Triangles, Congruence and Similarity MCQ Quiz - Objective Question with Answer for Triangles, Congruence and Similarity - Download Free PDF
Last updated on May 13, 2025
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
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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 =
Answer (Detailed Solution Below)
Triangles, Congruence and Similarity Question 4 Detailed Solution
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:
Answer (Detailed Solution Below)
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:
\(\frac{64}{25} = (\frac{AB}{QP})^2\)
⇒ \(\frac{64}{25} = (\frac{AB}{12})^2\)
⇒ \(\frac{64}{25} = \frac{AB^2}{144}\)
⇒ \(\frac{64}{25} × 144 = AB^2\)
⇒ \(\frac{64 × 144}{25} = AB^2\)
⇒ \(\frac{9216}{25} = AB^2\)
⇒ 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.
Answer (Detailed Solution Below)
Triangles, Congruence and Similarity Question 6 Detailed Solution
Download Solution PDFGiven:
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 = ?
Answer (Detailed Solution Below)
Triangles, Congruence and Similarity Question 7 Detailed Solution
Download Solution PDFGiven:
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:
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?
Answer (Detailed Solution Below)
Triangles, Congruence and Similarity Question 8 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.
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:
Answer (Detailed Solution Below)
Triangles, Congruence and Similarity Question 9 Detailed Solution
Download Solution PDFGiven:
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:
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.
Answer (Detailed Solution Below)
Triangles, Congruence and Similarity Question 10 Detailed Solution
Download Solution PDFGiven:
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,
(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,
(2.1) ∠BOC = 180∘ - ∠A
(2.2) ∠AOB = 180∘ - ∠C
(3.3) ∠AOC = 180∘ - ∠B
Calculation:
According to the question, the required image is:
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?
Answer (Detailed Solution Below)
Triangles, Congruence and Similarity Question 11 Detailed Solution
Download Solution PDFGiven:
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:
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 cmΔPQR inscribes in a circle with centre O. If PQ = 12 cm, QR = 16 cm and PR = 20 cm, find the circumradius of a triangle.
Answer (Detailed Solution Below)
Triangles, Congruence and Similarity Question 12 Detailed Solution
Download Solution PDFGiven:
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:
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:
Answer (Detailed Solution Below)
Triangles, Congruence and Similarity Question 13 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 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?
Answer (Detailed Solution Below)
Triangles, Congruence and Similarity Question 14 Detailed Solution
Download Solution PDFGiven:
AB : AC = 5 : 2
BC = 9 cm
Concept used:
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:
Answer (Detailed Solution Below)
Triangles, Congruence and Similarity Question 15 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 cm2