ABC is a triangle in which \((AB + BC)\) exceeds CA by \(10\ \text{cm}\), \((BC + CA)\) exceeds AB by \(8\ \text{cm}\) and \((CA + AB)\) exceeds BC by \(72\ \text{cm}\). What is the area of the triangle?
\(180\ \text{cm}^2\)
Let \(AB=x, BC=y, CA=z\). Then \(x+y-z=10\), \(y+z-x=8\), \(z+x-y=72\). Adding the first two: \(2y=18 \Rightarrow y=9\). Adding the last two: \(2z=80 \Rightarrow z=40\). Adding the first and third: \(2x=82 \Rightarrow x=41\). So sides are \(41, 9, 40\) with perimeter \(90\), semi-perimeter \(s=45\). By Heron's formula, Area \(= \sqrt{s(s-x)(s-y)(s-z)} = \sqrt{45 \times 4 \times 36 \times 5} = \sqrt{32400} = 180\ \text{cm}^2\).
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