Practice MCQ Questions and Answer on Triangles

71.

PQR is an equilateral triangle. MN is drawn parallel to QR such that M is on PQ and N is on PR. If PN = 6 cm, then the length of MN is:

(A) 3 cm
(B) 6 cm
(C) 12 cm
(D) 4.5 cm

72.

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

73.

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°

74.

An isosceles triangle ABC is right-angled at B. D is a point inside the triangle ABC. P and Q are the feet of the perpendiculars drawn from D on the side AB and AC respectively of ÃÂÃÂÃÂÃÂÃÂÃÂABC. If AP = a cm, AQ = b cm and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAD = 15ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, sin 75ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° = ?

(A) $\frac{2 b}{\sqrt{3} a}$
(B) $\frac{a}{2 b}$
(C) $\frac{\sqrt{3} a}{2 b}$
(D) $\frac{2 a}{\sqrt{3} b}$

Solution: According to question from ΔAQD, ∠A = $\frac{180 - 90}{2}$ ∠A = 45° ∠DAQ = 30° sin 60° = $\frac{A Q}{A D}$ $\frac{\sqrt{3}}{2}$ = $\frac{b}{A D}$ AD = $\frac{2 b}{\sqrt{3}}$ From ΔAPD $\begin{aligned} \sin 75^{\circ} = \frac{A P}{A D} \\ \sin 75^{\circ} = \frac{a}{2 b} \times \sqrt{3} \\ \sin 75^{\circ} = \frac{\sqrt{3} a}{2 b} \end{aligned}$

75.

In a right-angled triangle, the product of two sides is equal to half of the square of the third side i.e., hypotenuse. One of the acute angle must be

(A) 60°
(B) 30°
(C) 45°
(D) 15°

Solution: According to question, Given : ab = $\frac{C^{2}}{2}$ . . . . . . . . . . (i) ∴ In ΔABC Using Pythagoras theorem AC2 = AB2 + BC2 c2 = a2 + b2 . . . . . . . . . . . (ii) Put the value of C2 in equation (i) 2ab = a2 + b2 a2 + b2 - 2ab = 0 (a - b)2 = 0 ∴ a - b = 0 a = b If a = b means ABC is isosceles right angle triangle it means ∠A = 45° ∠B = 45°

76.

If the length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, then the length of the median to its greatest side is -

(A) 8 cm
(B) 6 cm
(C) 5 cm
(D) 4.8 cm

Solution: According to question, Length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, this is right angle triangle. Note: In right angle triangle median divides the hypotenuse in two equal parts ∴ BD = $\frac{H}{2}$ BD = $\frac{10}{2}$ BD = 5 cm

77.

ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12 $\sqrt{5}$ and BC = 24 cm then radius of circle is:

(A) 10 cm
(B) 15 cm
(C) 12 cm
(D) 14 cm

Solution: $\begin{aligned} R_{2} = \frac{a b c}{4 △} \\ △ = \sqrt{S (s - a) (s - b) (s - c)} \end{aligned}$ $= \sqrt{12 (\sqrt{5} + 1) (12) \times 12 \times 12 (\sqrt{5} - 1)}$ where a = 12 $\sqrt{5}$ , b = 12 $\sqrt{5}$ & c = 24 $\begin{aligned} S = \frac{a + b + c}{2} \\ S = \frac{24 \sqrt{5} + 24}{2} \\ S = 12 (\sqrt{5} + 1) \\ R_{2} = \frac{12 \sqrt{5} \times 12 \sqrt{5} \times 24}{4 \times 12 \times 12 \times 2} \\ R_{2} = \frac{30}{2} \\ R_{2} = 15 cm \end{aligned}$

78.

In a triangle ABC, the side BC is extended up to D such that CD = AC. If ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAD = 109ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ACB = 72ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° then the value of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ABC is

(A) 35°
(B) 60°
(C) 40°
(D) 45°

Solution: According to question, Given : ∠BAD = 109° ∠ACB = 72° ∴ ∠ACD = 180° - 72° ∠ACD = 108° ∴ AC = CD ∠CAD = ∠CDA In ΔCDA ∠CAD + ∠CDA + ∠DCA = 180° 2∠CAD + 108° = 180° 2∠CAD = 180° -108° 2∠CAD = 72° ∠CAD = $\frac{72^{\circ}}{2}$ ∠CAD = 36° ∴ ∠CAB = 109° - 36° ∠CAB = 73° In ΔABC ∠ABC + ∠ACB + ∠CAB = 180° ∠ABC + 72° + 73° = 180° ∠ABC + 145° = 180° ∠ABC = 180° - 145° ∠ABC = 35°

79.

In ÃÂÃÂÃÂÃÂÃÂÃÂABC, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A + ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B = 65ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B + ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ C = 140ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, then find ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B.

(A) 40°
(B) 25°
(C) 35°
(D) 20°

80.

Let ABC be an equilateral triangle and AD perpendicular to BC, then AB² + BC² + CA² = ?

(A) 3AD2
(B) 5AD2
(C) 2AD2
(D) 4AD2

Solution: AB2 = AD2 + BD2 . . . . . . . (i) AC2 = AD2 + CD2 . . . . . . . (ii) AB2 + AC2 = 2AD2 + BD2 + CD2 AB2 + AC2 + BC2 = 2AD2 + a2 + $\frac{a^{2}}{4}$ + $\frac{a^{2}}{4}$ AB2 + AC2 + BC2 = 4AD2 $(a^{2} - \frac{a^{2}}{4} = A D^{2})$

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Practice MCQ Questions and Answer on Triangles

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