Question Number 221117 by mr W last updated on 25/May/25
Commented by mr W last updated on 25/May/25
$${find}\:{shaded}\:{area}\:=? \\ $$
Answered by fantastic last updated on 25/May/25
Commented by fantastic last updated on 25/May/25
$${We}\:{have}\:{to}\:{find}\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\left({x}+{y}\right)\left(\sqrt{\mathrm{36}−{x}^{\mathrm{2}} }+{y}\right)−{x}\left(\sqrt{\left.\mathrm{36}−{x}^{\mathrm{2}} \right)}\right.\right. \\ $$$${or}=\frac{{xy}}{\mathrm{2}}+\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+\frac{{y}\sqrt{\mathrm{36}−{x}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${Now}\:{OD}=\left(\sqrt{\mathrm{36}−{x}^{\mathrm{2}} }+{y}\right)\:\left({O}\:{is}\:{the}\:{right}\:{angle}\:{vertex}\right) \\ $$$${but}\:{OD}=\sqrt{\mathrm{64}−\left({x}+{y}\right)^{\mathrm{2}} } \\ $$$${so}\:\left(\sqrt{\mathrm{36}−{x}^{\mathrm{2}} }+{y}\right)=\sqrt{\mathrm{64}−\left({x}+{y}\right)^{\mathrm{2}} } \\ $$$${or}\:\mathrm{36}−\cancel{{x}^{\mathrm{2}} }+{y}^{\mathrm{2}} +\mathrm{2}{y}\sqrt{\mathrm{36}−{x}^{\mathrm{2}} }=\mathrm{64}−\cancel{{x}^{\mathrm{2}} }−{y}^{\mathrm{2}} −\mathrm{2}{xy} \\ $$$${or}\:\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}{xy}\:+\mathrm{2}{y}\sqrt{\mathrm{36}−{x}^{\mathrm{2}} }=\mathrm{64}−\mathrm{36}=\mathrm{28} \\ $$$${or}\:{y}^{\mathrm{2}} +{xy}+{y}\sqrt{\mathrm{36}−{x}^{\mathrm{2}} }=\mathrm{14} \\ $$$${or}\:\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+\frac{{xy}}{\mathrm{2}}+\frac{{y}\sqrt{\mathrm{36}−{x}^{\mathrm{2}} }}{\mathrm{2}}=\mathrm{7} \\ $$$${so}\:{the}\:{area}\:{is}\:\mathrm{7}\:{sq}.{unit} \\ $$
Commented by Rasheed.Sindhi last updated on 25/May/25
$${fantastic}! \\ $$
Answered by Frix last updated on 25/May/25
$${AB}={CD}={x} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{64} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\left({a}−{x}\right)^{\mathrm{2}} +\left({b}−{x}\right)^{\mathrm{2}} =\mathrm{36} \\ $$$$\mathrm{Shaded}\:\mathrm{area} \\ $$$$\mathrm{A}=\frac{{ab}}{\mathrm{2}}−\frac{\left({a}−{x}\right)\left({b}−{x}\right)}{\mathrm{2}}=\frac{\left({a}+{b}\right){x}−{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{64} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\left({a}+{b}\right){x}−{x}^{\mathrm{2}} =\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\mathrm{36}}{\mathrm{2}}=\mathrm{2}{A}=\mathrm{14} \\ $$$${A}=\mathrm{7} \\ $$
Commented by Rasheed.Sindhi last updated on 25/May/25
$$\mathrm{sir},\:\:\vee.\:\mathbb{N}\boldsymbol{\mathrm{i}}\subset\in! \\ $$
Commented by Frix last updated on 25/May/25
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Answered by mr W last updated on 25/May/25
Commented by mr W last updated on 25/May/25
$$\mathrm{4}×\:{shaded}\:{area}\:=\mathrm{8}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} \\ $$$${shaded}\:{area}\:=\frac{\mathrm{8}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{7}\:\checkmark \\ $$
Commented by fantastic last updated on 25/May/25
$${Never}\:{thought}\:{like}\:{this}! \\ $$
Commented by Rasheed.Sindhi last updated on 25/May/25
$${e}^{\mathcal{X}} {cellent}\:{sir}! \\ $$