In triangle ABC,

AB = AC = x; BC = 10 cm and the area of the triangle is 60 cm². Find x.

In △ ABC,

Draw AD ⊥ BC

By formula,

Area of triangle = $\dfrac{1}{2} \times$ base × height

Given,

Area of triangle ABC = 60 cm²

$\Rightarrow \dfrac{1}{2} \times BC \times AD = 60 \\[1em] \Rightarrow \dfrac{1}{2} \times 10 \times AD = 60 \\[1em] \Rightarrow 5 \times AD = 60 \\[1em] \Rightarrow AD = \dfrac{60}{5} = 12 \text{ cm}.$

We know that,

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

∴ BD = CD = $\dfrac{BC}{2} = \dfrac{10}{2}$ = 5 cm.

In right-angled triangle ADB,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AD² + BD²

⇒ x² = 12² + 5²

⇒ x² = 144 + 25

⇒ x² = 169

⇒ x = $\sqrt{169}$ = 13 cm.

Hence, x = 13 cm.