Triangles, Congruence and Similarity MCQ Quiz - Objective Question with Answer for Triangles, Congruence and Similarity - Download Free PDF

Last updated on May 13, 2025

This Topic is generally considered an easy Topic to gain conceptual understanding but Triangles, Congruence and Similarity MCQs Quiz can be tricky to deal with. The trick here is to really understand and apply different properties and conditions of triangles. Practice Triangles, Congruence and Similarity objective questions with this curation of questions by Testbook and don’t miss out on any Triangles, Congruence and Similarity Question Answers. Also get some tips, tricks and shortcuts to solve them with ease.

Latest Triangles, Congruence and Similarity MCQ Objective Questions

Triangles, Congruence and Similarity Question 1:

In ΔABC, ∠C = 90° and CD ⊥ AB, also ∠A = 65°, then ∠CBA is equal to

22000

  1. 25°
  2. 35°
  3. 65°
  4. 40°
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 25°

Triangles, Congruence and Similarity Question 1 Detailed Solution

Given:

∠C = 90° 

CD ⊥ AB

∠A = 65°

Concept used:

The sum of the angles of a triangle is 180°.

Calculation:

In ΔABC,

∠BAC + ∠CBA  + ∠ACB  = 180°

⇒ 65° + 90° + ∠CBA  = 180°

⇒ ∠CBA  = 25°

Important Points

We can arrive at the solution based on the perpendicular as well. But it is more time-consuming. Since ∠A and ∠C are given, it is wise to directly apply them in the concept and get the solution.

Triangles, Congruence and Similarity Question 2:

In triangle ABC, points D and E are on AB and AC respectively such that DE is parallel to BC. If AD = 6 cm, DB = 4 cm, AE = 9 cm, then the length of EC (in cm) is

  1. 7
  2. 6.4
  3. 6
  4. 5.5

Answer (Detailed Solution Below)

Option 3 : 6

Triangles, Congruence and Similarity Question 2 Detailed Solution

Given:

In triangle ABC:

Points D and E lie on AB and AC respectively

DE ∥ BC

AD = 6 cm, DB = 4 cm ⇒ AB = AD + DB = 10 cm

AE = 9 cm

Concept used:

By Basic Proportionality Theorem (Thales’ Theorem):

If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides those sides in the same ratio.

Formula used:

AD / DB = AE / EC

Calculation:

6 / 4 = 9 / EC

⇒ 3 / 2 = 9 / EC

⇒ EC = (2 × 9) ÷ 3 = 18 ÷ 3 = 6 cm

∴ The length of EC is 6 cm

Triangles, Congruence and Similarity Question 3:

If the ratio of corresponding sides of two similar triangles is 1 ∶ 3, then what is the ratio of the areas of the two triangles?

  1. 1 ∶ 3
  2. 1 ∶ 9
  3. 1 ∶ 27
  4. 1 ∶ 81

Answer (Detailed Solution Below)

Option 2 : 1 ∶ 9

Triangles, Congruence and Similarity Question 3 Detailed Solution

Given:

If the ratio of corresponding sides of two similar triangles is 1 ∶ 3

Formula used:

Ratio of areas of two similar triangles = (Ratio of sides)2

Calculation:

Ratio of areas = (1 ∶ 3)2

⇒ Ratio of areas = 12 ∶ 32

⇒ Ratio of areas = 1 ∶ 9

∴ The correct answer is option (2).

Triangles, Congruence and Similarity Question 4:

If the bisectors of angle ∠ABC and ∠ACB of a triangle ABC meet at a point 0 and if ∠A = 60°, then ∠BOC =

  1. 135°
  2. 120°
  3. 105°
  4. 75°

Answer (Detailed Solution Below)

Option 2 : 120°

Triangles, Congruence and Similarity Question 4 Detailed Solution

- www.bijoux-oeil-de-tigre.com

In triangle \( ABC \), if the bisectors of \( \angle ABC \) and \( \angle ACB \) meet at point \( O \) and \( \angle A = 60^\circ \), then the angle \( \angle BOC \) is given by:

\[ \angle BOC = 90^\circ + \frac{\angle A}{2} \]

Substituting \( \angle A = 60^\circ \):

\[ \angle BOC = 90^\circ + \frac{60^\circ}{2} = 120^\circ \]

Thus, the measure of \( \angle BOC \) is:

\[ \boxed{120^\circ} \]

Triangles, Congruence and Similarity Question 5:

Let Δ ABC ∼ Δ QPR and \(\rm \frac{area \ of \ \Delta ABC}{area \ of \ \Delta PQR}=\frac{64}{25}\) If QP = 12 cm, PR = 10 cm and AC = 15 cm, then the length of AB is:

  1. 12.8 cm
  2. 24 cm
  3. 16 cm
  4. 19.2 cm

Answer (Detailed Solution Below)

Option 4 : 19.2 cm

Triangles, Congruence and Similarity Question 5 Detailed Solution

Given:

Δ ABC ∼ Δ QPR

\(\frac{area \ of \ Δ ABC}{area \ of \ Δ PQR}=\frac{64}{25}\)

Formula used:

In similar triangles, the ratio of the areas is equal to the square of the ratio of the corresponding sides.

\(\frac{area \ of \ Δ ABC}{area \ of \ Δ PQR} = (\frac{corresponding \ side \ of \ Δ ABC}{corresponding \ side \ of \ Δ PQR})^2\)

Calculation:

5-5-2025 IMG-1083 Prakhar -2

\(\frac{64}{25} = (\frac{AB}{QP})^2\)

⇒ \(\frac{64}{25} = (\frac{AB}{12})^2\)

⇒ \(\frac{64}{25} = \frac{AB^2}{144}\)

⇒ \(\frac{64}{25} × 144 = AB^2\)

⇒ \(\frac{64 × 144}{25} = AB^2\)

⇒ \(\frac{9216}{25} = AB^2\)

⇒ 368.64 = \(AB^2\)

⇒ AB = \(\sqrt{368.64}\)

⇒ AB = 19.2 cm

∴ The correct answer is option (4).

Top Triangles, Congruence and Similarity MCQ Objective Questions

ABC is a right-angled triangle. A circle is inscribed in it. The length of the two sides containing the right angle are 10 cm and 24 cm. Find the radius of the circle.

  1. 3 cm
  2. 5 cm
  3. 2 cm
  4. 4 cm

Answer (Detailed Solution Below)

Option 4 : 4 cm

Triangles, Congruence and Similarity Question 6 Detailed Solution

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Given:

ABC is a right-angled triangle. A circle is inscribed in it.

The length of the two sides containing the right angle are 10 cm and 24 cm

Calculations:

Hypotenuse² = 10² + 24²    (Pythagoras theorem)

Hypotenuse = √676 = 26

Radius of the circle (incircle) inside a triangle = ( Sum of sides containing right angle – Hypotenuse)/2

⇒ (10 + 24 - 26)/2

⇒ 8/2

⇒ 4

∴ The correct choice is option 4.

ΔABC ~ ΔPQR, the areas of ΔABC and ΔPQR are 64 cm2 and 81 cm2, respectively and AD and PT are the medians of ΔABC and ΔPQR, respectively. If PT = 10.8 cm, then AD = ?

  1. 9 cm
  2. 12 cm
  3. 8.4 cm
  4. 9.6 cm

Answer (Detailed Solution Below)

Option 4 : 9.6 cm

Triangles, Congruence and Similarity Question 7 Detailed Solution

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Given:

ΔABC ~ ΔPQR

Area of ΔABC = 64 cm2

Area of ΔPQR = 81 cm2

PT = 10.8 cm 

AD is the median of ΔABC.

PT is the median of ΔPQR

Concept: 

F1 Ashish Madhu 19.10.21 D2

The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides & medians. 

\(\text{Area of (ΔABC)}\over{\text{Area of (ΔPQR)}}\) = \(({AD\over PT})^2\)

Calculation:

\(\text{Area of (ΔABC)}\over{\text{Area of (ΔPQR)}}\) = \(({AD\over PT})^2\)

⇒ \(64\over81\) = \(({AD\over PT})^2\) 

​ \(AD\over PT\) = \(\sqrt{64\over81}\) = \(8\over9\)

​ \(AD\over 10.8\) = \(8\over9\)

​ AD = \({8\times10.8}\over9\) = 9.6 cm

∴ AD = 9.6 cm 

The sides of similar triangles ΔPQR and ΔDEF are in the ratio 5 ∶ 6. If area of ΔPQR is equal to 75 cm2, what is the area of ΔDEF?

  1. 150 cm2
  2. 90 cm2
  3. 108 cm2
  4. 120 cm2

Answer (Detailed Solution Below)

Option 3 : 108 cm2

Triangles, Congruence and Similarity Question 8 Detailed Solution

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Given:

ΔPQR ∼ ΔDEF

The sides of ΔPQR and ΔDEF are in the ratio 5 ∶ 6.

ar(PQR) = 75 cm2

Concepts used:

The ratio of area of similar triangles is equal to the square of the ratio of sides of corresponding triangles.

Calculation:

ΔPQR ∼ ΔDEF

ar(PQR)/ar(DEF) = (Side of ΔPQR/Side of ΔDEF)2

⇒ 75/ar(DEF) = (5/6)2

⇒ ar(DEF) = 108 cm2

∴ Area of ΔDEF is equal to 108 cm2.

In ΔABC, ∠A = 135°, CA = 5√2 cm and AB = 7 cm. E and F are midpoints of sides AC and AB, respectively. The length of EF (in cm) is:

  1. 5.5
  2. 6.5
  3. 6
  4. 5

Answer (Detailed Solution Below)

Option 2 : 6.5

Triangles, Congruence and Similarity Question 9 Detailed Solution

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Given:

AB (c)  = 7 cm, CA (b) = 5√2 cm, BC = a 

E and F are the mid-points of AC & AB.

∠A = 135°

Concept:

By cosine rule

Cos A = (b2 + c2 - a2)/ 2bc

Mid-point theorem

EF = BC/2

Calculation:

F1 Harshit 15-09-21 Savita D2

According to the cosine rule

\(cos135^0= \frac{(5\sqrt2)^2 + (7)^2 - (a)^2}{2\times7\times5\sqrt2}\)

⇒ \(\frac{-1}{\sqrt2}= \frac{(5\sqrt2)^2 + (7)^2 - (a)^2}{2\times7\times5\sqrt2}\)

⇒ \(-70= 50 + 49 - a^2 \)

⇒ a2 = 169

⇒ a = 13, a ≠ - 13

⇒ BC = 13

From formula used

EF = BC/2 = 13/2 cm 

∴ EF = 6.5 cm

In ΔABC, O is the orthocenter and I is the incenter for the given triangle, If ∠BIC - ∠BOC = 90, then find the ∠A.

  1. 120
  2. 140
  3. 90
  4. 180

Answer (Detailed Solution Below)

Option 1 : 120

Triangles, Congruence and Similarity Question 10 Detailed Solution

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Given:

In ΔABC, O is the orthocenter and I is the incenter for the given triangle,  

If ∠BIC - ∠BOC = 90.

Formula used:

(1) In ΔABC, I is the incenter for the given triangle,

F1 Madhuri Railways 22.06.2022 D12 

(1.1) ∠BIC = 90 + \(\frac{1}{2}\)∠A

(1.2) ∠AIC = 90 + \(\frac{1}{2}\)∠B

(1.3) ∠AIB = 90 + \(\frac{1}{2}\)∠C

(2) In ΔABC, O is the orthocenter for the given triangle,

F1 Madhuri Railways 22.06.2022 D13

(2.1) ∠BOC = 180∘ - ∠A

(2.2) ∠AOB = 180∘ - ∠C

(3.3) ∠AOC = 180∘ - ∠B

Calculation:

According to the question, the required image is:

F1 Madhuri Railways 22.06.2022 D14

As we know,

∠BOC = 180∘ - ∠A     ----(1)

∠BIC = 90 + \(\frac{1}{2}\)∠A    ----(2)

Now, subtract equation (1) from (2).

⇒ ∠BIC - ∠BOC = 90 + \(\frac{1}{2}\)∠A  - (180∘ - ∠A )

⇒ 90 = 90 + \(\frac{1}{2}\)∠A  - 180 + ∠A

⇒ 90 = \(\frac{3}{2}\)∠A  - 90

⇒ 180 = \(\frac{3}{2}\)∠A

⇒ ∠A = 120

∴ The required answer is 120.

Additional Information

(1) Incenter - It is the intersection point of all three angle bisectors of a triangle.

(1.1) Angle bisector cuts the angle into two equal half.

(2) Orthocenter - It is the intersection point of all three altitudes drawn from the vertex to the opposite side of the triangle.

(2.1) The altitude of a triangle is perpendicular to the opposite side.

In ΔABC, AB = 8 cm. ∠A is bisected internally to intersect BC at D. BD = 6 cm and DC = 7.5 cm. What is the length of CA?

  1. 12 cm
  2. 12.5 cm
  3. 10.5 cm
  4. 10 cm

Answer (Detailed Solution Below)

Option 4 : 10 cm

Triangles, Congruence and Similarity Question 11 Detailed Solution

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Given:

In ΔABC, AB = 8cm

∠A bisected internally to intersect BC at D.

AB = 8 cm, BD = 6 cm and DC = 7.5 cm

Concept used:

F1 SSC Priya 14-3-24 D22

In a triangle, the angle bisector of an angle is divides the opposite side to the angle in the ratio of the remaining two sides.

\(\frac{AB}{AC} = \frac{BD}{DC}\)

Calculation:

AB = 8 cm, ∠A is bisected internally to intersect BC at D,

⇒ AB/AC = BD/CD

⇒ 8/AC = 6/7.5

∴ AC = 10 cm

ΔPQR inscribes in a circle with centre O. If PQ = 12 cm, QR = 16 cm and PR = 20 cm, find the circumradius of a triangle.

  1. 10 cm
  2. 8 cm
  3. 6 cm
  4. 20 cm

Answer (Detailed Solution Below)

Option 1 : 10 cm

Triangles, Congruence and Similarity Question 12 Detailed Solution

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Given:

In ΔPQR,

PQ = 12 cm, QR = 16 cm and PR = 20 cm

Concept used:

a, b and c denote the sides of the triangle and A denotes the area of the triangle,

Then the measure of the circumradius(r) is.

r = [abc/4A]

If the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the biggest side is a right angle.

Hypotenuse2 = Perpendicular2 + Base2 

Calculation:

F6 Amar TTP 3.0 18-5-2021 Swati D2

Here, we can see that 

(20)2 = (16)2 + (12)2  = 400 

⇒ PR2 = QR2 + PQ2

So, ΔPQR is a right angle triangle.

Area of ΔPQR = (½ ) × base × Perpendicular

⇒ Area of ΔPQR = (½ ) × 16 × 12

⇒ Area of ΔPQR = 96 cm2

Circumradius (r) = [abc/4 × Area] 

Circumradius (r) = [(12 × 16 × 20)/4 × 96] = 10 cm

∴ The circumradius of a triangle is 10 cm.

 

For a right-angled triangle, the circumcenter lies at the midpoint of the hypotenuse. All the vertices of a triangle are at an equal distance from the circumcenter.

PO = QO = OR = r

⇒ PO = PR/2

⇒ PO = 20/2 = 10 cm

 The circumradius of a triangle is 10 cm.

If Δ ABC ∼ Δ QPR, \(\rm \frac{ar(\Delta ABC)}{ar(\Delta PQR)}=\frac{4}{9}\), AC = 12 cm, AB = 18 cm and BC = 10 cm, then PR (in cm) is equal to:

  1. 15
  2. 8
  3. 10
  4. 18

Answer (Detailed Solution Below)

Option 1 : 15

Triangles, Congruence and Similarity Question 13 Detailed Solution

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Given:

 Δ ABC ∼ Δ QPR, \(\rm \frac{ar(Δ ABC)}{ar(Δ PQR)}=\frac{4}{9}\),

AC = 12 cm, AB = 18 cm and BC = 10 cm

Concept used:

F1 Vinanti Teaching 13.10.22 D5

Δ ABC ∼ Δ QPR 

 Ratio of respective areas = Ratio of square of respective sides

that is, \(\rm \frac{ar(Δ ABC)}{ar(Δ QPR)}=\frac{AB^2}{QP^2}=\frac{BC^2}{PR^2}=\frac{AC^2}{QR^2}\)

Calculation:

⇒ 4/9 = (10)2/PR2

PR2 = 900/4

⇒ PR2 = 225

⇒ PR = 15 cm

Mistake Points In this question, it is given that ΔABC is similar to ΔQPR. Do not misread as ΔPQR.

Δ ABC ∼ Δ QPR, this relation matters while applying the similarity rule, 

What is given in the ratio form, is just for your confusion.

In a triangle ABC, AB : AC = 5 : 2, BC = 9 cm. BA is produced to D, and the bisector of the Angle CAD meets BC produced at E. What is the length (in cm) of CE?

  1. 9
  2. 6
  3. 10
  4. 3

Answer (Detailed Solution Below)

Option 2 : 6

Triangles, Congruence and Similarity Question 14 Detailed Solution

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Given:

F1 Arun Madhu 13.01.22  D1

AB : AC = 5 : 2

BC = 9 cm

Concept used:

F1 Arun Madhu 13.01.22  D1

By external angel bisector theorem,

AB/AC = BE/CE

Calculation:

According to the given question

AB/AC = BE/CE (Using external angle bisector theorem)

⇒ 5/2 = (9 + CE)/CE

⇒ 5CE = 18 + 2CE

⇒ 3CE = 18

⇒ CE = 6 cm

∴ The length of CE is 6

Additional InformationBisector angle theorem - The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.

In ΔABC, MN∥BC, the area of quadrilateral MBCN = 130 cm2. If AN : NC = 4 : 5, then the area of ΔMAN is:

  1. 45 cm2
  2. 65 cm2
  3. 40 cm2
  4. 32 cm2

Answer (Detailed Solution Below)

Option 4 : 32 cm2

Triangles, Congruence and Similarity Question 15 Detailed Solution

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F1 A.K 13.5.20 Pallavi D1

Given,

the area of quadrilateral MBCN = 130 cm2.

AN : NC = 4 : 5

Hence, AC = 4 + 5 = 9

In ΔABC, If MN∥BC, then

Area of ΔAMN/area of ΔABC = (AN/AC)2

Area of ΔAMN/area of ΔABC = (4/9)2 = 16/81

Area of ΔAMN = 16 unit and area of ΔABC = 81 unit

Now, Area of quadrilateral MBCN = area of ΔABC - area of ΔAMN

⇒ 81 - 16 unit = 130 cm2

⇒ 65 unit = 130 cm2

⇒ 1 unit = 130/65 = 2 cm2

∴ Area of ΔAMN = 16 × 2 = 32 cm2
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