The-perimeter-of-an-isosceles-right-angled-triangle-is-2p-Find-out-the-area-of-the-triangle-

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Question Number 99 by sagarwal last updated on 25/Jan/15

$$\mathrm{The}\mathrm{perimeter}\mathrm{of}\mathrm{an}\mathrm{isosceles}\mathrm{right}-\mathrm{angled}$$$$\mathrm{triangle}\mathrm{is}2p.\mathrm{Find}\mathrm{out}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle}.$$

Answered by 123456 last updated on 14/Dec/14

$$\mathrm{lets}\mathrm{\Delta }\mathrm{A}BC\mathrm{be}\mathrm{a}\mathrm{triangle}\mathrm{with}\mathrm{lenght}a,b,c$$$$\mathrm{by}\mathrm{it}\mathrm{isosceles}\mathrm{mean}\mathrm{that}\mathrm{two}\mathrm{or}\mathrm{more}\mathrm{angle}\mathrm{and}\mathrm{lenght}\mathrm{are}\mathrm{equal}$$$$\mathrm{since}\mathrm{its}\mathrm{right}\mathrm{angle}\mathrm{and}\mathrm{isosceles},\mathrm{we}\mathrm{have}\mathrm{a}\mathrm{angle}90°,\mathrm{then}\mathrm{the}\mathrm{other}\mathrm{will}\mathrm{be}45°$$$$\mathrm{then}\mathrm{we}\mathrm{gets}\mathrm{seting}a,b\mathrm{catetes}$$$$a=b$$$$a^2+b^2=2a^2=c^2$$$$\mathrm{the}\mathrm{perimet}\mathrm{is}$$$$\mathrm{a}+\mathrm{b}+\mathrm{c}=2\mathrm{a}+\mathrm{c}=2\mathrm{p}$$$$\mathrm{then}\mathrm{the}\mathrm{area}\mathrm{is}$$$$\mathbb{A}=\frac{ab}{2}=\frac{a^2}{2}$$$$\mathrm{solving}$$$${2\mathrm{a}}^2-{\mathrm{c}}^2=0$$$$2\mathrm{a}+\mathrm{c}=2\mathrm{p}$$$$\mathrm{c}=2(\mathrm{p}-\mathrm{a})$$$${\mathrm{c}}^2={4\mathrm{p}}^2-8\mathrm{ap}+{4\mathrm{a}}^2$$$$-{2\mathrm{a}}^2+8\mathrm{ap}-{4\mathrm{p}}^2=0$$$${2\mathrm{a}}^2-8\mathrm{ap}+{4\mathrm{p}}^2=0$$$$\mathrm{\Delta }={(-8p)}^2-4\left(2\right)\left(4p^2\right)$$$$=64p^2-16p^2=48p^2$$$$a=\frac{(8\pm 2\sqrt{12})p}{4}=(2\pm \sqrt{12})p$$$$c=2p-2a$$$$\mathrm{we}\mathrm{gets}$$$$\mathrm{a}=(2+\sqrt{12})p$$$$\mathrm{c}=2(\sqrt{12}-1)p$$$$\mathbb{A}=\frac{{(2+\sqrt{12})}^2{p}^2}{2}=\frac{(16+4\sqrt{12})p^2}{2}=(8+2\sqrt{12})p^2$$

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