Question

The perimeter of right angled triangle is $24cm$.If its hypotenuse is $10$cm the area of this triangle is :

• A. $24c{m}^2$
• B. $10c{m}^2$
• C. $12c{m}^2$
• D. $48c{m}^2$

Answer 1

Answer: (A)

$24c{m}^2$

Perimeter $=24$
Sum of other two sides $=24-10-14$
(as $10$ is hypotenuse)
So, applying pythagoras theorem, we get
${10}^2=x^2+(14-x{)}^2$

$\Rightarrow 100=x^2+196+x^2-28x$

$\Rightarrow 2x^2-28x+96=0$

$\Rightarrow x^2-14x+48=0$

$\Rightarrow x^2-8x-6x+48=0$

So, $x=8$ or $6$

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

$=\frac{1}{2}\times 6\times 8$

$=24c{m}^2$