Congruence and Similarity MCQ Quiz - Objective Question with Answer for Congruence and Similarity - Download Free PDF
Last updated on Jun 6, 2025
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:
Answer (Detailed Solution Below)
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:
\(\frac{64}{25} = (\frac{AB}{QP})^2\)
⇒ \(\frac{64}{25} = (\frac{AB}{12})^2\)
⇒ \(\frac{64}{25} = \frac{AB^2}{144}\)
⇒ \(\frac{64}{25} × 144 = AB^2\)
⇒ \(\frac{64 × 144}{25} = AB^2\)
⇒ \(\frac{9216}{25} = AB^2\)
⇒ 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?
Answer (Detailed Solution Below)
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 =
⇒ (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:
Answer (Detailed Solution Below)
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:
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
Congruence and Similarity Question 6 Detailed Solution
Download Solution PDFGiven:
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:
Answer (Detailed Solution Below)
Congruence and Similarity Question 7 Detailed Solution
Download Solution PDFGiven:
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:
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:
Answer (Detailed Solution Below)
Congruence and Similarity Question 8 Detailed Solution
Download Solution PDFGiven,
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 cm2In the given triangle, O is the incentre, AE = 4 cm, AC = 9 cm and BC = 10 cm. What is the length of side AB?
Answer (Detailed Solution Below)
Congruence and Similarity Question 9 Detailed Solution
Download Solution PDF∵ 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.
Answer (Detailed Solution Below)
Congruence and Similarity Question 10 Detailed Solution
Download Solution PDFGiven
AD = 12 cm, CD = 16 cm
Formula used
Angle Bisector theorem,
AD/CD = AB/BC
Perimeter of triangle = Sum of all sides
Calculation
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.
Answer (Detailed Solution Below)
Congruence and Similarity Question 11 Detailed Solution
Download Solution PDFGiven:
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
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
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.
Answer (Detailed Solution Below)
Congruence and Similarity Question 12 Detailed Solution
Download Solution PDFGiven :-
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.
(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:
Answer (Detailed Solution Below)
Congruence and Similarity Question 13 Detailed Solution
Download Solution PDFGiven:
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:
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.
Answer (Detailed Solution Below)
Congruence and Similarity Question 14 Detailed Solution
Download Solution PDFGIVEN:
∠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?
Answer (Detailed Solution Below)
Congruence and Similarity Question 15 Detailed Solution
Download Solution PDFGiven:
In a Δ ABC, ∠BAC = 90°, AD is drawn perpendicular from A on BC
Concept used:
Similarity of triangles
Calculation:
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.