Question-206111

Question Number 206111 by cortano21 last updated on 07/Apr/24

Commented by cortano21 last updated on 07/Apr/24

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Answered by nikif99 last updated on 07/Apr/24

Let a: side of square. PR=(a/4) In orthogonal axis system yAx: Equation of AS: y=2x Eq. of DB: y=−x+a (x, y)=(Q_x , Q_y )=((a/3), ((2a)/3)) QT=Q_y −T_y =((2a)/3)−(a/2)=(a/6) A_(PQR) =(1/2)(PR)(QT)=(1/2)∙(a/4)∙(a/6)=20 ⇒ a^2 =960

$${Let}\:{a}:\:{side}\:{of}\:{square}. \\ $$$${PR}=\frac{{a}}{\mathrm{4}} \\ $$$${In}\:{orthogonal}\:{axis}\:{system}\:{yAx}: \\ $$$${Equation}\:{of}\:{AS}:\:{y}=\mathrm{2}{x} \\ $$$${Eq}.\:{of}\:{DB}:\:{y}=−{x}+{a} \\ $$$$\left({x},\:{y}\right)=\left({Q}_{{x}} ,\:{Q}_{{y}} \right)=\left(\frac{{a}}{\mathrm{3}},\:\frac{\mathrm{2}{a}}{\mathrm{3}}\right) \\ $$$${QT}={Q}_{{y}} −{T}_{{y}} =\frac{\mathrm{2}{a}}{\mathrm{3}}−\frac{{a}}{\mathrm{2}}=\frac{{a}}{\mathrm{6}} \\ $$$${A}_{{PQR}} =\frac{\mathrm{1}}{\mathrm{2}}\left({PR}\right)\left({QT}\right)=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{{a}}{\mathrm{4}}\centerdot\frac{{a}}{\mathrm{6}}=\mathrm{20}\:\Rightarrow \\ $$$${a}^{\mathrm{2}} =\mathrm{960} \\ $$

Commented by nikif99 last updated on 07/Apr/24

Answered by mr W last updated on 07/Apr/24

Commented by mr W last updated on 07/Apr/24

FP=DE=((DC)/2) FR=((DE)/2)=RP ((RQ)/(QE))=((RP)/(DE))=(1/2) ⇒[PQR]=(1/3)[ERP]=(1/3)×(([ABCD])/(16)) ⇒[ABCD]=48×[PQR]=48×20=960

$${FP}={DE}=\frac{{DC}}{\mathrm{2}} \\ $$$${FR}=\frac{{DE}}{\mathrm{2}}={RP} \\ $$$$\frac{{RQ}}{{QE}}=\frac{{RP}}{{DE}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\left[{PQR}\right]=\frac{\mathrm{1}}{\mathrm{3}}\left[{ERP}\right]=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\left[{ABCD}\right]}{\mathrm{16}} \\ $$$$\Rightarrow\left[{ABCD}\right]=\mathrm{48}×\left[{PQR}\right]=\mathrm{48}×\mathrm{20}=\mathrm{960} \\ $$

Answered by A5T last updated on 07/Apr/24

Commented by A5T last updated on 07/Apr/24

DD′=x⇒RP=(x/2)⇒[DD′Q]=2^2 ×20=80 ⇒[QAB]=2^2 ×80=320 ⇒[ARPB]=300=2x^2 −((x^2 /4)+(x^2 /2))=((5x^2 )/4)⇒x^2 =240 [ABCD]=4x^2 =960

$${DD}'={x}\Rightarrow{RP}=\frac{{x}}{\mathrm{2}}\Rightarrow\left[{DD}'{Q}\right]=\mathrm{2}^{\mathrm{2}} ×\mathrm{20}=\mathrm{80} \\ $$$$\Rightarrow\left[{QAB}\right]=\mathrm{2}^{\mathrm{2}} ×\mathrm{80}=\mathrm{320} \\ $$$$\Rightarrow\left[{ARPB}\right]=\mathrm{300}=\mathrm{2}{x}^{\mathrm{2}} −\left(\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)=\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{4}}\Rightarrow{x}^{\mathrm{2}} =\mathrm{240} \\ $$$$\left[{ABCD}\right]=\mathrm{4}{x}^{\mathrm{2}} =\mathrm{960} \\ $$

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