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Question Number 209889 by Tawa11 last updated on 24/Jul/24
Answered by som(math1967) last updated on 25/Jul/24
let PQ=QS=QR=x cos∠PQS=((x^2 +x^2 −12^2 )/(2x^2 ))=((x^2 −72)/x^2 ) cos∠RQS=((x^2 +x^2 −8)/(2x^2 ))=((x^2 −4)/x^2 ) cos(90−∠PQS)=((x^2 −4)/x^2 ) ⇒sin∠PQS=((x^2 −4)/x^2 ) sin^2 ∠PQS+cos^2 ∠PQS=1 ⇒(((x^2 −72)/x^2 ))^2 +(((x^2 −4)/x^2 ))^2 =1 ⇒x^4 −144x^2 +5184+x^4 −8x^2 +16=x^4 ⇒x^4 −152x^2 +5200=0 ⇒(x^2 −100)(x^2 −52)=0 if x^2 −52=0 then cos∠PQS<0 ∴ x^2 −100=0⇒x^2 =100 cos∠PQS=((100−72)/(100))=((28)/(100))=(7/(25)) Green area=(1/2)×x^2 ×sin∠SQR =(1/2)×100×cos∠PQS =50×(7/(25))=14cm^2
$${let}\:{PQ}={QS}={QR}={x} \\ $$$${cos}\angle{PQS}=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{12}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} −\mathrm{72}}{{x}^{\mathrm{2}} } \\ $$$${cos}\angle{RQS}=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{8}}{\mathrm{2}{x}^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} } \\ $$$${cos}\left(\mathrm{90}−\angle{PQS}\right)=\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{sin}\angle{PQS}=\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} } \\ $$$${sin}^{\mathrm{2}} \angle{PQS}+{cos}^{\mathrm{2}} \angle{PQS}=\mathrm{1} \\ $$$$\Rightarrow\left(\frac{{x}^{\mathrm{2}} −\mathrm{72}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} +\left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{4}} −\mathrm{144}{x}^{\mathrm{2}} +\mathrm{5184}+{x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{16}={x}^{\mathrm{4}} \\ $$$$\Rightarrow{x}^{\mathrm{4}} −\mathrm{152}{x}^{\mathrm{2}} +\mathrm{5200}=\mathrm{0} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{100}\right)\left({x}^{\mathrm{2}} −\mathrm{52}\right)=\mathrm{0} \\ $$$$\:{if}\:{x}^{\mathrm{2}} −\mathrm{52}=\mathrm{0}\:{then}\:{cos}\angle{PQS}<\mathrm{0} \\ $$$$\therefore\:{x}^{\mathrm{2}} −\mathrm{100}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} =\mathrm{100} \\ $$$${cos}\angle{PQS}=\frac{\mathrm{100}−\mathrm{72}}{\mathrm{100}}=\frac{\mathrm{28}}{\mathrm{100}}=\frac{\mathrm{7}}{\mathrm{25}} \\ $$$$\:{Green}\:{area}=\frac{\mathrm{1}}{\mathrm{2}}×{x}^{\mathrm{2}} ×{sin}\angle{SQR} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{100}×{cos}\angle{PQS} \\ $$$$=\mathrm{50}×\frac{\mathrm{7}}{\mathrm{25}}=\mathrm{14}{cm}^{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 25/Jul/24
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 25/Jul/24
∠PSR=180−((90)/2)=135° PR=(√(12^2 +(2(√2))^2 +2×12×2(√2)×((√2)/2)))=10(√2) QP=QS=QR=((10(√2))/( (√2)))=10 Δ_(SQR) =((2(√2))/2)×(√(10^2 −(((2(√2))/2))^2 ))=14 ✓
$$\angle{PSR}=\mathrm{180}−\frac{\mathrm{90}}{\mathrm{2}}=\mathrm{135}° \\ $$$${PR}=\sqrt{\mathrm{12}^{\mathrm{2}} +\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}×\mathrm{12}×\mathrm{2}\sqrt{\mathrm{2}}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=\mathrm{10}\sqrt{\mathrm{2}} \\ $$$${QP}={QS}={QR}=\frac{\mathrm{10}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}=\mathrm{10} \\ $$$$\Delta_{{SQR}} =\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}×\sqrt{\mathrm{10}^{\mathrm{2}} −\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{14}\:\checkmark \\ $$
Commented by mr W last updated on 25/Jul/24
Commented by Tawa11 last updated on 25/Jul/24
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 25/Jul/24
Sir, where you wrote PQ. You mean PR? And only one thing is not clear sir. angle PSR = 180 − ((90)/2)
$$\mathrm{Sir},\:\mathrm{where}\:\mathrm{you}\:\mathrm{wrote}\:\:\mathrm{PQ}. \\ $$$$\mathrm{You}\:\mathrm{mean}\:\:\mathrm{PR}? \\ $$$$ \\ $$$$\mathrm{And}\:\mathrm{only}\:\mathrm{one}\:\mathrm{thing}\:\mathrm{is}\:\mathrm{not}\:\mathrm{clear}\:\mathrm{sir}. \\ $$$$\mathrm{angle}\:\mathrm{PSR}\:=\:\mathrm{180}\:−\:\frac{\mathrm{90}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 25/Jul/24
Commented by Tawa11 last updated on 25/Jul/24
Very clear sir. I really appreciate.
$$\mathrm{Very}\:\mathrm{clear}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$
Answered by A5T last updated on 26/Jul/24
Let PQ=s⇒PR=s(√2);∠PQS=x⇒∠SQR=90°−x 4=s^2 −s^2 cos(90−x)...(i); 72=s^2 −s^2 cos(x)...(ii) (((ii))/((i)))⇒18=((1−cosx)/(1−sinx))⇒18−18(√(1−cos^2 x))=1−cosx ⇒325sin^2 x+34sinx−35=0⇒cosx=(7/(25))⇒s^2 =100 ⇒[green]=((s^2 ×sin(90−x))/2)=50×(7/(25))=14
$$ \\ $$$$ \\ $$$$ \\ $$$${Let}\:{PQ}={s}\Rightarrow{PR}={s}\sqrt{\mathrm{2}};\angle{PQS}={x}\Rightarrow\angle{SQR}=\mathrm{90}°−{x}\: \\ $$$$\mathrm{4}={s}^{\mathrm{2}} −{s}^{\mathrm{2}} {cos}\left(\mathrm{90}−{x}\right)…\left({i}\right);\:\mathrm{72}={s}^{\mathrm{2}} −{s}^{\mathrm{2}} {cos}\left({x}\right)…\left({ii}\right) \\ $$$$\frac{\left({ii}\right)}{\left({i}\right)}\Rightarrow\mathrm{18}=\frac{\mathrm{1}−{cosx}}{\mathrm{1}−{sinx}}\Rightarrow\mathrm{18}−\mathrm{18}\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} {x}}=\mathrm{1}−{cosx} \\ $$$$\Rightarrow\mathrm{325}{sin}^{\mathrm{2}} {x}+\mathrm{34}{sinx}−\mathrm{35}=\mathrm{0}\Rightarrow{cosx}=\frac{\mathrm{7}}{\mathrm{25}}\Rightarrow{s}^{\mathrm{2}} =\mathrm{100} \\ $$$$\Rightarrow\left[{green}\right]=\frac{{s}^{\mathrm{2}} ×{sin}\left(\mathrm{90}−{x}\right)}{\mathrm{2}}=\mathrm{50}×\frac{\mathrm{7}}{\mathrm{25}}=\mathrm{14} \\ $$

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