Given: right-angled $\triangle ABC$, the length of the hypotenuse $c = 5$ and the sum of the legs $a+b = 6$. Find the area of the triangle. I tried creating the following system $a+b = 6; a^2+b^2 =25$ however it got me nowhere, I get a really ugly quadratic equation .
$\endgroup$ 22 Answers
$\begingroup$We have \begin{align*} \text{Area }&=\frac12ab\sin \angle C\\[3pt] &=\frac12ab\sin\frac{\pi}2\\[3pt] &=\frac12ab\\[3pt] &=\frac14\left[(a+b)^2-(a^2+b^2)\right]\\[3pt] &=\frac14(36-25)\\[3pt] &=\frac{11}4 \end{align*}
$\endgroup$ $\begingroup$Squaring $a+b$
$(a+b)^2=a^2+b^2+2ab$
Putting $a+b = 6; a^2+b^2 =25$
$\Rightarrow 36=25+2ab$
$\Rightarrow ab=11/2$
Area is half of $ab$ = $11/4$
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