Question

The difference between the sides at right angles in a right angled triangle is 14 cm. The area of the triangle is 120 cm${}^2$. Calculate the perimeter of the triangle.

Answer 1

Given that, in a right angled triangle, the difference between the sides at right angle is $14cm$. The area of the triangle is $120c{m}^2$.

To find out: The perimeter of the triangle.

Let one of the sides at the right angle of the triangle be $xcm$.

Hence, the other side will be $(x+14)cm$.

We know that, area of a triangle is $\frac{1}{2}\times b\times h$

In a right angled triangle, the sides at the right angle forms the base and altitude of the triangle.

Hence, $b$ and $h$ can interchangeably be $x$ and $x+14$.

$\therefore$ The area of the triangle $=\frac{1}{2}\times x\times (x+14)$

The area is given as $120c{m}^2$

Hence, $\frac{1}{2}\times x\times (x+14)=120$

$\Rightarrow x(x+14)=240$

$\Rightarrow x^2+14x-240=0$

$\Rightarrow x^2+24x-10x-240=0$

$\Rightarrow x(x+24)-10(x+24)=0$

$\Rightarrow (x+24)(x-10)=0$

Hence, $x=-24$ or $x=10$

The length of the side cannot be negative. Hence, $-24$ cannot be the value of $x$.

$\therefore x=10$

and $x+14=10+14$

$=24$

Now, the third side of the triangle will be the side opposite to the right angle, which is the hypotenuse.

We know that, for a right angled triangle, $(\text{Hypotenuse}{)}^2=(\text{Base}{)}^2+(\text{Perpendicular}{)}^2$

Let the hypotenuse of the triangle be $Hcm$

$\therefore H^2={10}^2+{24}^2$

$\Rightarrow H=\sqrt{{10}^2+{24}^2}$

$\Rightarrow H=\sqrt{100+576}$

$\Rightarrow H=\sqrt{676}$

$\therefore H=26cm$

Now, the perimeter of a triangle is the sum of all sides.

Hence, the perimeter of the given triangle $=10+24+26$

$=60cm$

Hence, the perimete of the triangle is $60cm$.