Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm.

Given,

base = 6 cm and perimeter = 16 cm

Consider △ABC as an isosceles triangle in which,

Let, AB = AC = x cm.

So, BC = 6 cm.

We know that,

Perimeter of △ABC = AB + BC + AC

Substituting the values we get,

⇒ 16 = x + 6 + x

⇒ 16 = 2x + 6

⇒ 16 – 6 = 2x

⇒ 10 = 2x

⇒ x = $\dfrac{10}{2}$ = 5 cm.

Area of an isosceles triangle = $\dfrac{1}{4}b\sqrt{4a^2 - b^2}$, where a is length of equal sides and b is the length of other side.

Substituting values we get,

$A = \dfrac{1}{4} \times 6 \times \sqrt{4(5)^2 - 6^2} \\[1em] = \dfrac{1}{4} \times 6 \times \sqrt{4 \times 25 - 36} \\[1em] = \dfrac{1}{4} \times 6 \times \sqrt{100 - 36} \\[1em] = \dfrac{1}{4} \times 6 \times \sqrt{64} \\[1em] = \dfrac{1}{4} \times 6 \times 8 \\[1em] = 12 \text{ cm}^2.$

Hence, area of isosceles triangle = 12 cm².