Practice MCQ Questions and Answer on Triangles

11.

In ÃÂÃÂÃÂÃÂÃÂÃÂABC, two points D and E are taken on the lines AB and BC respectively in such a way that AC is parallel to DE. Then ÃÂÃÂÃÂÃÂÃÂÃÂABC and ÃÂÃÂÃÂÃÂÃÂÃÂDBE are :

(A) Similar only if D lies outside the line segment AB
(B) Congruent only If D lies out side the line segment AB
(C) Always similar
(D) Always congruent

12.

The sum of three altitudes of a triangle is

(A) Equal to the sum of three sides
(B) Less than the sum of sides
(C) Greater than the sum of sides
(D) Twice the sum of sides

13.

If the sides of a right angled triangle are three consecutive integers, then the length of the smallest side is

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

14.

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 $\begin{aligned} = \frac{a + b - c}{2} \\ = \frac{8 + 6 - 10}{2} \\ = \frac{4}{2} \\ = 2 c m \end{aligned}$

15.

If the median drawn on the base of a triangle is half of its base the triangle will be

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

16.

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°

17.

In a ÃÂÃÂÃÂÃÂÃÂÃÂABC, AB = AC and BA is produced to D such that AC = AD. Then the ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BCD is :

(A) 100°
(B) 60°
(C) 80°
(D) 90°

18.

Given that the ratio of altitudes of two triangles is 4 : 5, ratio of their areas is 3 : 2, the ratio of their corresponding bases is :

(A) 5 : 8
(B) 15 : 8
(C) 8 : 5
(D) 8 : 15

Solution: $\begin{aligned} \frac{area of triangle 1}{area of triangle 2} = \frac{3}{2} \\ \Rightarrow \frac{\frac{1}{2} \times B_{1} \times 4}{\frac{1}{2} \times B_{2} \times 5} = \frac{3}{2} \\ \Rightarrow \frac{B_{1}}{B_{2}} = \frac{3}{2} \times \frac{5}{4} \\ \Rightarrow \frac{B_{1}}{B_{2}} = \frac{15}{8} \\ ∴ B_{1} : B_{2} = 15 : 8 \end{aligned}$

19.

In ÃÂÃÂÃÂÃÂÃÂÃÂABC, AD ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ¥ BC, then

(A) AB2 - BD2 = AC2 - CD2
(B) AB2 + BD2 = AC2 - CD2
(C) AB2 - BD2 = AC2 + CD2
(D) AB2 - AC2 = BD2 + CD2

20.

If ABC is an equilateral triangle and D is a point of BC such that AD ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ¥ BC, then

(A) AB : BD = 1 : 1
(B) AB : BD = 1 : 2
(C) AB : BD = 2 : 1
(D) AB : BD = 3 : 2

Solution: According to question, In equilateral triangle 'AD' bisects the BC in two equal parts. Let side of equilateral triangle is 2 cm $\begin{aligned} ∴ \frac{A B}{B D} = \frac{2}{1} \\ A B : B D = 2 : 1 \end{aligned}$

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Practice MCQ Questions and Answer on Triangles

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