Practice MCQ Questions and Answer on Triangles

In a ÃÂÃÂÃÂÃÂÃÂÃÂPQR, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ Q = 55ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ R = 35ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°. Find the ratio of angles subtended by side QR on circumcenter, incenter and orthocenter of the triangle.

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

Solution: Circumcenter at the mid point of QR hence angle made by QR = 2 × 90° = 180° Angle made by QR at In center = 90° + $\frac{1}{2}$ × ∠P = 135° Ortho center is at point 'P' Hence angle made by QR = 90 Then ration C : I : O = 180 : 135 : 90 = 4 : 3 : 2

In a ÃÂÃÂÃÂÃÂÃÂÃÂABC, BC is extended upto D; ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ACD = 120ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B = $\frac{1}{2}$ ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A, then ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A is:

(A) 60°
(B) 75°
(C) 80°
(D) 90°

Solution: ∠C = 180° - 120° = 60° ∠A + ∠B + ∠C = 180° ∠A + $\frac{1}{2}$ ∠A + 60° = 180° $\frac{3}{2}$ ∠A = 120° ∠A = 80°

G is the centroid of the equilateral ÃÂÃÂÃÂÃÂÃÂÃÂABC. If AB = 10 cm then length of AG is ?

(A) $\frac{5 \sqrt{3}}{3} c m$
(B) $\frac{10 \sqrt{3}}{3} c m$
(C) $5 \sqrt{3} c m$
(D) $10 \sqrt{3} c m$

Solution: According to question, Given : AB = BC = CA = 10 cm G = Centroid AG = 2 units GD = 1 unit AD = 3 units = Height As we know that the height of the equilateral triangle is $\begin{aligned} = \frac{\sqrt{3}}{2} \times 10 = 5 \sqrt{3} \\ ∴ 3 units = 5 \sqrt{3} \\ 1 unit = \frac{5 \sqrt{3}}{3} \\ 2 units = \frac{5 \sqrt{3}}{3} \times 2 \\ 2 units = \frac{10 \sqrt{3}}{3} \\ ∴ AG = \frac{10 \sqrt{3}}{3} c m \end{aligned}$

AD is the median of a triangle ABC and O is the centroid such that AO = 10 cm. The length of OD (in cm) is

(A) 4 cm
(B) 5 cm
(C) 6 cm
(D) 8 cm

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

In ÃÂÃÂÃÂÃÂÃÂÃÂPQR, S and T are point on sides PR and PQ respectively such that ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ PQR = ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ PST, If PT = 5 cm, PS = 3 cm and TQ = 3 cm, then length of SR is

(A) 5 cm
(B) 6 cm
(C) $\frac{31}{3}$ cm
(D) $\frac{41}{3}$ cm

Solution: According to question, Given : PT = 5 cm PS = 3 cm TQ = 3 cm SR = ? ΔPQR ∼ ΔPST $\begin{aligned} \frac{P R}{P T} = \frac{P Q}{P S} \\ \frac{P R}{5} = \frac{8}{3} \\ P R = \frac{40}{3} \\ ∴ S R = P R - P S \\ S R = \frac{40}{3} - 3 \\ S R = \frac{40 - 9}{3} \\ S R = \frac{31}{3} cm \end{aligned}$

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°

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

I is the incentre of a triangle ABC. If ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ACB = 55ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ABC = 65ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° then the value of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BIC is

(A) 130°
(B) 120°
(C) 140°
(D) 110°

Solution: According to question, Given : ∠ACB = 55° ∠ABC = 65° ∠BIC = ? ∴ ∠ACB + ∠ABC + ∠BAC = 180° ∠BAC = 180° - 55° - 65° ∠BAC = 60° We know that ∠BIC = 90 + $\frac{1}{2}$ ∠A ∠BIC = 90 + $\frac{1}{2}$ × 60 ∠BIC = 90 + 30 ∠BIC = 120° Alternate: In ΔBIC, $\frac{1}{2}$ ∠B + $\frac{1}{2}$ ∠C + ∠BIC = 180° $\frac{1}{2}$ (65° + 55°) + ∠BIC = 180° ∠BIC = 180° - 60° ∠BIC = 120°

10.

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

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Practice MCQ Questions and Answer on Triangles

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