Practice MCQ Questions and Answer on Triangles

81.

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

82.

In a triangle ABC, AB + BC = 12 cm, BC + CA = 14 cm and CA + AB = 18 cm. Find the radius of the circle (in cm) which has the same perimeter as the triangle

(A) $\frac{5}{2}$
(B) $\frac{7}{2}$
(C) $\frac{9}{2}$
(D) $\frac{11}{2}$

Solution: According to question, AB + BC = 12 cm BC + CA = 14 cm CA + AB = 18 cm 2(AB + BC + CA) = 44 cm AB + BC + CA = $\frac{44}{2}$ cm AB + BC + CA = 22 cm Perimeter of triangle = 22 cm Perimeter of triangle = Perimeter of circle 22 = 2πr 2 × $\frac{22}{7}$ × r = 22 r = $\frac{7}{2}$ cm

83.

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.

84.

In an isosceles triangle, if the unequal angle is twice the sum of the equal angles, then each equal angle is

(A) 120°
(B) 60°
(C) 30°
(D) 90°

Solution: According to question, Given : AB = AC y = 2(x + x) y = 4x ∴ x + x + y = 180° 2x + 4x = 180° 6x = 180° x = $\frac{180^{\circ}}{6}$ x = 30°

85.

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

86.

ÃÂÃÂÃÂÃÂÃÂÃÂABC is similar to ÃÂÃÂÃÂÃÂÃÂÃÂDEF. If the sides of ÃÂÃÂÃÂÃÂÃÂÃÂABC, that is AB, BC and CA, are 3, 4 and 5 cms respectively, what would be the perimeter of the ÃÂÃÂÃÂÃÂÃÂÃÂDEF, if the side DE measures 12 cms ?

(A) 24 cms
(B) 30 cms
(C) 36 cms
(D) 48 cms

Solution: Perimeter of ΔABC = 3 + 4 + 5 = 12 $\begin{aligned} \frac{Perimeter of △ A B C}{Perimeter of △ D E F} = \frac{A B}{D E} \\ \frac{12}{Perimeter of △ D E F} = \frac{3}{12} \end{aligned}$ Perimeter of ΔDEF = 48 cms

87.

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

88.

If two angles of a triangle are 21ÃÂÃÂÃÂÃÂ° and 38ÃÂÃÂÃÂÃÂ°, then the triangle is :

(A) Right-angled triangle
(B) Acute-angled triangle
(C) Obtuse-angled triangle
(D) Isosceles triangle

89.

PQR is an equilateral triangle. MN is drawn parallel to QR such that M is on PQ and N is on PR. If PN = 6 cm, then the length of MN is:

(A) 3 cm
(B) 6 cm
(C) 12 cm
(D) 4.5 cm

90.

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 $\begin{aligned} (x - 2) x (x + 2) \\ ↓ ↓ ↓ \\ 6 8 10 \\ ↓ \\ Triplet \\ x = 8 \end{aligned}$

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Practice MCQ Questions and Answer on Triangles

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