Sushila Devi Bansal College

An Institute of Bansal Group of Institutes

Umaria, A. B. Road, Near Rau, Indore -
453331

Mensuration
Q1. What is the area of an equilateral triangle of side 16 cm?

a) 48√3 sq.cm

b) 128√3 sq.cm

c) 96√3 sq.cm

d) 64√3 sq.cm

Answer: 64√3 sq.cm

Explanation:
Area of an equilateral triangle = √34S2.If S = 16, Area of triangle = √34×16×16 = 64√3 cm2

Q2. If the sides of a triangle are 26 cm, 24 cm and 10 cm, what is its area?

a) 120sq.cm

b) 130 sq.cm

c) 312 sq.cm

d) 315 sq.cm

Answer: 120sq.cm

Explanation:
The triangle with sides 26cm, 24cm and 10cm is right angled, where the hypotenuse is 26cm.Are
a of the triangle = 12×24×10 = 120cm2

Q3. The perimeter of a triangle is 28 cm and the inradius of the triangle is 2.5 cm. What is the
area of the triangle?

a) 35 sq.cm

b) 42 sq.cm

c) 49 sq.cm

d) 70 sq.cm

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 Sushila Devi Bansal College
An Institute of Bansal Group of Institutes
Umaria, A. B. Road, Near Rau, Indore -
453331

Answer: 35 sq.cm

Explanation:
Area of a triangle = r * sWhere r is the inradius and s is the semi perimeter of the triangle.Area of
triangle = 2.5×282 = 35 cm2

Q4. Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance
between them is 15 cm.

a) 225 sq.cm

b) 275 sq.cm

c) 285 sq.cm

d) 315 sq.cm

Answer: 285 sq.cm

Explanation:
Area of a trapezium = 12(sum of parallel sides)×(perpendicular distance between them) = 12 (20
+18)×(15) = 285 cm2

Q5.Find the area of a parallelogram with base 24 cm and height 16 cm.

a) 262 sq.cm

b) 384 sq.cm

c) 192 sq.cm

d) 131 sq.cm

Answer: 384 sq.cm

Explanation:

Area of a parallelogram = base×height = 24×16 = 384 cm2

Q6. The ratio of the length and the breadth of a rectangle is 4 : 3 and the area of the rectangle is
6912 sq cm. Find the ratio of the breadth and the area of the rectangle?

a) 1 : 96

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 Sushila Devi Bansal College
An Institute of Bansal Group of Institutes
Umaria, A. B. Road, Near Rau, Indore -
453331

b) 1 : 48

c) 1 : 84

d) 1 : 68

Answer: 1 : 96

Explanation:

Solution:Let the length and the breadth of the rectangle be 4x cm and 3x respectively.(4x)×(3x)
= 691212x2 = 6912 x2 = 576 = 4×144 = 22×122(x>0)⇒ x = 2×12 = 24Ratio of the breadth and t
he areas = 3x : 12x2 = 1 : 4x = 1 : 96.

Q7.A wire in the form of a circle of radius 3.5 m is bent in the form of a rectangle, whose length
and breadth are in the ratio of 6 : 5. What is the area of the rectangle?

a) 60 sq.cm

b) 30 sq.cm

c) 45 sq.cm

d) 15 sq.cm

Answer: 30 sq.cm

Explanation:

Let l = 6x and b = 5x 2(6x+5x) = 2×227×3.5 ⇒

The circumference of the circle is equal to the permeter of the rectangle.
x = 1 Therefore l = 6 cm and b = 5 cm .Area
of the rectangle = 6×5 = 30 cm2

Q8. A cube of side one meter length is cut into small cubes of side 10 cm each. How many such
small cubes can be obtained?

a) 10

b) 100

c) 1000

d) 10000

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 Sushila Devi Bansal College
An Institute of Bansal Group of Institutes
Umaria, A. B. Road, Near Rau, Indore -
453331

Answer: 1000

Explanation:
Along one edge, the number of small cubes that can be cut = 10010 = 10.Along each edge 10 cub
es can be cut. (Along length, breadth and height).Total number of small cubes that can be cut = 1
0×10×10 = 1000

Q9. The radius of a wheel is 22.4 cm. What is the distance covered by the wheel in making 500
revolutions?

a) 252 m

b) 704 m

c) 352 m

d) 808 m

Answer: 704 m

Explanation:

Solution:
In one resolution, the distance covered by the wheel is its own circumference. Distance covered i
n 500 revolutions. = 500×2×227×22.4 = 70400 cm = 704 m

Q10. The ratio of the volumes of two cubes is 729 : 1331. What is the ratio of their total surface
areas

a) 81 : 121

b) 9 : 11

c) 729 : 1331

d) 27 : 121

Answer: 81 : 121

Explanation:

Ratio of the sides = 3√729 : 3√1331 = 9 : 11

Ratio of surface areas = 92 : 112 = 81 : 121

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 Sushila Devi Bansal College
An Institute of Bansal Group of Institutes
Umaria, A. B. Road, Near Rau, Indore -
453331

Q11. An order was placed for the supply of a carpet whose length and breadth were in the ratio
of 3 : 2. Subsequently, the dimensions of the carpet were altered such that its length and breadth
were in the ratio 7 : 3 but were was no change in its perimeter. Find the ratio of the areas of the
carpets in both the cases.

a) 4 : 3

b) 8 : 7

c) 4 : 1

d) 6 : 5

Answer: 8 : 7

Explanation:
Let the length and breadth of the carpet in the first case be 3x units and 2x units respectively.Let

2x) = 2(7y+3y)⇒ 5x = 10y ⇒

the dimensions of the carpet in the second case be 7y, 3y units respectively.From the data,. 2(3x+
x = 2yRequired ratio of the areas of the carpet in both the c
ases = 3x×2x : 7y×3y = 6x2 : 21y2= 6×(2y)2 : 21y2= 6×4y2 : 21y2= 8 : 7

Q12. A park in the form of a right-angled triangle has a base and height of 10 m and 15 m
respectively. Find the area of the park?

a) 45

b) 75

c) 60

d) 150

Answer: 75

Explanation: Base = 10mHeight = 15mArea of the triangle = 12×10×15 = 75m2

Q13. The percentage increase in the area of a rectangle, if each of its sides is increased by 20%
is:

a) 40%

b) 42%

c) 44%

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 Sushila Devi Bansal College
An Institute of Bansal Group of Institutes
Umaria, A. B. Road, Near Rau, Indore -
453331

d) 46%

Answer: 44%

Explanation:

Let original length = x metres and original breadth = y metres.Original area = (xy) m2.
New length = 120100×x m=65×x m.New breadth = 120100×y m=65ym.
New Area = (65)x×(65)y = (3625)xy.
The difference between the original area = xy and new area 3625xy is= (3625)xy−xy
= xy(3625−1) = xy(1125) or (1125)xy Increase % = ((1125)xy(xy))×100
% = 44%.

Q14. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the
middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq.
m, then what is the width of the road?

a) 2.91 m

b) 3 m

c) 5.82 m

d) 4 m

Answer: 3 m

Explanation: Area of the park = (60×40)m2 = 2400m2.

Area of the lawn = 2109m2.Area of the crossroads = (2400−2109)m2 = 291m2.Let the width of t
he road be x metres. Then,60x+40x−x2 = 291 x2−100x+291 = 0 (x−97)(x−3) = 0
x = 3.

Q15. 50 men took a dip in a water tank 40 m long and 20 m broad on a religious day. If the
average displacement of water by a man is 4 cubic meter , then the rise in the water level in the
tank will be:

a) 20 cm

b) 25 cm

c) 35 cm

d) 50 cm

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 Sushila Devi Bansal College
An Institute of Bansal Group of Institutes
Umaria, A. B. Road, Near Rau, Indore -
453331

Answer: 25 cm

Explanation:

Total volume of water displaced = (4×50)m3 = 200m3.Rise in water level = 20040×20m = 0.25
m = 25cm.

Q16. There is a rectangular Garden whose length and width is 60m X 20m.There is a walkway
of uniform width around the garden. Area of the walkway is 516 sq.m. Find the width of the
walkway?

a) 1

b) 2

c) 3

d) 4

Answer: 3

Explanation:
Let the width of rectangle be x.So the length & breath is increased by 2x.So new total area along

⇒
with walkway is (60+2x)×(20+2x)So (60+2x)×(20+2x)60×20 = 516⇒(60+2x)×(20+2x) = 1716
(30+x)×(10+x) = 429

⇒ (30+x)×(10+x) = 33×13

⇒ x=3

Q17. If a circle and a square have the same area, then what will be the perimeter of the square, if
the diameter of the circle is 8 cm.

a) 4π

b) 32

c) 4√π

d) 16√π

Answer: 16√π

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 Sushila Devi Bansal College
An Institute of Bansal Group of Institutes
Umaria, A. B. Road, Near Rau, Indore -
453331

Explanation:
We know,Area of circle = Area of Square.πR2 = (side)2Side = √(π)×R= √(π)×Diameter2 = 4√(π)
Perimeter of square = 4×side = 16√(π)

Q18. A field with equal sides of 40m is full of grass. One cow is tied to each corner of the field
by a rope of 14m. How much grass will not be eaten by those 4 cows?

a) 984 sq.m

b) 1056 sq.m

c) 1224 sq.m

d) 856 sq.m

Answer: 984 sq.m

Explanation:

One cow at each corner So, 4 cows = 4 corners = 4 sides.

Field with 4 sides that are equal = Square.
So, we can draw it as followsArea of square = 40×40 = 1600 sq.m.
The part eaten by the cow ie. light green part; are 4 parts of an entire circle.The radius of this circ
le = length of rope = 14m.

∴Area not eaten by cow = 1600−πr2 = 1600−(227)×14×14

Area not eaten by cows =( Area of square)minus(Area of circle)

Area mt eaten by cow = 984 sq.m

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 Sushila Devi Bansal College
An Institute of Bansal Group of Institutes
Umaria, A. B. Road, Near Rau, Indore -
453331

Q19. Poles are to be created along the boundary of a rectangular field in such a way that distance
between any two adjacent poles is 1.5 metres. The perimeter of the field is 21 metres and length
and the breadth are in the ratio of 4:3 respectively. How many poles will be required?

a) 14

b) 16

c) 15

d) 20

Answer: 14

Explanation:

Let the length and breadth be 4x and 3x metres respectively.2(4x+3x) = 21⇒ 14x = 21 ⇒ x
= 2114 = 1.5 Length = 6m and Breadth = 4.5m.So number of poles = 14

Q20. Find the length of the longest pole that can be placed in a room 12 m long, 8m broad and 9
m high.

a) 16 m

b) 17 m

c) 18 m

d) 19 m

Answer: 17 m

Explanation:
The length of the longest pole can be calculated by calculating the value of the diagonal of a roo
mwhich is considered to be in the shape of a cuboid.

Hence, The Diagonal of the cuboid = √l2+b2+h2 = √122+82+92 = √289=17m.

Q21. If the radius of a cylinder is doubled and the height remains same, the volume will be

a) Doubled

b) Halved

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 Sushila Devi Bansal College
An Institute of Bansal Group of Institutes
Umaria, A. B. Road, Near Rau, Indore -
453331

c) Same

d) Four times

Answer: Four times

Explanation:

If r be the radius and h be the height, then volume = πr2h.

If radius is doubled and height remain same,
The volume will be = π(2r)2h = π×4r2h = 4πr2h = 4×Volume.
The volume is four times more than the original volume.

Q22. An ant wants to go from one corner of the cube to the diagonally opposite corner by
crawling [can't fly]. Minimum distance it has to travel to reach there if the side of the cube is
10 ?

a) 10√3

b) √10+30

c) 30

d) 10√5

Answer: 10√5

Explanation:
Path for D to F is = DI+IFSay the distance AI = xThe total path length, L = 102+x2+(10−x)2+1
2By maxima or minima dLdx = 0⇒x = 102So L=10√5DI is in the ABCD plane & IF is in the A
BFE plane.

Q23. If the sum of the interior angles of a regular polygon measures up to 144 degrees, how
many sides does the polygon have?

a) 10 sides

b) 8 sides

c) 12 sides

d) 9 sides

Answer: 10 sides

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 Sushila Devi Bansal College
An Institute of Bansal Group of Institutes
Umaria, A. B. Road, Near Rau, Indore -
453331

Explanation:
Exterior angle = 180−Interior angle = 180−144 = 36.Total sum of the exterior angles of a polygo

l have the sum of the interior angles = (2N−4)×right angles. ∴ 144N = (2N−4)×90144N−180
n = 360°.Therefore no.of sides N = 36036 = 10.It’s a decagon.Aliter :A polygon with N sides wil

N = −360 36N = 360 N = 10 It’s a polygon with 10 sides.

Q24. A polygon has 27 diagonals. The number of sides of the polygon is

a) 9

b) 10

c) 11

d) 12

Answer: 9

Explanation:
We know that,Number of diagonals of a polygon = n(n−3)2According to the question,n(n−3)2 =
27 n(n−3) = 54 n2−3n−54 = 0n2×9n+6n−54 = 0n(n9)+6(n9) = 0(n+6)(n−9) = 0n = 9, − 6 (Negle
cting negative value)n = 9. Hence, sides of polygon are 9.

Q25. Consider Square S inscribed in circle C, what is the ratio of the areas of S and C? And,
Consider Circle Q inscribed in Square S, what is the ratio of the areas of S and Q?

a) 2:π, 4:π

b) 4:π, 2:π

c) 1:π, 4:π

d) 2:π, 1:π

Answer: 2:π, 4:π

Explanation: Go about it by focussing on the radius of the Circle or the side of the Square.
If square is inside the circle, ratio of areas of square to that of circle is 2 : π.
If circle is inside square, ratio of areas of square to that of circle is 4 : π.
Remember, circle area goes with π, square area goes with number.
The question is "What is the ratio of the areas of S and C? and what is the ratio of the areas of S
and Q?" Hence, the answer is 2:π, 4:π. Choice A is the correct answer.

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 Sushila Devi Bansal College
An Institute of Bansal Group of Institutes
Umaria, A. B. Road, Near Rau, Indore -
453331

Website: https://sdbc.ac.in Contacts: +91 9669662255, 9669660055, E-Mail ID: director@sdbc.ac.in