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.

Answer 1

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.......(1)

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

$$\begin{gathered} \Rightarrow 24=\frac{1}{2}\times b\times (2+b) \\ \Rightarrow 48=b+\frac{1}{2}{b}^2 \\ \Rightarrow 48=2b+b^2 \\ \Rightarrow b^2+2b-48=0 \\ \Rightarrow (b+8)(b-6)=0 \\ \Rightarrow b=-8\mathrm{or}6 \end{gathered}$$
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

$$\begin{gathered} (\mathrm{Hyp}{)}^2=(\mathrm{Oneside}{)}^2+(\mathrm{Otherside}{)}^2 \\ \Rightarrow {\mathrm{Hyp}}^2=8^2+6^2 \\ \Rightarrow {\mathrm{Hyp}}^2=64+36 \\ \Rightarrow {\mathrm{Hyp}}^2=100 \\ \Rightarrow \mathrm{Hyp}=10\mathrm{cm} \end{gathered}$$

So, the third side is 10 cm.

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