Question

The lengths of the two sides of a right triangle containing the right angle differ by 2 cm. If the areah of the triangle is 24 $c{m}^2$, find the perimeter of the triangle.

Answer 1

Let the base of the triangle be b cm and its height or perpendicular be p cm.

Given, p - b = 2 ... (1)

and area of triangle = 24 $c{m}^2$

⇒$\frac{1}{2}\times b\times p=24$ ​​​​​​​

⇒ $b\times p=48$​​​​​​​​​​​​​​

⇒ $b=\frac{48}{p}$ ​​​​​​​

On putting the value of b in (1), we get

$p-\frac{48}{p}=2$ ​​​​​​​

⇒$p^2-2p-48=0$ ​​​​​​​

⇒ $p^2-8p+6p-48=0$ ​​​​​​​

⇒ $(p-8)(p+6)=0$ ​​​​​​​

⇒ $p=8,-6$ ​​​​​​​

p ≠ -6. So, p = 8 cm

On putting p in (1), we get

⇒ b = 6 cm

So, Hypotenuse (h) =$\sqrt{p^2+b^2}=\sqrt{8^2+6^2}=\sqrt{64+36}=\sqrt{100}=10$ ​​​​​​​ ​​​​​​​

Hence, perimeter of triangle = p + b + h = 8 cm + 6 cm + 10 cm = 24 cm