Practice MCQ Questions and Answer on Triangles

31.

For a triangle ABC, D and E are two points on AB and AC such that AD = $\frac{1}{4}$ AB, AE = $\frac{1}{4}$ AC. If BC = 12 cm, then DE is :

(A) 5 cm
(B) 4 cm
(C) 3 cm
(D) 6 cm

Solution: According to question, By using B.P.T $\begin{aligned} \frac{A D}{A B} = \frac{A E}{A C} = \frac{D E}{B C} \\ \frac{A D}{A B} = \frac{D E}{B C} \\ \Rightarrow \frac{1}{4} = \frac{D E}{12} \\ \Rightarrow D E = 3 cm \end{aligned}$

32.

ABC is an isosceles triangle with AB = AC, A circle through B touching AC at the middle point intersects AB at P. Then AP : AB is:

(A) 4 : 1
(B) 2 : 3
(C) 3 : 5
(D) 1 : 4

Solution: According to question, Let AB = AC = 2x ∵ AQ = QC = x ∴ AB is a secant ∴ AP × AB = AQ2 AP × 2x = x2 $\begin{aligned} A P = \frac{x}{2} \\ \frac{A P}{A B} = \frac{x}{2 \times 2 x} = \frac{1}{4} \\ \frac{A P}{A B} = \frac{1}{4} \\ A P : A B = 1 : 4 \end{aligned}$

33.

ABC is a triangle, PQ is line segment intersecting AB is P and AC in Q and PQ || BC. The ratio of AP : BP = 3 : 5 and length of PQ is 18 cm. The length of BC is:

(A) 28 cm
(B) 48 cm
(C) 84 cm
(D) 42 cm

Solution: ∵ PQ || BC So ∠AQP = ∠ACB = α and ∠APQ = ∠ABC = β So, ΔABC and ΔAPQ $\begin{aligned} \frac{A P}{A B} = \frac{P Q}{B C} \\ \frac{3}{8} = \frac{P Q}{B C} \\ \frac{3}{8} = \frac{18}{B C} \\ B C = 48 cm \end{aligned}$

34.

In an isosceles triangle ÃÂÃÂÃÂÃÂÃÂÃÂABC, AB = AC and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A = 80ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°. The bisector of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ C meet at D. The ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BDC is equal to.

(A) 90°
(B) 100°
(C) 130°
(D) 80°

Solution: ∵ AB = AC Point D is the incenter ∴ ∠BDC = 90° + $\frac{1}{2}$ ∠A = 90° + $\frac{1}{2}$ × 80° = 90° + 40° = 130°

35.

D is any point on side AC of ÃÂÃÂÃÂÃÂÃÂÃÂABC. If P, Q, X, Y are the mid-point of AB, BC, AD and DC respectively, then the ratio of PX and QY is

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

Solution: According to question, PX || BD [mid point theorem] ∴ PX = $\frac{1}{2}$ BD Similarly, QY || BD ∴ QY = $\frac{1}{2}$ BD ∴ PX : QY, $\frac{1}{2}$ BD : $\frac{1}{2}$ BD PX : QY = 1 : 1

36.

In a triangle ABC, AB + BC = 12 cm, BC + CA = 14 cm and CA + AB = 18 cm. Find the radius of the circle (in cm) which has the same perimeter as the triangle

(A) $\frac{5}{2}$
(B) $\frac{7}{2}$
(C) $\frac{9}{2}$
(D) $\frac{11}{2}$

Solution: According to question, AB + BC = 12 cm BC + CA = 14 cm CA + AB = 18 cm 2(AB + BC + CA) = 44 cm AB + BC + CA = $\frac{44}{2}$ cm AB + BC + CA = 22 cm Perimeter of triangle = 22 cm Perimeter of triangle = Perimeter of circle 22 = 2πr 2 × $\frac{22}{7}$ × r = 22 r = $\frac{7}{2}$ cm

37.

In triangle ABC a straight line parallel to BC intersects AB and AC at D and E respectively. If AB = 2AD, then DE : BC is

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

Solution: According to question, Given : AB = 2AD $\frac{A B}{A D} = \frac{2}{1}$ By applying B. P. T $\begin{aligned} \frac{A D}{A B} = \frac{D E}{B C} = \frac{A E}{A C} \\ \frac{D E}{B C} = \frac{1}{2} \\ ∴ D E : B C = 1 : 2 \end{aligned}$

38.

In a ÃÂÃÂÃÂÃÂÃÂÃÂABC, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A + ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B = 118ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A + ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ C = 96ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°. Find the value of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A.

(A) 36°
(B) 40°
(C) 30°
(D) 34°

39.

BL and CM are medians of ÃÂÃÂÃÂÃÂÃÂÃÂABC right-angled at A and BC = 5 cm. If BL = $\frac{3 \sqrt{5}}{2}$ cm, then the length of CM is

(A) $2 \sqrt{5}$ cm
(B) $5 \sqrt{2}$ cm
(C) $10 \sqrt{2}$ cm
(D) $4 \sqrt{5}$ cm

Solution: According to question, According to figure, when two medians intersect each other in a right angled triangle then we use this equation. ⇒ 4 (BL2 + CM2) = 5BC2 ⇒ 4 × ${(\frac{3 \sqrt{5}}{2})}^{2}$ + 4CM2 = 5BC2 ⇒ 45 + 4CM2 = 125 ⇒ CM2 = $\frac{125 - 45}{4}$ ⇒ CM2 = 20 ⇒ CM = $2 \sqrt{5}$ cm

40.

The side QR of an equilateral triangle PQR is produced to the point S in such a way that QR = RS and P is joined to S. Then the measure of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ PSR is

(A) 30°
(B) 15°
(C) 60°
(D) 45°

Solution: According to question, Given : PQR is an equilateral triangle QR = RS PR = RS ∠SRP = 180° - 60° (Exterior ∠) ∠SRP = 120° ∴ ∠RPS = ∠RSP ∴ ∠RPS + ∠PRS + ∠RSP = 180° 2∠PSR = 180° - 120° ∠PSR = $\frac{60^{\circ}}{2}$ ∠PSR = 30°

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Practice MCQ Questions and Answer on Triangles

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