The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle.

Let ABC be an isosceles triangle with AB = AC = a cm and BC = b cm.

Altitude (AD) = 8 cm

Perimeter = 32 cm

Perimeter = sum of all sides of a triangle

⇒ 32 = a + a + b

⇒ 32 = 2a + b

⇒ b = 32 - 2a ………(1)

In an isosceles triangle, the altitude drawn from the common vertex bisects the base.

Thus, AD bisects BC.

So, BD = DC = $\dfrac{b}{2}$

∴ ∠ADC = ∠ADB = 90°.

In triangle ADB,

By pythagorean theorem,

$\Rightarrow AB^2 = AD^2 + DB^2 \\[1em] \Rightarrow a^2 = 8^2 + \Big(\dfrac{b}{2}\Big)^2 \\[1em] \Rightarrow a^2 = 64 + \dfrac{b^2}{4} \\[1em] \Rightarrow a^2 = \dfrac{256 + b^2}{4} \\[1em] \Rightarrow 4a^2 = 256 + b^2$

Substituting the value of b from equation (1) in above equation, we get :

$\Rightarrow 4a^2 = 256 + (32 - 2a)^2 \\[1em] \Rightarrow 4a^2 = 256 + 1024 -128a + 4a^2 \\[1em] \Rightarrow 0 = 1280 - 128a \\[1em] \Rightarrow 128a = 1280 \\[1em] \Rightarrow a = \dfrac{1280}{128} \\[1em] \Rightarrow a = 10 \text{ cm}.$

∴ b = 32 - 2(10)

⇒ b = 32 - 20

⇒ b = 12 cm.

Area of triangle ABC = $\dfrac{1}{2}$ × Base × Height

= $\dfrac{1}{2}$ × BC × AD

= $\dfrac{1}{2}$ × 12 × 8

= 6 × 8 = 48 cm².

Hence, area of triangle = 48 cm².