Practice MCQ Questions and Answer on Triangles

61.

Length of the sides of a triangle are a, b and c respectively. If a² + b² + c² = ab + bc + ca then the triangle is:

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

62.

In a triangle, if three altitudes are equal, then the triangle is

(A) Obtuse
(B) Equilateral
(C) Right
(D) Isosceles

63.

The orthocenter of a triangle is the point where?

(A) The medians meet
(B) The altitudes meet
(C) The right bisectors of the sides of
(D) The bisectors of the angles

64.

If the circumcentre of a triangle lies outside it, then the triangle is

(A) Equilateral
(B) Acute angled
(C) Right angled
(D) Obtuse angled

65.

In triangle PQR, points A, B and C are taken on PQ, PR and QR respectively such that QC = AC and CR = CB. If ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ QPR = 40ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, then ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ACB is equal to:

(A) 140°
(B) 40°
(C) 70°
(D) 100°

66.

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°

67.

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°

68.

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 : $\begin{aligned} A B^{2} = B D . B C \\ {(\sqrt{B D^{2} + A D^{2}})}^{2} = B D . B C \\ {(\sqrt{4^{2} + 6^{2}})}^{2} = 4. B C \\ \frac{52}{4} = B C, \\ ∴ B C = 13 c m \end{aligned}$

69.

For a triangle ABC, D and E are two points on AB and AC such that AD = $\frac{1}{4}$ AB, AE = $\frac{1}{4}$ AC. If BC = 12 cm, then DE is :

(A) 5 cm
(B) 4 cm
(C) 3 cm
(D) 6 cm

Solution: According to question, By using B.P.T $\begin{aligned} \frac{A D}{A B} = \frac{A E}{A C} = \frac{D E}{B C} \\ \frac{A D}{A B} = \frac{D E}{B C} \\ \Rightarrow \frac{1}{4} = \frac{D E}{12} \\ \Rightarrow D E = 3 cm \end{aligned}$

70.

If angle bisector of a triangle bisects the opposite side, then what type of triangle is it?

(A) Right angled
(B) Equilateral
(C) Isosceles and equilateral
(D) Isosceles

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Practice MCQ Questions and Answer on Triangles

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