# The perimeter of a right triangle is 40 cm and its hypotenuse measures 17 cm.

Question:

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

Solution:

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:

$a^{2}+b^{2}=c^{2}$

$\Rightarrow a^{2}+(23-a)^{2}=17^{2}$

$\Rightarrow a^{2}+529-46 a+a^{2}=289$

$\Rightarrow 2 a^{2}-46 a+529-289=0$

$\Rightarrow 2 a^{2}-46 a+240=0$

$\Rightarrow a^{2}-23 a+120=0$

$\Rightarrow(a-15)(a-8)=0$

$\Rightarrow a=15$ or $a=8$

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

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

In any case,

Area of a triangle $=\frac{1}{2} \times$ base $\times$ height

$=\frac{1}{2} \times 8 \times 15$

$=60 \mathrm{~cm}^{2}$