ABC is an isosceles triangle with AB = AC = 12 cm and BC = 8 cm. The area of the triangle is :

1. $32\sqrt{2}$ cm²

2. $16\sqrt{2}$ cm²

3. $8\sqrt{2}$ cm²

4. $12\sqrt{2}$ cm²

Let AD be the altitude.

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

∴ BD = CD = $\dfrac{BC}{2} = \dfrac{8}{2}$ = 4 cm.

In right angle triangle ABD,

By pythagoras theorem,

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

⇒ AB² = AD² + BD²

⇒ 12² = AD² + 4²

⇒ 144 = AD² + 16

⇒ AD² = 144 - 16

⇒ AD² = 128

⇒ AD = $\sqrt{128} = 8\sqrt{2}$ cm.

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

From figure,

⇒ Area of △ ABC = Area of △ ABD + Area of △ ACD

⇒ Area of △ ABC = $\dfrac{1}{2} \times BD \times AD + \dfrac{1}{2} \times CD \times AD$

⇒ Area of △ ABC

$= \dfrac{1}{2} \times AD \times (BD + CD) = \dfrac{1}{2} \times 8\sqrt{2} \times (4 + 4) = 4\sqrt{2} \times 8 = 32\sqrt{2} \text{ cm}^2.$

Hence, Option 1 is the correct option.