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

Last updated on Jun 6, 2025

Congruence and Similarity MCQs Quiz are framed in a confusing manner and require practice and patience to solve them. Generally considered an easy Topic to understand, Congruence and Similarity holds great importance in Geometry and can be tricky to deal with. Practice this set of Congruence and Similarity objective questions to improve your grasp on the Topic. The Congruence and Similarity Question Answers would be solved even more quickly with the tips and tricks given in this article.

Latest Congruence and Similarity MCQ Objective Questions

Congruence and Similarity Question 1:

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

Congruence and Similarity Question 1 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).

Congruence and Similarity Question 2:

If the ratio of corresponding sides of two similar triangles is 5 : 7, then what is the ratio of the areas of the two triangles?

  1. 125 : 49
  2. 25 : 49
  3. 25 : 7
  4. 5 : 7

Answer (Detailed Solution Below)

Option 2 : 25 : 49

Congruence and Similarity Question 2 Detailed Solution

Given:

The ratio of corresponding sides of two similar triangles = 5 : 7

Formula Used:

The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides.

Calculation:

Let the ratio of the corresponding sides be 5/7.

The ratio of the areas of the two triangles = (Ratio of sides)2" id="MathJax-Element-9-Frame" role="presentation" style="position: relative;" tabindex="0">(Ratio of sides)2

⇒ (5/7)2

⇒ 25/49

The ratio of the areas of the two triangles is 25 : 49.

Congruence and Similarity Question 3:

In Δ PQR, a line joining point X on side PQ and point Y on side PR is parallel to side QR. If the ratio PY ∶ YR is 3 ∶ 5 and length PX = 7 cm, then the length of side PQ is:

  1. 14.66 cm
  2. 18 cm
  3. 18.66 cm
  4. 14 cm

Answer (Detailed Solution Below)

Option 3 : 18.66 cm

Congruence and Similarity Question 3 Detailed Solution

Given:

In ΔPOR, a line joining point X on side PQ and point Y on side PR is parallel to side QR.

The ratio PY ∶ YR is 3 ∶ 5.

Length PX = 7 cm.

Formula used:

By the basic proportionality theorem (Thales' theorem), if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

Calculations:

Let the length of PQ be x cm.

According to the basic proportionality theorem,

\(\dfrac{PX}{XQ} = \dfrac{PY}{YR}\)

Given PY : YR = 3 : 5,

\(\dfrac{PX}{XQ} = \dfrac{3}{5}\)

Let XQ = y cm,

\(\dfrac{7}{y} = \dfrac{3}{5}\)

⇒ 7 × 5 = 3 × y

⇒ 35 = 3y

⇒ y = \(\dfrac{35}{3} cm\)

Therefore,

PQ = PX + XQ = 7 cm + \(\dfrac{35}{3} cm\)

⇒ PQ = \(\dfrac{21}{3} + \dfrac{35}{3}\)

⇒ PQ = \(\dfrac{56}{3} cm\)

⇒ PQ = 18.66 cm

∴ The correct answer is option (3).

Congruence and Similarity Question 4:

ΔABC ∼ ΔQRP and PQ = 6 cm, QR = 8 cm and PR = 10 cm. If ar (ΔABC) : ar (ΔPQR) = 1 : 4, then AB is equal to:

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

Answer (Detailed Solution Below)

Option 4 : 4 cm

Congruence and Similarity Question 4 Detailed Solution

Given:

ΔABC ~ ΔQRP

QR = 8 cm

PR = 10 cm

PQ = 6 cm

ar(ΔABC) : ar(ΔQRP) = 1 : 4

Formula used:

If ΔABC ~ ΔQRP,

Then ar(ΔABC) / ar(ΔQRP) = (AB/QR)2 = (BC/RP)2 = (AC/QP)2

Calculation:

ar(ΔABC) / ar(ΔQRP) = 1/4

(AB/QR)2 = 1/4

⇒ (AB/8)2 = 1/4

⇒ AB/8 = √(1/4)

⇒ AB/8 = 1/2

⇒ AB = 8/2

⇒ AB = 4 cm

∴ AB = 4 cm

Congruence and Similarity Question 5:

In ΔPQR and ΔXYZ,  the given side lengths are PQ = 5 cm, QR = 6 cm, PR = 7 cm and XY = 5 cm, YZ = 6 cm, XZ = 7 cm. Which of the following statements is true?

  1. ΔPQR ≅ ΔYXZ
  2. ΔPRQ ≅ ΔXYZ
  3. ΔPQR ≅ ΔXYZ
  4. ΔQPR ≅ ΔXYZ

Answer (Detailed Solution Below)

Option 3 : ΔPQR ≅ ΔXYZ

Congruence and Similarity Question 5 Detailed Solution

Given:

In ΔPQR:

PQ = 5 cm, QR = 6 cm, PR = 7 cm

In ΔXYZ:

XY = 5 cm, YZ = 6 cm, XZ = 7 cm

Formula used:

SSS (Side-Side-Side) Congruence Rule:

If three sides of one triangle are equal to three sides of another triangle, then the triangles are congruent.

Calculation:

Comparing side lengths:

PQ = XY = 5 cm

QR = YZ = 6 cm

PR = XZ = 7 cm

Since all corresponding sides are equal, by the SSS Congruence Rule:

⇒ ΔPQR ≅ ΔXYZ

∴ The correct statement is ΔPQR ≅ ΔXYZ.

Top Congruence and Similarity MCQ Objective Questions

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

Congruence and Similarity Question 6 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.

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

Congruence and Similarity Question 7 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 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

Congruence and Similarity Question 8 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

In the given triangle, O is the incentre, AE = 4 cm, AC = 9 cm and BC = 10 cm. What is the length of side AB?

19.11.2018.002

  1. 12 cm
  2. 8 cm
  3. 10 cm
  4. 14 cm

Answer (Detailed Solution Below)

Option 2 : 8 cm

Congruence and Similarity Question 9 Detailed Solution

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∵ AE + EC = AC

⇒ EC = 5 cm

According to angle bisector theorem,

So, AE/EC = AB/BC

⇒ 4/5 = AB/10

∴ AB = 8 cm

In a ΔABC right angled at B, D is a point on AC such that BD is an angle bisector of B. If AD = 12 cm, CD = 16 cm then find the perimeter of triangle ABC.

  1. 49.6 cm
  2. 67.2 cm
  3. 56.4 cm 
  4. 48 cm

Answer (Detailed Solution Below)

Option 2 : 67.2 cm

Congruence and Similarity Question 10 Detailed Solution

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Given

AD = 12 cm, CD = 16 cm 

Formula used 

Angle Bisector theorem,

AD/CD = AB/BC

Perimeter of triangle = Sum of all sides

Calculation

F1 Ashish Shraddha 06.08.2021 D2 1

In Δ ABC, BD is an angle bisector of ∠B

AD/CD = AB/BC

⇒ 12/16 = AB/BC

⇒ AB : BC = 3 : 4

From the triplets of right angle triangle

If AB = 3x and BC = 4x

then AC = 5x

AC = 12 + 16 = 28 cm

⇒ 5x = 28 cm

⇒ x = 5.6 cm

Perimeter of triangle = 5x + 3x + 4x = 12x

⇒ 67.2 cm

∴ The required answer is 67.2 cm

In ΔDEF, DE = 9 cm, EF = 12 cm and DF = 7 cm. If DO is the angle bisector of ∠EDF meets EF at O, then find the length of OF.

  1. 5.25 cm
  2. 4.35 cm
  3. 4.00 cm
  4. 4.25 cm

Answer (Detailed Solution Below)

Option 1 : 5.25 cm

Congruence and Similarity Question 11 Detailed Solution

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Given:

DE = 9, EF = 12 cm, and DF = 7 cm

DO is the angle bisector of ∠EDF

Concept used:

In a triangle, If AD is the angle bisector of ∠BAC intersect the opposite side BC at D then

F1 Amar Kumar Anil 19-05.21 D26

By angle bisector theorem

AB/AC = BD/CD

Calculation:

DE = 9, EF = 12 cm, and DF = 7 cm

DO is the angle bisector of ∠EDF

F1 Amar Kumar Anil 19-05.21 D27

By angle bisector theorem,

DE/DF = EO/OF

⇒ 9/7 = (EF - OF)/OF

⇒ 9/7 = (12 - OF)/OF

⇒ 9 × OF = 84 - 7 × OF

⇒ 16 × OF = 84

⇒ OF = 84/16

⇒ OF = 5.25 cm

∴ The value of OF is 5.25 cm.

If ΔABC and ΔPQR are similar. AB = 8 cm, PQ = 12 cm, QR = 18 cm and RP = 24 cm, then the perimeter of ΔABC is _________ cm.

  1. 36
  2. 54
  3. 27
  4. 42

Answer (Detailed Solution Below)

Option 1 : 36

Congruence and Similarity Question 12 Detailed Solution

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Given :-

ΔABC and ΔPQR are similar

AB = 8 cm

PQ = 12 cm

QR = 18 cm 

RP = 24 cm

Concept Use :-

Two triangles are similar if they have the same ratio of corresponding sides and equal pair of corresponding angles. 

browse

(1) AB/PQ = BC/QR = AC/PR

(2) angle A = angle P , angle B = angle Q , angle C = angle R

Calculation :-

⇒  AB/PQ = BC/QR = AC/PR

 8/12 = BC/18 = AC/24

⇒ BC = (18 × 8) /12 = 12cm

and,

⇒ AC = (12 × 24) /18 = 16 cm

Now,

 Perimeter of triangle ABC = AB +BC + AC = 8 + 12 + 16

⇒ Perimeter of triangle ABC = 36cm

The area of ΔABC is 44 cm2. If D is the midpoint of BC and E is the midpoint of AB, then the area (in cm2) of ΔBDE is:

  1. 44
  2. 22
  3. 11
  4. 5.5

Answer (Detailed Solution Below)

Option 3 : 11

Congruence and Similarity Question 13 Detailed Solution

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Given:

Area of ΔABC = 44 cm2

BD = CD

AE = BE

Concept used:

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion), and hence the two triangles are similar.

The ratio of areas of two similar triangles is proportional to the squares of the corresponding sides of both triangles.

Calculation:

F1 Ashish 07-04-2020 Savita D2

According to the concept,

DE ∥ AC, and AC = 1/2 × DE

Let BC = 2 unit, and BD = 1 unit

Area of ΔBDE/Area of ΔBCA = (BD/BC)2

⇒ Area of ΔBDE/44 = (1/2)2

⇒ Area of ΔBDE = (1/4) × 44 = 11 cm2

∴ The area of ΔBDE is 11 cm2

In the figure given below, ∠AED = ∠BAC, AD || BC, AD : BC = 7 : 9 and AC = 36 cm, then find the length of EC.

F1 Shraddha Ashish 05.11.2020 D11

  1. 10 cm
  2. 12 cm
  3. 8 cm
  4. 16 cm

Answer (Detailed Solution Below)

Option 3 : 8 cm

Congruence and Similarity Question 14 Detailed Solution

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GIVEN:

F1 Shraddha Ashish 05.11.2020 D11

∠AED = ∠BAC

AD || BC

AD : BC = 7 : 9

AC = 36 cm

CONCEPT:

Similarity of triangle

CALCULATION:

Since ∠AED = ∠BAC  

∠EAD = ∠ACB (alternate interior angle)

Rule of similarity AA(Angle-Angle)

∆EDA and ∆ABC are similar.

Now,

AD/BC = AE/AC

⇒ 7/9 = AE/36

⇒ AE = 28 cm

Now,

EC = AC – AE = 36 – 28 = 8 cm

∴ The value of EC is 8 cm 

In a Δ ABC, ∠BAC = 90°, A perpendicular AD is drawn from point A on line BC. Which among the following is the mean proportional between BD and BC?

  1. CD
  2. AB
  3. AD
  4. AC

Answer (Detailed Solution Below)

Option 2 : AB

Congruence and Similarity Question 15 Detailed Solution

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Given:

In a Δ ABC, ∠BAC = 90°, AD is drawn perpendicular from A on BC

Concept used:

Similarity of triangles

Calculation:

 F1 Arun k 23-2-22 Savita D10

Since AD⊥BC, ΔBAC ~ ΔBDA

So, we know,

BC/AB = AB/BD

⇒ BC × BD = AB2 

⇒ BC : AB :: AB : BD

∴ AB is the following is the mean proportional between BD and BC.

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