Practice MCQ Questions and Answer on Triangles

If two angles of a triangle are 21ÃÂÃÂÃÂÃÂ° and 38ÃÂÃÂÃÂÃÂ°, then the triangle is :

(A) Right-angled triangle
(B) Acute-angled triangle
(C) Obtuse-angled triangle
(D) Isosceles triangle

In a triangle ABC, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAC = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and AD is perpendicular to BC. If AD = 6 cm and BD = 4 cm then the length of BC is:

(A) 8 cm
(B) 10 cm
(C) 9 cm
(D) 13 cm

Solution: According to question, Given: BAC is a right angle triangle AD ⊥ BC AD = 6 cm BD = 4 cm BC = ? In ΔBAD $\begin{aligned} A B = \sqrt{B D^{2} + A D^{2}} \\ A B = \sqrt{4^{2} + 6^{2}} \\ A B = \sqrt{52} c m \end{aligned}$ ΔBAC ∼ ΔBDA $\begin{aligned} ∴ \frac{B C}{A B} = \frac{A B}{B D} \\ ∴ \frac{B C}{\sqrt{52}} = \frac{\sqrt{52}}{4} \\ B C = \frac{52}{4} \\ B C = 13 c m \end{aligned}$ Alternate : $\begin{aligned} A B^{2} = B D . B C \\ {(\sqrt{B D^{2} + A D^{2}})}^{2} = B D . B C \\ {(\sqrt{4^{2} + 6^{2}})}^{2} = 4. B C \\ \frac{52}{4} = B C, \\ ∴ B C = 13 c m \end{aligned}$

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

In a ÃÂÃÂÃÂÃÂÃÂÃÂABC, AB = BC, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B = xÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A = (2x - 20)ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, Then ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B is :

(A) 54°
(B) 30°
(C) 40°
(D) 44°

In a ÃÂÃÂÃÂÃÂÃÂÃÂABC, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A + ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B = 75ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B + ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ C = 140ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, then ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B is:

(A) 40°
(B) 35°
(C) 50°
(D) 45°

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

In ÃÂÃÂÃÂÃÂÃÂÃÂABC, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B = 60ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ C = 40ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°. If AD and AE be respectively the internal bisector of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A and perpendicular on BC, then the measure of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ DAE is

(A) 5°
(B) 10°
(C) 40°
(D) 60°

Solution: According to question, Given : ∠B = 60° ∠C = 40° As we know that ∠A + ∠B + ∠C = 180° ∠A = 180° - 60° - 40° ∠A = 80° ∴ ∠BAD = $\frac{80^{\circ}}{2}$ = 40° In ΔAEB ∠A + ∠B + ∠E = 180° ∠A = 180° - 60° - 90° ∠A = 30° Then, ∠DAE = ∠DAB - ∠EAB ∠DAE = 40° - 30° ∠DAE = 10° By Trick $\begin{aligned} ∠ D A E = \frac{∠ B - ∠ C}{2} \\ = \frac{60^{\circ} - 40^{\circ}}{2} \\ = 10^{\circ} \end{aligned}$

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°

In ÃÂÃÂÃÂÃÂÃÂÃÂABC, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAC = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and AB = $\frac{1}{2}$ BC, Then the measure of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ACB is :

(A) 60°
(B) 30°
(C) 45°
(D) 15°

Solution: According to question, Given : BAC is right angle triangle AB = $\frac{1}{2}$ BC $\begin{aligned} \frac{A B}{B C} = \frac{1}{2} \frac{P}{H} = \frac{1}{2} \\ \sin θ = \frac{P}{H} = \frac{1}{2} \\ \sin 30^{\circ} = \frac{1}{2} \\ ∴ θ = ∠ A C B = 30^{\circ} \end{aligned}$

10.

Let O be the in-centre of a triangle ABC and D be a point on the side BC of ÃÂÃÂÃÂÃÂÃÂÃÂABC, such that OD ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ¥ BC. If ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BOD = 15ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, then ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ABC = ?

(A) 75°
(B) 45°
(C) 150°
(D) 90°

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Practice MCQ Questions and Answer on Triangles

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