SSC CHSL 20 Dec 2015 Shift 1 Paper

Question : 124 of 200

Marks: +1, -0

The hypotenuse of a right-angled triangle is 39 cm and the difference of other two sides is 21 cm. Then the area of the triangle is

Solution:

Let vertices of the triangle be A,B,C

AB^2 + BC^2 = AC^2 = 39^ 2 = 1521

........(1)
Let

(AB − BC)^ 2 = AB^2 + BC^2 − 2(AB)(BC)

.....(2)
Substitute equation (1) in equation (2)

21 ^2 = 1521 − 2(AB)(BC)

( \Given AB − BC = 21)

441 = 1521 − 2(AB)(BC)

⇒ 1080 = 2(AB)(BC)

⇒ (AB)(BC) = 540

Area of the triangle =

1/2

x base x height =

1/2(AB)(BC) = 1/ 2 (540) = 270

Hence, option B is the correct answer.

Go to Question:

Prev Question

Next Question