Triangles

Similar Triangles
Similar Figures

Similar figures are the figures having the same shape but not necessarily the same size.
For example in the figure given below, the two triangles have the same shape, but their
perimeter and area are different.

Similar Triangles

Congruent Figures

Two figures are said to be congruent if they have the same shape as well as the same size.
Congruent figures are exactly the same. They have the same perimeter, area and can be
superposed on each other.

Difference between Congruency and Similarity

Congruent figures are exactly the same, with the same size, shape and dimensions.

Similar figures are scaled up or scaled down versions of each other. They have the same
shape but their sizes need not be the same.

Similar Polygons

Two polygons are similar if their corresponding angles are equal and their corresponding
sides are in the same ratio.

Two polygons cannot be similar if their number of sides are different.

Similar Triangles

Two triangles are similar if their

 corresponding angles are equal
corresponding sides are in the same ratio.

△ABC & △DEF are similar triangles

ΔABC ∼ ΔDEF if and only if

∠A = ∠D, ∠B = ∠E, ∠C = ∠F

AB BC CA
= =
DE EF FD

Basic Proportionality Theorem

Basic Proportionality Theorem

Basic Proportionality Theorem (BPT) states that if a line is drawn parallel to one
side of a triangle to intersect the other two sides in distinct points, the other two
sides are divided in the same ratio.

In the above figure DE| |BC. Then, BPT says that,

AD AE
=
DB EC

Converse of Basic Proportionality Theorem

 The converse of Basic Proportionality Theorem is true. It states that if a line
divides two sides of a triangle in the same ratio, then the line is parallel to the
third side.

In the above figure, if AD

DB
=
AE

EC
,

then, the converse of BPT states that DE| |BC.

Criteria for Similarity of Triangles

Criteria for Similarity of Triangles

Two triangles are said to be similar if their corresponding angles are equal and their
corresponding sides are in the same ratio. However, we need not check for all angles and
sides to ensure similarity. There are certain criteria to confirm the similarity of two
triangles by comparing a lesser number of corresponding parts of a triangle. These are

AAA similarity
SSS similarity
SAS similarity

AAA Similarity

According to AAA similarity criterion, if the corresponding angles of two

triangles are equal, then the corresponding sides are in the same ratio and the
triangles are similar.

If ∠A = ∠D, ∠B = ∠E and ∠C = ∠F , then ΔABC ~ ΔDEF

If two angles of a triangle are respectively equal to two angles of another triangle, then by
angle sum property, the third angle of the triangles are equal and the triangles are similar.
This is called AA similarity criterion.
SSS Similarity

According to SSS similarity criterion, if the sides of one triangle are proportional
to the corresponding sides of another triangle, then their corresponding angles
are equal and the triangles are similar.

△ABC & △DEF are similar by SSS criterion

If AB

DE
=
BC

EF
=
AC

DF
, then ΔABC ~ ΔDEF

SAS Similarity

According to SAS similarity criterion, if two sides of one triangle are proportional to two
sides of another triangle and the corresponding included angles are equal, then, the
triangles are similar.
 △ABC & △DEF are similar by SAS criterion

If AB

DE
=
AC

DF
and ∠A = ∠D, then ΔABC ~ ΔDEF

Areas of Similar Triangles

Relation between Areas and Sides of Similar Triangles

In the given figure, ΔABC is similar to ΔDEF , the ratio of their areas is given by,
area(ΔABC)
AB 2 BC 2 AC 2
= ( ) = ( ) = ( )
area(ΔDEF ) DE EF DF

Pythagoras Theorem
Perpendicular from Right Angle to Hypotenuse Divides the Triangle
into Two Similar Triangles

In a right-angled triangle, if a perpendicular is drawn from the vertex of the right

angle to the hypotenuse, then the triangles on both sides of the perpendicular
are similar to the whole triangle and to each other.
 CD is the perpendicular from the vertex to hypotenuse AB

In the above figure, CD is the perpendicular drawn from the vertex C on the hypotenuse AB
of ΔABC .
So, ΔABC ~ ΔCBD ~ ΔACD

Pythagoras Theorem

Pythagoras Theorem states that in a right-angled triangle, the square of the

hypotenuse is equal to the sum of the squares of the other two sides.

Conversely, In a triangle, if square of one side is equal to the sum of the squares
of the other two sides, then the angle opposite the first side is a right angle.
 CBSE Board Class 10 Maths Chapter 6- Triangles
Objective Questions

Areas of Similar Triangles

1. If △ ABC ~ △ DEF such that AB = 12 cm and DE = 14 cm. Find the ratio of areas
of △ ABC and △ DEF.

(A) 49/9
(B) 36/49
(C) 49/16
(D) 25/49

Answer: (B) 36/49

Solution: We know that the ratio of areas of two similar triangles is equal
to the ratio of the squares.
Of any two corresponding sides,
area of △ ABC / area of △ DEF = (AB/DE) 2= (12/14) 2= 36/49

2. D and E are points on the sides AB and AC respectively of a △ABC such that DE ||
BC. Which of the following statement is true?

(i) △ ADE ~ △ ABC

(ii) (area of △ ADE/ area of △ ABC) = (AD2/AB2)
(iii) (area of △ ADE/ area of △ ABC)= (AB2/ AD2)

(A) only (iii)

(B) only (i)
(C) only (i) and (ii)
(D) all (i) , (ii) and (iii)

Answer: (i) △ ADE ~ △ ABC and (ii) (area of △ ADE/ area of △ ABC) = (AD2/AB2)

Solution:

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 In △ ADE and △ ABC, we have

∠ ADE = ∠ B

[Since DE || BC ∠ ADE = ∠ B (Corresponding angles)]

and, ∠ A = △ A [Common]

△ ADE ~ △ ABC

Therefore, (area of △ ADE / area of △ ABC) = (AD2/AB2)

3.

In the figure, PB and QA are perpendicular to segment AB. If OA = 5 cm, PO = 7cm

and area (ΔQOA) = 150 cm2, find the area of ΔPOB.

(A) 233 cm2

(B) 294 cm2
(C) 300 cm2
(D) 420 cm2

Answer: (B) 294 cm2

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 Solution: Consider Δ~QOA and Δ POB

QA || PB,

Therefore, ∠ AQO = ∠ PBO [Alternate angles]

∠ QAO = ∠ BPO [Alternate angles]

and

∠ QOA = ∠ BOP [Vertically opposite angles]

Δs QOA ~ BOP [by AAA similarity]

Therefore, (OQ/ OB) = (OA/OP)

Now, area (POB)/ area (QOA) = (OP) 2/ (OA) 2= 72/ 52

Since area (QOA) =150cm2

⇒area (POB) =294cm2

4. Two isosceles triangles have equal angles and their areas are in the ratio 16: 25. The
ratio of corresponding heights is:

(A) 4:5
(B) 5:4
(C) 3:2
(D) 5:7

Answer: (A) 4:5

Solution: For similar isosceles triangles,

Area (Δ1) / Area (Δ2) = (h1)2 / (h2)2

(h1 / h2) = 4/5

Basic Proportionality Theorem

5. In △ABC, AB = 3 and, AC = 4 cm and AD is the bisector of ∠A. Then, BD : DC is —

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 (A) 9: 16
(B) 4:3
(C) 3:4
(D) 16:9

Answer: (C) 3:4

Solution:

The Angle-Bisector theorem states that if a ray bisects an angle of a triangle, then
it divides the opposite side into segments that are proportional to the other two
sides (It may be similar or may not depending on type of triangle it divides)

In △ABC

as per the statement AB/ AC= BD/DC i.e. a/b= c/d

So, BD/ DC= AB/AC= ¾

So, BD: DC = 3: 4

6. ABCD is a parallelogram with diagonal AC If a line XY is drawn such that XY ∥ AB.

BX/XC=?

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 (A) (AY/AC)
(B) DZ/AZ
(C) AZ/ZD
(D) AC/AY

Answer: (C) AZ/ZD

Solution: In the Δ ABC,

AB ∥ XZ

AB ∥ XY

∴ BX/ XC= AY/YC…. (By BPT)..... (1)

In parallelogram ABCD,
AB ∥ CD

AB ∥ CD ∥ XZ

In the Δ ACD,
CD ∥ YZ

∴ AY/YC= AZ/ZD ... (By BPT)...... (2)

From 1 & 2,

BX/XC= AY/YC= AZ/ZD

BX/XC= AZ/ZD

7.

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In ABC, Given that DE//BC, D is the midpoint of AB and E is a midpoint of AC.

The ratio AE: EC is ____.

(A) 1: 3
(B) 1:1
(C) 2:1
(D) 1:2

Answer: (B) 1:1

Solution:

DE is parallel to BC

So, In triangles ABC, ADE

∠DAE = ∠ECF {Alternate angles}

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 ∠ADE = ∠EFC {Alternate angles}

∠BAC = ∠DAE
By A.A.A similarity ABC≡ADE

⇒ AD/DB= AE/EC (Basic Proportionality Theorem)

Since, D is midpoint of AB.

AD=DB

⇒ AD/DB= 1/1=AE/EC

⇒ AE/EC= 1 /1

∴ AE: EC= 1:1

8. In ΔABC, AC = 15 cm and DE || BC. If AB/AD=3, Find EC.

(A) 5cm
(B) 10 cm
(C) 2.5cm
(D) 9cm

Answer: (B) 10cm

Solution: Given: DE∥BC

From basic proportionality theorem

AD/DB=AE/EC

Now, AB/AD= (AD+DB) /AD= 1 + (DB/AD)

⇒1+ (DB/AD) = 3

⇒ (DB/AD) = 2

⇒ AD/DB=AE/EC = ½

⇒2AE=EC⇒AC=AE+EC−−−−−−− (1)
On substituting value of EC in (1), we get
15=3AE⇒5=AE⇒EC=10cm

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Criteria for Similarity of Triangles

9. △ ABC is an acute angled triangle. DE is drawn parallel to BC as shown. Which of the following
are always true?

i) △ ABC ∼ △ ADE

ii) AD/BD= AE/EC

iii) DE= BC/2

(A) Only (i)

(B) (i) and (ii) only
(C) (i), (ii) and (iii)
(D) (ii) and (iii) only

Answer: (B) (i) and (ii) Only

Solution: Since DE || BC, AD/BD=AE/EC and hence △ABC ∼ △ADE

DE = BC/2 only if D and E are the mid points of AB and AC respectively. So this may not be
true always.

10. The triangles ABC and ADE are similar

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 Which of the following is true?

(A) EC/AC=AD/DE
(B) BC/BD=CE/DE
(C) AB/AD=BC/DE
(D) All of the Above

Answer: (C) AB/AD=BC/DE

Solution: Since the given triangles are similar, the ratios of corresponding sides are
equal.

So, AB/AD=BC/DE=AC/AE

11. If in △ CAB and △ FED, AB/ EF=BC/FD=AC/ED, then:

(A) △ ABC∼△ DEF

(B) △ CAB∼△ DEF
(C) △ ABC∼△ EFD
(D) △ CAB∼△ EFD

Answer: (C) △ ABC∼△ EFD

Solution: If two triangles are similar, corresponding sides are proportional.

Therefore, △ABC∼△EFD.

12. A tower of height 24m casts a shadow 50m and at the same time, a girl of height
1.8m casts a shadow. Find the length of the shadow of girl.

(A) 3.75m
(B) 3.5m
(C) 3.25m
(D) 3m

Answer: (A) 3.75m

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 Solution: In △ABC and △DEC
∠ABC=∠DEC=90°
∠C=∠C (common)
Therefore, △ABC∼△DEC [by AA similarity]

So, DE/AB=EC/BC

EC=DE × (BC/AB)

EC= 1.8 × (50/24) ⇒EC=3.75 m

Pythagoras Theorem

13. In the adjoining figure, if BC = a, AC = b, AB = c and ∠CAB = 120°,

then the correct relation is-

(A) a2 = b2 + c2 – bc
(B) a2 = b2 + c2 + bc
(C) a2 = b2 + c2 - 2bc
(D) a2 = b2 + c2 + 2bc

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Answer: (B) a2 = b2 + c2 + bc

Solution: In △CDB,

BC2 = CD2 + BD2 [By Pythagoras Theorem]

BC2 = CD2 + (DA+AB)2

BC2 = CD2 + DA2 + AB2 + (2×DA×AB) (i)

In △ADC,

CD2 + DA2 = AC2 (ii) [By Pythagoras Theorem]

Also, Cos60∘ = AD/AC

AC = 2AD (iii)

Putting the values from (ii) and (iii) in (i), we get

BC2 = AC2 + AB2 + (AC×AB)

a2 = b2 + c2 + bc

Alternatively,

Since ∠A is an obtuse angle in ΔABC, so

BC2 = AB2 + AC2 + 2AB . AD

= AB2 + AC2 + 2×AB× ½ ×AC

[∵ AD = AC cos 60∘ = 1/2AC]

= AB2 + AC2 + AB × AC

a2 = b2 + c2 + bc.

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14. If the distance between the top of two trees 20 m and 28 m tall is 17 m, then the
horizontal distance between the trees is :

(A) 11m
(B) 31m
(C) 15m
(D) 9m

Answer: (C) 15m

Solution: Let AB and CD be two trees such that AB = 20 m, CD = 28 m & BD = 17 m

Draw BE parallel to CD. Then, ED = 8 m.

By applying Pythagoras theorem:

BE2+DE2=BD2

=15m

∴ AC = BE = 15 m

15.

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In the figure △ABC is a right angled triangle with right angle at B. BD is perpendicular to AC.
Then which of the following options will hold true?

(A) AD2=DC×AC
(B) AB2=AD×AC
(C) AB2=AD×DC
(D) AB2=DC2+AD2

Answer: (B) AB2=AD×AC

Solution:

In △ABC and △ADB

∠ABC=∠ADB=90°
∠A=∠A (common angle)
Therefore, △ABC∼△ADB [by AA similarity]

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 AB/AD= AC/ AB

AB2=AC×AD

16. In a right △ABC, a perpendicular BD is drawn on to the largest side from the
opposite vertex. Which of the following does not give the ratio of the areas of △ABD
and △ACB?

(A) (AB/AC)2
(B) (AD/AB)2
(C) (AB/AD)2
(D) (BD/CB)2

Answer: (C) (AB/AD)2

Solution:

Consider ΔABD and ΔACB:

∠BAD = ∠BAC [common angle]

∠BDA = ∠ABC [ 90°]

By AA similarity criterion, △ABD ~ △ACB

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 Hence,

ar (ΔABD)/ ar(ΔACB) = (AB/AC)2= (AD/AB)2= (BD/CB)2

Similar Triangles

17. △ ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm.△ DEF is similar to △ABC.
If EF = 4 cm, then the perimeter of △DEF is –

(A) 7.5 cm
(B) 15cm
(C) 30cm
(D) 22.5cm

Answer: (B) 15cm

Solution: AB/DE= AC/DF=BC/EF=2/4=1/2

DE = 2×AB = 6 cm, DF = 2×AC = 5 cm

∴ Perimeter of △DEF = (DE + EF + DF) = 15 cm.

18. In △ ABC and △ DEF, ∠A = ∠E = 40∘ and AB/ED=AC/EF. Find ∠B if ∠F is 65°

(A) 85°
(B) 75°
(C) 35°
(D)65°

Answer: (B) 75°

Solution:

AB/ED= AC/EF (Given)

∠A = ∠E = 40°

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 Since, the ratio of adjacent sides and the included angles are equal.
∴△ABC is similar to △EDF by SAS similarity criterion.

Now, ∠C = ∠F = 65° [Corresponding angles of a similar triangles are equal]

∴ ∠B = 180°− (∠A+∠C)
=180° - (40° + 65°) = 75°

19. The ratio of the corresponding sides of two similar triangles is 1: 3. The ratio of their
corresponding heights is _________

(A) 1:3
(B) 3:1
(C) 1:9
(D) 9:1

Answer: (A) 1:3

Solution: Ratio of heights = Ratio of sides = 1: 3.

20. If Δ ABC and Δ DEF are similar such that 2AB = DE and BC = 8 cm, then Find EF.

(A) 16 cm
(B) 12 cm
(C) 8 cm
(D) 4 cm

Answer: (A) 16 cm

Solution:

2AB = DE

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2BC = EF

⇒ 2×8 = EF

⇒ EF = 16 cm

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(E)

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 POST CLASS NOTES

Triangles
Topics
 Similar Triangles
A
P

B C Q

∠A = ∠P k
∠B = ∠Q
∠C = ∠R

AB BC CA
= = =k
PQ QR RP
 Criteria for Similarity
of Triangles
P
A

14 10
AB BC CA 7 5
= =
PQ QR RP
B C Q R
6 12

X
P
70°
70°

60° 50° 60°

50° R Y Z
Q

P
A
15
10
AB BC 2
= = 60° 60°
PQ QR 3 B C Q R
8 12
∠B = ∠Q
 Ratio of Areas of
Similar Triangles
A

B C Q R

2 2 2
Area(∆ABC) AB BC CA
= = =
Area(∆PQR) PQ QR RP

∆ABC~ ∆ADC ~ ∆ADB AA D

A C
 Basic Proportionality Theorem

1 N M
Area of ∆APQ = × AP × QN
2
1
Area of ∆PBQ = × PB × QN P Q
2
1
Area of ∆APQ = × AQ × PM
2
1
Area of ∆QCP = × QC × PM
2
B C
1
× AP × QN
Area of ∆ APQ 2 AP
= = …………(1)
Area of ∆ PBQ 1 PB
× PB × QN
2
1
× AQ × PM
Area of ∆ APQ 2 AQ
= = ………(2)
Area of ∆ QCP 1 QC
× QC × PM
2

∴ ∆PBQ = ∆QCP …………(3)

(1), (2) (3)

AP AQ
=
PB QC
 Converse of Basic
Proportionality Theorem

A
AD AE
= , DE || BC.
DB EC E′
D E
DE

AD AE
= ……(1) B
DB EC C
DE BC DE’ BC

AD AE’
= ……(2)
DB E’C
1 2
AE AE′
=
EC E′C

AE AE′ AE+EC AE′+E′C

+1= +1 =
EC E′C EC E′C

AC AC
= , EC = E’C
EC E′C

E E’
DE’ || BC
∴ DE || BC.
 Properties of
Right-Angled Triangles

B
D
△ADB ~ △ABC
AD AB
∴ = (
AB AC )
A C
AB2 = AD × AC …………(1)

, △ADC ~△ABC

CD BC
∴ =
BC AC
BC2= CD × AC …………(2)

(1) + (2)

AB2 + BC2 = AD × AC + CD × AC
AB2 + BC2 = AC (AD + CD)
AD + CD = AC

∴ AC2 = AB2 + BC2

 △EGF, B
AC = EG BC = FG.
c
a
△EGF,
EF2 = EG2 + FG2 = b2 + a2 ……(1)
C b A

△ABC,
AB2 = AC2 + BC2 = b2 + a2 ……(2)

(1) (2) F
EF2 = AB2
c
a
EF = AB
⇒ △ ACB ≅ △EGF SSS
G b E
⇒ ∠C
∴ △ABC
 Important Theorems and Formulae

P
A

SSS
B C Q R
P X

AAA/AA
Q R Y Z
P
A

SAS
B C Q R

a c

a2 + b2 = c 2
b

P Q

AP AQ
PQ || BC, =
PB QC
B C
Mind Map
 Practice Questions - Term 1
Date: 19/11/2021
Subject: Mathematics
Topic : Triangles Class: X

1.
If AC , then AE _____.
AD 3
I n ΔABC, DE || BC and = . = 5.6 cm =
DB 5

A. 2.1 cm

B. 2.4 cm

C. 3.2 cm

D. 3.6 cm

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 Practice Questions - Term 1

2. In the given figure, if DE || BC then find AD.

A. 30 units

B. 50 units

C. 40 units

D. 10 units

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 Practice Questions - Term 1

3. Find CD, if AC = 4 cm, AB = 3 cm and AD = 2 cm.

A. 8
cm
3

B. 3
cm
8

C. 4
cm
3

D. 3
cm
4

4.
In ΔABC , point D and E lies on the line AB and AC respectively as shown
in the figure. Find the measure of ∠AED.

A. 65
∘

B. 40
∘

C. 75
∘

D. 70
∘

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 Practice Questions - Term 1

5.
Observe the given triangles and find the

value of ∠P .

A. ∘
60

B. ∘
40

C. ∘
50

D. ∘
65

6. In the figure, AC = 10cm, PC = 15cm, PQ = 12 cm, find PB.

A. 6 cm

B. 7 cm

C. 8 cm

D. 9 cm

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 Practice Questions - Term 1

7. In a △ABC , points P and Q are on sides AB and AC respectively. If AP = 3

cm, PB = 6 cm. AQ = 5 cm and QC = 10 cm, then BC = ____.

A. 4PQ

B. PQ

C. 3PQ

D. PQ
2

8.
Two isosceles triangles have equal angles and their areas are in the ratio 16
: 25. The ratio of corresponding heights is :

A. 4: 5

B. 5: 4

C. 3: 2

D. 5: 7

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 Practice Questions - Term 1

9.
In the figure given below, sides PB and QA are perpendiculars drawn to the
line segment AB.

If PO = 6 cm, QO = 9 cm and area of ΔP OB = 120 cm , then the area of

2

ΔQOA is

A. 360 cm
2

B. 270 cm
2

C. 240 cm
2

D. 290 cm
2

10. If D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC

then, CA = _____.
2

A. BC. CD

B. BD. DC

C. BC. BC

D. AD. DC

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 Practice Questions - Term 1

11.

In the given figure, DE || BC. If AD = 6 cm, AB = 24 cm and DE = 5 cm, then

BC = ____ cm.

A. 5

B. 10

C. 20

D. 24

12. In the given figure, DE||AC, DC||AP ,

BC = 4 cm and BP = 6 cm.

BE
Find the value of .
EC

A. 1:2

B. 2:1

C. 1:3

D. 3:1

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 Practice Questions - Term 1

13.
The line segment XY is parallel to side AC of Δ ABC and it divides the
triangle into two parts of equal areas. Find the ratio .
BX

AB

A. 1

√2

B. 1

C. 4

D. √2

14.
A tower of height 24 m casts a shadow 50 m and at the same time, a girl of
height 1.8 m casts a shadow. Find the length of her shadow.

A. 3m

B. 3.25 m

C. 3.5 m

D. 3.75 m

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 Practice Questions - Term 1

15. If ABC is an equilateral triangle of side 4a, then the length of its altitude is
________.

A. 2√3a

B. 7√9a

C. 4√3a

D. 5√2a

16.
In the adjoining figure, AB = 10 cm, BC =15 cm AD : DC = 2 : 3, then ∠ABC
is equal to -

A. ∘
30

B. ∘
40

C. ∘
45

D. 110
∘

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 Practice Questions - Term 1

17. In ΔABC , AC = 15 cm and DE || BC. If AD

DB
= , then EC = ______.
2

A. 5 cm

B. 10 cm

C. 7.5 cm

D. 12.5 cm

18.
△ ABC is a right angled triangle, right angled at B. BD is perpendicular to
AC. What is AC . DC?

A. BC. AB

B. BC
2

C. BD
2

D. AB. AC

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 Practice Questions - Term 1

19.
△ABC is an isosceles triangle in which AB = AC = 13 cm. If area of △ADC
is 169 cm , then area of △EF B is equal to
2

A. 196 cm
2

B. 324 cm
2

C. 169 cm
2

D. 396 cm
2

20.
If BC ||EF and F G||CD then,
AE

AB
= _____.

A.
AG

CD

B. GD

AC

C. AG

AD

D. CF

AF

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 Practice Questions - Term 1
Date: 19/11/2021
Subject: Mathematics
Topic : Triangles Class: X

1.
If AC , then AE _____.
AD 3
I n ΔABC, DE || BC and = . = 5.6 cm =
DB 5

 A. 2.1 cm

 B. 2.4 cm

 C. 3.2 cm

 D. 3.6 cm

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 Practice Questions - Term 1

I n ΔABC,

DE || BC (Given)

AD AE
= [Basic P roportionality T heorem]
AB AC

∵ AB = (3 + 5), AD = 3

AC = 5.6 cm (Given)

3 AE
=
3+5 5.6

3 AE
=
8 5.6

3×5.6
AE = = 3 × 0.7 = 2.1 cm
8

⇒ AE = 2.1 cm

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 Practice Questions - Term 1

2. In the given figure, if DE || BC then find AD.

 A. 30 units

 B. 50 units

 C. 40 units

 D. 10 units
By Basic Proportionality Theorem,
AD AE
=
DB EC

AD 20
⇒ =
45 30

20×45
⇒ AD = = 30 units
30

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 Practice Questions - Term 1

3. Find CD, if AC = 4 cm, AB = 3 cm and AD = 2 cm.

 A. 8
cm
3

 B. 3
cm
8

 C. 4
cm
3

 D. 3
cm
4

Perpendicular from right angle to hypotenuse divides the triangle into two
similar triangles and these are similar to the whole triangle also.

∴ △CAB ∼ △CDA

AB CA
⇒ =
DA CD

3 4
⇒ =
2 CD

8
⇒ CD = cm
3

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 Practice Questions - Term 1

4.
In ΔABC , point D and E lies on the line AB and AC respectively as shown
in the figure. Find the measure of ∠AED.

 A. 65
∘

 B. 40
∘

 C. 75
∘

 D. 70
∘

In ΔABC,

AD 4 AE 4.8 4
= and = =
DB 3 EC 3.6 3

AD AE 4
∴ = =
DB EC 3

We know that in a triangle, if a line segment intersects two sides and divides
them in the same ratio, then it will be parallel to the third side.

∴ DE || BC
∘
So, ∠AED = ∠ACB = 40

(Corresponding angles are equal)

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 Practice Questions - Term 1

5.
Observe the given triangles and find the

value of ∠P .

 A. ∘
60

 B. ∘
40

 C. ∘
50

 D. ∘
65

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 Practice Questions - Term 1
In ΔABC and ΔRQP ,

AB 3.8 1
= =
RQ 7.6 2

BC 6 1
= =
QP 12 2

3√3 1
CA
and = =
PR 2
6 √3

AB BC CA
⇒ = =
RQ QP PR

So, ΔABC ∼ ΔRQP

(By SSS similarity criterion)

∴ ∠C = ∠P

(Corresponding angles of similar

triangles are equal)

∘
But, ∠C = 180 − ∠A − ∠B

(By angle sum property)

∘ ∘ ∘ ∘
∠C = 180 − 80 − 60 = 40

∘
So, ∠P = 40

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 Practice Questions - Term 1

6. In the figure, AC = 10cm, PC = 15cm, PQ = 12 cm, find PB.

 A. 6 cm

 B. 7 cm

 C. 8 cm

 D. 9 cm
I n ΔABC and ΔP QC,
∘
∠P QC = ∠ABC = 90

∠C = ∠C (common angle)

T heref ore, ΔABC ∼ ΔP QC By AA similarity

AC AB BC 10 AB BC
= = ⇒ = =
PC PQ QC 15 12 QC

⇒ AB = 8cm

I n ΔABC,

2 2 2
AB + BC = AC

2 2 2
⇒ 8 + BC = 10

⇒ BC = 6cm

P B = P C − BC = 15 − 6 = 9cm

∴ P B = 9cm.

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 Practice Questions - Term 1

7. In a △ABC , points P and Q are on sides AB and AC respectively. If AP = 3

cm, PB = 6 cm. AQ = 5 cm and QC = 10 cm, then BC = ____.

 A. 4PQ

 B. PQ

 C. 3PQ

 D. PQ
2

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 Practice Questions - Term 1
Given: AP = 3 cm, PB = 6 cm,
AQ = 5 cm and QC = 10 cm

We have, AB = AP + PB = (3 + 6) cm = 9 cm
and, AC = AQ + QC = (5 + 10) cm = 15 cm

AP 3 1
∴ = =
AB 9 3

and
AQ 5 1
= =
AC 15 3

AP AQ
⇒ =
AB AC

Thus, in △P AQ and △BAC ,

AP AQ
=
AB AC

∠P AQ = ∠BAC

(Common in both triangles)

∴ △AP Q ∼ △ABC (by SAS similarity criterion)

PQ AQ
AP
⇒ = =
AB BC AC

PQ AQ 1
⇒ = =
BC AC 3

⇒ BC = 3P Q

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 Practice Questions - Term 1

8.
Two isosceles triangles have equal angles and their areas are in the ratio 16
: 25. The ratio of corresponding heights is :

 A. 4: 5

 B. 5: 4

 C. 3: 2

 D. 5: 7

For similar triangles,

2
(Corresponding length of 1st triangle)
=
Area of 1st triangle

Area of 2nd triangle 2

(Corresponding length of 2nd triangle)

2
Area (Δ ) (h )
1 1
=
Area (Δ2 ) 2
(h2 )

2
(h1 )
16
=
25 2
(h )
2

h 4
1
=
h 5
2

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 Practice Questions - Term 1

9.
In the figure given below, sides PB and QA are perpendiculars drawn to the
line segment AB.

If PO = 6 cm, QO = 9 cm and area of ΔP OB = 120 cm , then the area of

2

ΔQOA is

 A. 360 cm
2

 B. 270 cm
2

 C. 240 cm
2

 D. 290 cm
2

In ΔP OB and ΔQOA,
∘
∠P BO = ∠QAO = 90

∠P OB = ∠QOA

(Since vertically opposite angles are equal)

ΔP OB ∼ ΔQOA

(by AA similarity criterion)

ar(ΔP OB) 2
PO
⇒ =
ar(ΔQOA) 2
QO

(The ratio of the areas of two similar triangles is equal to the square of the
ratio of their corresponding sides).
2 2
120 cm 6
⇒ =
2
ar(QOA) 9

2
120 cm ×81
2
⇒ ar(QOA) = = 270 cm
36

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 Practice Questions - Term 1

10. If D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC

then, CA = _____. 2

 A. BC. CD

 B. BD. DC

 C. BC. BC

 D. AD. DC

Given: ∠ADC = ∠BAC

In △BAC and △ADC ,

∠ACB = ∠ACD .... (common angles)

∠BAC = ∠ADC .... (Given)

∴ △BAC ∼ △ADC .... (AA simlilarity)

BA AC BC
⇒ = =
AD CD AC

AC BC
⇒ =
CD AC

2
⇒ AC = BC. CD

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 Practice Questions - Term 1

11.

In the given figure, DE || BC. If AD = 6 cm, AB = 24 cm and DE = 5 cm, then

BC = ____ cm.

 A. 5

 B. 10

 C. 20

 D. 24

Consider △ADE and △ABC ,

Given, DE || BC

∠ ADE = ∠ ABC
∠ AED =∠ ACB

(Since corresponding angles are equal)

∴ △ ADE ∼ △ ABC (by AA similarity criterion)

DE AD
⇒ =
BC AB

(corresponding sides of similar triangles are in same ratio)

DE×AB
⇒ BC =
AD

5×24
⇒ BC =
6

⇒ BC = 20 cm

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 Practice Questions - Term 1

12. In the given figure, DE||AC, DC||AP ,

BC = 4 cm and BP = 6 cm.

BE
Find the value of .
EC

 A. 1:2

 B. 2:1

 C. 1:3

 D. 3:1
Given: DE || AC.

Applying BPT to △BAC we get ,

BE BD
= ⋯ (i)
EC DA

Also, it is given that DC || AP .

Applying BPT to △BAP we get ,

BC BD
= ⋯ (ii)
CP DA

From (i) and (ii), we get

BE BC
=
EC CP

We know that, CP = BP - BC
BE BC
⇒ =
EC BP −BC

Given: BC = 4 cm and BP = 6 cm .
BE 4 2
⇒ = =
EC 6−4 1

∴ BE : EC = 2 : 1

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 Practice Questions - Term 1

13.
The line segment XY is parallel to side AC of Δ ABC and it divides the
triangle into two parts of equal areas. Find the ratio .
BX

AB

 A. 1

√2

 B. 1

 C. 4

 D. √2

We have XY || AC (given)

So, ∠ BXY = ∠ A and ∠ BYX = ∠ C (since corresponding angles are equal)

∴ △ ABC ∼ △ XBY (AA similarity criterion)

The ratio of the area of two similar triangles are equal to the ratio of the
squares of their corresponding sides.
2
ar(△ABC)
So, ar(△XBY )
= (
AB

XB
) ​

Also, ar(△ABC) = 2 × ar(△XBY )

ar(△ABC)
So, =
2

1
ar(△XBY )

Therefore, (
AB 2
) =
XB 1

√2
AB
⇒ =
XB 1

Taking reciprocal on both sides, we get

BX 1
=
AB √2

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 Practice Questions - Term 1

14.
A tower of height 24 m casts a shadow 50 m and at the same time, a girl of
height 1.8 m casts a shadow. Find the length of her shadow.

 A. 3m

 B. 3.25 m

 C. 3.5 m

 D. 3.75 m

Let AB be the height of tower, BC be the length of the shadow of the tower,
ED be the height of the girl and EC = x be the length of her shadow.

In △ABC and △DEC ,

∘
∠ABC = ∠DEC = 90

∠BCA = ∠ECD (common in both triangles)

Therefore, △ABC ∼ △DEC (by AA similarity criterion)

⇒
DE

AB
=
EC

BC
(corresponding sides of similar triangles are in same ratio)

BC
⇒ EC = DE ×
AB

50
⇒ EC = 1.8 ×
24

⇒ EC = 3.75 m

Hence, length of the her shadow is 3.75 m.

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 Practice Questions - Term 1

15. If ABC is an equilateral triangle of side 4a, then the length of its altitude is
________.

 A. 2√3a

 B. 7√9a

 C. 4√3a

 D. 5√2a

Given: Length of each side of the equilateral triangle = 4a

Construction: Draw a perpendicular from A to BC. Let this meeting point be
D.

In △ABD and △ACD,

AB = AC
(∵ All sides of an equilateral triangle are equal)
AD = AD (Common)
∘
∠ADB = ∠ADC = 90
∴ △ABD ≅ △ACD

(By RHS congruency)

⇒ BD = DC = 2a (by cpct)

In △ABD,
AB
2
= AD
2
+ BD
2
(By pythagoras theorm)
2 2 2
⇒ AD = AB − BD
2 2 2
⇒ AD = (4a) − (2a)
2 2 2
⇒ AD = 16a − 4a
2 2
⇒ AD = 12a

⇒ AD = 2√3a

∴ Altitude of an equilateral triangle with side 4a will be 2√3a.

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 Practice Questions - Term 1

16.
In the adjoining figure, AB = 10 cm, BC =15 cm AD : DC = 2 : 3, then ∠ABC
is equal to -

 A. 30
∘

 B. 40
∘

 C. 45
∘

 D. 110
∘

Clearly, = and = =
AD 2 AB 10 2

DC 3 BC 15 3

So, AD
= AB

DC BC

Thus, BD divides AC in the ratio of the other two sides.

By Converse of Angular Bisector theorem,

BD is the bisector of ∠B
∴ ∠ ABC = 2 ( ∠CBD)

Now, ∠CBD = 180 - (130 + 30 ) .... ∘ ∘ ∘

(Angle Sum
Property)
= 20 ∘

But , ∠ABC = 2 ( ∠CBD)

= 2 x 20
= 40 ∘

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 Practice Questions - Term 1

17. In ΔABC , AC = 15 cm and DE || BC. If AD

DB
= , then EC = ______.
2

 A. 5 cm

 B. 10 cm

 C. 7.5 cm

 D. 12.5 cm
Given: AC = 15 cm
DE∥ BC
AD 2
=
DB 1

In Δ ADE and Δ ABC,

∠DAE = ∠BAC [ Common ]

∠ADE = ∠ABC and ∠AED = ∠ACB

[ ∵ DE∥ BC, corresponding angles are equal ]

⇒ ΔADE ∼ ΔABC [AA similarity]

From Basic Proportionality theorem,

AD AE
=
DB EC
AE 2
⇒ =
EC 1

⇒ AE = 2EC

Given : AC = 15cm

⇒ AE + EC = 15cm

⇒ 3EC = 15cm

⇒ EC = 5cm

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 Practice Questions - Term 1

18.
△ ABC is a right angled triangle, right angled at B. BD is perpendicular to
AC. What is AC . DC?

 A. BC. AB

 B. BC
2

 C. BD
2

 D. AB. AC

Consider △ADB and △ABC

∠BAD = ∠BAC [common angle]

∠BDA = ∠ABC [ 90 ]∘

Therefore by AA similarity criterion, △ADB and △ABC are similar.

So, AB

AC
=
AD

AB
⇒ AB
2
= AC.AD ---(I)

Similarly, △BDC and △ABC are similar.

So, = = AC.DC ---(II)

BC DC 2

AC BC
⇒ BC

Dividing (I) and (II) and cancelling out AC, we get, (

AB

BC
) = AD

DC
----(III)

Also, In △ADB, AB = AD + DB and in 2 2 2

△BDC , CB = CD + DB
2 2 2

[Pythagoras theorem]

Subtracting these two equations above and cancelling off DB on both 2

sides, we get

AB
2
- BC = AD - CD ⇒
2 2 2
AB
2
+ CD = AD + BC
2 2 2
--------------(IV)
2 2 2 2

Dividing this equation with BC on both sides, + = +

2 AB CD AD BC

2 2 2 2
BC BC BC BC
2 2 2

+ = +1
AB CD AD
⇒
2 2 2
BC BC BC

2 2 2

-1= -
AB AD CD
⇒
2 2 2
BC BC BC

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 Practice Questions - Term 1
2

From (III), we know that (

AB
)
BC
= AD

DC

2 2
AD −CD
⇒
AD

DC
-1= 2
[Using (III)]
BC

(AD−CD)(AD+CD)
⇒
AD−DC

DC
= 2
BC

⇒
DC
1
= AC

2
BC

2
⇒ BC = AC. DC

Alternatively,

Consider △ABC and △BDC ,

∘
∠ABC = ∠BDC = 90

∠C = ∠C [Common angle]

Therefore by AA similarity criterion, △ABC and △BDC are similar.

AC

BC
= BC

DC

BC
2
= AC × DC

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 Practice Questions - Term 1

19.
△ABC is an isosceles triangle in which AB = AC = 13 cm. If area of △ADC
is 169 cm , then area of △EF B is equal to
2

 A. 196 cm
2

 B. 324 cm
2

 C. 169 cm
2

 D. 396 cm
2

In ΔADC and ΔEF B,

∠ACD = ∠EBF

(Since base angles of an isosceles triangle are equal)

∘
∠ADC = ∠EF B (given 90 )

ΔADC ∼ ΔEF B (by AA similarity criterion)

2
ar(ΔADC)
So,
AC
=
2
ar(ΔEF B) (EB)

(Since the ratio of the areas of two similar triangles is equal to the square of
the ratio of their corresponding sides).
2 2
169 cm 13
⇒ =
2
ar(ΔEF B) (EC+CB)

2 2
169 cm 13
⇒ =
2
ar(ΔEF B) (8+10)

(18×18)×169
2
⇒ ar(ΔEF B) = = 324 cm
2
(13)

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 Practice Questions - Term 1

20.
If BC ||EF and F G||CD then,
AE
= _____.
AB

 A. AG

CD

 B. GD

AC

C.
AG
 AD

D.
CF
 AF

In ΔABC,

EF ∥ BC

AE AF
∴ = . . . . (1)
AB AC

(Basic proportionality theorem)

In ΔACD,

F G ∥ CD

AF AG
∴ = . . . . (2)
AC AD

(Basic proportionality theorem)

From 1 & 2,

AE AF AG
= =
AB AC AD

AE AG
⇒ =
AB AD

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