(√3−1)
ð 5(1.73−1)

ð 5×0.73
ð 3.65 m

19) Two vertical buildings which are 200 m apart and the height of one
is double that of the other. When a cat is in exactly in between the two
buildings and the angle of elevation made by the cat, finds the angular
elevations of their tops to be complementary. Find the heights of the
buildings.
a) 65 m and 130 m
b) 130 m and 260 m
c) 141 m and 282 m
d) 70.5 m and 141 m
Answer: d
Explanation:
Let AB and CD be the buildings with heights h and 2h respectively.

Given that distance between the heights, BD = 200 m

Let E be the middle point of BD,
AEB = θ
CED = (90-θ) (∵ given that angular elevations are complementary)
Since E is the middle point of BD, we have BE = ED = 100 m
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Height and Distance
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From the right ABE,

ð tanθ=AB/BE
ð tanθ=h/100 ……...... (1)
ð h= 100 tanθ
From the right ∆ EDC,
ð tan(90−θ)=CD/ED
ð cotθ=2h/100
=> 2h= 100 cotθ ……….. (2)
(1) × (2)
=> [2h]2 =[100]2
[∵tanθ×cotθ=tanθ×1tanθ=1][∵tanθ×cotθ=tanθ × 1tanθ=1]
=>√2h=100
=>h = 100/√2
=>2h=2×70.5=141
i.e., the heights of the buildings are 70.5 m and 141 m.

20) The angles of depression and elevation of the top of a tower 11 m

high from top and bottom of a building are 60° and 30° respectively.
What is the height of the building?
a) None of these
b) 22 m
c) 33 m
d) 44m

Answer: d
Explanation:
Let DC be the tower, AB be the building.

Given that DBC = 30°, DAE = 60°, DC = 11 m

ð tan 30°=DC/BC
ð 1/√3=11/BC
ð BC = 11√3 m
ð AE = BC =11√3 m ⋯ (1)
ð tan60°=ED/AE
ð √3=ED/11√3
ð ED=11√3 x √3 [∵ Substituted value of AE from (1)]
ð 11×3=33
Height of the building
ð AB = EC = (ED + DC)
ð (33 + 11) = 44 m

Practice Questions
1) Raj is flying a kite starting from point A. At the end of 2 minutes, Raj
finds the angular elevation of the kite as 60°. If the point at which Raj is
standing is 150 m away from point A, what is the speed of the Kite?
a) 2.16 meter/sec
b) 0.72 meter/sec
c) 0.63 meter/sec
d) 3.87 meter/sec

2) A vertical tower stands on ground and is surmounted by a vertical

Poster of height 18 m. At a point on the ground, the angle of elevation
of the bottom and the top of the poster are 30° and 60° respectively.
What is the height of the tower?

a) 10.40 m

b) 15.57 m

c) 9 m
d) 12 m

3) From the top of a building 100 m high, the angles of depression of

the top and bottom of a small building are 30° and 60° respectively.
What is the height of the small building?

a) 52 m

b) 66.67 m
c) 50 m

d) 33.33 m

4) The angle of elevation of the top of a lighthouse 60 m high, from two

Ships sailing on its opposite sides are 45° and 60°. What is the distance
between these two ships?

a) 30m

b) 94.6m

c) 45m

d) 103.8m

5) From a tower of 80 m high, the angle of depression of a man is

30°.How far is the Man from the tower?
a) 40m

b) 138.4m

c) 46.24m

d) 160m

6) A wire 10 m long just reaches the top of a Pole and makes an angle of
60° with the ground after it broke. Find the distance of the foot of the
Pole from the wire touching the ground (√3=1.73)

a) 5 m
b) 17.3 m

c) 8.65 m

d) 4.32 m

7) On the same side of a lighthouse, two ships are located. Observed

from the top of the lighthouse, their angles of depression are 45° and
60°. If the height of the lighthouse is 600 m, the distance between the
ships is approximately equal to:

a) 272 m

b) 254 m

c) 288 m

d) 284 m

8) The man observing the top of the tower at an angle of elevation of

the of 30°. If the man moves 40 m towards the tower, the angle of
elevation of the top of the tower increases by 15°. The height of the
tower is:

a) 62.2 m

b) 64.2 m
c) 52.2 m

d) 54.6 m

9) A kite is flying at height of a 15 metre high makes an angle of

elevation of 60° with the man standing at bottom of an building and
angle of elevation of 30° with man standing at the top of the building.
What is the height of the building?

a) 12m

b) 5m

c) 8m

d) 10m

10) A man is watching from the top of a tower a car moving away from
the tower. The car makes an angle of depression of 45° with the man's
eye when at a distance of 100 metres from the tower. After 10
seconds, the angle of depression becomes 30°. What is the
approximate speed of the car?

a) 32.42 km/hr.

b) 26.28 km/hr.

c) 24.22 km/hr.
d) 31.25 km/hr.

11) A man on the top of a tower observes a car moving at a uniform

speed coming directly towards it. If it takes 12 minutes for the angle of
depression to change from 30° to 45°, how soon after this will the car
reach the observation tower?

a) 12min 23 second
b) 15 min 24 seconds
c) 16 min 24 seconds
d) 8 min

12) Raj looks at the top of a pole. The angle made is 45°. The angle
made reduces to 30°, if he moves back by 10 m. How high is the pole?

a) 10/(√3- 1) m

b) 10/( √3+ 1) m

c) 20 m

d) 203 m

13) A tree of height 24 cm stands in my vicinity. On a stormy night, the

tree broke and fell in such a way that the upper part remained attached
to the stem, forming an angle of 30° with the ground. Find the height
from where it broke.

a) 7.5 cm

b) 8 cm

c) 9.5 cm

d) 12 cm

14) The actual height of a child is 3 feet but due to the position of Sun
he casts a shadow of 2 feet. He is standing next to an pole and notices
that it casts a shadow of 32 feet. Find the height of the pole.

a) 48 ft

b) 54 ft
c) 63 ft

d) 72 ft

15) Two friends Ajay and Vijay are having two sticks of heights of
height 8 cm and 15 cm respectively which are placed on ground. The
distance between the two Sticks was 24 cm. Find the distance between
the tops of two Sticks.

a) 24cm

b) 24.5cm

c) 25cm

d) 31cm

16) A building is under construction. The top of the building forms 30°
angle of elevation from a point on the adjoining plot that is 150 m.
After a month, the angle of elevation formed by the top of the building
from the same point increased to 60°. How much was the building
constructed in this 1 month.

a) 200/√3 m

b) 100√3 m

c) 200√3 m
d) 3001√3 m

17) When looking at the top of a pole from two different points m and
n on the ground, the angles of elevation formed are 60° and 45°
respectively. Find the height of the pole.

a) O𝐦𝐧√𝟑𝐮𝐧𝐢𝐭𝐬

b)√𝟑 𝐦𝐧 units

c) mn units

d) 3mn units

18) From a point 450 meters away from the foot of a tower, the top of
the tower is observed at an angle of elevation of 45°, then the height
(in meters) of the tower is?

a) 375

b) 450
c) 225

d) 250

19) The angle of elevation of a tower at a point 90 m from it is

cot-1(3/5).Then the height of the tower is

a) 45

b) 150

c) 112.5

d) 90

20) On the level ground, the angle of elevation of the top of a tower is
30°.on moving 20 meters nearer, the angle of elevation is 60°.Then the
height of the tower is

a) 10

b) √3

c) 10√3

d) 20√3

Answer Key
1(a) 2(c) 3(b) 4(b) 5(b)
6(c) 7(b) 8(d) 9(d) 10(b)
11(c) 12(a) 13(b) 14(a) 15(c)
16(b) 17(a) 18(b) 19(b) 20(c)

Solutions
1) Answer- A

Explanation:

Let C be the position of Raj. Let A be the position

at which Kite leaves the earth and B be the position of the Kite after 2
minutes.Given that CA = 150 m, BCA = 60°tan60°=BA/CA
ð √3 = BA/150
ð BA=150√3
i.e, the distance travelled by the Kite =150√3=15√3 meterstime taken
= 2 min = 2 × 60 = 120 secondsSpeed =Distance/Time
1#$√&
=>
1)$
=> 1.25√3

=> 1.25×1.73
=> 2.16 meter/second

2) Answer- C
Explanation:

Let DC be the vertical tower and AD be the vertical Poster. Let B be the
point of observation. Given that AD = 18 m, ABC = 60°, DBC = 30°Let
DC be h.=> tan30°=DC/BC
1 3
ð =
√& 45

45
ð h= ⋯ (1)
√&
=> tan60°=AC/BC
16+3
ð √3=
45
ð 18+h=BC x√3 …….(2)

From (1) and (2)=> 3h=18+h

ð 2h=18
=> h= 9m i.e., the height of the tower = 9 m

3) Answer- B
Explanation:

Consider the diagram shown above. AC represents the bigger building

and DE represents the Smaller building. Given that AC = 100 m XAD =
ADB = 30° (∵ AX || BD ) XAE = AEC = 60° (∵ AX || CE)Let DE = h, Then,
BC = DE = h, AB = (100-h) (∵ AC=100 and BC = h),BD = CEtan60°=AC/CE
=>√3=100/CE
=>CE = 100/√3 (1) tan30°=AB/BD
1 1$$23
=> =
√& 47
1 1$$23
=> = "## (∵ BD = CE and substituted the value of CE from
√&
√%
1 1$$
equation 1)=>(100−h)= x
√& √&
= 100-33.33= 66.67m
i.e., the height of the smaller building = 66.67 m

4) Answer- B
Explanation:

Let BD be the lighthouse and A and C be the two points on ground. Then,
BD, the height of the lighthouse = 60 m BAD = 45° , BCD =
60°tan45°=BD/BA

⇒1=60/BA

⇒BA=60 m ⋯ (1)

tan60° = BD/BC
8$
⇒ √3 =
45

8$
⇒ BC=
√&

BC= 20√3 = 20 x 1.73

=34.6m

Distance between A and C

=AC = BA+ BC = 60 + 34.6 = 94.6m

5) Answer- B

Explanation:

Let AC be the tower and B be the position of the Man.Then BC = the

distance of the man from the foot of the tower.Given that height of the
tower, AC = 80 m and the angle of depression, DAB = 30° ABC = DAB =
30° (because DA || BC)tan30°=AC/BC
=>tan30°=80/BC
6$
=>BC = 80/tan30° = "
√%
= 80×1.73=138.4m i.e., Distance of the bus from the foot of the tower =
138.4 m

6) Answer- C

Explanation: A

Wire

Pole

C B

Let BA be the Wire and AC be the pole as shown above.Then the distance
of the foot of the wire from the pole = BCGiven that BA = 10 m , BAC =
60°sin 60°=BC/BA

√& 45
=
) 1$

√&
ð BC = 10 x
)

ð 5×1.73

ð 8.65 m

7) Answer- B

Explanation:

Let DC be the lighthouse and A and B be the ships as shown above.Given

that DC = 600 m , DAC = 45°, DBC = 60°tan 60°=DC/BC
√3 = 600/BC
8$$
BC= ⋯(1) tan 45°=DC/AC
√&
1=600/AC
AC= 600 ⋯(2) Distance between the objects= AB = (AC - BC)
1
= 600 (1-W X) [∵ from (1) and (2)]
√&
= 200√3(√3 − 1)
= 200 (3-1.73)
= 254 m

8) Answer- D
Explanation:

Let DC be the tower and A and B be the positions of the man such that AB
= 40m, We have DAC = 30°, DBC = 45°Let DC = htan30°=DC/AC
1 ,
=> =
√& '(
=>𝐴𝐶 = ℎ√3 ⋯ (1)
=> 𝑡𝑎𝑛45° = 𝐷𝐶/𝐵𝐶
1 = DC/BC
ð BC=h …. (2)
AB = (AC-BC)
40 = (h√3 –h)
40 = h (√3 -1)
9$
h=
√& 21

9$ ;√&+1<
h=
)

h= 20 (1.73-1)
h= 20 x 0.73
h= 54.6m

9) Answer- D

Explanation:

Consider the diagram shown above. AC represents the height of kite and
DE represents the Building. Given that AC = 15 m, ADB = 30°, AEC =
60°Let DE = h, Then, BC = DE = h, AB = (15-h) (∵ AC=15 and BC = h),BD =
CEtan60°=AC/CE
=> √3 = 15/CE
=> CE = 15/√3 ….. (1) => tan30°=AB/BD
1
=> = (15−h)/BD
√&
1 1#2,
=> = "& (∵ BD = CE and substituted the value of CE from
√&
√%
equation1)=> h= 15-5 =10 units
i.e., height of the Building = 10 m

10) Answer- B

Explanation:

Consider the diagram shown above.Let AB be the tower. Let C and D be

the positions of the carThen, ACB = 45° , ADC = 30°, BC = 100 m
=> tan45°=AB/BC
=>1=AB/100
=> AB = 100 ⋯(1)=> tan30°=AB/BD

1
=> =100/BD (∵ Substituted the value of AB from equation 1)=>
√&
BD =100√3CD = (BD - BC) =(100√3−100)
=100(√3−1)
It is given that the distance CD is covered in 10 seconds.i.e., the
distance 100(√3−1) is covered in 10 seconds.Required
speed=Distance/Time
1$$(√&21)
=>
1$
ð 10(1.73-1)
ð 7.3 meter/seconds
ð 7.3 × (18/5) km/hr = 26.28 km/hr

11) Answer- C

Explanation:

Consider the diagram shown above. Let AB be the tower. Let D and C be
the positions of the car.Then, ADC = 30° , ACB = 45°Let AB = h, BC = z,
CD = ytan45°=AB/BC=h/z

=>1=h/z

=>h=z ⋯ (1) => tan30° =

'" '" ,
= =
"* "(+(* ?+@

1 ,
=> =
√& ?+@

=>z + y = √3h

=>y = √3h – z (∵ Substituted the value of x from equation 1 )=> y

= √3h – hGiven that distance y is covered in 12 minutes.i.e, distance √3h
– h is covered in 8 minutes.Time to travel distance z= Time to travel
distance h (∵ Since x = h as per equation 1).Let distance h is covered in t
minutes.Since distance is proportional to the time when the speed is
constant, we have h (√3 − 1) ∝ 12 ….(A)
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Height and Distance
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h ∝ 12 …..(B)

Dividing (A) and (B)

,(√&21) 1)
= =
, -

1)
ð √3 − 1) =
-
1)
ð t=
√&21)
ð t =1200/73
Time = 16.392 minutes
Time = 16 + (0.392 x 60)
= 16 minutes 24 seconds

12) Answer- A

Explanation:
Let PQ be the pole and M and N be positions where Raj stands.
Tip:
P
tan θ =
B
Where, P = Perpendicular; B = Base

1
ð tan30° = ; tan 45° = 1;
&
ð tan 60° = 3 ; tan 90° = Not defined
ð Now , tan45° = 1 = PQ/MQ
∴ PQ = MQ
1 PQ PQ
tan 30° = = =
√3 NQ 10+MQ
∴ 10 + MQ = √3 PQ∴ 10 + PQ = √3 PQ --------> As, PQ = MQ
10
∴ PQ = M
√3 - 1

13) Answer- B
Explanation:

Let the tree break at height h cm from ground at point M.The broken part
makes angle of 30°∴ Broken Part MP = MN = 24 - h (Orange line in
diagram)
Tip:
P
Sin θ =
H
Where, P = Perpendicular;H = Hypotenuse
1 1
Sin 30° = ; Sin 45° = ;
2 2

3
Sin 60° = ; Sin 90° = 1
2
In △MNQ,
1 MQ h
Sin 30° = = =
2 MN 24-h
∴ 24 - h = 2h∴ h = 8cm = Tree breaks at this height

14) Answer - A
Explanation:
Both the child and pole are near each other and are
illuminated by same sun from same direction.
So angle of elevation for sun is same for both.
So ratio of object to shadow will be same for all
objects. (Proportionality Rule)
Object height 3 H
= =
Shadow length 2 32
∴ H = 48 ft = Height of pole

15) Answer - C
Explanation: Let MN = Ajay's Stick &
PQ = Vijay's Stick
From the diagram we can see that
MR = 24cm & PR = 15 - 8 = 7cmBy Pythagoras
Theorem,Hypotenuse2 = (side1)2 + (side2)2∴
MP = 242 + 72∴ MP = 25cm

16) Answer: B
Explanation:
Tip:
P
Tan θ =
B
Where, P = Perpendicular ; B = Base
1
Tan 30° = ; Tan 45° = 1;
√3
Tan 60° = √3; Tan 90° = Not defined

Original Building height = h = MQNew building height = PQ

1 MQ
In △MQN, tan 30° = =
√3 NQ

150
∴ MQ =
√3

PQ
In △PQN, tan 60° = √3=
NQ
∴ PQ = 150√3Building grew = PQ – MQ
1#$ ) )
Building grew = 150√3 − = 150 = 3 x 50 x = 100√3
√& √& √&

17) Answer - A
Explanation:
Tip:
P
Tan θ =
B
Where, P = Perpendicular ; B = Base
1
Tan 30° = ; Tan 45° = 1;
√3
Tan 60° =√3 ; Tan 90° = Not defined

Height of pole = PQ
ð tan60° = √3 = PQ/m
ð tan45° = 1 = PQ/n
Multiply both equations
PQ PQ
√3x 1 = x
M N
∴ PQ = Omn√3 units

18) Answer – B
Explanation

From the right angled triangle

tan(45°)= X/450
=> X = 450 m

19) Answer - B

Let cot-1(3/5) = x
=> cot (x) = 3/5
=> tan(x) = 5/3
From the right angled triangle
tan(x) = h/90
=> h = 5/3*90 =150 m

20) Answer – C
Explanation:

Let h be the height of tower

From figure.
20 =h (cot30 - cot60)
=> 20 = h (√3 − 1/√3)
=> 20√3 = h (3 − 1)
=> h = 10√3.

What are CATKing Bible LOD 1 Books?

How to make best use of CATKing Bible LOD 1 Books?

i. Attend the CATKing Concept Builder Classes to gain an idea of what all
are the basic pointers of the chapters.
ii. Go through that chapter in the CATKing Bible LOD 1 Books and read all
the Theory and Formulae provided in the Introduction of the chapter.
iii. Make a note of all the formulae and try to learn them by Heart, this
will be your ready reckoner to revise before exams.
iv. Go through all the Solved Examples and solve them simultaneously
while referring the solutions provided to understand the best way to
solve each type of questions.
v. Solve all the Practice Questions provided on your own and then refer
to the solutions at the end so as to verify if you have solved the questions
correctly or is there a better smarter approach for the same question.
vi. If you are able to solve majority questions correctly, then move to the
next step of preparation by taking the Topicwise Tests.
vii. Once you are done with good set of 4-5 Topics, give the Sectional and
Full length Mocks and see where you stand.
viii. While you analyze the mocks see to it that you highlight your weak
and strong areas. For all weak areas, you must get back to these CATKing
Bible LOD 1 Books, read up your formulae, check out the Solved Exams
and your concepts will be revised.
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Mensuration
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Table of Contents

Introduction 2
Solved Examples 4
Practice Questions 19
Answer Key 25
Solutions 25

Let’s get started...

Introduction
Mensuration: Mensuration is the branch of Geometry which deals with
the study of 2D and 3D Geometric shapes. It mainly deals with the
measurement of lengths, areas and volumes of these shapes.

Important Formulae
1) Rectangle
• Area = lb

• Perimeter = 2(l+b)

2) Square
• Area = a×a

• Perimeter = 4a

3) Parallelogram
• Area = l × h

• Perimeter = 2(l+b)

4) Triangle
• Area = b×h/2 or √s(s-a)(s-b)(s-c)…………….where s = (a+b+c)/2

5) Right Angled Triangle

• Area =1/2(bh)

• Perimeter = b+h+d

6) Isosceles Right Angled Triangle

• Area = a /2
2

• Perimeter = 2a+d……………………….where d=a√2

7) Equilateral Triangle
• Area = √3(a /4) or ½(ah)….where h = √3a/2
2

• Perimeter = 3a

8) Trapezium
• Area = 1/2h(a + b), (a and b are parallel sides)

• Perimeter = Sum of all sides

9) Rhombus
• Area = (d1 × d2)/2

• Perimeter = 4a

• Area = 1/2 × Diagonal × (Sum of offsets)

11) Kite
• Area = (d1×d2)/2

• Perimeter = 2 × Sum on non-adjacent sides

12) Circle
• Area = πr or πd /4
2 2

• Circumference = 2πr or πd

• Area of sector of a circle = (θπr )/360

13) Cube
• Volume: V = a
3

• Lateral surface area = 4a

• Surface Area: S = 6a
2

• Longest Diagonal (d) = √3a

14) Cuboid
• Volume of cuboid: lbh

• Total surface area = 2(lb + bh + hl)

• Length of diagonal =√(l +b +h )

2 2 2

15) Right Circular Cylinder

• Volume of Cylinder = πr h
2

• Lateral Surface Area (LSA or CSA) = 2πrh

• Total Surface Area = TSA = 2πr(r + h)

• Volume of hollow cylinder = πh(R – r ), (‘R’ is the bigger radius,

2 2

and ‘r’ is smaller one)

16) Right Circular cone

• Volume = 1/3πr h
2

• Curved surface area: CSA= πrl

• Total surface area = TSA = πr(r + l)

17) Sphere
• Volume: V = 4/3 πr
3

• Surface Area: S = 4πr

18) Hemisphere
• Volume = 2/3πr
3

• Curved surface area (CSA) = 2πr

• Total surface area = TSA = 3πr

19) Prism
• Volume = Base area x h

• Lateral Surface area = perimeter of the base x h

20) Pyramid
• Volume of a right pyramid = (1/3) × area of the base × height.

• Area of the lateral faces of a right pyramid = (1/2) × perimeter of

the base x slant height.

• Area of whole surface of a right pyramid = area of the lateral faces

+ area of the base.

21) Tetrahedron
• Area of its slant sides = 3a √3/4
2

• Area of its whole surface = √3a

• Volume of the tetrahedron = (√2/12)a

22) Regular Hexagon

• Area = 3√3a /2
2

• Perimeter = 6a

23) Some other Formulae

• Area of Pathway running across the middle of a rectangle = w(l+b-w)

• Perimeter of Pathway around a rectangle field = 2(l+b+4w)

• Area of Pathway around a rectangle field =2w(l+b+2w)

• Perimeter of Pathway inside a rectangle field =2(l+b-4w)

• Area of Pathway inside a rectangle field =2w(l+b-2w)
• Area of four walls = 2h(l+b)

Solved Examples
1. 12 laddoos are tightly packed into a sweet box in one layer. Each row
has 4 laddoos and each column has 3 laddoos. What part of the box is
empty?
A) 10/21

B) 15/19
C) 11/23
D) 9/17
Answer - A
Explanation:
Let the diameter of laddoo be 2r
Length of the box = 4*2r = 8r
Breadth = 3*2r = 6r
Height = 2r
Volume of the box = 8r x 6r x 2r = 96r3
Volume of 12 laddoos = 12 * (4/3) * (22/7) * r3 = (352 r 3/7)
The volume of empty space = 96 r3 - (352 r 3/7)
= (672 r3 - 352 r3)/7 = (320 r3)/7
The required fraction = ((320 r3/7)/96 r3) = 10/21

2. A wire in form of a circle of radius 24.5 cm is cut made into a

rectangular shape whose sides are in the ratio of 3:8. What is the larger
side of the rectangle?
A) 58 cm
B) 52 cm
C) 56 cm
D) 64 cm

Answer – C
Explanation:
circumference = 2*(22/7)*24.5 = 154 cm
length of rectangle sides are 3x, 8x.
circumference = 2*(3x+8x)
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Mensuration
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154 = 22x
X = (154/22) = 7
Therefore, the larger side = 8x = 8x7 = 56 cm

3. A stick is resting against a tree trunk. The height of the stick is 10m
and distance from the bottom end of stick to tree is 5m. How high is the
upper end of stick from the ground?
A) 5 m
B) 5√3 m
C) 7√3 m
D) 7 m
Answer - B
Explanation:
Hypotenuse2 = (base)² + (altitude)²
102 = 52 + altitude2
Altitude2 = 102 - 52 = 75
Altitude = √75 = 5√3

4. Length of diagonal of a parallelepiped is 8cm and the sum of length,

breadth and height is 12cm. What is the total surface area of the
parallelepiped?
A) 75 cm2
B) 80 cm2
C) 85 cm2
D) 52 cm2
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Answer - B
Explanation:
l2 + b2 + h2 = 82 = 64
l + b + h = 12
(l+b+h)2 = l2 + b2 + h2 + 2(lb+bh+hl)
122 = 64 + 2(lb+bh+hl)
2(lb+bh+hl) = 144-64 = 80 cm2

5. The adjacent three faces of a cuboid have areas 324 cm2, 100 cm2 and
49 cm2. Then what is the volume of the box?
A) 1245 cm3
B) 1250 cm3
C) 1200 cm3
D) 1260 cm3
Answer - D
Explanation:
Area of the box = lbh
Area of three faces = lb x bh x lh = (lbh)2
(lbh)2= (324 x 100 x 49)
Taking square root on both sides
lbh = 18 x 10 x 7
Therefore, the volume of the box = 1260 cm3

6. An architect draws a plan of a rectangular bedroom and shows it to

his client, the client asks him to reduce the length of the bedroom by
30% but he should keep the area of the bedroom unaltered. Then by
what percent would the architect have to increase the width of the
bedroom?
A) 41.25%
B) 42.86%
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C) 43.76%
D) 40.50%
Answer – B
Explanation:
Suppose the new width is X.
=> L x B = (L – (30L/100)) x X
=> L x B = ((100L – 30L)/100) x X
=> X = (L x B x 100)/70L
=> X = B x 10/7
=> X = 1.4286 B
Therefore the width is to be increased by 42.86% to keep the area same.

7. The boundary rope of a circular cricket ground of radius 49 m is

converted into a rectangular football turf whose sides are in the ratio of
4:7. The smaller side of the rectangular football turf is?
A) 56 m
B) 58 m
C) 45 m
D) 52 m

Answer – A
Explanation:
circumference = 2*(22/7)*49 = 308 m
length of rectangular sides are 4x, 7x.
circumference = 2*(4x + 7x)
308 = 22x
X= 308/22 = 14
smaller side of rectangle = 4x = 4*14 = 56 m
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8. Santa and Banta started walking from a point at the same time. One
was walking towards north and the other was walking towards east.
One of them covered thrice the distance than the other. After stopping,
when they looked at each other, the distance between both of them
was 12 m. The area of the triangle created by them is?
A) 23.4 m2
B) 26.6 m2
C) 22.5 m2
D) 21.6 m2
Answer – D
Explanation :
x2 + (3x)2 = (12)2
10 x2= 144
x2 = 14.4 m2
Area = 1/2 (x *3x)
= 3(14.4)/2
= 21.6 m2

9. A conical vessel of height 21cm is filled upto the brim with milk. The
milk is later poured into cylindrical vessel, whose radius is one third of
the radius of the conical vessel. Then the height of milk in the new
vessel is?
A) 67 cm
B) 60 cm
C) 63 cm
D) 65 cm

Answer – C
Explanation :
1/3*(3.14*r2*h1) = 3.14*(r/3)2*h2
(r2*21)/3 = (r2/9)*h2
h2 =7*9 = 63cm

10. The base of a right angled triangle is 20 cm and its area is 80 cm2.
What is the length of its Hypotenuse?(approximate)
A) 22 cm
B) 16 cm
C) 18 cm
D) 24 cm
Answer – A
Explanation :
Area = [h*(20)]/2
80 = h*10
h = 80/10 = 8 cm
x2 = 82+202 = (64+400)
x = 21.5 = 22cm

11. The perimeter of a rectangle is half the circumference of a circle.

The area of the circle is 3850 cm2. What is the area of the rectangle if
the length of the rectangle is 25 cm?
A) 750 cm2
B) 725 cm2
C) 700 cm2
D) 730 cm2

Answer – A
Explanation :
Area of circle = (22/7)*r2
3850 = (22/7)*r2
r2= 3850*7/22 = 175*7 = 1225
r = 35
circumference of the circle = 2*(22/7)*35 = 220 cm
perimeter of the rectangle = 2(25+b)
220/2 = 50+2b
110 = 50+2b
2b = 60
b = 30 cm
Area of the rectangle = 25*30 = 750 cm2

12. A circle is drawn inside a square such that the circle exactly fits in
the square and the diameter of the circle is equal to the side of the
square, which is 24 cm. What is the area of the space left out in the
square which is not covered by the circle? (approximate)
A) 124 cm2
B) 115 cm2
C) 117 cm2
D) 120 cm2

Answer – A
Explanation :
(24*24) – [22/7 (12*12)]
= 576 – 452.16
= 123.84m2

13. The ratio of the side of the room is 4:3, the cost of whitewashing the
ceiling of the room at 40 paise per square metre is Rs.2800 and the cost
of preparing the walls at 25 paise per square metre is Rs.2500. The
height of the room is?
A) 20 m
B) 25 m
C) 30 m
D) 35 m
Answer – C
Explanation:
Area of ceiling = 2800/0.40 = 7000 m2
4x*3x = 7000
12x2 = 7000
X2 =583.33
X=24.15 = 24
Area of wall = 2500/0.25
= 10000 m2
H = 10000/2(96+72)
= 10000/336
= 29.77
= 30 m

14. There are 2 rectangular tanks A and B with lengths 6 m and 9 m in a

square field. If the total area of the square field excluding the
rectangular tanks is 300 m2 and the breadth of both the rectangular
tanks is 1/3 of the side of the square field, find the perimeter of the
square field?
A) 72 m
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B) 78 m
C) 84 m
D) 80 m
Answer – D
Explanation :
Area A = 6*a/3 = 2a, (‘a’ is the side of the square)
Area B = 9*a/3 = 3a
a2-5a-300 = 0
(a-20)(a+15) = 0
a=20
Perimeter of the square = 4a = 4*20 = 80 m

15. Suhas installs fencing around his 78 m2 rectangular garden. Since he

has only 55 m of fence, he fences three sides of the garden letting his
house compound wall act as the fourth side fencing. The dimensions of
the garden are (All sides are having length in integers)?
A) 12m X 3m
B) 26m X 3m
C) 6m X 13m
D) 39m X 2m

Answer – B
Explanation :
L =55 – 2b
area of the garden = 78 sq m,
L * b =78
b*(55-2b) = 78
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55b – 2b2 = 78
2b2 -55b+ 78 =0
b= 3/2, b=26
b = 3/2 then L = 52
b=26 then L = 3
Dimension = 26m * 3 m

16. In a rectangle the ratio of the length and breadth is 3:1. If each of
the length and breadth is increased by 4m their ratio becomes 10:7. The
area of the original rectangle in m² is?
A) 4m²
B) 3m²
C) 2.7m²
D) 3.5m²
Answer – B
Explanation:
[3x + 4 / x + 4] = 10/7
x = 12/11 = 1.09 i.e = approx. 1
Area of the original rectangle = 3x * x = 3x²
Area of the original rectangle = 3 * 12 = 3m²

17. The length of a rectangle wall is 3/2 times of its height. If the area of
the wall is 600 m². What is the sum of the length and height of the wall?
A) 30
B) 40
C) 50
D) 80
Answer – C
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Explanation:
length = 3x
height = 2x
Area of the wall = 3x * 2x = 6x² = 600
x = 10; Sum of the length and height of the wall = 50

18. From a rectangle ABCD of area 768 sq cm, a semi-circular part with
diameter AB and area 72π sq cm is removed. The perimeter of the
leftover portion, in cm, is?
A) 82 + 24π
B) 86 + 8π
C) 80 + 16π
D) 88 + 12π
Answer – D
Explanation:
Given that ABCD be the rectangle of area of 768 sq cm. A semi-circular
part with diameter AB and area 72π sq cm is removed
From the area given we can find the radius r as
πr2/2 = 72π
r2 = 144
r = 12 cm

The diameter AB is 24 since r = 12

So, we have a rectangle ABCD.

AB × BC = 768
24 × BC = 768
BC = 768/24
BC = 32
Now we have to find the perimeter of the leftover portion (shaded area)
Perimeter = AD + DC + BC + 1/2 perimeter of the circle
Perimeter = 32 + 24 + 32 + 2πr/2 (r = 12 cm)
Perimeter = 88 + 12π cm
The perimeter of the leftover portion, in cm, is 88 + 12π

19. Two mutually perpendicular chords AB and CD meet at a point P

inside the circle such that AP = 6 cms, PB = 4 cms and DP = 3 cms. What
is the area of the circle?
A) 125π/4 sq cms
B) 120π/4 sq cms
C) 122π/4 sq cms
D) 135π/4 sq cms

Answer – A
Explanation:
We construct following diagram to get the answer quickly,

As AB and CD are two chords that intersect at P, AP * PB = CP * PD

6 * 4 = CP * 3
CP = 8
From center O draw OM ⊥r AB and ON ⊥r CD.
From the center a line ⊥r to a chord bisects the chord.
So, we have AM = MB = 5 cm
MP = 1 cm, ON = 1 cm, CD = 11 cm, CN = 5.5 cm
ON2 + CN2 = OC2
12 + 5.52 + r2
1 + 30.25 = r2
Area = πr2
= π * 31.25
31.25π = 125π/4 sq cms

20. ABC grows potatoes in his backyard which is in the shape of a

square. Each potato takes 1 cm2 in his backyard. This year, he has been
able to grow 131 more potatoes than last year. The shape of the

backyard remained a square. How many potatoes did ABC produce this
year?
A) Data Insufficient
B) 4563
C) 4356
D) 4366
Answer – C
Explanation:
Let the area of backyard be x2 this year and y2 last year
∴ X2 - Y2 = 131
=> (X+Y) * (X-Y) = 131
Now, 131 is a prime number.
=> (X+Y) * (X-Y) = 131 x 1
=> X+Y = 131
X-Y = 1
=> 2X = 132 => X = 66
and Y = 65
∴ Number of potatoes produced this year = 662 = 4356

Practice Questions
1. Madhur has 60 m of barbed wire to be used for fencing the garden.
The rectangular garden is enclosed from one side by the compound wall
and three other sides need to be fenced using the wire. If the area of
the garden is 400 m2, what can be the dimensions of the garden?
A) 25 m x 16 m
B) 40 m x 10 m
C) 50 m x 8 m
D) 80 m x 5 m

2. Find the cost of carpeting a hexagonal room with each side of

hexagon being 15 m and the perpendicular distance between the center
and the edge is 20m, given the rate of carpeting is Rs.120 per m2?
A) Rs. 1,12,000
B) Rs. 1,08,000
C) Rs. 1,05,500
D) Rs. 1,10,000

3. Find the perimeter of a rectangle under following conditions:

The area of a square and rectangle is same. The difference between
length and breadth of the rectangle is 21m and the length of diagonal of
the square is 14√2 m.
A) 70 m
B) 84 m
C) 48 m
D) 56 m

4. The length, breadth and height of a room are in the ratio 7:2:5. If the
length and breadth are halved while the height is doubled, then the
total area of the four walls of the room will _________?
A) get doubled
B) get halved
C) remain same
D) Cannot be determined

5. There are two cones whose heights are in the ratio 1:2 and the radii
of their bases in the ratio 2:1. Find the ratio of their volumes.
A) 1:4
B) 4:1
C) 1:2
D) 2:1

6. Circumference of a circle-A is 3/2 times perimeter of a square. Area

of the square is 900 cm2. What is the area of another circle-B whose
diameter is equal to the radius of the circle-A?
A) 635.47 cm2
B) 641.50 cm2
C) 642.68 cm2
D) 651.65 cm2

7. A cow is tethered to a peg in the corner of a rectangular grass field

with a 20 m long rope. The dimensions of the field are 50 m long and 35
m wide. How much area of the field can the cow graze?
A) 314.28 m2
B) 315.51 m2
C) 310.14 m2
D) 340.27 m2

8. If a circle is inscribed in a rectangle such that it fits perfectly inside

the rectangle of length 12 cm and breadth 10 cm. Then the area of circle
is?
A) 73.81 cm2
B) 68.54 cm2
C) 70.50 cm2
D) 78.57 cm2

9. If one man takes a dip in a swimming pool, the average displacement

of water is 4 cm3. Suppose that 150 men dip in a swimming pool of
dimension 90 cm x 40 cm, how much water will be displaced?
A) 1/3rd
B) 1/5th
C) 1/4th
D) 1/6th

10. A cylindrical tank of diameter 50 cm and height 100 cm is full of

water. If 10 litres of water is drawn off from the tank, then the
remaining water level in the tank will become?
A) 93.6 cm
B) 95.5 cm
C) 94.9 cm
D) 96.4 cm

11. The length of a rectangular wall is 5/2 times of its height. If the area
of the wall is 1000 m². What is the sum of the length and height of the
wall?
A) 70 m
B) 80 m
C) 60 m
D) 50 m

12. Suppose the area of a circle is 550 cm². It’s diameter is equal to
radius of a semicircle. What is the perimeter of the semicircle?
A) 153.12 cm
B) 124.54 cm
C) 135.9 cm
D) 141.25 cm

13. One side of rectangular ground is 10 m and its diagonal is 15 m. Find

the area of ground?
A) 110.5 m2
B) 115.2 m2
C) 117.1 m2
D) 111.8 m2

14. Phil wants to go to point B from point A which are located at the
opposite corners of a square garden of side 10√2 m. He decides to take
the shortest route and runs diagonally through the garden. How much
distance did he run for?
A) 15 m
B) 17 m
C) 20 m
D) 22 m

15. A rectangular ground of 20 m length and 12 m breadth has a gravel

path 3 m wide all around it on the outside. What is the area of the
path?
A) 225 m²
B) 228 m²
C) 230 m²
D) 235 m²

16. A spherical rubber ball of radius 14 cm is cut by a cutter at a

distance of “x” cm from its centre, into 2 different pieces. What should
be the value of “x” such that the cumulative surface area of the newly
formed pieces is 3/28 more than the rubber ball’s original surface area?
A) 12.4 cm
B) 13.4 cm
C) 15.4 cm
D) 15.8 cm

18. A slice from a circular pancakes of diameter 14 inches is cut in a such

a way that each slice of pancake has a central angle of 45°. What is the
area of each slice of Pancakes (in square inches)?
A) 13.25
B) 15.25
C) 19.25
D) 18.25

19. The length and the breadth of a rectangular window are increased
by 1 m each and due to this the area of the window increased by 21 sq.
m. But if the length is increased by 1 m and breadth decreased by 1 m,
area is decreased by 5 sq. m. Find the perimeter of the window.
A) 20 m
B) 25 m
C) 35 m
D) 40 m

Answer Key
1(B) 2(B) 3(A) 4(C) 5(D)
6(C) 7(A) 8(D) 9(D) 10(C)
11(A) 12(C) 13(D) 14(C) 15(B)
16(A) 17(D) 18(C) 19(D) 20(B)

Solutions
1. Answer - B
Explanation:
Area = 400
l x b = 400 … (1)
Let the side having compound wall be the length of the rectangle
therefore,
l + 2b = 60 => b = (60 – l)/2
Substituting the value of b in equation 1
=> l x (60 - l)/2 = 400
=> 60l - l2 = 800
=> l2 - 60l + 800 = 0
=> l2 – 40l – 20l + 800 = 0
=> l(l – 40) – 20(l – 40) = 0
=> (l – 40) (l – 20) = 0
=> l = 40 or l = 20
Case i: If l = 40m
=> b = (60 – l)/2 = (60 – 40)/2 = 20/2 = 10m
Case ii: If l = 20m
=> b = (60 – l)/2 = (60 – 20)/2 = 40/2 = 20m
Hence the dimensions of the Garden can either be 40m x 10m or 20m x
20m.

2. Answer – B
Explanation:
The hexagon can be considered as 6 equal triangles, because it is a
regular hexagon.
Area of the floor = 6 x ½(bh)
=6 x ½(15 x 20) = 900 m2
Cost of carpeting = 120 x 900 =Rs. 1,08,000.

3. Answer – A
Explanation:
For square,
Diagonal = side√2
So, side of square is 14m
Area of square = 142 = 196
Therefore for rectangle,
l x b = 196 … (1)
l-b = 21 ; l = 21+b
Putting l in (1)
(21+b) x b = 196
21b + b2 = 196
b2 +21b - 196 = 0
b = 7 or -28
So, b = 7 and l = 28
So perimeter of rectangle = 2(l+b) = 2 x 35 = 70 m.

4. Answer – C
Explanation:
Length = l = 7x, Breadth = b = 2x, Height = h = 5x
Area of four walls = 2 (l + b) h = 2(7x + 2x) 5x = 90x2
Length gets halved = 7x/2, Breadth halved = x, Height doubled = 10x
New area of four walls = 2 (7x/2 + x) 10x = 90x2
Hence the area remains the same

5. Answer – D
Explanation:
Ratio of volume = (1/3πr12h1) / (1/3πr22h2)
We know that,
Height = 1 : 2
Radius = 2 : 1
Ratio of volume = ( r12h1)/( r22h2) = 4x1/1x2 =2/1
Therefore, ratio is 2:1

6. Answer – C
Explanation :
The side of the square = a = √900 = 30 cm.
Perimeter = 4 X 30 = 120 cm
The circumference of the circle = 3/2 x 120 = 22/7 x 2 x r
Radius = 45*7/11 = 28.6 cm
Diameter of Circle-B = Radius of Circle-A = 28.6 cm.
Radius of Circle-B = 28.6/2 = 14.3 cm
The area = 22/7 x 14.3 x 14.3 = 642.68 cm2

7. Answer – A
Explanation :
Area = [22*20*20/7] / 4
= 8800/28 = 314.28 m2

8. Answer – D
Explanation :
Radius = Breadth/2 = 10/2 = 5
Area of circle = 22*5*5/7 = 550/7 = 78.57 cm2

9. Answer – D
Explanation:
Total volume displaced by 150 men= 150×4 cm3
However, volume = lxbxh = 90x40xh
90x40xh = 150×4
h=150*4/90*40 = 600/3600 = 1/6.
so, the water level rises by 1/6 cm.

10. Answer – C
Explanation :
1 litres = 1000 cm3
10 litres = 10000 cm3
Volume of cylinder = πr2h = 22*25*25*100/7 = 196428.57 cm3
Remaining volume of water = 196428.57 – 10000 = 186428.57 cm3
Remaining Volume = πr2hr
186428.57 = 22*25*25*hr/7
hr = 186428.57*7/22*25*25 = 1305000/13750 = 94.9 cm

11. Answer - A
Explanation:
Area of the wall = 5x * 2x = 10x² = 1000
x = 10; Sum of the length and height of the wall = 5x+2x = 50+20 = 70 m

12. Answer – C
Explanation:
Area of a circle = 550 cm²
πr² = 550
22r2/7 = 550
r2 = 550*7/22 = 175
r = 13.22; D = 26.44
D = Radius of Semi Circle = 26.44
Perimeter of the semicircle = πr + 2r
= (22*26.44/7)+(2*26.44)
= 135.9 cm

13. Answer – D
Explanation :
d = √(l² + b²)
15 = √(l² + 10²)
l² = 15² – 10² => l2 = 225 – 100 = 125
l = 11.18
Area = 11.18 * 10 = 111.8 m²

15. Answer – B
Explanation :
Area of ground = 20 * 12 = 240
Total area (ground + path) = (20 + 6)*(12 + 6) = 468
Area of path = 468 – 240 = 228 m²

16. Answer – A
Explanation:

Area of the ball = 4πr2

= 4π * 14 * 14
= 784π
Cumulative area of the 2 pieces = 3/28 more than the ball's surface area.
Therefore, extra area = 3/28 * 784π = 84π
Now this extra area = areas of 2 new circles of those 2 pieces
Let area of each new circle = πr12
πr12 = 84π / 2 = 42π
Or r12 = 42
Now, r1, x and r form a right-angled triangle.

r2 = x2 + r12
x2 = 14*14 - 42
= 196 – 42 = 154
x = 12.4 cm

18. Answer – C
Explanation:
Diameter = 14
R = D/2 = 14/2 =7
Area of each slice of Pancake =πr² * Θ/360°
= (22/7) * 7 * 7 * (45°/360°)
=19.25

19. Answer – D
Explanation:
Let original length = l, breadth = b, so area = lb
When l and b increased by 1:
(l+1) (b+1) = lb + 21
Solve, l + b = 20
When l increased by 1, b decreased by 1:
(l+1) (b-1) = lb – 5
Solve, l – b = 4
Now solve both equations, l = 12, b = 8
Perimeter = 2(12 + 8) = 40 m

What are CATKing Bible LOD 1 Books?

CATKing Bible LOD 1 Books are specially designed books which are useful in getting
students Boosted Up and Ready for All Management Entrance Tests (CAT/ CET/
NMAT/ CMAT/ SNAP/ TISSNET/ MICAT/ IIFT). These books cover all the topics and
are distributed section wise: Quantitative Ability, Data Interpretation, Logical
Reasoning and Verbal Ability. They are recommended for all students who wish to
get their Basics clear in any section for any Management Entrance Test.
Understand the Basics concepts from the theory section, to better relate to all
video lessons and live classes.

How to make best use of CATKing Bible LOD 1 Books?

i. Attend the CATKing Concept Builder Classes to gain an idea of what all are the
basic pointers of the chapters.
ii. Go through that chapter in the CATKing Bible LOD 1 Books and read all the
Theory and Formulae provided in the Introduction of the chapter.
iii. Make a note of all the formulae and try to learn them by Heart, this will be your
ready reckoner to revise before exams.
iv. Go through all the Solved Examples and solve them simultaneously while
referring the solutions provided to understand the best way to solve each type of
questions.
v. Solve all the Practice Questions provided on your own and then refer to the
solutions at the end so as to verify if you have solved the questions correctly or is
there a better smarter approach for the same question.
vi. If you are able to solve majority questions correctly, then move to the next step
of preparation by taking the Topicwise Tests.
vii. Once you are done with good set of 4-5 Topics, give the Sectional and Full
length Mocks and see where you stand.
viii. While you analyze the mocks see to it that you highlight your weak and strong
areas. For all weak areas, you must get back to these CATKing Bible LOD 1 Books,
read up your formulae, check out the Solved Exams and your concepts will be
revised.

Table of Contents

Introduction 1
Solved Examples 10
Practice Questions 24
Answer Key 36
Solutions 35

Let’s get started..

Introduction
Lines and Angles:
• Sum of all angles on a straight line is 180°.
• Vertically opposite angles are congruent (equal).
• If any point is equidistant from the end points of a segment, then it
must lie on a perpendicular bisector.
• When 2 parallel lines are intersected by a transversal, corresponding
angles are equal, alternate angles are equal and co-interior angles
supplementary. (All acute angles formed are equal to each other and
all obtuse angles are equal to each other)
Concept: The ratio of intercepts formed by a transversal intersecting 3
parallel lines is equal to the ratio of corresponding intercepts formed by
any other transversal.

ab=cd=ef

Triangles:
• Sum of interior angles of a triangle is 180° and sum of exterior angles is
360°.
• Exterior Angle= Sum of remote interior angles.
• Sum of two sides is always greater than the third side and the
difference between the two sides is always lesser than the third side.
• Side opposite to the biggest angle is longest and the side opposite to
the smallest angle is shortest.

Area of a triangle:

!
Area = x base x height
"
!
Area = x Product of sides x Sine of included angle
"
#$%$&
Area = √s(s-a) x (s-b) x (s-c), where s (semi-perimeter)=
"
Area = r x s [r is the radius of the incircle)
#(%(&
Area = [R is radius of circumcircle]
)*

A Median of a triangle is a line segment joining a vertex to the midpoint

of the opposing side. The three medians intersect in a single point, called
the Centroid of the triangle. Centroid divides the median in the ratio of
2:1

An Altitude of a triangle is a straight line through a vertex and

perpendicular to the opposite side or an extension of the opposite side.
The three altitudes intersect in a single point, called the Orthocenter of
the triangle.

A Perpendicular Bisector is a line that forms a right angle with one of the
triangle's sides and intersects that side at its midpoint. The three
perpendicular bisectors intersect in a single point, called the
Circumcentre of the triangle. It is the center of the circumcircle which
passes through all the vertices of the triangle.

An Angle Bisector is a line that divides the angle at one of the vertices in
two equal parts. The three angle bisectors intersect in a single point,
called the Incentre of the triangle. It is the center of the incircle which
touches all sides of a triangle.

Points to note:
Centroid and incentre will always lie inside the triangle.
• For an acute angled triangle, the Circumcenter and the Orthocenter

will lie inside the triangle.

• For an obtuse angled triangle, the Circumcenter and the Orthocenter
will lie outside the triangle.

• For a right-angled triangle, the Circumcenter will lie at the mid-point

of the hypotenuse and the orthocenter will lie at the vertex at which
the angle is 90°
Points to note:
The orthocenter, centroid, and circumcenter always lie on the same line
known as Euler line.
• The orthocenter is twice as far from the centroid as the circumcenter

is.
• If the triangle is isosceles, then the incentre lies on the same line.

• If the triangle is equilateral, all four are the same point.

Theorems:
Mid-Point Theorem:

The line joining the midpoint of any two sides is parallel to the third side
and is half the length of the third side.
Basic proportionality theorem:

If DE∥BC. Then:
+, +.
=
,- ./

Apollonius’ Theorem:

AB + AC = 2 (AD + BD )
2 2 2 2

Interior Angle Bisector Theorem:

AE BA
=
ED BD

Special Triangles:
1) Right angled triangle:

ΔABC ∼ ΔADB ∼ ΔBDC

BD2= AD x DC and AB x BC=BD x AC

2) Equilateral Triangle:

All angles are equal to 60°. All sides are also equal.
√1
a) Height = x side
"
√1
b) Area = x side2
)
!
c) Inradius = x Height
1
"
d) Circumradius = x Height
1

3) Isosceles Triangle:

Angles opposite to equal sides are equal.

&
Area = x √(4a2-c2)
)

4) 30°-60°-90° Triangle:

√1
Area = x (X)2
"

5) 45°-45°-90° Triangle

Area = (X) 2/2

6) 30°-30°-120° Triangle:

√1
Area = x (X)2
)

Similarity of triangles:
Two triangles are similar if their corresponding angles are congruent and
corresponding sides are in proportion.
Tests of similarity:
1. AA
2. SSS
3. SAS
For similar triangles, if the sides are in the ratio a:b
• Corresponding heights are in the ratio a:b

• Corresponding medians are in the ratio a:b

• Circumradii are in the ratio a:b

• Inradii are in the ratio a:b

• Perimeters are in the ratio a:b

• Areas are in the ratio a :b

2 2

Solved Examples
Q.1) Find the area of a triangle whose sides are 5 cms, 6 cms and 7 cms.
a) 6 sq. cms
b) 6√𝟑𝐬𝐪. 𝐜𝐦𝐬
c) 6√𝟔𝐬𝐪. 𝐜𝐦𝐬
d) 6√𝟐𝐬𝐪. 𝐜𝐦𝐬

ANSWER: C
Area of triangle = √s(s − a)(s − b)(s − c)
Where a, b and c are the sides of the triangle and
#$%$&
s= is the semi perimeter of the triangle.
"
a = 5, b = 5, c = 7
2$3$4
s= = 9 cm
"
Area of triangle = √9(9 − 5)(9 − 6)(9 − 7)
= √9(4)(3)(2)
= 6√6 cm
Hence, [C].

Q.2) In ∆PQR, QS and RS are angle bisectors. If ∠QPR= 80o, find the
measure of ∠𝐐𝐒𝐑. (In degrees)

a) 90
b) 75
c) 120
d) 130
ANSWER: D
In ∆PQR, ∠P + ∠Q + ∠R = 180. Therefore, ∠Q + ∠R = 100.
Thus, ∠SQR + ∠SRQ = 50.
Therefore, ∠QSR = 130o
Hence, [D].

Q.3) All possible obtuse-angled triangles with sides of integer lengths

are constructed, such that two of the sides have lengths 7 and 14. How
many such triangles exist?
a) 7
b) 9
c) 10
d) 11

ANSWER: C
Let x be the length of the third side. So, we have x < 7 + 14 and x + 7 > 14.
Therefore, 7 < x < 21.
Now for the triangle to be obtuse,
Case 1:
x2 + 72 < 142
This gives possible values as 8, 9, 10, 11 and 12.
Case 2:
72 + 142 < x2
This gives possible values as 16, 17, 18, 19 and 20.
Therefore, total 10 such triangles are possible.
Hence, [C].

Q.4) Find the area of triangle XYZ where XY = 17.29, XZ = 18 and YK = 12,
where YK is the altitude.
a) 90
b) 108
c) 120
d) 135
ANSWER: B

!
Area= x 18 x 12= 108
"
Hence, [B].

Q.5) Find the area of triangle PQR where, PQ = 15, PR = 15, QR=18 and
Median PK = 12.
a) 90
b) 108
c) 120
d) 135
ANSWER: B
In an isosceles triangle, median on a non-equal side= Altitude

!
Therefore, Area= x 18 x 12= 108 .
"
Hence, [B].

Q.6) In triangle QPR, QS is an altitude of the side PR, which meets side
PR in S such that P – S – R. Suppose side PQ = 8, side QR = 12 and angle
R = 30 degrees. Calculate lengths of QS, SR and PS respectively.
a) 𝟔√𝟑, 𝟔, 𝟐√𝟕
b) 6, 6√𝟑, 𝟐√𝟕
c) 𝟐√𝟕, 𝟔, 𝟔√𝟑
d) 𝟐√𝟕, 𝟔√𝟑, 𝟔

ANSWER: B
QS= 6 (side opposite to 30o is half of side opposite to 90o in a 30o-60o-90o
triangle)

SR= 6√3 (side opposite to 60o is √3 times the side opposite to 30o)
PS= √PQ2-QS2 = √82-62
= 2√7,
Hence [B].

Q.7) Calculate area of parallelogram PQRS (Double Triangle), if SR = 30

cm, PS = 18 and angle Q = 60°.
a) 270√𝟑.
b) 540
c) 270
𝟐𝟕𝟎
d)
√𝟑

ANSWER: A
Area of parallelogram = a x b x sin𝜃
Where a & b are the sides & 𝜃 is the angle between these sides.
Area of parallelogram, PQRS= 30 x 18 x sin 60o
√1
= 30 x 18 x = 270√3.
"
Hence, [A].

Q.8) ABC is a right-angled triangle with ∠𝐁 = 90o, AB = 21 units and BC =

20 units, then find the circumradius.
a) 13.5 units
b) 14.5 units
c) 14 units
d) 15 units
ANSWER: B

By Pythagoras theorem,
CA2 = AB2 + BC2
= 212 + 202= 441 + 400 = 841 = 29 units.
Length of Hypotenuse (CA) = 29 units.
In a right-angled triangle, circumcentre is the midpoint of the
hypotenuse.
!
Therefore, circumradius= x 29 = 14.5 units.
"
Hence, [B].

Q.9) In PQR, PQ = 8 cms, PR = 6 cms and QR = 10 cms. Find the length

of the median PS.
a) 5
b) 15
c) 10
d) None of the above

ANSWER: A

Since PS is the median

Therefore QS = SR = 5 cm
By Apollonius theorem,
82 + 62 = 2(PS2 + 52)
50 = PS2 + 25
PS2= 25, PS = 5 cm
Hence, [A].

Q.10) AE is the median of ∆𝐀𝐁𝐂. If the area of ∆𝐀𝐁𝐂 is 40 sq. cm, then
find the area of triangle ∆𝐀𝐁𝐄.
a) 10 sq cm
b) 20 sq cm
c) 40 sq cm
d) 80 sq cm

ANSWER: B
AE is the median of ∆ABC.
Area of ∆ABE = Area of ∆AEC
!
= x Area of ∆ABC = Area of ∆ABE
"
!
= x 40 sq cm = 20 sq cm.
"
Hence, [B].

Q.11) In ABC, DE || BC and AB = 5 cms, BC = 10 cms and AD = 3 cms.

Find (DE).
a) 4
b) 2
c) 8
d) 6

ANSWER: D

∠DAE ≅ ∠BAC ( Common Angle)

∵ DE ∥ BC
∠ADE = ∠ABC (Corresponding angle)
+, ,. 1 ,.
∴ = ⇒ =
+- -/ 2 !9
1
∴ DE = x 10= 6 cm.
2
Hence, [D].

Q.12) In an equilateral triangle the circumradius is 12cm. Find out the

area of the triangle.
a) 216√𝟑
b) 108√𝟑
c) 100√𝟑
𝟏𝟎𝟖
d)
√𝟑

ANSWER: B

From the above figure

1
AD = x 12= 18 cm.
"
From the property of equilateral triangles
√1
AD = a, where ‘a’ is the side of the triangle.
"
"
Hence, a = x 18 = 12√3cm.
√1
√1
Area of an equilateral triangle = x a2
)
√1
x (12√3)2 = 108√3 sq cm.
)
Hence, [B].

Q.13) A triangle of area 60 sq. cms has two of its sides as 8 cms and 15
cms. If the inradius of the triangle is 3 cms, then find the circumradius.
a) 17 cms
b) 8.5 cms
c) 9.5 cms
d) 10 cms

ANSWER: B
Since it is a right angle triangle with the perpendicular sides 8 and 15.
Thus, hypotenuse = 17.
Circumradius of a right angle triangle is half of its hypotenuse. Thus,
circumradius = 8.5 cms.
Hence, [B].

Q.14) If the area of an equilateral triangle is 25√𝟑 sq cm, then find the
inradius of the triangle.
𝟓
a) 𝐜𝐦
√𝟑
b) 𝟓√𝟑 cm
𝟓
c) 𝐜𝐦
𝟑
d) 15 cm

ANSWER: A
√1
Area of an equilateral triangle = x a2, where ‘a’ is the side of the
)
triangle.
√1
∴ x a2= 25√3
)
Side = 10 cm
+=># ?@ A=B#CDE>
Inradius =
F
1 ( GBH> 1 ( !9
= = = 15 cm.
" "
+=># ?@ I=B#CDE> "2√1 2
Inradius = = = cm.
F !2 √1
Hence, [A].

Q.15) In the figure, AD is the external bisector of EAC, which intersects

BC produced in D. If AB = 12 units, AC = 6 units and CD = 8 units, find BC.

a) 4 units
b) 6 units
c) 8 units
d) 3 units

ANSWER: C
By exterior angle bisector theorem,
AB BD
=
AC DC
Let BC = x
12 x + 8
=
6 8
x = 8 units.
Hence, [C].

Q.16) The sides of ABC are 8 cms, 10 cms and 12 cms with the smallest
angle ‘C’. Find the length of the altitude from the vertex C.
a) 15 √𝟑 𝐜𝐦
𝟏𝟓
b) √𝟕 cm
𝟒
c) 15 cm
d) 4√𝟕 cm

ANSWER: B

Let the height of ∆ABC be h = CD

!
Area of ∆ABC = x Base x Height
"
!
= x 8 x h = 4h -------- i)
"
Area = √s(s − a)(s − b)(s − c)
#$%$&
s= is the semi perimeter of the triangle.
"
K$!9$!"
= = 15 cm.
"
= √15(15 − 8)(15 − 10)(15 − 12)
= 15√7 sq cm -------- ii)
From i) and ii),
4h = 15√7
!2
h = √7 cm
)
Hence, [B].

Q.18) Choose the correct alternative.

In ∆𝐏𝐐𝐑, PX = PY, RY = RZ and ∠𝐏𝐐𝐑 = 90o
If ∠𝐘𝐗𝐐 = 𝟏𝟎𝟓o , find the measurement of ∠𝐘𝐙𝐐(𝐢𝐧 𝐝𝐞𝐠𝐫𝐞𝐞𝐬).

a) 105
b) 120
c) 135
d) 90

ANSWER: B
Let ∠PXY = ∠PYX = a and ∠RYZ = ∠RZY = b
P+ Q + R = 180o. Therefore, P + R = 90o
P + 2a = 180o and R + 2b = 180o
Therefore, 2a + 2b + P + R = 360o. Therefore, a + b = 135o.
But a + b + ∠XYZ = 180o
Therefore, ∠XYZ = 45o
Now in quadrilateral XYZQ, ∠XQZ + ∠YXQ + ∠YZQ + ∠XYZ = 360o
Therefore, ∠YZQ= 120o
Hence, [B].

Q.19) AB = AC = 15. BC = 18. AD ⊥ BC. F is the midpoint of AC and DE =

9. What is the length of segment DF?

a) 9
b) 7.5
c) 12
d) None of these

ANSWER: B
D is the midpoint of BC and F is the midpoint of AC. Also, ∆CFD and
∆CAB are similar.
Therefore, DF = 0.5 x AB = 7.5
Hence, [B].

Q.20) Find the area of triangle XYZ where XY + YZ + XZ = 28 and in-radius

= 4.
a) 45
b) 51
c) 56
d) 60

ANSWER: C

Area of triangle = Inradius x semi-perimeter

"K
= 4 x = 56
"
Hence, [C].

Practice Questions

Q.1) Find the area of triangle PQR where. PQ = QR = 15 and PR = 24 and

circumradius = 12.5
a) 90
b) 102
c) 108
d) 112

Q.2) Choose the correct alternative.

In the given figure, CE = 7 cm, AF = 5 cm, CF = 10 cm. Find the area of
ABCD.

a) 160 sq. cm
b) 270 sq. cm
c) 200 sq. cm
d) 240 sq. cm

Q.3) ABC is a triangle and AD, BE and CF are the medians from the
vertices A, B and C, respectively. If the area of ∆𝐀𝐁𝐂 is 72, then find the
sum of areas of the following:
∆𝐀𝐎𝐁, ∆𝐀𝐁𝐃, ∆𝐀𝐎𝐅, ∆𝐀𝐄𝐅, ∆𝐃𝐄𝐅, ∆𝐀𝐄𝐃, ∆𝐄𝐎𝐅 𝐚𝐧𝐝 𝐪𝐮𝐚𝐝𝐫𝐢𝐥𝐚𝐭𝐞𝐫𝐚𝐥 𝐀𝐄𝐎𝐅.
(O is the centroid of the triangle ABC.)
a) 145
b) 152
c) 156
d) 160

Q.4) a & b are the lengths of the base & height of a right angled triangle
whose hypotenuse is ‘h’. If the value of a & b are positive integers,
which of the following cannot be a value of the square of the
hypotenuse?
a) 13
b) 23
c) 37
d) 41

Q.5) In a triangle PQR, angle P = 60°. PS is an angle bisector of angle P.

Side PQ = 12 cm and side PR = 20 cms. Calculate the length of PS.
𝟏𝟎√𝟑
a)
𝟐
𝟏𝟓√𝟑
b)
𝟐
c) 𝟓√𝟑
𝟐𝟎√𝟑
d)
𝟐

Q.6) Choose the correct alternative.

In the given figure, ∠EDB = 115o, ∠BCE = 65o, AC = 15 units, EC = 10
units and AD = 5 units. Find BD.

a) 10
c) 15
c) 20
d) 25

Q.7) Choose the correct alternative.

If in triangle ABC, AD is the median, AB = 6 cms, AC = 8 cms and AD = 5
cms, then find the area of ABD.
a) 11 sq. cms
b) 12 sq. cms
c) 24 sq. cms
d) 16 sq. cms

Q.8) Choose the correct alternative.

In ∆𝐀𝐁𝐂, AD is the altitude. P is the midpoint of BD. Q is the midpoint
of AC. BC= 12 and AD = 16.

Find the length of segment PQ.

a) 8
b) 10
c) 12
d) 15

Q.9) Choose the correct alternative.

XY = XZ. XC ⊥ YZ. ∆𝐀𝐁𝐂 is an equilateral triangle. V is midpoint of XZ.

AB = XB.
Find the measurement of ∠𝐁𝐗𝐀. (in degrees)

a) 30
b) 45
c) 36
d) 42

Q.10) Find the measurement of ∠LOK.

AB‖CD. (Angle measurements are in degrees)

a) 18
b) 24
c) 54
d) 108

Q.11) A, B and C are the measures of angles of ΔABC. Also, a, b and c are
lengths of the triangle.

If B = 110° and C = 35°, then what kind of triangle is ΔABC?

a) Equilateral
b) Obtuse
c) Isosceles
d) Both 2 and 3

Q.12) In a ΔABC, measurements of sides of the triangle are 10, 24 and

26. What kind of a triangle is ΔABC?
a) Isosceles
b) Right-angled
c) Scalene
d) Cannot be determined

Q.13) In the given figure, PQ‖EF. Then find x

a) 120⁰
b) 180⁰
c) 200⁰
d) 240⁰

Q.14) How many triangles can be formed by joining the vertices of a

regular octagon such that at least one side of the triangle is same as the
side of the octagon?
a) 40
b) 32
c) 36
d) 48

Q.15) ABCD is a square with point P on side AB such that AP:PB = 1:3. X
is the point of intersection of segments PC and BD. If area of triangle
BXC is 24 sq. units, what is area of quadrilateral APXD? (in sq. units)
a) 24
b) 38
c) 19
d) Data insufficient

Q.16) The three sides of a right angled triangle have integral lengths and
also form arithmetic progression. A possible length of one of the sides
is:
a) 22
b) 91
c) 82
d) 56

Q.17) A regular pentagon ABCDE is fitted inside a regular hexagon

APQRST such that P-B-Q, Q-C-R, R-D-S and S-E-T. Find ∠SED
a) 18
b) 30
c) 42
d) 54

Q.18) ∆ABC is right-angled at B with AB = 6 cm and BC = 8 cm. A square

PQRS is inscribed in ∆ABC such that the point P and Q lie on AC, point S
on AB and point R on BC. Find the approximate length of the sides of
the square PQRS.
a) 4.6 cm
b) 3.2 cm
c) 4.2 cm
d) 3.8 cm

Q.19) Two chords of length 12 cm and 16 cm, of a circle of radius 10 cm,

are perpendicular to each other. Find the distance (in cm) between the
midpoints of these two chords.
a) 18
b) 6
c) 8
d) 10

Q.20) What is the ratio of the sides AB to AC to CB in an isosceles right–

angled triangle ABC, right–angled at A?
a) 𝟏: 𝟏: √𝟐
b) 𝟏: √𝟑: 𝟐
c) √𝟐: 𝟏: 𝟏
d) 𝟏: 𝟐: √𝟑

Answer Key
1(C) 2(B) 3(C) 4(B) 5(B)
6(A) 7(B) 8(B) 9(A) 10(C)
11(D) 12(B) 13(D) 14(A) 15(B)
16(D) 17(A) 18(B) 19(D) 20(A)

Solutions
Q.1) ANSWER: C
#(%(&
Area of triangle =
)*
We know that, a = 15, b =15, c = 24, R = 12.5
!2 ( !2 ( ")
Area of triangle =
) ( !".2
Hence, 108.

Q.2) ANSWER: B
!
Area of ABCD = x (diagonal) x (sum of perpendiculars)
"
Where DE = √CD2-CE2
(∵ ∆CDE is a right-angled triangle)
= √252-72 = 24 cm
BF = √AB2-AF2
(∵ ∆ADF is a right angled triangle)
√169 − 25 = 12 cm
!
Area of ABCD = x (15) x (12 + 24)
"
= 270 sq. cm
Q.3) ANSWER: C

!
The area of 6 small triangles formed = x 72 = 12
3
After joining E to F, F to D and E to D, we get four congruent triangles.
!
Therefore, (AEF) = (ECD) = (DBF) = (EFD) = x 72 = 18
)
!
Area of ∆AOB = x 72 = 24
1
!
Area of ∆ABD = x 72 = 36
"
!
Area of ∆AOF = x 72 = 12
3
!
Area of ∆AEF = x 72 = 18
)
!
Area of ∆DEF = x 72 = 18
)
!
Area of ∆AED = x 72 = 24
)
! !
Area of ∆EOF = x area of ∆DEF = x 72 = 6
1 !"
!
Area of AEOF = x 72 = 24
1
24 + 36 + 12 + 18 + 18 + 18 + 6 + 24 = 156.

Q.4) ANSWER: B

The value of the square of the hypotenuse = h2 = a2 + b2. As the problem

states that ‘a & b’ are positive integers’ the values of a2 and b2 will have
to be perfect squares. Hence, we need to find out the value amongst the
four answer choices which can’t be expressed as sum of two perfect
squares.
Choice 1 is 13 = 9 + 4 = 32 + 22, so not the answer.
Choice 2 is 23 cannot be expressed as a sum of two numbers each of
which in turn happen to be perfect squares.
Therefore, choice 2 is the answer.
Choice 3 is 37 = 36 + 1 = 62 + 12
Choice 4 is 41 = 16 + 25 = 42 + 52 .
Hence, [B].

Q.5) ANSWER: B

Area of ∆PQR= Area of ∆PQS + ∆PSR

! ! !
x PQ x PR x sin60o = x PQ x PS x sin30o + x PS x PR x sin30o
" " "
! √1 ! ! ! !
x 12 x 20 x = x 12 x PS x + x PS x 20 x
" " " " " "
60√3 = 3PS + 5PS
60√3 = 8PS
39√1 !2√1
PS = =
K "
Hence, [B].
Q.6) ANSWER: A
∠ADE = 180 − 115 = 65o [Linear pair]
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Triangles
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∠ADE = ∠ACB
And ∠DAE = ∠BAC
⟹ ∆DAE~∆CAB [AA test]
AD AE
∴ =
AC AB
2 2
= [AE = AC - EC]
!2 +-
⟹ DB = 15 – 5 = 10 units.
Hence, [A].

Q.7) ANSWER: B
By Apollonius theorem,

(AB)2 + (AC)2 = 2[(AD)2 + (BD)2]

(6)2 + (8)2 = 2[(5)2 + (BD)2]
(BD)2 = 25 ⟹ BD = 5
Therefore, the sides of ∆ABD are 6, 5, 5
3$2$2
s= =8
"
Area = √s(s − a)(s − b)(s − c)
√8(2)(3)(3) = 12 sq cm.
Hence, [B].

Q.8) ANSWER: B
Drop a perpendicular from Q on DC. Let that point be T. Considering
∆ADC, T will be midpoint of segment DC. Therefore, PT = 0.5 x BC = 6

Also, QT = 0.5 x AD = 8
In right angled triangle QTP, QT = 8 and TP = 6. Therefore, PQ = 10
Hence, [B].

Q.9) ANSWER: A
∠ABC = 60o. Therefore, ∠ABX = 120o.
But ∆ABX is an isosceles triangle as AB = XB.
∴ ∠BAX = ∠BXA = 30o
Hence, [A].

Q.10) ANSWER: C
10x = 180. Therefore, x = 18°
∴∠LOK=3x =3*18
∴ ∠LOK = 54⁰.
Hence, [C].

Q.11) ANSWER: D
As B=110 > 90⁰, ΔABC is an obtuse triangle.
Also, A+B+C = A+110+35 = 180.
∴ A=C=35
∴ ΔABC is also an isosceles triangle.
Hence, [D].

Q.12) ANSWER: B
102 + 242 = 100 + 576 = 676 = 262
Thus, 10, 24 and 26 is a Pythagorean triplet. Therefore, the given triangle
is a right-angled triangle.
Hence, [B].

PQ∥EF
Now, construct a line GH, which is parallel to both the lines, PQ and EF.
∵PQ∥GH and AB is transversal, then
∠GBA + ∠BAC = 180o (Interior angles)
∠GBA = 180o-50o = 130o----------------------i)
∵ GH ∥ EF and BC is transversal, then
∠GBC + ∠BCE = 180o (Interior angles)
∠GBC = 180o-70o = 110o ----------------------ii)
From i) and ii)
∠GBA + ∠GBC = 240o
x = 240o
Hence, [D].

Q.14) ANSWER: A
At least one side of the triangle is same as the side of the octagon.
Therefore we have to consider two cases: first is when only one side of
the triangle is same as the side of the octagon and the second is when

two sides of the triangle are same as the sides of the octagon.
Case 1: When only one side of the triangle is same as the side of the
octagon
Number of triangles that can be formed: 8 × 4 = 32
Case 2: When two sides of the triangle are same as the sides of the
octagon
Total number of ways in which two adjacent sides of an octagon can be
chosen is: 8
Therefore total number of triangles that can be formed: 32 + 8=40.
Hence, [A].

Q.15) ANSWER: B

∆BXP~∆DXC
BX PB 3
∴ = =
DX CD 4
Perpendicular dropped from point C on the line BD will be common
height for triangles BXC and DXC.
Therefore, area of triangles BXC and DXC will be in ratio of their bases,
i.e. BX and DX.
Area of ΔDXC DX
=
Area of ΔBXC BX
)
Area of ΔDXC = x 24= 32 sq. units.
1
Since, ∆BXP~∆DXC
+=># ?@ M-NO -N
= ( )2
+=># ?@ M,N/ ,N
1
Area of ΔBXP = ( )2 x 32 = 18 sq. units.
)

Now, Area of ΔABD = Area of ΔBCD = 32 + 24 = 56 sq. units.

Area of quadrilateral APXD = Area of ΔABD – Area of ΔBXP
= 56 – 18 = 38 sq. units.
Hence, [B].

Q.16) ANSWER: D
Suppose three sides are (a – x), a and (a + x). Clearly, the length of the
hypotenuse : a + x
Therefore, (a + x)2 = (a – x)2 + a2
⇒ a = 4x
Thus, sides of triangle are 3x, 4x and 5x, respectively. This is a multiple of
basic Pythagorean triplet of 3, 4 and 5.
Out of the given alternatives, only 56 is divisible by either 3, 4 0r 5.
Hence, [D].

Q.17) ANSWER: A

Since, a regular pentagon is fitted inside a regular hexagon.

ΔDRC will be isosceles triangle.

∠RDC = 180-120/2 = 30⁰
m ∠SDE = 180-108-30= 42⁰
m ∠SED = 180-120-42= 18⁰
Hence, [A].

Q.18) ANSWER: B
Based on the given information, we get the following figure:

AB = 6 cm and BC = 8 cm, AC = 10 cm (Pythagorean triplet)

Since, SP is perpendicular to AP.
ΔAPS~ΔABC (by AA test)
+O +- 3 1
= = ⇒ AP = PS
OF -/ K )
Now, RQ is perpendicular to QC
Therefore, ΔCQR~ΔCBA (by AA test)
/P /- K )
= = ⇒ CQ = RQ
*P +- 3 1
Let the side of square PQRS be x cm.
Now, AC = AP + PQ + QC
3 4
⇒ 10 = x + x + x
4 3
37
⇒ 10 = x
12
⇒ x ≈ 3.2cm
Hence, [B].

Let P be the centre of the circle. M & N are mid-points of chords AB & CD
respectively.

AB=16 cm & CD=12 cm

In ΔPMA, PM2 = PA2 – AM2
= 100 – 64 = 36 cm
In ΔPNC, PN2 = PC2 – CN2
= 100-36 = 64 cm
ΔPMN is right angled triangle
∴ MN2 = PM2 + PN2 = 36 + 64 = 100
∴ MN = 10 cm
Hence, [D].

Q.20) ANSWER: A
In an isosceles right-angled triangle ABC,
∠A = 90o and ∠B = ∠C = 45o
Therefore, AB = AC
Let AC = AC = 1 cm
∴ BC2 = AB2 + AC2 = 1 + 1 = 2
BC = √2
∴ AB ∶ AC ∶ CB = 1 ∶ 1: √2.
Hence, [A].

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Tests (CAT/ CET/ NMAT/ CMAT/ SNAP/ TISSNET/ MICAT/ IIFT). These books
cover all the topics and are distributed section wise: Quantitative Ability,
Data Interpretation, Logical Reasoning and Verbal Ability. They are
recommended for all students who wish to get their Basics clear in any
section for any Management Entrance Test.

How to make best use of CATKing Bible LOD 1 Books?

i. Attend the CATKing Concept Builder Classes to gain an idea of what all
are the basic pointers of the chapters.
ii. Go through that chapter in the CATKing Bible LOD 1 Books and read all
the Theory and Formulae provided in the Introduction of the chapter.
iii. Make a note of all the formulae and try to learn them by Heart, this will
be your ready reckoner to revise before exams.
iv. Go through all the Solved Examples and solve them simultaneously
while referring the solutions provided to understand the best way to solve
each type of questions.
v. Solve all the Practice Questions provided on your own and then refer to
the solutions at the end so as to verify if you have solved the questions
correctly or is there a better smarter approach for the same question.
vi. If you are able to solve majority questions correctly, then move to the
next step of preparation by taking the Topicwise Tests.
vii. Once you are done with good set of 4-5 Topics, give the Sectional and
Full length Mocks and see where you stand.
viii. While you analyze the mocks see to it that you highlight your weak and
strong areas. For all weak areas, you must get back to these CATKing Bible
LOD 1 Books, read up your formulae, check out the Solved Exams and your
concepts will be revised.
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Trigonometry
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Table of Contents

Introduction 1
Solved Examples 3
Practice Questions 20
Answer Key 26
Solutions 26

Let’s get started...

Introduction
The word ‘trigonometry’ is derived from the Greek words ‘tri’ (meaning
three), ‘gon’ (meaning sides) and ‘metron’ (meaning measure). In fact,
trigonometry is the study of relationships between the sides and angles of
a triangle. The earliest known work on trigonometry was recorded in Egypt
and Babylon. Early astronomers used it to find out the distances of the
stars and planets from the Earth. Even today, most of the technologically
advanced methods used in Engineering and Physical Sciences are based on
trigonometrical concepts.
As we know, the three sides of a right-angled triangle are:
Ø Base: The side that is horizontal to the plane.
Ø Perpendicular: The side making an angle of 90 degree with
the Base.
Ø Hypotenuse: The longest side of the triangle.

The Trigonometric Ratios are:

sin θ = Opposite Side/Hypotenuse
sec θ = Hypotenuse/Adjacent Side
cos θ = Adjacent Side/Hypotenuse
tan θ = Opposite Side/Adjacent Side
cosec θ = Hypotenuse/Opposite Side
cot θ = Adjacent Side/Opposite Side

The Reciprocal Identities are:

cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ
sin θ = 1/cosec θ
cos θ = 1/sec θ
tan θ = 1/cot θ

Trigonometric Table:
It’s a table that you can refer to if you are not so sure about the values of
different angles. Below is the table for trigonometry formulas of different
angles which are commonly used for solving an ample number of
problems.
Angles (In
0° 30° 45° 60° 90° 180° 270° 360°
Degrees)
Angles (In
0° π/6 π/4 π/3 π/2 π 3π/2 2π
Radians)
sin 0 1/2 1/√2 √3/2 1 0 -1 0
cos 1 √3/2 1/√2 1/2 0 -1 0 1
tan 0 1/√3 1 √3 ∞ 0 ∞ 0
cot ∞ √3 1 1/√3 0 ∞ 0 ∞
cosec ∞ 2 √2 2/√3 1 ∞ -1 ∞

sec 1 2/√3 √2 2 ∞ -1 ∞ 1

Solved Examples
Q.1. What is the maximum value of x if 99 cos B + x ≤ 20 sin B?
A) – 101
B) 101
C) – 100
D) – 25
Answer – A
Explanation:
We have Cos(A ± B) = CosACosB ∓ SinASinB and
−1 ≤ Cosθ ≤ 1
99 20
If CosA = then SinA =
101 101
−1 ≤ Cos(A + B) ≤ 1
99 20
−1 ≤ CosB − SinB ≤ 1
101 101
−101 ≤ 99CosB − 20SinB ≤ 101
99CosB − 101 ≤ 20SinB
Therefore x = −101

Q.2. For how many values of x ∈ (−10π, 10π) does the equation
Sin!"!# x − Sin! x = Cos ! x − Cos !"!# x?
A) 32
B) 36
C) 39
D) 40

Answer- C
Explanation:
Sin!"!# x − Sin! x = Cos ! x − Cos !"!# x
Sin!"!# x + Cos !"!# x = Sin! x + Cos ! x = 1
We know that Sin! x + Cos ! x = 1 for all values of x.
Also, if Cosx < 1 then Cos !"!# x < Cos ! x and If Sinx < 1 then
Sin!"!# x < Sin! x.
Therefore, only possibility for Sin!"!# x + Cos !"!# x = 1 is either
Sin!"!# x = 1 or Cos !"!# = 1.
Now, Cosx = 1 for x = −9π, −8π, … ,0, π, … ,9π and at these 19 values of
x, Sinx = 0.
Also, Sinx = 1 for x = −9.5π, −8.5π, … 0.5π, 1.5π, . . ,9.5π and at these
20 values of x, Cosx = 0.
Therefore, at total at 19+20=39 values of x the given equations will be
hold.

Q.3. For the real number x what is the value of

Max{5Sinx + 12Cosx − 15} − Min{11Cosx − 60Sinx + 70}?
A) 11
B) 7
C) 13
D) -11

Answer- D
Explanation:
5 12
5Sinx + 12Cosx − 15 = 13 R Sinx + CosxS − 15
13 13
= 13TSin(A + x)U − 15
#! %
Where SinA = , CosA =
#$ #$
13TSin(A + x)U − 15 ≤ 13(1) − 15 = −2
Max{5Sinx + 12Cosx − 15} = −2
11 60
11Cosx − 60Sinx + 70 = 61 R Cosx − SinxS + 70
61 61
11 60
= 61TSin(A − x)U + 70 where SinA = and CosA =
61 61
61(−1) + 70 ≤ 61TSin(A − x)U + 70
Min{11Cosx − 60Sinx + 70} = 9
Therefore, the required answer =−2 − 9 = −11

Q.4. If Sec ! A = tanA and aSec#& A + bSec ' A + cSec & A + dSec ( A +
eSec ! A + f = 0 then what is the value of ab! + c ! d + ef ! ?
A) -13
B) 21
C) -55
D) 54
Answer- C
Explanation:
Sec ! A = tanA
Sec ( A = tan! A = 1 − Sec ! A
Sec ' A = 1 − 2Sec ! A + Sec ( A
Sec ' A − 1 = Sec ( A − 2Sec ! A
Sec#& A − 2Sec ' A + 1 = Sec ' A − 4Sec & A + 4Sec ( A
Sec#& A − 3Sec ' A + 4Sec & A − 4Sec ( A + 1 = 0
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By Comparing with
aSec#& A + bSec ' A + cSec & A + dSec ( A + eSec ! A + f = 0
a = 1, b = −3, c = 4, d = −4, e = 0, f = 1
ab! + c ! d + ef ! = 1 × 9 + 16 × −4 + 0 = 9 − 64 = −55

Q.5. What is the value of ∑." - -

+/"(log )*+ $"! Sin n − log )*+ $"! Cos n ) ?
A) 1
B) -2
C) 0
D) 91
Answer- B
Explanation:
."
∑ logSin n- − ∑ log Cos n-
^(log )*+ $"" Sin n- − log )*+ $"" Cos n- ) =
∑ log tan 30"
+/"
Consider the numerator, ∑ logSin n- − ∑ log Cos n-
:= log Sin 0- + log Sin 1- + ⋯ + log Sin 90- − log Cos 0- −
log Cos 1- − ⋯ − log Cos 90- = 0
Since we have Sin(90- − x) = Cosx
Therefore, the entire summation is zero

Q.6. Sundar is standing between two buildings of unequal height. Sundar

is 1 unit close to the shorter building and the distance between two
buildings is 5 units. If Sundar observe the top of the short building the
angle of elevation is 45- and for the other building the angle of elevation
will be 60- , then what will be the length of the wire between the top of
the two buildings?
A) 3`2T4 − √3U

B) 2`3T4 + √3U

C) 3`2T4 + √3U

D) 2`T14 − √3U
Answer- D
Explanation:

B
A
E

D X C

Let AD and BC represent two buildings and Sundar is standing at point X.

Since Sundar is one unit close to AD, DX = 2, and XC = 3

Also, from the given data ∠DXA = 45- and ∠CXB = 60-
01 01
Therefore,tan45- = => 1 = => AD = 2
12 !

Therefore, BE=BC − AD = 3√3 − 2 and AE = 5

Therefore, by Pythagoras Theorem,
AB ! = AE ! + EB !
AB ! = 25 + 27 − 12√3 + 4
AB ! = 56 − 12√3 = 4T14 − √3U

AB = 2`T14 − √3U

Q.7. Which of the following cannot be the measure of the hypotenuse if

the height and the base of the right-angle triangle is positive integer?
A) 13
B) 24
C) 25
D) 29
Answer- B
Explanation:
Pythagoras Triplet (a, b, c) where c is the hypogenous of the triangle and
a and b are two sides of the right-angle triangle.
Student should know some standard Pythagoras triplet with integer sides
and some of them has been given below make a note and verify them:
(3,4,5)
(5,12,13)
(8,15,17)
(7,24,25)
(20,21,29)

(9,40,41)
(28,45,53)
(11,60,61)
(33,56,65)
(16,63,65)
Hence in our example, 13,25 and 29 can be the value of the hypogenous.

56+ 57+
Q.8. If CosecA = , CosA = and
! !
am( + bm$ n + cm! n! + dmn$ + en( + fm! + gn! = −16
then what is the value of abc − def − g?
A) 2
B) -2
C) 4
D) -4
Answer- C
Explanation:
56+ 57+
CosecA = , CosA =
! !
2
SinA =
m+n
We have, Sin! A + Cos ! A = 1
4 (m − n)!
+ =1
(m + n)! 4
16 + (m − n)! (m + n)! = 4(m + n)!
16 + (m! − n! )! = 4m! + 8mn + 4n!
16 + m( − 2m! n! + n( = 4m! + 8mn + 4n!
m( − 2m! n! − 4m! − 4n! − 8mn + n( + 16 = 0
Comparing with am( + bm$ n + cm! n! + dmn$ + en( + fm! + gn! =
−16
a = 1, b = 0, c = −2, d = 0, e = 1, f = −4, g = −4

abc − def − g = 0 − 0 − (−4) = 4

Q.9. Which of the following can be the value of the hypogenous where
the side of the right-angle triangle are of integer length?
A) 16
B) 17
C) 18
D) 19
Answer- B
Explanation:
We have Pythagoras Triplet (8,15,17). Hence 17 can be the side of the
hypogenous with all side of integer length.

Q.10. In ΔABC if a, b, c are the length of the side opposite to angles A, B

and C respectively then what is the value of
(b − c)SinA + a(SinB − SinC) + (a − c)SinB + b(SinA − SinC) +
c(SinA − SinB) + (a − b)SinC ?
A) 0
B) 4 a Sin B − 4 b SinC
C) 4 b Sin A + 4 a SinB
D) None of these

Answer- B
Explanation:
NOTE:
* : ;
Sine Rule we will have: = = =k
89+0 89+3 89+4
aSinB = bSinA
bSinC = cSinB
aSinC = cSinA
(b − c)SinA + a(SinB − SinC) + (a − c)SinB + b(SinA − SinC)
+ c(SinA − SinB) + (a − b)SinC
= bSinA − cSinA + aSinB − aSinC + aSinB − cSinB + bSinA − bSinC +
cSinA − cSinB + aSinC − bSinC
= 2bSinA + 2aSinB − 2cSinB − 2bSinC
= 4aSinB − 4bSinC

(+ (5
Q.11. If (cosθ + cotθ) = # # and (cosθ − cotθ) = # #, then what is
+ 75 + 75
the value of mn?
A) 1
B) 2
C) 6
D) None of these

Answer- D
Explanation:
(+
From the given condition, we have (cosθ + cotθ) = # # … . (1)
+ 75
4m
(cosθ − cotθ) = ! !
… (2)
n −m
Adding (1) and (2) we get,
2(m + n)
Cosθ = !
n − m!
! +75
Cosθ = ⇒ Secθ =
+75 !

Subtracting (2) from (1) we get,

! +65
Cotθ = ⇒ tanθ =
+65 !
We have 1 + tan θ = Sec ! θ
!

(n + m)! (n − m)!
1+ =
4 4
4 + n + 2nm + m = n! − 2mn + m!
! !

4mn = −4
mn = −1

4-<=789+=
Q.12.If Cosx = then what is the value of tan 2x?
√!
A) -1
B) 0
C) 1
D) 2√3

Answer- A
Explanation:
From given condition, we have√2Cosx = Cosx − Sinx
Sinx = T1 − √2 UCosx
tanx = T1 − √2U
2tanx 2T1 − √2U 2T1 − √2U
tan2x = = ! = = −1
1 − tan! x 1 − T1 − √2U −2T1 − √2U

((#74-<@; =)
Q.13. If t = ≠ 2 then what is the value of Cosec x + Cotx
#74-<@; =64-) =
?
A) √t
B) t !
C) 2t
!
D)
!7)
Answer- D
Explanation:
((#74-<@; =)
Given t =
#74-<@; =64-) =
4(1 − Cosec x) 1 + Cot x + Cosec x
t= ×
1 − Cosec x + Cot x 1 + Cot x + Cosec x
4(1 + Cot x + Cosec x − Cosec x − Cosecx Cot x − Cosec ! x)
t=
(1 + Cot x)! − Cosec ! x
Note that we have use the formula: 1 + Cot ! x = Cosec ! x

4Cot x(1 + Cot x − Cosec x)

t=
2Cot x
t
= (1 + Cot x − Cosec x)
2

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t 2−t
Cosec x − Cot x = 1 − =
2 2
! !
Since Cosec x − Cot x = 1
(Cosecx − Cotx)(Cosec x + Cotx) = 1
2
Cosec x + Cot x =
2−t
##B
Q.14. What is the value of Sin ?
(
A) 1
#
B) −
√!
#
C)
√!
√$
D)
!
Answer- C
Explanation:
NOTE: Add-Co Cut-Co Rule:
Step (1) Convert the given angle in the form 90k ± θ
Step (2) Check the quadrant for sign for trigonometric function
Step (3) If k is even then same function and if k is odd then add co or cut
the co whichever is possible and then find the value of the resulting
function at θ

Sin Cosec + All +

tan cot + Cos Sec +

##B #
Sin = Sin 495- = Sin (5 × 90- + 45- ) = Cos 45- =
( √!

Since 1st quadrant where Sin is positive and Since 5 is odd in Sin we added
Co to get Cosine=Cos

(B
Q.15. What is the value of Cos ?
$
A) 1
#
B)
!
√$
C) −
!
√$
D)
!
Answer- C
Explanation:
4π - - - "
√3
Cos ( )
= Cos 210 = Cos 2 × 90 + 30 = −Cos 30 = −
3 2
Since 3 quadrant sign of Cos is negative and since 2 is even Cos is not
rd

changed.

Q.16. If Sinx + Cosx = 1 then what is the range of Cosx − Sinx

A) [0,1]
B) [−2,2]
C) (−2,2)
D) [−1,1]

Answer- D
Explanation:
Since we have Cos ! x − Sin! = Cos2x
(Cosx − Sinx)(Cosx + Sinx) = Cos2x
1(Cosx − Sinx) = Cos2x
And Range of Cos function is [−1,1] implies (Cosx − Sinx) has the range
[-1,1]

!"
Q.17. If Cosx = then find the possible value of Sinx tanx +
!.
Secx Cosecx?
(&$#
A)
&"."
#"&(.
B)
#!#'"
#"&(.
C)
&"."
(&$#
D)
#!#'"
Answer- A
Explanation:
-CC-<9)@ <9D@ *DH*;@+)
We have Sin x = and Cosx =
EFC-)@+-G< EFC-)@+-G<
∴ + x! 20!
= 29!

∴ x = 21
21
Sinx =
29
89+$ =6# !#$ 6# (&$#
Consider Sin x tanx + Sec x Cosec x = = #% #" =
89+ = 4-< = !.$ I#&JI#&J &"."

Q.18. If Sinx − 2 = 3 Cos x then which of the following is the possible

value of 10Cos x + 6?
√$
A)
!
#
B)
!
C) 1
D) √6
Answer- D
Explanation:
Sin x = 2 + 3Cos x
Sin! x = (2 + 3Cos x)!
1 − Cos ! x = 4 + 12Cos x + 9 Cos ! x
10 Cos ! x + 12Cos x + 3 = 0
−12 ± √12! − 4 × 10 × 3 −12 ± √24 −6 ± √6
Cosx = = =
20 20 10
10Cosx + 6 = ±√6

4-<$""! ×)*+$$"! ×8@;#$%!

Q.19.What is the value of ?
4-)$#%! ×89+#$%! ×4-<@;!#"!
A) 1
B) 0
#
C)
!√$
D) √3

Answer- C
Explanation:
By using our add-co cut-co rule we have,
- -
1
Cos 300 = Sin30 =
2
" -
1
tan 330 = −Cot60 = −
√3
Sec 135- = −Cosec45- = −√2
Cot 315- = −tan 45- = −1
1
Sin 135- = Cos 45- =
√2
Cosec 210 = −Cosec 30- = −2
-

Cos300- × tan330- × Sec135- 2√2 1

= =
Cot315- × Sin135- × Cosec210- 2√3 × 2√2 2√3

Q.20. What is the value of 109(Sin! 31- ) + Sin15- Sin75- + Cos ! 15- −
Cos ! 75- + 109(Sin! 59- )?
($L7!√$
A)
(
($L6!
B)
(√$
($L6!√$
C)
(
($L7!
D)
(√$

Answer- C
Explanation:
We use the properties:
Sin(90- − θ) = Cosθ
Cos(90- − θ) = Sinθ
tan(90- − θ) = Cotθ
Sin2x = 2Sinx Cosx
Cos2x = Cos ! x − Sin! x
109(Sin! 31- ) + Sin15- Sin75- + Cos ! 15- − Cos ! 75- + 109(Sin! 59- )
= 109(Sin! 31- + Cos ! 31- ) + Sin15- Cos15" + Cos ! 15- − Sin! 15-
Sin30-
= 109(1) + + Cos30"
2
# √$ ($L6!√$
= 109 + + =
( ! (

Practice Questions
5
Q.1. If SinxCosx = and Cos ! x = Sin! x + n then what is the value of
!
! !
m +n ?
A) 1
B) 2
C) 0
D) 3

Q.2. If 2nTtanx + Cos(−x)U − 4 = mn and 2nTtanx − Cos(−x)U + 4 =

mn then what is the value of n! − m! ?
A) 1
B) 0
C) 2
D) 4

Q.3. If Cosx + Sinx = √2Sinx then which of the following is the possible
value of Cot2x?
A) −√3
B) 1
C) √3
D) −1

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'(#7)*+=)
Q.4. If t = ≠ 8 then which of the following is the
4-)=(#6)*+=68@;=)
4-<=
possible value of ?
8@;=7)*+=6#
()7')
A)
(
(
B)
)7'
)
C)
!)7#
D) t !
B
Q.5. What is the value of Sin7 ?
$
#
A)
√!
√$
B)
!
#
C) −
√!
7√$
D)
!

Q.6. What is the value of Cot 2670- ?

A) 1
B) −1
#
C) −
√$
D) −√3

B B
Q.7. What is the value of (Cosx + Sinx) RCos o + xp + Sin o − xpS?
! !
A) −1
B) 0
C) 1
D) None of these

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%&
Q.8. If Cotx = then what is the value of Sin2x?
$$
$&.&
A)
(!!%
$&.&
B)
%&!%
#'('
C)
(!!%
D) Cannot be determined

Q.9. If tanx − 4Secx = 6 then what is the possible value of 15secx + 24?
A) 39
B) 2(5√3 + 12)
C) −√21
D) None of these

Q.10. What is the value of Cosec(765- ) + Sin(660- ) − Cos(480- ) −

tan(1920- )?
A) 0
#6!√$
B)
!
#7!√!6√$
C)
!
#6!√!6√$
D)
!

)+ %"! 6)+ #"!

Q.11. What is the value of + Cot11- Cot 79- −
#7)*+%"! )*+ #"!
#64-) &"" 4-) $""
4-) &"! 74-) $"!
A) 1
B) 1 + 2√3
C) 2 + √3
7!√$6#
D)
√$

Q.12. Which of the following can be the side of the hypogenous where
the side of the right-angle triangle are of integer length?
A) 59
B) 60
C) 61
D) 62

Q.13. What is the maximum value of x if 9CosP + x ≤ 40SinP?

A) −36
B) −49
C) −42
D) −41

Q.14. For the real number x, what is the value of

Max{7Sinx + 24Cosx − 27} − Min{20Cosx − 21Sinx + 40}
A) 11
B) 9
C) −13
D) 13

Q.15. If Cosec ! A = CotA and

pCosec#& A − qCosec ' A − rCosec & A − sCosec ( A − tCosec ! A − u = 0
then what is the value of qp! + tr ! + s ! u?
A) −13
B) −21
C) 53
D) 54

Q.16. Which of the following cannot be the measure of the hypotenuse if

the height and the base of the right-angle triangle is positive integer?
A) 41
B) 61
C) 15
D) 65

Q.17. What is the value of

^[tan x - cot(90 − x)- + Cosec(x − 90)- Sec(−x)- ] ?

=/#
A) −1
B) 90
C) −89
D) Cannot be determined

Q.18. If Cos3x + 3Cosx = a, then which of the following is the value of

Cosx?
$ *
A) `
!
$ (
B) `
*
$ !
C) `
*
$ *
D) `
(

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C6M C7M
Q.19. If Sin A = , Sec A = and
! !
ep( + fp! q! + gp! + hp! + ipq + jq( = −64, then what is the value of
fg + i + ehj?
A) 4
B) −4
C) 2
D) −2

Answer Key
1 (A) 2 (D) 3 (D) 4 (B) 5 (B)
6 (D) 7 (D) 8 (A) 9 (C) 10 (D)
11 (B) 12 (C) 13(D) 14 (C) 15 (A)
16 (C) 17 (C) 18 (D) 19 (B) 20 (A)
Solutions
Answer 1 -A
Explanation:
From given data we have 2SinxCosx = m and Cos ! x − Sin! x = n
Sin2x = m and Cos2x = n
Sin! 2x + Cos ! 2x = 1
m! + n! = 1

Answer 2 - D
Explanation:
Note: Cos(-x)= Cos(x) and Sec(-x)=Sec(x) and for the other trigonometric
function negative sign come out of the function.
From the given conditions we will have,
5+6(
tanx + Cosx = …(1)
!+
5+7(
tanx − Cosx = … (2)
!+
By solving (1) and (2) we get ,
m 2 n
tanx = and Cosx = implies Secx =
2 n 2
! !
And we have tan x + 1 = sec x
m! n !
1+ =
4 4
! !
n −m
=1
4
n ! − m! = 4
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Trigonometry
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Answer 3 – D
Explanation:
Cosx + sinx = √2Sinx
Cosx = T√2 − 1USinx
Cotx = T√2 − 1U
1 √2 + 1
tanx = × = √2 + 1
√2 − 1 √2 + 1
2tanx
tan2x =
1 − tan! x
2T√2 + 1U
tan2x = !
1 − T√2 + 1U
!N√!6#O
tan2x = = −1
7!N√!6#O
Cot2x = −1

Answer 4 - B
Explanation:
8(1 − tanx) 8tanx(1 − tanx) 1 + tanx − secx
t= = ×
Cotx(1 + tanx + secx) 1 + tanx + secx 1 + tanx − secx
8tanx(1 + tanx − secx − tanx − tan! x + tanxsecx)
t=
(1 + tanx)! − sec ! x
8tanx(1 − secx − sec ! x + 1 + tanx secx)
t=
2tanx
t = 4T2 − secx(Secx − tanx + 1)U
t
2 − = secx(secx − tanx + 1)
4
8−t
= secx(secx − tanx + 1)
4
;-<= (
=
<@;=7)*+=6# '7)
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Trigonometry
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Answer 5 – B
Explanation:
By add-co cut-co rule,
7π - - - -
√3
Sin = Sin 420 = Sin (4 × 90 + 60 ) = Sin 60 =
3 2

Answer 6 – D
Explanation:
By add-co cut-co rule,
Cot 2670" = Cot(90- × 29 + 60- ) = − tan tan 60" = −√3

Answer 7 – D
Explanation:
π π
(Cosx + Sinx) vCos o + xp + Sin o − xpw
2 2
= (Cosx + Sinx)(−Sinx + Cosx) = Cos ! x − Sin! x = Cos2x
Which will change as the value of x changes. Hence the answer is none of
these

Answer 8 - A
Explanation:
*DH*;@+)
Since Cotx = and hence hypo! = 33! + 56! = 4225 = 65!
-CC-<9)@
opp 33 adj 56
Sinx = = and Cosx = =
hypo 65 hypo 65
33 56 3696
Sin2x = 2Cosx Sinx = 2 × × =
65 65 4225

Answer 9 - C
Explanation:
tanx − 4secx = 6
tanx = 6 + 4secx
tan! x = (6 + 4secx)!
x − 1 = 36 + 48secx + 16x
15x + 48secx + 37 = 0
T−48 ± √48! − 4 × 15 × 37U
Secx =
2 × 15
−48 ± √84
Secx =
2 × 15
−24 ± √21
Secx =
15
15Sec x + 24 = −√21

Answer 10 - D
Explanation:
By using add co cut co rule we have,
Cosec(765- ) = Cosec(8 × 90- + 45- )
= Cosec45" = √2
√3
Sin 660- = (7 × 90- + 30- ) = −Cos(30- ) = −
2
1
Cos(480- ) = Cos(5 × 90- + 30- ) = −Sin30" = −
2
tan (1920" ) = tan (21 × 90- + 30- ) = −cot 30" = −√3
Now Cosec(765- ) + Sin(660- ) − Cos(480- ) − tan(1920- )
√3 1
= √2 − + + √3
2 2

2√2 √3 1
= + +
2 2 2
1 + 2√2 + √3
=
2

Answer 11 – B
Explanation:
tan 50- + tan 10- - -
1 + Cot 60" Cot 30"
+ Cot11 Cot 79 +
1 − tan50- tan 10- −(Cot 60- − Cot 30- )
= tan (50- + 10- ) + tan79- cot 79- + Cot(60- − 30- )
4-)* 4-):7# (4-)*4-):6#)
Note: Cot(a + b) = and Cot(a − b) =
4-):64-)* 4-):74-)*
- -
= tan 60 + 1 + Cot30
= √3 + 1 + √3
= 1 + 2√3

Answer 12 - C
Explanation:
We have (11,60,61) is a Pythagoras Triplet

Answer 13 - D
Explanation:
Cos(P + Q) = CosPCosQ − SinPSinQ
. ("
If CosQ = then SinQ =
(# (#
−1 ≤ Cos(P + Q) ≤ 1
9 40
−1 ≤ CosP × − SinP × ≤1
41 41
−41 ≤ 9CosP − 40SinP ≤ 41
9CosP − 41 ≤ 40SinP
x = −41

Answer 14 - C
Explanation:
7Sinx + 24Cosx − 27
7 24
= 25 R sin x + cosxS − 27
25 25
= 25Sin(A + x) − 27
!( L
Where SinA = and Cos A =
!% !%
25Sin(A + x) − 27 ≤ 25(1) − 27 = −2
Max{7Sinx + 24Cosx − 27} = −2
Now, 20Cosx − 21Sinx + 40
20 21
= 29 R Cosx − SinxS + 40
29 29
= 29TSin(A − x)U + 40
!" !#
Where SinA = and CosA =
!. !.
29(−1) + 40 ≤ 29TSin(A − x)U + 40
11 ≤ 29TSin(A − x)U + 40
Min{20Cosx − 21Sinx + 40} = 11
Therefore, the required answer is −2 − 11 = −13

Answer 15 - A
Explanation:
Cosec ! A = CotA
Cosec ( A = Cot ! A
Cosec ( A = 1 − Cosec ! A
Cosec ' A = 1 − 2Cosec ! A + Coesc ( A
Cosec ' A − 1 = Cosec ( A − 2Cosec ! A
Cosecc#& A − 2Cosec ' A + 1 = Cosec ' A − 4Cosec & A + 4Cosec ( A
Cosec#& A − 3Cosec ' A + 4Cosec & A − 4Cosec ( A + 1 = 0
By Comparing with

pCosec#& A − qCosec ' A − rCosec & A − sCosec ( A − tCosec ! A − u = 0

p = 1, q = 3, r = −4, s = 4, t = 0, u = −1
qp! + tr ! + s ! u = 3 × 1! + 0 + 4! (−1) = 3 − 16 = −13

Answer 16- C
Explanation:
We have following Pythagoras Triplet,
(9,40,41)
(11,60,61)
(16,63,65)
And 15 is not hypotonus for triplet

Answer 17 - C
Explanation:
'.

^[tan x - cot(90 − x)- + Cosec(x − 90)- Sec(−x)- ]

=/#
= ∑[tan x - tan x - − Cosec(90 − x)- Secx - ]
'.

= ^[tan! x - − Sec ! x - ]
=/#
We have used tan(90 − x) = cotx and Cosec(90 − x) = sec x and
Cosec(−x) = −Cosecx and Sec(−x) = Sec x
Now we have 1 + tan! x = Sec ! x
tan! x − Sec ! x = −1
'. '.

^[tan! x - − Sec ! x - ] = ^ −1
=/# =/#
= −1 × 89 = −89

Answer 18 - D
Explanation:
NOTE the Formula: Sin 3x = 3Sinx − 4Sin$ x
Cos3x = 4Cos $ x − 3Cosx
3tanx − 1x
tan 3x =
1 − 3x
Therefore, a = 4Cos $ x
a
= Cos $ x
4
$ a
Cosx = `
4

Answer 19 - B
Explanation:
C6M (
SinA = and CosA =
! C7M
Sin! A + Cos ! A=1
(p + q) !
16
+ =1
4 (p − q)!
(p + q)! (p − q)! + 64 = 4(p − q)!
64 + (p! − q! )! = 4p! − 8pq + 4q!
p( − 2p! q! + q( + 64 = 4p! − 8pq + 4q!
p( − 2p! q! − 4p! − 4p! − 8pq + q( + 64 = 0
Comparing, e = 1, f = −2, g = −4, h = −4, i = −8, j = 1
fg + i − ehj = (−2)(−4) + (−8) + (−4)
= 8 − 8 − 4 = −4

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