Correct Answer: 120

Solution :
Let the base of the isosceles triangle as $b\operatorname{ cm }$ and each of the equal sides as $a\operatorname{ cm }$.
The altitude drawn to the base = $8\operatorname{ cm }$
The perimeter of the isosceles triangle = $64\operatorname{ cm }$
In an isosceles triangle, the altitude bisects the base, dividing it into two equal parts.
Let each part as $\frac{b}{2}$.
We can use the Pythagorean theorem,
$⇒a = \sqrt{(\frac{b}{2})^2 + 8^2}$ ____(i)
Given that the perimeter of the triangle is $64\operatorname{ cm }$.
$⇒2a + b = 64$ ____(ii)
From equation (i) and equation (ii),
$⇒2\sqrt{(\frac{b}{2})^2 + 8^2} + b = 64$
$⇒\sqrt{ b^2 + 256}=64-b$
$⇒b^2 + 256=4096-128b+b^2$
$⇒128b=3840$
$⇒b = 30\operatorname{ cm }$
From equation (ii),
$⇒a = 17\operatorname{ cm }$.
The area of the triangle $=\frac{1}{2}bh=\frac{1}{2} \times 30 \times 8 = 120\operatorname{ cm^2 }$
Hence, the correct answer is 120.