Each of the equal sides of an isosceles triangle is 2 cm more than its height and the base of the triangle is 12 cm. Find the area of the triangle.

Let ABC be an isosceles triangle in which AB = AC, AD ⊥ BC and BC is the base.

Given,

Each of the equal sides of the isosceles triangle is 2 cm more than its height.

Let the height of the triangle be h cm.

Equal sides: AB = AC = h + 2

Base: BC = 12 cm

In Δ ABD and Δ ACD,

AD = AD [Common Side]

∠ADB = ∠ADC [Both equal to 90°]

AB = AC [Δ ABC is an isosceles triangle]

∴ Δ ABD ≅ Δ ACD [By R.H.S. axiom]

∴ BD = CD [C.P.C.T.C.]

∴ BD = CD = $\dfrac{BC}{2} = \dfrac{12}{2}$ = 6 cm.

By using the Pythagoras theorem in Δ ABD,

⇒ BD² + AD² = AB²

⇒ 6² + h² = (h + 2)²

⇒ 36 + h² = h² + 4 + 2 × h × 2

⇒ 36 = 4 + 4h

⇒ 4h = 32

⇒ h = $\dfrac{32}{4}$

⇒ h = 8 cm.

$\Rightarrow \text{Area of triangle ABC} = \dfrac{1}{2} \times \text{ base } \times \text{ height } \\[1em] = \dfrac{1}{2} \times BC \times AD \\[1em] = \dfrac{1}{2} \times 12 \times 8 \\[1em] = 6 \times 8 \\[1em] = 48 \text{ cm}^2.$

Hence, area of the triangle = 48 cm².