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Question-148858




Question Number 148858 by Tawa11 last updated on 31/Jul/21
Answered by EDWIN88 last updated on 01/Aug/21
Commented by EDWIN88 last updated on 01/Aug/21
shaded area =2×(1/4)π(2^2 ) +16−(1/4)π(4^2 )+2(4−(1/4)π(2^2 ))  =2π+16−4π+8−2π  =24−4π
$${shaded}\:{area}\:=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{2}^{\mathrm{2}} \right)\:+\mathrm{16}−\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{4}^{\mathrm{2}} \right)+\mathrm{2}\left(\mathrm{4}−\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{2}^{\mathrm{2}} \right)\right) \\ $$$$=\mathrm{2}\pi+\mathrm{16}−\mathrm{4}\pi+\mathrm{8}−\mathrm{2}\pi \\ $$$$=\mathrm{24}−\mathrm{4}\pi\: \\ $$
Commented by Tawa11 last updated on 01/Aug/21
Thanks sir. God bless you.
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

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