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Question Number 67775 by TawaTawa last updated on 31/Aug/19

Answered by MJS last updated on 31/Aug/19

ABCD is a square with side s=(√(196))=14 ⇒ the yellow sector′s area =(π/8)s^2 =((49π)/2) minus the white segment which intersects the diagonal in the center of the square the area of the white segment = quarter circle minus half square =((49π)/4)−((49)/2) ⇒ the yellow area =((49)/4)π+((49)/2)≈62.9845

$${ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square}\:\mathrm{with}\:\mathrm{side}\:{s}=\sqrt{\mathrm{196}}=\mathrm{14} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{yellow}\:\mathrm{sector}'\mathrm{s}\:\mathrm{area}\:=\frac{\pi}{\mathrm{8}}{s}^{\mathrm{2}} =\frac{\mathrm{49}\pi}{\mathrm{2}}\:\mathrm{minus} \\ $$$$\mathrm{the}\:\mathrm{white}\:\mathrm{segment}\:\mathrm{which}\:\mathrm{intersects}\:\mathrm{the} \\ $$$$\mathrm{diagonal}\:\mathrm{in}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square} \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{white}\:\mathrm{segment}\:=\:\mathrm{quarter} \\ $$$$\mathrm{circle}\:\mathrm{minus}\:\mathrm{half}\:\mathrm{square}\:=\frac{\mathrm{49}\pi}{\mathrm{4}}−\frac{\mathrm{49}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{yellow}\:\mathrm{area}\:=\frac{\mathrm{49}}{\mathrm{4}}\pi+\frac{\mathrm{49}}{\mathrm{2}}\approx\mathrm{62}.\mathrm{9845} \\ $$

Commented by MJS last updated on 31/Aug/19

corrected my mistake

$$\mathrm{corrected}\:\mathrm{my}\:\mathrm{mistake} \\ $$

Commented by TawaTawa last updated on 31/Aug/19

God bless you sir.

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

Answered by mr W last updated on 31/Aug/19

a^2 =196 ⇒a=(√(196))=14 R=a=14 r=(a/2)=7 A_(shaded) =((πR^2 )/8)−(r^2 /2)((π/2)−sin (π/2)) =((π 14^2 )/8)−(7^2 /2)((π/2)−1) =((49(π+2))/4) =63 cm^2

$${a}^{\mathrm{2}} =\mathrm{196} \\ $$$$\Rightarrow{a}=\sqrt{\mathrm{196}}=\mathrm{14} \\ $$$${R}={a}=\mathrm{14} \\ $$$${r}=\frac{{a}}{\mathrm{2}}=\mathrm{7} \\ $$$${A}_{{shaded}} =\frac{\pi{R}^{\mathrm{2}} }{\mathrm{8}}−\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{sin}\:\frac{\pi}{\mathrm{2}}\right) \\ $$$$=\frac{\pi\:\mathrm{14}^{\mathrm{2}} }{\mathrm{8}}−\frac{\mathrm{7}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{49}\left(\pi+\mathrm{2}\right)}{\mathrm{4}} \\ $$$$=\mathrm{63}\:{cm}^{\mathrm{2}} \\ $$

Commented by TawaTawa last updated on 31/Aug/19

God bless you sir. Thanks for your time

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

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