Question-150150

Question Number 150150 by Tawa11 last updated on 09/Aug/21

Commented by MJS_new last updated on 10/Aug/21

6×6−[4quarter circles r=2]−[3circles r=1]= =36−7π

$$\mathrm{6}×\mathrm{6}−\left[\mathrm{4}{quarter}\:{circles}\:{r}=\mathrm{2}\right]−\left[\mathrm{3}{circles}\:{r}=\mathrm{1}\right]= \\ $$$$=\mathrm{36}−\mathrm{7}\pi \\ $$

Commented by Tawa11 last updated on 09/Aug/21

$$\mathrm{I}\:\mathrm{got}\:\:\mathrm{14}\:???? \\ $$

Commented by Tawa11 last updated on 10/Aug/21

Thank you sir. God bless you. Thanks for clarification.

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}.\:\mathrm{Thanks}\:\mathrm{for}\:\mathrm{clarification}. \\ $$

Commented by Tawa11 last updated on 10/Aug/21

Thanks sir . I follow the same thing you did for me in Q149510

$$\mathrm{Thanks}\:\mathrm{sir}\:.\:\mathrm{I}\:\mathrm{follow}\:\mathrm{the}\:\mathrm{same}\:\mathrm{thing}\:\mathrm{you}\:\mathrm{did}\:\mathrm{for}\:\mathrm{me}\:\mathrm{in}\:\:\mathrm{Q149510} \\ $$

Answered by maged last updated on 10/Aug/21

Blue Area=2(2×4−π)+(4×4−π×2^2 )+(2×2−π) Blue Area=(16−2π)+(16−4π)+(4−π) Blue Area=16+16+4−2π−4π−π Blue Area=36−7π Blue Area=36−7×((22)/7)=36−22 Blue Area=14 cm^2

$$\mathrm{Blue}\:\mathrm{Area}=\mathrm{2}\left(\mathrm{2}×\mathrm{4}−\pi\right)+\left(\mathrm{4}×\mathrm{4}−\pi×\mathrm{2}^{\mathrm{2}} \right)+\left(\mathrm{2}×\mathrm{2}−\pi\right) \\ $$$$\mathrm{Blue}\:\mathrm{Area}=\left(\mathrm{16}−\mathrm{2}\pi\right)+\left(\mathrm{16}−\mathrm{4}\pi\right)+\left(\mathrm{4}−\pi\right) \\ $$$$\mathrm{Blue}\:\mathrm{Area}=\mathrm{16}+\mathrm{16}+\mathrm{4}−\mathrm{2}\pi−\mathrm{4}\pi−\pi \\ $$$$\mathrm{Blue}\:\mathrm{Area}=\mathrm{36}−\mathrm{7}\pi \\ $$$$\mathrm{Blue}\:\mathrm{Area}=\mathrm{36}−\mathrm{7}×\frac{\mathrm{22}}{\mathrm{7}}=\mathrm{36}−\mathrm{22} \\ $$$$\mathrm{Blue}\:\mathrm{Area}=\mathrm{14}\:\mathrm{cm}^{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 10/Aug/21

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

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