Find the area (in square cm) of a right angled triangle whose altitude is 7 cm less than its base and its hypotenuse is 17 cm.

Correct option is D

Given:

Hypotenuse of a triangle = 17cm

Altitude of triangle = 7 less than base

Formula Used:

Pythagoras’ Theorem

Area of triangle =$$\frac{1}{2} \times b \times h$$​​

Solution:

Let the base of triangle be x

Then altitiude = x-7

Using Pythagoras’ Theoram

​$$(H)^2 = (P)^2 +(B)^2$$​ 

​$$(17)^2 = (x-7)^2 +(x)^2$$​ 

​$$289 = x^2 + 49 - 14x + x^2$$​ 

​$$2x^2 -14x - 240 =0$$​​

​$$x^2 -7x -120 = 0$$​​

Solving we get:

x = 15 and -8

Neglecting negative value

Base = 15cm

Height = (x-7) = 15-7 = 8cm

Area of triangle =$$\frac{1}{2} \times 15 \times 8 = 60cm^2$$​​