Question

The difference between the sides at right angle in a right-angled triangle is 7 cm. The area of the triangle is 60 cm². Find its perimeter.

Answer 1

Given:
Area of the triangle = $60{\mathrm{cm}}^2$
Let the sides of the triangle be a, b and c, where a is the height, b is the base and c is hypotenuse of the triangle.
$a-b=7\mathrm{cm}$
a = 7 + b.......(1)
$\mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\times b\times h$
$$\begin{gathered} \Rightarrow 60=\frac{1}{2}\times b\times (7+b) \\ \Rightarrow 120=7b+b^2 \\ \Rightarrow b^2+7b-120=0 \\ \Rightarrow (b+15)(b-8)=0 \\ \Rightarrow b=-15\mathrm{or}8 \end{gathered}$$

Side of a triangle cannot be negative.
Therefore, b = 8 cm.

Substituting the value of b = 8 cm, in equation (1):
a = 7+8 = 15 cm

Now, a = 15 cm, b = 8 cm

Now, in the given right triangle, we have to find third side.
$$\begin{gathered} (\mathrm{Hyp}{)}^2=(First\mathrm{side}{)}^2+(Second\mathrm{side}{)}^2 \\ \Rightarrow {\mathrm{Hyp}}^2=8^2+{15}^2 \\ \Rightarrow {\mathrm{Hyp}}^2=64+225 \\ \Rightarrow {\mathrm{Hyp}}^2=289 \\ \Rightarrow \mathrm{Hyp}=17\mathrm{cm} \end{gathered}$$
So, the third side is 17 cm.

Perimeter of a triangle = $a+b+c$.
Therefore, required perimeter of the triangle $=15+8+17=40\mathrm{cm}$.