Practice MCQ Questions and Answer on Triangles

71.

If in a triangle ABC, D and E are on the sides AB and AC, such that, DE is parallel to BC and $\frac{A D}{B D}$ = $\frac{3}{5}$ . If AC = 4 cm, then AE is

(A) 1.5 cm
(B) 2.0 cm
(C) 1.8 cm
(D) 2.4 cm

Solution: According to question, Given : AD = 3 BD = 5 AB = 8 AC = 4 AE = ? By applying B. P. T $\begin{aligned} \frac{A D}{A B} = \frac{A E}{A C} = \frac{D E}{B C} \\ \frac{3}{8} = \frac{A E}{4} \\ A E = \frac{3}{2} = 1.5 cm \end{aligned}$

72.

Let ABC be an equilateral triangle and AD perpendicular to BC, then AB² + BC² + CA² = ?

(A) 3AD2
(B) 5AD2
(C) 2AD2
(D) 4AD2

Solution: AB2 = AD2 + BD2 . . . . . . . (i) AC2 = AD2 + CD2 . . . . . . . (ii) AB2 + AC2 = 2AD2 + BD2 + CD2 AB2 + AC2 + BC2 = 2AD2 + a2 + $\frac{a^{2}}{4}$ + $\frac{a^{2}}{4}$ AB2 + AC2 + BC2 = 4AD2 $(a^{2} - \frac{a^{2}}{4} = A D^{2})$

73.

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

74.

In a ÃÂÃÂÃÂÃÂÃÂÃÂABC ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A : ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B : ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ C = 2 : 3 : 4. A line CD drawn || to AB, then the ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ACD is :

(A) 40°
(B) 60°
(C) 80°
(D) 20°

75.

ABC is a triangle in which ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°. Let P be any point on side AC. If BC = 10 cm, AC = 8 cm and BP = 9 cm, then AP = ?

(A) $2 \sqrt{5}$ cm
(B) $3 \sqrt{5}$ cm
(C) $2 \sqrt{3}$ cm
(D) $3 \sqrt{3}$ cm

Solution: According to question, BC = 10 cm, AC = 8 cm, BP = 9 cm In ΔCAB, BC2 = AB2 + AC2 102 = AB2 + 82 AB2 = 100 - 64 AB2 = 36 AB = 6 cm In ΔPAB BP2 = AB2 + AP2 92 = 62 + AP2 AP2 = 81 - 36 AP2 = 45 AP = $3 \sqrt{5}$ cm

76.

In ÃÂÃÂÃÂÃÂÃÂÃÂABC, AB = BC = K, AC = $\sqrt{2}$ k, then ÃÂÃÂÃÂÃÂÃÂÃÂABC is a :

(A) Right isosceles triangle
(B) Isosceles triangle
(C) Right-angled triangle
(D) Equilateral triangle

Solution: ∵ AB = BC = K ⇒ AC = $\sqrt{2}$ K ⇒ (AC)2 = (AB)2 + (BC)2 ⇒ ( $\sqrt{2}$ K)2 = K2 + K2 ⇒ 2K2 = 2K2 ⇒ Therefore ΔABC will be a right isosceles triangle.

77.

In ÃÂÃÂÃÂÃÂÃÂÃÂABC, AD ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ¥ BC and AD = BD ÃÂÃÂÃÂÃÂÃÂÃÂ DC. The measure of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BAC is : 2

(A) 75°
(B) 90°
(C) 45°
(D) 60°

78.

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°

79.

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

80.

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°

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Practice MCQ Questions and Answer on Triangles

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