Question Number 224324 by fantastic last updated on 03/Sep/25
Answered by fantastic last updated on 03/Sep/25
Commented by fantastic last updated on 05/Sep/25
![Area EFHJ=AGHJE−AGHF−AEF AEF=AED−AFD =AEFGD−(AFHD−DHFG) ∴EFGH =AGHJE−AGHF−AEFGD+AFHD−DHFG =AFHD−AEFGD−AGHF Eq.of two quartercircles x^2 +(y−a)^2 =a^2 (x−a)^2 +(y−a)^2 =a^2 [A is (0,0)] ⇒x=(a/2) y=a(1+((√3)/2))[other one is −ve.here it will not apply] We got E((a/2),((2+(√3))/2)) ED=(√((0−(a/2))^2 +(a−a−a((√3)/2))^2 )) =(√((a^2 /4)+((3a^2 )/4)))=a ∴DEC is eqi.triangle AEFGD=((30^0 )/(360^0 ))πa^2 −(((60^0 )/(360^0 ))πa^2 −((√3)/4)a^2 ) =−(1/(12))πa^2 +((√3)/4)a^2 AGHF=2{((1/4)π×(a^2 /4))−((1/2)×(a^2 /4))} =(1/8)πa^2 −(a^2 /4) AFHD=(1/2)π(a^2 /4) =(1/8)πa^2 ∴EFHJ=(1/8)πa^2 −(((√3)/4)a^2 −(1/(12))πa^2 )−((1/8)πa^2 −(a^2 /4)) =(1/8)πa^2 −((√3)/4)a^2 +(1/(12))πa^2 −(1/8)πa^2 +(a^2 /4) =(a^2 /4)(1−(√3))+(1/(12))πa^2 =(a^2 /4)(1−(√3)+(1/3)π) ≈0.07878668590693a^2 ≈0.0788a^2](https://www.tinkutara.com/question/Q224333.png)
$${Area}\:{EFHJ}={AGHJE}−{AGHF}−{AEF} \\ $$$${AEF}={AED}−{AFD} \\ $$$$={AEFGD}−\left({AFHD}−{DHFG}\right) \\ $$$$\therefore{EFGH} \\ $$$$=\cancel{{AGHJE}}−{AGHF}−{AEFGD}+{AFHD}\cancel{−{DHFG}} \\ $$$$={AFHD}−{AEFGD}−{AGHF} \\ $$$${Eq}.{of}\:{two}\:{quartercircles} \\ $$$${x}^{\mathrm{2}} +\left({y}−{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\left({x}−{a}\right)^{\mathrm{2}} +\left({y}−{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \left[{A}\:{is}\:\left(\mathrm{0},\mathrm{0}\right)\right] \\ $$$$\Rightarrow{x}=\frac{{a}}{\mathrm{2}}\:{y}={a}\left(\mathrm{1}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left[{other}\:{one}\:{is}\:−{ve}.{here}\:{it}\:{will}\:{not}\:{apply}\right] \\ $$$${We}\:{got}\:{E}\left(\frac{{a}}{\mathrm{2}},\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$${ED}=\sqrt{\left(\mathrm{0}−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({a}−{a}−{a}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{4}}}={a} \\ $$$$\therefore{DEC}\:{is}\:{eqi}.{triangle} \\ $$$${AEFGD}=\frac{\mathrm{30}^{\mathrm{0}} }{\mathrm{360}^{\mathrm{0}} }\pi{a}^{\mathrm{2}} −\left(\frac{\mathrm{60}^{\mathrm{0}} }{\mathrm{360}^{\mathrm{0}} }\pi{a}^{\mathrm{2}} −\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{a}^{\mathrm{2}} \right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{12}}\pi{a}^{\mathrm{2}} +\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{a}^{\mathrm{2}} \\ $$$${AGHF}=\mathrm{2}\left\{\left(\frac{\mathrm{1}}{\mathrm{4}}\pi×\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right)−\left(\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\pi{a}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${AFHD}=\frac{\mathrm{1}}{\mathrm{2}}\pi\frac{{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\pi{a}^{\mathrm{2}} \\ $$$$\therefore{EFHJ}=\frac{\mathrm{1}}{\mathrm{8}}\pi{a}^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{a}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{12}}\pi{a}^{\mathrm{2}} \right)−\left(\frac{\mathrm{1}}{\mathrm{8}}\pi{a}^{\mathrm{2}} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$=\cancel{\frac{\mathrm{1}}{\mathrm{8}}\pi{a}^{\mathrm{2}} }−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{12}}\pi{a}^{\mathrm{2}} \cancel{−\frac{\mathrm{1}}{\mathrm{8}}\pi{a}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{12}}\pi{a}^{\mathrm{2}} \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\left(\mathrm{1}−\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\pi\right) \\ $$$$\approx\mathrm{0}.\mathrm{07878668590693}{a}^{\mathrm{2}} \\ $$$$\approx\mathrm{0}.\mathrm{0788}{a}^{\mathrm{2}} \\ $$
Commented by fantastic last updated on 03/Sep/25
$${Mr}\:{W}\:{I}\:{am}\:{requesting}\:{you} \\ $$$${to}\:\:{check}\:{my}\:{answer}\:{in}\:{geometry} \\ $$$${expressions}\:{app}. \\ $$$$\left.:\right) \\ $$
Commented by mr W last updated on 04/Sep/25
$${i}\:{think}\:{your}\:{answer}\:{is}\:{right}. \\ $$
Commented by fantastic last updated on 04/Sep/25
$${thanks}\:{for}\:{the}\:{information}\:{sir} \\ $$