One of the equal sides of an isosceles triangle is 13 cm and its perimeter is 50 cm. Find the area of the triangle.

Given:

One of the equal sides of an isosceles triangle = 13 cm

Perimeter of the triangle = 50 cm

Let a be the length of the third side of the triangle.

∵ Perimeter of triangle = sum of sides

⇒ 13 + 13 + a = 50 cm

⇒ 26 + a = 50 cm

⇒ a = 50 - 26 cm

⇒ a = 24 cm

Let a = 13 m, b = 13 m and c = 24 m.

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{13 + 13 + 24}{2}\\[1em] = \dfrac{50}{2}\\[1em] = 25$

∵ Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

= $\sqrt{25(25 - 13)(25 - 13)(25 - 24)}$ cm²

= $\sqrt{25 \times 12 \times 12 \times 1}$ cm²

= $\sqrt{3,600}$ cm²

= 60 cm²

Hence, the area of the triangle is 60 cm².