Question

The perimeter of triangle is 50 cm. One side of the triangle is 4 cm longer than the smallest side and the third side is 6 cm less than twice the smallest side. Find the area of the triangle.

Answer 1

Let ABC be any triangle with perimeter 50 cm.
Let the smallest side of the triangle be x.
Then the other sides be x + 4 and 2x − 6.

Now,
x + x + 4 + 2x − 6 = 50 (∵ perimeter is 50 cm)
⇒ 4x − 2 = 50
⇒ 4x = 50 + 2
⇒ 4x = 52
⇒ x = 13

∴ The sides of the triangle are of length 13 cm, 17 cm and 20 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{13+17+20}{2}=\frac{50}{2}=25\mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ABC=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{25\left(25-13\right)\left(25-17\right)\left(25-20\right)} \\ =\sqrt{25\left(12\right)\left(8\right)\left(5\right)} \\ =20\sqrt{30}{\mathrm{cm}}^2 \end{gathered}$$

Hence, the area of the triangle is $20\sqrt{30}{\mathrm{cm}}^2$.