# The lengths of the two sides of a right triangle containing the right angle differ by 2 cm.

Question:

The lengths of the two sides of a right triangle containing the right angle differ by 2 cm. If the area of the triangle is 24 cm2, find the perimeter of the triangle.

Solution:

Given:

Area of triangle $=24 \mathrm{~cm}^{2}$

Let the sides be a and b, where a is the height and b is the base of triangle.

$a-b=2 \mathrm{~cm}$

$a=2+b \ldots \ldots .(1)$

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

$\Rightarrow 24=\frac{1}{2} \times b \times(2+b)$

$\Rightarrow 48=b+\frac{1}{2} b^{2}$

$\Rightarrow 48=2 b+b^{2}$

$\Rightarrow b^{2}+2 b-48=0$

$\Rightarrow(b+8)(b-6)=0$

$\Rightarrow b=-8$ or 6

Side of a triangle cannot be negative.

Therefore, b = 6 cm.

Substituting the value of b = 6 cm in equation(1), we get:
a = 2+6 = 8 cm

Now,  a = 8 cm, b = 6 cm

In the given right triangle we have to find third side. Using the relation

$(\mathrm{Hyp})^{2}=(\text { Oneside })^{2}+(\text { Otherside })^{2}$

$\Rightarrow \mathrm{Hyp}^{2}=8^{2}+6^{2}$

$\Rightarrow \mathrm{Hyp}^{2}=64+36$

$\Rightarrow \mathrm{Hyp}^{2}=100$

$\Rightarrow \mathrm{Hyp}=10 \mathrm{~cm}$

So, the third side is 10 cm.

So, perimeter of the triangle = b + c
= 8+6+10
​=24 cm