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Question Number 161733 by Tawa11 last updated on 21/Dec/21

Answered by FongXD last updated on 21/Dec/21

let O be the point where the 2 intersect let T∈(QR), where (PT)⊥(QR) • in the right triangle PQT we have: PQ^2 =PT^2 +TQ^2 ⇒ PT=(√((3+1)^2 −(3−1)^2 ))=2(√3) • cos(∠OQR)=((TQ)/(PQ))=((3−1)/(3+1))=(1/2), ⇒ ∠OQR=(π/3) • ∠QPT=(π/2)−(π/3)=(π/6), ⇒ ∠OPS=(π/2)+(π/6)=((2π)/3) • Area of PQRS=(((PS+QR)PT)/2)=(((1+3)×2(√3))/2)=4(√3) (1) • Area of sector OQR=((∠OQR)/2)×R^2 =(π/6)×3^2 =((3π)/2) (2) • Area of sector OPS=((∠OPS)/2)×r^2 =(π/3) (3) ⇒ Area of the shaded region=(1)−(2)−(3) =4(√3)−((3π)/2)−(π/3)=4(√3)−((11π)/6)

$$\:\:\:\mathrm{let}\:\mathrm{O}\:\mathrm{be}\:\mathrm{the}\:\mathrm{point}\:\mathrm{where}\:\mathrm{the}\:\mathrm{2} \:\mathrm{intersect} \\ $$$$\:\:\:\mathrm{let}\:\mathrm{T}\in\left(\mathrm{QR}\right),\:\mathrm{where}\:\left(\mathrm{PT}\right)\bot\left(\mathrm{QR}\right) \\ $$$$\bullet\:\mathrm{in}\:\mathrm{the}\:\mathrm{right}\:\mathrm{triangle}\:\mathrm{PQT} \\ $$$$\mathrm{we}\:\mathrm{have}:\:\mathrm{PQ}^{\mathrm{2}} =\mathrm{PT}^{\mathrm{2}} +\mathrm{TQ}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{PT}=\sqrt{\left(\mathrm{3}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{3}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\bullet\:\mathrm{cos}\left(\angle\mathrm{OQR}\right)=\frac{\mathrm{TQ}}{\mathrm{PQ}}=\frac{\mathrm{3}−\mathrm{1}}{\mathrm{3}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}},\:\Rightarrow\:\angle\mathrm{OQR}=\frac{\pi}{\mathrm{3}} \\ $$$$\bullet\:\angle\mathrm{QPT}=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}=\frac{\pi}{\mathrm{6}},\:\Rightarrow\:\angle\mathrm{OPS}=\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{6}}=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\bullet\:\mathrm{Area}\:\mathrm{of}\:\mathrm{PQRS}=\frac{\left(\mathrm{PS}+\mathrm{QR}\right)\mathrm{PT}}{\mathrm{2}}=\frac{\left(\mathrm{1}+\mathrm{3}\right)×\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{4}\sqrt{\mathrm{3}}\:\:\:\left(\mathrm{1}\right) \\ $$$$\bullet\:\mathrm{Area}\:\mathrm{of}\:\mathrm{sector}\:\mathrm{OQR}=\frac{\angle\mathrm{OQR}}{\mathrm{2}}×\mathrm{R}^{\mathrm{2}} =\frac{\pi}{\mathrm{6}}×\mathrm{3}^{\mathrm{2}} =\frac{\mathrm{3}\pi}{\mathrm{2}}\:\:\:\left(\mathrm{2}\right) \\ $$$$\bullet\:\mathrm{Area}\:\mathrm{of}\:\mathrm{sector}\:\mathrm{OPS}=\frac{\angle\mathrm{OPS}}{\mathrm{2}}×\mathrm{r}^{\mathrm{2}} =\frac{\pi}{\mathrm{3}}\:\:\:\left(\mathrm{3}\right) \\ $$$$\Rightarrow\:\mathrm{Area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{shaded}\:\mathrm{region}=\left(\mathrm{1}\right)−\left(\mathrm{2}\right)−\left(\mathrm{3}\right) \\ $$$$=\mathrm{4}\sqrt{\mathrm{3}}−\frac{\mathrm{3}\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{3}}=\mathrm{4}\sqrt{\mathrm{3}}−\frac{\mathrm{11}\pi}{\mathrm{6}} \\ $$

Commented by Tawa11 last updated on 21/Dec/21

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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