Question

In a right angled triangle one of the perpendicular sides is $4cm$ greater than the other and $4cm$ less than the hypotenuse. Find the area of that triangle.

• A. $72{cm}^2$
• B. $48{cm}^2$
• C. $36{cm}^2$
• D. $96{cm}^2$

Answer 1

The correct option is D $96{cm}^2$

Let PQR be the right angled triangle. Now,

$a=b+4\Rightarrow b=a-4$

$c=a+4$

Using pythayoras theorem,

$a^2+b^2=c^2\Rightarrow a^2+{(a-4)}^2={(a+4)}^2$

$\Rightarrow a^2+a^2+16-8a=a^2+16+8a\Rightarrow a^2=16a\Rightarrow a=16cm$

$\therefore b=16-4=12cm$

Hence, ar $△PQR=\frac{1}{2}\times a\times b=\frac{1}{2}\times 16\times 12=96{cm}^2$