Question Number 224702 by gregori last updated on 28/Sep/25
Commented by gregori last updated on 28/Sep/25
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Answered by mehdee7396 last updated on 28/Sep/25
$${cosA}_{\mathrm{1}} =\frac{\mathrm{10}}{\:\mathrm{5}\sqrt{\mathrm{5}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}={sinA}_{\mathrm{2}} \\ $$$${s}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×\frac{\mathrm{5}\sqrt{\mathrm{5}}}{\mathrm{2}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}=\mathrm{25} \\ $$$${s}=\mathrm{100}−\mathrm{25}−\mathrm{25}−\frac{\mathrm{25}}{\mathrm{2}}=\frac{\mathrm{75}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by mehdee7396 last updated on 28/Sep/25
Answered by mr W last updated on 28/Sep/25
Commented by mr W last updated on 28/Sep/25
$$\Box={area}\:{of}\:{square} \\ $$$${A}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{\Box}{\mathrm{2}}=\frac{\Box}{\mathrm{4}} \\ $$$${A}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{\Box}{\mathrm{4}}=\frac{\Box}{\mathrm{8}} \\ $$$${shaded}\:{area}={A}_{\mathrm{1}} +{A}_{\mathrm{2}} =\frac{\Box}{\mathrm{4}}+\frac{\Box}{\mathrm{8}}=\frac{\mathrm{3}\Box}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}×\mathrm{100}^{\mathrm{2}} }{\mathrm{8}}=\mathrm{37}.\mathrm{5}\:{cm}^{\mathrm{2}} \\ $$
Answered by fantastic last updated on 28/Sep/25
