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Question Number 149510 by Tawa11 last updated on 05/Aug/21

Commented by Ar Brandon last updated on 06/Aug/21

36−4×(9/2)=18cm^3

$$\mathrm{36}−\mathrm{4}×\frac{\mathrm{9}}{\mathrm{2}}=\mathrm{18cm}^{\mathrm{3}} \\ $$

Commented by Tawa11 last updated on 06/Aug/21

cm^2 ? And sir, how did you get 4 × 5

$$\mathrm{cm}^{\mathrm{2}} \:? \\ $$$$\mathrm{And}\:\mathrm{sir},\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\:\:\mathrm{4}\:×\:\mathrm{5} \\ $$

Commented by Ar Brandon last updated on 06/Aug/21

Just some guess work, hahaha ! Maybe Mr W has a better method.

$$\mathrm{Just}\:\mathrm{some}\:\mathrm{guess}\:\mathrm{work},\:\mathrm{hahaha}\:! \\ $$$$\mathrm{Maybe}\:\mathrm{Mr}\:\mathrm{W}\:\mathrm{has}\:\mathrm{a}\:\mathrm{better}\:\mathrm{method}. \\ $$

Commented by Tawa11 last updated on 06/Aug/21

Hahahaha ... Thanks sir. I appreciate.

$$\mathrm{Hahahaha}\:...\:\:\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by MJS_new last updated on 06/Aug/21

counting the white areas (1) left upper corner 2×2=4 (2) middle square + right lower corner 2×2=4 (3) the symmetric small “triangles” 2×(2×2−2^2 π/4)=8−2π (4) the symmetric big “triangles” 2×(4×4−4^2 π/4)=32−8π sum of these =48−10π = white area 36−(48−10π)=10π−12 = blue area

$$\mathrm{counting}\:\mathrm{the}\:\mathrm{white}\:\mathrm{areas} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{left}\:\mathrm{upper}\:\mathrm{corner}\:\mathrm{2}×\mathrm{2}=\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{middle}\:\mathrm{square}\:+\:\mathrm{right}\:\mathrm{lower}\:\mathrm{corner}\:\mathrm{2}×\mathrm{2}=\mathrm{4} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{the}\:\mathrm{symmetric}\:\mathrm{small}\:``\mathrm{triangles}'' \\ $$$$\:\:\:\:\:\mathrm{2}×\left(\mathrm{2}×\mathrm{2}−\mathrm{2}^{\mathrm{2}} \pi/\mathrm{4}\right)=\mathrm{8}−\mathrm{2}\pi \\ $$$$\left(\mathrm{4}\right)\:\mathrm{the}\:\mathrm{symmetric}\:\mathrm{big}\:``\mathrm{triangles}'' \\ $$$$\:\:\:\:\:\mathrm{2}×\left(\mathrm{4}×\mathrm{4}−\mathrm{4}^{\mathrm{2}} \pi/\mathrm{4}\right)=\mathrm{32}−\mathrm{8}\pi \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{these}\:=\mathrm{48}−\mathrm{10}\pi\:=\:\mathrm{white}\:\mathrm{area} \\ $$$$\mathrm{36}−\left(\mathrm{48}−\mathrm{10}\pi\right)=\mathrm{10}\pi−\mathrm{12}\:=\:\mathrm{blue}\:\mathrm{area} \\ $$

Commented by Tawa11 last updated on 06/Aug/21

Thanks sir. God bless you.

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

Commented by Tawa11 last updated on 07/Aug/21

Sir Mjs and ajfour. Thanks, I understand your solutions very well now. God bless you.

$$\mathrm{Sir}\:\mathrm{Mjs}\:\:\mathrm{and}\:\:\mathrm{ajfour}.\:\mathrm{Thanks},\:\mathrm{I}\:\mathrm{understand}\:\mathrm{your}\:\mathrm{solutions}\:\mathrm{very}\:\mathrm{well}\:\mathrm{now}. \\ $$$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}. \\ $$

Answered by ajfour last updated on 06/Aug/21

B=(π/2)(4cm)^2 −2{4cm^2 −(π/4)(2cm)^2 } −4cm^2 =(10π−12)cm^2

$${B}=\frac{\pi}{\mathrm{2}}\left(\mathrm{4}{cm}\right)^{\mathrm{2}} −\mathrm{2}\left\{\mathrm{4}{cm}^{\mathrm{2}} −\frac{\pi}{\mathrm{4}}\left(\mathrm{2}{cm}\right)^{\mathrm{2}} \right\} \\ $$$$\:\:\:\:\:−\mathrm{4}{cm}^{\mathrm{2}} \\ $$$$\:\:=\left(\mathrm{10}\pi−\mathrm{12}\right){cm}^{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 06/Aug/21

Thanks sir. God bless you.

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

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