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Question-138299

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Question Number 138299 by cherokeesay last updated on 12/Apr/21
Answered by bemath last updated on 12/Apr/21
BD = r_1 +2 ⇔ r_1 ((√2)−1)=2 r_1 = (2/( (√2)−1)) = ((2((√2)+1))/(2−1))=2(√2)+2 shaded area=(2(√2)+2)^2 −(1/4)π(2(√2)+2)^2 −(1/4)π(4) =4(3+2(√2))(((4−π)/4))−π = (3+2(√2))(4−π)−π
$${BD}\:=\:{r}_{\mathrm{1}} +\mathrm{2}\:\Leftrightarrow\:{r}_{\mathrm{1}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=\mathrm{2} \\ $$$${r}_{\mathrm{1}} =\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\:=\:\frac{\mathrm{2}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\mathrm{2}−\mathrm{1}}=\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2} \\ $$$${shaded}\:{area}=\left(\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{4}\right) \\ $$$$=\mathrm{4}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\left(\frac{\mathrm{4}−\pi}{\mathrm{4}}\right)−\pi \\ $$$$=\:\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\left(\mathrm{4}−\pi\right)−\pi \\ $$
Answered by nadovic last updated on 12/Apr/21
2r_1 ^2 = (r_1 + 2)^2 r_1 (√2) − r_1 = 2 r_1 = (2/( (√2) − 1)) r_1 = 2 + 2(√2) Area of ABCD, A = (2 + 2(√2))^2 _ A = 12 + 8(√2) Area of ADC, A_1 = (1/2)r_1 ^2 θ_1 A_1 = (1/2)(2 + 2(√2))^2 (π/2) _ A_1 = π(3 + 2(√2)) Area of r_2 sector, A_2 = (1/2)r_2 ^2 θ_2 A_2 = (1/2)(2)^2 (π/2) _ A_2 = π Area of hatched part = A − (A_1 + A_2 ) = (12 + 8(√2)) − (3π + 2π(√2) + π) = 4(3 + 2(√2)) − (3 + 2(√2))π − π = (3 + 2(√2))(4 − π) − π
$$\mathrm{2}{r}_{\mathrm{1}} ^{\mathrm{2}} \:=\:\left({r}_{\mathrm{1}} \:+\:\mathrm{2}\right)^{\mathrm{2}} \\ $$$${r}_{\mathrm{1}} \sqrt{\mathrm{2}}\:−\:{r}_{\mathrm{1}} \:=\:\mathrm{2} \\ $$$${r}_{\mathrm{1}} \:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}\:−\:\mathrm{1}} \\ $$$${r}_{\mathrm{1}} \:=\:\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{Area}\:\mathrm{of}\:{ABCD},\:{A}\:=\:\left(\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{} {\:}\:{A}\:=\:\mathrm{12}\:+\:\mathrm{8}\sqrt{\mathrm{2}} \\ $$$$\mathrm{Area}\:\mathrm{of}\:\mathrm{ADC},\:{A}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{r}_{\mathrm{1}} ^{\mathrm{2}} \theta_{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{A}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:_{} {A}_{\mathrm{1}} \:=\:\pi\left(\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{Area}\:\mathrm{of}\:{r}_{\mathrm{2}} \:{sector},\:{A}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{r}_{\mathrm{2}} ^{\mathrm{2}} \theta_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{A}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\right)^{\mathrm{2}} \frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:_{} \:\:{A}_{\mathrm{2}} \:=\:\pi \\ $$$$\mathrm{Area}\:\mathrm{of}\:\mathrm{hatched}\:\mathrm{part}\:=\:{A}\:−\:\left({A}_{\mathrm{1}} \:+\:{A}_{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{12}\:+\:\mathrm{8}\sqrt{\mathrm{2}}\right)\:−\:\left(\mathrm{3}\pi\:+\:\mathrm{2}\pi\sqrt{\mathrm{2}}\:+\:\pi\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4}\left(\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right)\:−\:\left(\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right)\pi\:−\:\pi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right)\left(\mathrm{4}\:−\:\pi\right)\:−\:\pi\: \\ $$$$ \\ $$

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