Question-145701

Question Number 145701 by alcohol last updated on 07/Jul/21

Commented by MJS_new last updated on 07/Jul/21

1. square−2×(square−quarter circle)= =−square+2×quarter circle= =half circle−square=4^2 π/2−4^2 = =8π−16 2. trapezoid−rectangular−half circle= =((6+8)/2)×4−4×2−2^2 π/2=20−2π

$$\mathrm{1}. \\ $$$$\mathrm{square}−\mathrm{2}×\left(\mathrm{square}−\mathrm{quarter}\:\mathrm{circle}\right)= \\ $$$$=−\mathrm{square}+\mathrm{2}×\mathrm{quarter}\:\mathrm{circle}= \\ $$$$=\mathrm{half}\:\mathrm{circle}−\mathrm{square}=\mathrm{4}^{\mathrm{2}} \pi/\mathrm{2}−\mathrm{4}^{\mathrm{2}} = \\ $$$$=\mathrm{8}\pi−\mathrm{16} \\ $$$$\mathrm{2}. \\ $$$$\mathrm{trapezoid}−\mathrm{rectangular}−\mathrm{half}\:\mathrm{circle}= \\ $$$$=\frac{\mathrm{6}+\mathrm{8}}{\mathrm{2}}×\mathrm{4}−\mathrm{4}×\mathrm{2}−\mathrm{2}^{\mathrm{2}} \pi/\mathrm{2}=\mathrm{20}−\mathrm{2}\pi \\ $$

Answered by iloveisrael last updated on 07/Jul/21

(1) shaded area = 2((1/4)π(4)^2 −(1/2)(4)^2 ) = 2(4π−8)=8π−16

$$\left(\mathrm{1}\right)\:{shaded}\:{area}\:=\:\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{4}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}\right)^{\mathrm{2}} \right) \\ $$$$\:=\:\mathrm{2}\left(\mathrm{4}\pi−\mathrm{8}\right)=\mathrm{8}\pi−\mathrm{16} \\ $$

Answered by iloveisrael last updated on 07/Jul/21

(2) shaded area=(((4+6)/2)).4−(1/2)π(2)^2 = 20−2π

$$\left(\mathrm{2}\right)\:{shaded}\:{area}=\left(\frac{\mathrm{4}+\mathrm{6}}{\mathrm{2}}\right).\mathrm{4}−\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\mathrm{2}\right)^{\mathrm{2}} \\ $$$$=\:\mathrm{20}−\mathrm{2}\pi\: \\ $$

Commented by alcohol last updated on 07/Jul/21

$${please}\:{explain}\:{this}\: \\ $$$${thanks} \\ $$

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Question-145701

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