In an isosceles triangle, the altitude drawn to the base is 8 cm and the total perimeter of the triangle is 32 cm. Assuming the altitude is drawn from the vertex between the equal sides, find the area of the triangle in square centimetres (cm^2).

Introduction / Context:
This problem checks your understanding of isosceles triangles and how the altitude to the base behaves. In an isosceles triangle, the altitude from the vertex to the base bisects the base into two equal parts, creating two identical right triangles. We use perimeter information to find the side lengths and then compute area.

Given Data / Assumptions:

• Altitude to the base, h = 8 cm
• Perimeter, P = 32 cm
• Let each equal side be a cm
• Let base be b cm
• Altitude bisects base, so each half is b/2

Concept / Approach:
Use two equations: (1) Perimeter: 2a + b = 32. (2) Pythagoras in one right triangle: a^2 = h^2 + (b/2)^2. Solve for a and b, then area = (1/2) * b * h.

Step-by-Step Solution:
Perimeter: 2a + b = 32 Right triangle relation: a^2 = 8^2 + (b/2)^2 = 64 + (b^2/4) From perimeter: b = 32 - 2a Substitute into a^2 equation (solving) gives a = 10 and b = 12 Area = (1/2) * b * h = (1/2) * 12 * 8 = 48 cm^2

Verification / Alternative check:
Check Pythagoras: half-base = 12/2 = 6. Then a^2 = 8^2 + 6^2 = 64 + 36 = 100, so a = 10. Perimeter becomes 10 + 10 + 12 = 32, correct.

Why Other Options Are Wrong:
44 cm^2: would not match base and side lengths that satisfy both altitude and perimeter. 52 cm^2: implies base would be 13 cm if h=8, but then perimeter and right triangle relation break. 56 cm^2: implies base 14 cm with h=8, which conflicts with the required equal sides. 60 cm^2: implies base 15 cm with h=8, not consistent with perimeter 32 cm.

Common Pitfalls:
Not using the fact that the altitude bisects the base in an isosceles triangle. Using area formulas before finding the base correctly. Forgetting that perimeter includes both equal sides plus the base.

Final Answer:
Area of the triangle = 48 cm^2