D) False — area is $ \frac{1}{2}ab $, and $ r s = r \cdot \frac{a+b+c}{2} $, and $ r = \frac{a+b-c}{2} $, so $ r s = \frac{a+b-c}{2} \cdot \frac{a+b+c}{2} = \frac{(a+b)^2 - c^2}{4} = \frac{a^2 + 2ab + b^2 - c^2}{4} $. Since $ c^2 = a^2 + b^2 $, this becomes $ \frac{2ab}{4} = \frac{ab}{2} $ — yes.

Sat, 09 May 2026 01:06:30 +0000

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