Question:medium

The altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32cm. Find the area of the triangle?

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Always draw a diagram for geometry problems. Use algebraic equations for lengths and apply theorems like Pythagoras. Double-check calculations and compare with options. If discrepancies arise, highlight them.
Updated On: Jul 14, 2026
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Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Let the equal side be \(s\) and half of the base be \(k\). From the perimeter, \(2s+2k=32\), so \(s+k=16\). From the Pythagorean theorem using the altitude of 8, \(s^2=k^2+8^2\), which rearranges to \(s^2-k^2=64\).

Step 2: Factor the left side as a difference of squares: \((s-k)(s+k)=64\). Since \(s+k=16\), substitute to get \((s-k)(16)=64\), so \(s-k=4\).

Step 3: Solve the pair \(s+k=16\) and \(s-k=4\) by adding and subtracting: \(2s=20 \Rightarrow s=10\), and \(2k=12 \Rightarrow k=6\). So the base is \(2k=12\) cm.

Step 4: Compute the area: \( \text{Area} = \frac{1}{2}\times12\times8 = 48 \).
\[ \boxed{48 \text{ cm}^2} \]
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