Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

The sides of a triangle are 35 cm, 54 cm and 61 cm respectively. Find the length of its longest altitude.

Answer» Let a = 35 cm, b = 54 cm and c = 61 cm. Then,
`s=(1)/(2)(a+b+c)=(1)/(2)(35+54+61)cm=75cm`.
` therefore" "(s-a)=(75-35)cm = 40 cm,(s-b)=(75-54)cm = 21cm and (s-c)=(75-61)cm = 14 cm`.
`therefore" ""area of the triangle"= sqrt(s(s-a)(s-b)(s-c))cm^(2)`
`= sqrt(75 xx 40 xx 21 xx 14)cm^(2)`
`= sqrt(15 xx 15 xx 14 xx 14 xx 4 xx 5)cm^(2)`
`=(15 xx 14 xx 2) sqrt(5)cm^(2) = 420 sqrt(5) cm^(2)`.
The longest altitude will be on smallest base.
Let the longest altitude be h cm. Then,
base = 35 cm and height = h cm.
`therefore" "(1)/(2) xx 35 xx h = 420 sqrt(5) rArr h = ((2 xx 420 xx sqrt(5))/(35))cm = 24 sqrt(5)` cm.
Hence, the longest altitude is `24 sqrt(5)` cm.
Discussion
2.

Find the area of a triangle having base 25 cm and height 10.8 cm .

Answer» Here, base = 25 cm and height = 10.8 cm.
Area of the triangle `= ((1)/(2) xx "base" xx "height")`sq units
`= ((1)/(2) xx 25 xx 10.8)cm^(2) = 135 cm^(2)`.
Hence, the area of the given triangle is 135 `cm^(2)`.
Discussion
3.

The perimeter of an equilateral triangle is 60 cm. Find its (i) area and (ii) height. (Given, `sqrt(3) = 1732`.)

Answer» Perimeter of the given equilateral triangle = 60 cm.
Length of each of its sides `=(60)/(3) cm = 20 cm`.
(i) Area of the triangle `=((sqrt(3))/(4)xx a^(2))`sq units
`=((sqrt(3))/(4)xx 20 xx 20)cm^(2)=(100 xx sqrt(3))cm^(2)`
`=(100 xx 1.732)cm^(2) = 173.2 cm^(2)`.
Hence, the area of the given triangle is 173.2 `cm^(2)`.
(ii) Let the height of the given triangle be h cm. Then,
`"its area"=((1)/(2)xx "base" xx "height")=((1)/(2)xx 20 xx h)cm^(2)`.
`therefore" "(1)/(2)xx 20 xx h=173.2 rArr h=(173.2)/(10) rArr "height" = 17.32 cm`.
Hence, the height of the given triangle is 17.32 cm.
Discussion
4.

The area of a trapezium is 475 `cm^(2)` and its height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.

Answer» Correct Answer - 23 cm, 27 cm
Let the parallel sides be x cm and (x + 4) cm. Then,
`(1)/(2)xx[x+(x+4)]xx 19 = 475 rArr 2x + 4 = (475 xx 2)/(19) rArr 2x = 46 rArr x = 23`.
Discussion
5.

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find the area of the triangle. Hint Ratio of sides = 3 : 3 : 2.

Answer» Correct Answer - `32 sqrt(3) cm^(2)`
Discussion
6.

The perimeter of an isosceles triangle is 42 cm and its base is `1(1)/(2)` times each of the equal sides. Find (i) the length of each side of the triangle, (ii) the area of the triangle, and (iii) the height of the triangle. (Given, `sqrt(7) = 2.64`.)

Answer» Correct Answer - (i) 12 cm, 12 cm, 18 cm (ii) 71.28 `cm^(2)` (iii) 7.92 cm
(i) `x + x + (3)/(2)x = 42 rArr (7x)/(2) = 42 rArr x = (84)/(7)=12`.
So, the sides are 12 cm, 12 cm, 18 cm.
(ii) Area of the triangle `= sqrt(s(s-a)(s-b)(s-c))=sqrt(21 xx 9 xx 9 xx 3)cm^(2)`
`" "=(27 xx sqrt(7))cm^(2)=(27 xx 2.64)cm^(2)=71.28 cm^(2)`.
(iii) `(1)/(2) xx 18 cm xx h = 71.28 cm^(2) rArr h=(71.28)/(9)cm = 7.92 cm`.
Discussion
7.

The sides of a quadrilateral ABCD taken in order are 6 cm, 8 cm, 12 cm and 14 cm respectively and the angle between the first two sides is a right angle. Find its area. (Given, `sqrt(6) = 2.45`.)

Answer» Correct Answer - 82.8 `cm^(2)`
Discussion
8.

The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle isA. `16 sqrt(5) cm^(2)`B. `8 sqrt(5) cm^(2)`C. `16 sqrt(3) cm^(2)`D. `8 sqrt(3) cm^(2)`

Answer» Correct Answer - B
Discussion
9.

Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find (i) the area of the field and (ii) the length of the perpendicular from the opposite vertex on the side measuring 154 m.

Answer» Correct Answer - (i) 2772 `m^(2)` (ii) 36 m
Discussion