Practice MCQ Questions and Answer on Triangles

31.

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 $\begin{aligned} \frac{A P}{A B} = \frac{P Q}{B C} \\ \frac{3}{8} = \frac{P Q}{B C} \\ \frac{3}{8} = \frac{18}{B C} \\ B C = 48 cm \end{aligned}$

32.

ABC is an isosceles triangle such that AB = AC and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B = 35ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, AD is the median to the base BC. Then ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAD is

(A) 70°
(B) 35°
(C) 110°
(D) 55°

Solution: According to question, AB = AC, ∠B = ∠C ∠A + ∠B + ∠C = 180° ∠A + 2∠B = 180° ∠A = 180° - 70° ∠A = 110° Note : In isosceles triangle median bisects the opposite side and make angle 90° on opposite side. It also bisects the vertex angle. ∠BAD = $\frac{∠ A}{2}$ ∠BAD = $\frac{110^{\circ}}{2}$ ∠BAD = 55°

33.

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 $\begin{aligned} \frac{A D}{A B} = \frac{D E}{B C} = \frac{A E}{A C} \\ \frac{D E}{B C} = \frac{1}{2} \\ ∴ D E : B C = 1 : 2 \end{aligned}$

34.

In case of an acute angled triangle, its orthocenter lies:

(A) Inside the triangle
(B) Outside the triangle
(C) On the triangle
(D) On one of the vertex of the triangle

35.

If two medians BE and CF of a triangle ABC, intersect each other at G and if BG = CG, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BGC = 60ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, BC = 8 cm, then area of the triangle ABC is:

(A) $96 \sqrt{3}$ cm2
(B) $48 \sqrt{3}$ cm2
(C) 48 cm2
(D) $54 \sqrt{3}$ cm2

Solution: According to question, ⇒ ∵ ∠BGC = 60° (given) ⇒ ∠GBC = ∠GCB = x° ⇒ x° + x° + 60° = 180° ⇒ x = 60° ⇒ So ΔBGC is an equilateral triangle with side 8 cm each Then, Area pf triangle ΔBGC = $\frac{\sqrt{3}}{4}$ a2 = $\frac{\sqrt{3}}{4}$ 82 = 16 $\sqrt{3}$ cm2 ⇒ Area of ΔABC = Area (ΔBGC + ΔAGC + ΔAGB) ⇒ Area of ΔABC = 3 × 16 $\sqrt{3}$ ⇒ Area of ΔABC = 48 $\sqrt{3}$ cm2 {∵ ΔBGC = ΔAGC = ΔAGB}

36.

If ÃÂÃÂÃÂÃÂÃÂÃÂABC is an isosceles triangle with ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ C = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and AC = 5 cm then AB is:

(A) 5 cm
(B) 10 cm
(C) 5 $\sqrt{2}$ cm
(D) 2.5 cm

Solution: According to question, Given: ∠C = 90° BC = AC = 5 cm (Isosceles triangle) By Pythagoras theorem AB2 = AC2 + BC2 AB2 = 52 + 52 AB2 = 25 + 25 AB2 = 50 AB = 5 $\sqrt{2}$ cm

37.

In ÃÂÃÂÃÂÃÂÃÂÃÂABC, AD ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ¥ BC and AD = BD ÃÂÃÂÃÂÃÂÃÂÃÂ DC. The measure of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAC is : 2

(A) 75°
(B) 90°
(C) 45°
(D) 60°

38.

ABC is a right angled triangled, right angled at C and P is the length of the perpendicular from C on AB. If a, b and c are the length of the sides BC, CA and AB respectively, then

(A) $\frac{1}{p^{2}} = \frac{1}{b^{2}} - \frac{1}{a^{2}}$
(B) $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$
(C) $\frac{1}{p^{2}} + \frac{1}{a^{2}} = - \frac{1}{b^{2}}$
(D) $\frac{1}{p^{2}} = \frac{1}{a^{2}} - \frac{1}{b^{2}}$

Solution: According to question, ACB is a right angle triangle ∴ area of ΔACB $\frac{1}{2}$ × AC × BC = $\frac{1}{2}$ × AB × PC $\frac{1}{2}$ × b × a = $\frac{1}{2}$ × c × p c = $\frac{a b}{p}$ . . . . . . . (i) By using pythagoras theorem AB2 = AC2 + BC2 c2 = b2 + a2 . . . . . . . . . . (ii) Put the value of C in equation (ii) $\begin{aligned} {(\frac{a b}{p})}^{2} = a^{2} + b^{2} \\ \Rightarrow \frac{a^{2} b^{2}}{p^{2}} = a^{2} + b^{2} \\ \Rightarrow \frac{1}{p^{2}} = \frac{a^{2}}{a^{2} b^{2}} + \frac{b^{2}}{a^{2} b^{2}} \\ \Rightarrow \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} \end{aligned}$ Alternate : From figure, $\begin{aligned} P = \frac{a b}{c} \\ P = \frac{a b}{\sqrt{a^{2} + b^{2}}} (∵ a^{2} + b^{2} = c^{2}) \\ P^{2} = \frac{a^{2} b^{2}}{a^{2} + b^{2}} \\ \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} \end{aligned}$

39.

If the three angles of a triangle are: ${(x + 15)}^{\circ},$ ÃÂÃÂ ${(\frac{6 x}{5} + 6)}^{\circ}$ ÃÂÃÂ and ${(\frac{2 x}{3} + 30)}^{\circ}$ ÃÂÃÂ then the triangle is:

(A) Isosceles
(B) Equilateral
(C) Right angled
(D) Scalene

Solution: According to question, $\Rightarrow (x + 15^{\circ}) + {(\frac{6 x}{5} + 6)}^{\circ} +$ ${(\frac{2 x}{3} + 30)}^{\circ} =$ $180^{\circ}$ ${∠ A + ∠ B + ∠ C = 180^{\circ}}$ $\Rightarrow x + \frac{6 x}{5} + \frac{2 x}{3} =$ $180^{\circ} - (15 + 6 + 30)$ $\begin{aligned} \Rightarrow \frac{15 x + 18 x + 10 x}{15} = 180 - 51 \\ \Rightarrow 43 x = 129 \times 15 \\ \Rightarrow x = 45^{\circ} \\ \Rightarrow each angle \\ \Rightarrow {(x + 15)}^{\circ} = 45 + 15 = 60^{\circ} \\ \Rightarrow {(\frac{6 x}{5} + 6)}^{\circ} = 60^{\circ} \\ \Rightarrow {(\frac{2 x}{3} + 30)}^{\circ} = 60^{\circ} \end{aligned}$ ∵ All three angles are equal 60° ⇒ Triangle will be equilateral triangle

40.

In a ÃÂÃÂÃÂÃÂÃÂÃÂABC, AB = AC and BA is produced to D such that AC = AD. Then the ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BCD is :

(A) 100°
(B) 60°
(C) 80°
(D) 90°

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Practice MCQ Questions and Answer on Triangles

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