Practice MCQ Questions and Answer on Triangles

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

In a triangle ABC, the side BC is extended up to D such that CD = AC. If ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAD = 109ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ACB = 72ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° then the value of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ABC is

(A) 35°
(B) 60°
(C) 40°
(D) 45°

Solution: According to question, Given : ∠BAD = 109° ∠ACB = 72° ∴ ∠ACD = 180° - 72° ∠ACD = 108° ∴ AC = CD ∠CAD = ∠CDA In ΔCDA ∠CAD + ∠CDA + ∠DCA = 180° 2∠CAD + 108° = 180° 2∠CAD = 180° -108° 2∠CAD = 72° ∠CAD = $\frac{72^{\circ}}{2}$ ∠CAD = 36° ∴ ∠CAB = 109° - 36° ∠CAB = 73° In ΔABC ∠ABC + ∠ACB + ∠CAB = 180° ∠ABC + 72° + 73° = 180° ∠ABC + 145° = 180° ∠ABC = 180° - 145° ∠ABC = 35°

If I be the incentre of ÃÂÃÂÃÂÃÂÃÂÃÂABC and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B = 70ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ C = 50ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, then the magnitude of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BIC is

(A) 130°
(B) 60°
(C) 120°
(D) 105°

Solution: According to question, As we know that ∠A + ∠B + ∠C = 180° ∴ ∠A = 180° -70° - 50° ∠A = 60° ∴ ∠BIC = 90° + $\frac{1}{2}$ × 60° ∠BIC = 120°

The length of the three sides of a right angled triangle are (x - 2) cm, (x) cm and (x + 2) cm respectively. Then the value of x is

(A) 10
(B) 8
(C) 4
(D) 0

Solution: According to question, ABC is a right angle triangle ∴ Apply Pythagoras theorem AC2 = AB2 +BC2 (x + 2)2 = (x - 2)2 + x2 x2 + 4 + 4x = x2 + 4 - 4x + x2 x2 = 8x x = 8 Alternate : From option approach $\begin{aligned} (x - 2) x (x + 2) \\ ↓ ↓ ↓ \\ 6 8 10 \\ ↓ \\ Triplet \\ x = 8 \end{aligned}$

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

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

In a right angled triangle ÃÂÃÂÃÂÃÂÃÂÃÂDEF, if the length of the hypotenuse EF is 12 cm, then the length of the median DX is:

(A) 3 cm
(B) 4 cm
(C) 6 cm
(D) 12 cm

Solution: Median of right angle $\begin{aligned} = \frac{E F}{2} \\ = \frac{12}{2} \\ = 6 cm \end{aligned}$

O is the incentre of ÃÂÃÂÃÂÃÂÃÂÃÂABC and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A = 30ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, then ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BOC is

(A) 100°
(B) 105°
(C) 110°
(D) 90°

Solution: According to question, Given: ∴ ∠BOC = 90° + $\frac{1}{2}$ ∠A ∠BOC = 90° + $\frac{1}{2}$ × 30° ∠BOC = 90° + 15° ∠BOC = 105°

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

10.

ABC is an isosceles triangle such that AB = AC and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B = 35ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, AD is the median to the base BC. Then ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAD is

(A) 70°
(B) 35°
(C) 110°
(D) 55°

Solution: According to question, AB = AC, ∠B = ∠C ∠A + ∠B + ∠C = 180° ∠A + 2∠B = 180° ∠A = 180° - 70° ∠A = 110° Note : In isosceles triangle median bisects the opposite side and make angle 90° on opposite side. It also bisects the vertex angle. ∠BAD = $\frac{∠ A}{2}$ ∠BAD = $\frac{110^{\circ}}{2}$ ∠BAD = 55°

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Practice MCQ Questions and Answer on Triangles

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