Question

The difference between the sides at right angles in a right angled triangle is $14$ cm. The area of the triangle is $120$ cm${}^2$. The perimeter of the triangle

• A. $68cm$
• B. $64cm$
• C. $60cm$
• D. $58cm$

Answer 1

Answer: (C)

$60cm$

Let the sides at right angles in the given right-angled triangle be $a$ and $b$.

Without loss of generality, we let $a>b$.

The by the problem $a-b=14\Rightarrow a=(b+14)$........(1).

Now the are of the right-angled triangle be $\frac{1}{2}\times a\times b=\frac{b(b+14)}{2}$ cm${}^2$. [Using (1)]

According to the problem

$\frac{b(b+14)}{2}=120$

$b^2+14b-240=0$

$(b-10)(b+24)=0$

This gives $b=10\Rightarrow a=24$.

Now the hypotenuse of the right-angled triangle be $\sqrt{{10}^2+{24}^2}=26$ cm.

Now the perimeter of the triangle be $(10+24+26)=60$ cm.