Question

The perimeter of a right triangle is 40 cm and its hypotenuse measures 17 cm. Find the area of the triangle.

Answer 1

The perimeter of a right-angled triangle = 40 cm
Therefore , a+b+c= 40 cm
Hypotenuse = 17 cm
Therefore, c = 17 cm
a+b+c= 40 cm
$\Rightarrow$a+b+17 = 40
$\Rightarrow$a+b = 23
$\Rightarrow$b = 23 $-$ a...........(i)
Now, using Pythagoras' theorem, we have:
$$\begin{gathered} a^2+b^2=c^2 \\ \Rightarrow a^2+(23-a{)}^2={17}^2 \\ \Rightarrow a^2+529-46a+a^2=289 \\ \Rightarrow 2a^2-46a+529-289=0 \\ \Rightarrow 2a^2-46a+240=0 \\ \Rightarrow a^2-23a+120=0 \\ \Rightarrow (a-15\left)\right(a-8)=0 \\ \Rightarrow a=15ora=8 \end{gathered}$$

Substituting the value of a=15, in equation(i) we get:
b = 23-a
= 23 - 15
= 8 cm

If we had chosen $a=8\mathrm{cm}$, then, $b=23-8=15\mathrm{cm}$

In any case,
​ $$\begin{gathered} \mathrm{Area}\mathrm{of}\text{a t}\mathrm{riangle}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height} \\ =\frac{1}{2}\times 8\times 15 \\ =60{\mathrm{cm}}^2 \end{gathered}$$