Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

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

$$\mathrm{2}{r}_{\mathrm{1}} ^{\mathrm{2}} \:=\:\left({r}_{\mathrm{1}} \:+\:\mathrm{2}\right)^{\mathrm{2}} \\ $$$${r}_{\mathrm{1}} \sqrt{\mathrm{2}}\:−\:{r}_{\mathrm{1}} \:=\:\mathrm{2} \\ $$$${r}_{\mathrm{1}} \:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}\:−\:\mathrm{1}} \\ $$$${r}_{\mathrm{1}} \:=\:\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{Area}\:\mathrm{of}\:{ABCD},\:{A}\:=\:\left(\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{} {\:}\:{A}\:=\:\mathrm{12}\:+\:\mathrm{8}\sqrt{\mathrm{2}} \\ $$$$\mathrm{Area}\:\mathrm{of}\:\mathrm{ADC},\:{A}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{r}_{\mathrm{1}} ^{\mathrm{2}} \theta_{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{A}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:_{} {A}_{\mathrm{1}} \:=\:\pi\left(\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\mathrm{Area}\:\mathrm{of}\:{r}_{\mathrm{2}} \:{sector},\:{A}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}{r}_{\mathrm{2}} ^{\mathrm{2}} \theta_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{A}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\right)^{\mathrm{2}} \frac{\pi}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:_{} \:\:{A}_{\mathrm{2}} \:=\:\pi \\ $$$$\mathrm{Area}\:\mathrm{of}\:\mathrm{hatched}\:\mathrm{part}\:=\:{A}\:−\:\left({A}_{\mathrm{1}} \:+\:{A}_{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{12}\:+\:\mathrm{8}\sqrt{\mathrm{2}}\right)\:−\:\left(\mathrm{3}\pi\:+\:\mathrm{2}\pi\sqrt{\mathrm{2}}\:+\:\pi\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{4}\left(\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right)\:−\:\left(\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right)\pi\:−\:\pi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\right)\left(\mathrm{4}\:−\:\pi\right)\:−\:\pi\: \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com