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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 .

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  • $\begingroup$ Continue: so $(a+b)^2=36$, so $2ab=11$. In other words, the area $\frac{ab}{2}=\frac{11}{4}$ $\endgroup$ – almagest Apr 17 '16 at 18:20
  • $\begingroup$ All you need is $\frac{1}{2}ab$ and $(ab)=((a+b)^2-(a^2+b^2))/2$ $\endgroup$ – Anurag A Apr 17 '16 at 18:21
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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*}

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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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