Practice MCQ Questions and Answer on Triangles

11.

ABC is a triangle in which ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°. Let P be any point on side AC. If BC = 10 cm, AC = 8 cm and BP = 9 cm, then AP = ?

(A) $2 \sqrt{5}$ cm
(B) $3 \sqrt{5}$ cm
(C) $2 \sqrt{3}$ cm
(D) $3 \sqrt{3}$ cm

Solution: According to question, BC = 10 cm, AC = 8 cm, BP = 9 cm In ΔCAB, BC2 = AB2 + AC2 102 = AB2 + 82 AB2 = 100 - 64 AB2 = 36 AB = 6 cm In ΔPAB BP2 = AB2 + AP2 92 = 62 + AP2 AP2 = 81 - 36 AP2 = 45 AP = $3 \sqrt{5}$ cm

12.

In ÃÂÃÂÃÂÃÂÃÂÃÂPQR, straight line parallel to the base QR cuts PQ at X and PR at Y. If PX : XQ = 5 : 6, then XY : QR will be

(A) 5 : 11
(B) 6 : 5
(C) 11 : 6
(D) 11 : 5

Solution: ∵ ΔPQR ∼ ΔPXY $\begin{aligned} \frac{P X}{P Q} = \frac{X Y}{Q R} \\ \frac{5}{(5 + 6)} = \frac{X Y}{Q R} \\ X Y : Q R = 5 : 11 \end{aligned}$

13.

Possible length of the sides of a triangle are:

(A) 2cm, 3cm, 6cm
(B) 3cm, 4cm, 5cm
(C) 2.5cm, 3.5cm, 6cm
(D) 4cm, 4cm, 9cm

14.

In a triangle ABC, if ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A + ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ C = 140ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A + 3ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B = 180ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, then ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A is equal to:

(A) 80°
(B) 40°
(C) 60°
(D) 20°

15.

The centroid of a triangle is G. If area of ÃÂÃÂÃÂÃÂÃÂÃÂABC = 72 sq. unit, then the area of ÃÂÃÂÃÂÃÂÃÂÃÂBGC is?

(A) 16 sq. units
(B) 24 sq. units
(C) 36 sq. units
(D) 48 sq. units

Solution: G is centroid Area of ΔBGC = $\frac{1}{3}$ area of ΔABC ΔBGC = $\frac{1}{3}$ × 72 ΔBGC = 24 sq. units

16.

A point D is taken on the side BC of a right-angled triangle ABC, where AB is hypotenuse. Then

(A) AB2 + CD2 = AD2 + BC2
(B) CD2 + BD2 = 2AD2
(C) AB2 + AC2 = 2AD2
(D) AB2 = AD2 + BC2

17.

Angle between the internal bisectors of two angles of a triangle ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ C is 120ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, then ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A is :

(A) 20°
(B) 30°
(C) 60°
(D) 90°

Solution: According to question, Given : ∠BIC = 120° ∠BIC = 90° + $\frac{1}{2}$ ∠A $\frac{∠ A}{2}$ = (120° - 90°) $\frac{∠ A}{2}$ = 30° ∠A = 60°

18.

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 $\begin{aligned} S = \frac{a + b + c}{2} \\ S = \frac{24 \sqrt{5} + 24}{2} \\ S = 12 (\sqrt{5} + 1) \\ R_{2} = \frac{12 \sqrt{5} \times 12 \sqrt{5} \times 24}{4 \times 12 \times 12 \times 2} \\ R_{2} = \frac{30}{2} \\ R_{2} = 15 cm \end{aligned}$

19.

In a triangle ABC, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ C = 55ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, $A D$ ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ¥ $B C$ . What is the value of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAD ?

(A) 35°
(B) 60°
(C) 45°
(D) 55°

20.

In ÃÂÃÂÃÂÃÂÃÂÃÂABC ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and AD ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ¥ BC where D lies on BC. If BC = 8 cm, AD = 6 cm, then arÃÂÃÂÃÂÃÂÃÂÃÂABC : arÃÂÃÂÃÂÃÂÃÂÃÂACD = ?

(A) 4 : 3
(B) 25 : 16
(C) 16 : 9
(D) 25 : 9

Solution: According to question, ΔABC ∼ ΔACD $\begin{aligned} ∴ \frac{area of Δ A B C}{area of Δ A C D} = \frac{B C^{2}}{A D^{2}} \\ \frac{area of Δ A B C}{area of Δ A C D} = \frac{8^{2}}{6^{2}} \\ \frac{area of Δ A B C}{area of Δ A C D} = \frac{64}{36} \\ \frac{area of Δ A B C}{area of Δ A C D} = \frac{16}{9} \end{aligned}$ area ΔABC : area ΔACD = 16 : 9

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Practice MCQ Questions and Answer on Triangles

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