Practice MCQ Questions and Answer on Triangles

121.

In a triangle ABC, AB = AC, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAC = 40ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° then the external angle at B is :

(A) 90°
(B) 70°
(C) 110°
(D) 80°

122.

For a triangle base is 6 $\sqrt{3}$ cm and two base angles are 30ÃÂÃÂÃÂÃÂ° and 60ÃÂÃÂÃÂÃÂ°. Then height of the triangle is

(A) 3 $\sqrt{3}$ cm
(B) 4.5 cm
(C) 4 $\sqrt{3}$ cm
(D) 2 $\sqrt{3}$ cm

Solution: According to question, Given : ABC is a right angle triangle BC = 6 $\sqrt{3}$ ∴ sin 30° = $\frac{P}{H}$ = $\frac{A B}{6 \sqrt{3}}$ AB = 3 $\sqrt{3}$ sin 60° = $\frac{P}{H}$ = $\frac{A N}{A B}$ $\frac{\sqrt{3}}{2}$ = $\frac{A N}{3 \sqrt{3}}$ AN = $\frac{9}{2}$ AN = 4.5 cm

123.

The length of the two sides forming the right angle of a right angled triangle are 6 cm and 8 cm. The length of its circum-radius is :

(A) 5 cm
(B) 7 cm
(C) 6 cm
(D) 10 cm

Solution: According to question, ABC is a right angle triangle, ∴ AB = 8 cm BC = 6 cm ∴ AC2 = AB2 + BC2 AC = 64 + 36 AC = $\sqrt{100}$ AC = 10 cm In right triangle Circum Radius IR = $\frac{A C}{2}$ = $\frac{10}{2}$ = 5 cm

124.

BL and CM are medians of ÃÂÃÂÃÂÃÂÃÂÃÂABC right-angled at A and BC = 5 cm. If BL = $\frac{3 \sqrt{5}}{2}$ cm, then the length of CM is

(A) $2 \sqrt{5}$ cm
(B) $5 \sqrt{2}$ cm
(C) $10 \sqrt{2}$ cm
(D) $4 \sqrt{5}$ cm

Solution: According to question, According to figure, when two medians intersect each other in a right angled triangle then we use this equation. ⇒ 4 (BL2 + CM2) = 5BC2 ⇒ 4 × ${(\frac{3 \sqrt{5}}{2})}^{2}$ + 4CM2 = 5BC2 ⇒ 45 + 4CM2 = 125 ⇒ CM2 = $\frac{125 - 45}{4}$ ⇒ CM2 = 20 ⇒ CM = $2 \sqrt{5}$ cm

125.

In an equilateral triangle ABC, G is the centroid. Each side of the triangle is 6 cm. The length of AG is:

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

Solution: In Equilateral triangle $\begin{aligned} A G : G D = 2 : 1 \\ A D = \frac{\sqrt{3}}{2} a \\ A D = \frac{\sqrt{3}}{2} \times 6 \\ A D = 3 \sqrt{3} \\ 3 \to 3 \sqrt{3} \\ 1 \to \sqrt{3} \\ A G = 2 unit = 2 \sqrt{3} cm \end{aligned}$

126.

If two medians BE and CF of a triangle ABC, intersect each other at G and if BG = CG, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BGC = 60ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, BC = 8 cm, then area of the triangle ABC is:

(A) $96 \sqrt{3}$ cm2
(B) $48 \sqrt{3}$ cm2
(C) 48 cm2
(D) $54 \sqrt{3}$ cm2

Solution: According to question, ⇒ ∵ ∠BGC = 60° (given) ⇒ ∠GBC = ∠GCB = x° ⇒ x° + x° + 60° = 180° ⇒ x = 60° ⇒ So ΔBGC is an equilateral triangle with side 8 cm each Then, Area pf triangle ΔBGC = $\frac{\sqrt{3}}{4}$ a2 = $\frac{\sqrt{3}}{4}$ 82 = 16 $\sqrt{3}$ cm2 ⇒ Area of ΔABC = Area (ΔBGC + ΔAGC + ΔAGB) ⇒ Area of ΔABC = 3 × 16 $\sqrt{3}$ ⇒ Area of ΔABC = 48 $\sqrt{3}$ cm2 {∵ ΔBGC = ΔAGC = ΔAGB}

127.

Consider the following statements :
I. Three sides of a triangle are equal to three sides of another triangle, then the triangles are congruent.
II. If three angles of a triangle are respectively equal to three angles of another triangle, then the two triangles are congruent.
Of these statements :

(A) I and II both are true
(B) I is true and II is false
(C) I is false and II is true
(D) None of these

128.

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°

129.

In ÃÂÃÂÃÂÃÂÃÂÃÂABC, AC = BC and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ABC = 50ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, the side BC is produced to D so that BC = CD then the value of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAD is:

(A) 80°
(B) 40°
(C) 90°
(D) 50°

130.

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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