Practice MCQ Questions and Answer on Triangles

121.

ÃÂÃÂÃÂÃÂÃÂÃÂABC be a right-angled triangle where ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and AD ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ¥ BC. If ar (ÃÂÃÂÃÂÃÂÃÂÃÂABC) = 40 cm, ar (ÃÂÃÂÃÂÃÂÃÂÃÂACD) = 10 cm and AC = 9 cm, then the length of BC is22

(A) 12 cm
(B) 18 cm
(C) 4 cm
(D) 6 cm

Solution: According to question, Given: AC = 9 cm area of ΔABC = 40 cm2 area of ΔADC = 10 cm2 ΔABC ∼ ΔADC $\frac{area of Δ A B C}{area of Δ A D C} = \frac{A B^{2}}{A D^{2}} = \frac{B C^{2}}{A C^{2}}$ (In similar Δratio of their area is square of ratio of corresponding sides) $\begin{aligned} \frac{40}{10} = \frac{B C^{2}}{{(9)}^{2}} \\ \frac{40}{10} \times 81 = B C^{2} \\ B C = 18 cm \end{aligned}$

122.

ABC is an isosceles triangle with AB = AC, A circle through B touching AC at the middle point intersects AB at P. Then AP : AB is:

(A) 4 : 1
(B) 2 : 3
(C) 3 : 5
(D) 1 : 4

Solution: According to question, Let AB = AC = 2x ∵ AQ = QC = x ∴ AB is a secant ∴ AP × AB = AQ2 AP × 2x = x2 $\begin{aligned} A P = \frac{x}{2} \\ \frac{A P}{A B} = \frac{x}{2 \times 2 x} = \frac{1}{4} \\ \frac{A P}{A B} = \frac{1}{4} \\ A P : A B = 1 : 4 \end{aligned}$

123.

If the length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, then the length of the median to its greatest side is -

(A) 8 cm
(B) 6 cm
(C) 5 cm
(D) 4.8 cm

Solution: According to question, Length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, this is right angle triangle. Note: In right angle triangle median divides the hypotenuse in two equal parts ∴ BD = $\frac{H}{2}$ BD = $\frac{10}{2}$ BD = 5 cm

124.

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 $\begin{aligned} = \frac{E F}{2} \\ = \frac{12}{2} \\ = 6 cm \end{aligned}$

125.

Consider the following statements :
I. Every equilateral triangle is necessarily an isosceles triangle.
II. Every right-angled triangle is necessarily an isosceles triangle.
III. A triangle in which one of the median is perpendicular to the side it meets, is necessarily an isosceles triangle.
The correct statements are:

(A) I and II
(B) II and III
(C) I and III
(D) I, II and III

126.

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

127.

ABC is an isosceles triangle with AB = AC. The side BA is produced to D such that AB = AD. If ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ABC = 30ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, then ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BCD is equal to

(A) 45°
(B) 90°
(C) 30°
(D) 60°

Solution: According to question, In ΔABC Exterior angle CAD = ∠ABC + ∠ACB CAD = 2∠ABC (∵ ∠ABC = ∠ACB) CAD = 2 × 30° CAD = 60° In ΔCAD, ∠ACD = ∠ADC = $\frac{180 - ∠ CAD}{2}$ = 60° ⇒ ∠BCD = ∠ACD + ∠BCA ⇒ ∠BCD = 60° + 30° ⇒ ∠BCD = 90°

128.

ABC is a right-angled triangle with AB = 6 cm and BC = 8 cm. A circle with center O has been inscribed inside ÃÂÃÂÃÂÃÂÃÂÃÂABC. The radius of the circle is

(A) 1 cm
(B) 2 cm
(C) 3 cm
(D) 4 cm

Solution: According to question, Given : AB = 6 cm, BC = 8 cm In right angle ΔABC By using Pythagoras theorem AC2 = AB2 + BC2 AC2 = 62 + 82 AC2 = 36 + 64 AC2 = 100 AC = 10 cm In radius $\begin{aligned} = \frac{a + b - c}{2} \\ = \frac{8 + 6 - 10}{2} \\ = \frac{4}{2} \\ = 2 c m \end{aligned}$

129.

The orthocentre of a right angled triangle lies

(A) Outside the triangle
(B) At the right angular vertex
(C) On its hypotenuse
(D) Within the triangle

130.

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 $\begin{aligned} \frac{A D}{A B} = \frac{A E}{A C} = \frac{D E}{B C} \\ \frac{A D}{A B} = \frac{D E}{B C} \\ \Rightarrow \frac{1}{4} = \frac{D E}{12} \\ \Rightarrow D E = 3 cm \end{aligned}$

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Practice MCQ Questions and Answer on Triangles

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