In an isosceles triangle, the altitude drawn to the base is 8 cm and the perimeter is 32 cm. Find the area of the triangle (in sq cm).

Introduction / Context:
This question tests properties of an isosceles triangle combined with right-triangle geometry. In an isosceles triangle, the altitude from the vertex to the base bisects the base into two equal halves. That creates two identical right triangles. With altitude given, if we can find the base length, the area becomes straightforward because Area = (1/2)*base*height. The perimeter gives a relation between the equal sides and the base. Using these facts together allows us to solve for the side length and base length without guessing.

Given Data / Assumptions:

• Altitude to base (height) = 8 cm
• Perimeter = 32 cm
• Let equal sides = a cm each
• Let base = b cm
• Altitude bisects base => half-base = b/2

Concept / Approach:
Use perimeter: 2a + b = 32. Use Pythagoras in the right triangle: a^2 = 8^2 + (b/2)^2. Solve these equations to find b, then area = (1/2)*b*8.

Step-by-Step Solution:
Perimeter: 2a + b = 32 => b = 32 - 2a Right triangle: a^2 = 8^2 + (b/2)^2 Substitute b: a^2 = 64 + ((32 - 2a)/2)^2 = 64 + (16 - a)^2 Expand: a^2 = 64 + (a^2 - 32a + 256) = a^2 - 32a + 320 Cancel a^2: 0 = -32a + 320 => a = 10 b = 32 - 2*10 = 12 Area = (1/2)*b*height = (1/2)*12*8 = 48

Verification / Alternative check:
Half-base is 6 cm. Check side: sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10 cm, matching a. Perimeter becomes 10+10+12=32, consistent. Hence area is correct.

Why Other Options Are Wrong:
24 results from halving incorrectly or using base/2 as the base. 60 and 72 imply a larger base than what the perimeter allows. 36 comes from using wrong height or wrong base after solving.

Common Pitfalls:
Forgetting the altitude bisects the base in an isosceles triangle, using b instead of b/2 in Pythagoras, or making algebra errors when expanding (16 - a)^2.

Final Answer:
The area of the triangle is 48 sq cm.