Question Number 27055 by Joel578 last updated on 01/Jan/18 | ||
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Commented by Joel578 last updated on 01/Jan/18 | ||
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$${EB}\:=\:\mathrm{4} \\ $$$${ED}\:=\:\mathrm{8} \\ $$$${EC}\:=\:\mathrm{7} \\ $$$$\mathrm{Find}\:{EA} \\ $$ | ||
Answered by mrW1 last updated on 01/Jan/18 | ||
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Commented by mrW1 last updated on 01/Jan/18 | ||
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$${a}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} \:\:\:...\left({i}\right) \\ $$$${a}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}} \:\:...\left({ii}\right) \\ $$$${b}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} \:\:\:...\left({iii}\right) \\ $$$$\left({i}\right)+\left({iii}\right)−\left({ii}\right): \\ $$$${a}^{\mathrm{2}} +{c}^{\mathrm{2}} +{b}^{\mathrm{2}} +{d}^{\mathrm{2}} −{a}^{\mathrm{2}} −{d}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}} =\mathrm{1}={x}^{\mathrm{2}} \\ $$$$\Rightarrow{EA}={x}=\sqrt{\mathrm{1}}=\mathrm{1} \\ $$$$ \\ $$$${generally}: \\ $$$${EA}^{\mathrm{2}} +{ED}^{\mathrm{2}} ={EB}^{\mathrm{2}} +{EC}^{\mathrm{2}} \\ $$ | ||
Commented by Rasheed.Sindhi last updated on 01/Jan/18 | ||
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$$\mathcal{E}{xcellent}! \\ $$ | ||
Commented by Joel578 last updated on 02/Jan/18 | ||
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$${thank}\:{you}\:{very}\:{much} \\ $$ | ||