Isosceles triangle with altitude and perimeter\nIn an isosceles triangle, the altitude drawn to the base is 8 cm. The perimeter of the triangle is 32 cm. Find the area of the triangle (in cm²).

Introduction / Context:
For an isosceles triangle, the altitude to the base bisects the base and forms two congruent right triangles. With perimeter given, we can determine the base and hence the area quickly.

Given Data / Assumptions:

• Altitude to base = 8 cm
• Perimeter = 32 cm
• Let equal sides be a and base be b

Concept / Approach:
The altitude splits the base into b/2 and forms a right triangle with legs 8 and b/2 and hypotenuse a. So a = sqrt(8^2 + (b/2)^2). Also 2a + b = 32. Area = (1/2)*b*8 = 4b.

Step-by-Step Solution:

Verification / Alternative check:
a = sqrt(64 + 36) = sqrt(100) = 10 cm. Perimeter = 10 + 10 + 12 = 32 cm (consistent).

Why Other Options Are Wrong:
24, 60, and 72 do not match the derived base and the formula (1/2)*b*h with h = 8 cm.

Common Pitfalls:
Forgetting that the altitude bisects the base in an isosceles triangle, or confusing perimeter with semiperimeter while substituting values leads to errors.

Final Answer:
48