An altitude of \(12\sqrt{2}\) cm is drawn to the base of an isosceles triangle. The perimeter of this triangle is 48 cm. Find the area of the triangle.
\(72\sqrt{2}\) cm2
Let the base be \(2b\) and each equal side be \(a\). The altitude to the base bisects it, forming a right triangle with legs \(b\) and \(12\sqrt{2}\) and hypotenuse \(a\).
So \(a^2 = b^2 + (12\sqrt{2})^2 = b^2 + 288\).
The perimeter gives \(2a + 2b = 48\), so \(a + b = 24\), giving \(a = 24 - b\).
Substitute: \((24-b)^2 = b^2 + 288\), so \(576 - 48b + b^2 = b^2 + 288\).
Then \(576 - 48b = 288\), giving \(48b = 288\) and \(b = 6\).
So the base is \(2b = 12\) cm and the equal side is \(a = 18\) cm.
Area \(= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 12\sqrt{2} = 72\sqrt{2}\) cm2.
Hence, the area of the triangle is \(72\sqrt{2}\) cm2.
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