Question #175612

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Rectangular diagonals intersection point coincides with the center of the circle.
The length of the rectangle is equal to 8 and the width is equal to 2√2. The length of the circle radius is 2. Calculate the rectangle and
the area of the common part (area of circle inside rectangle) of the circle.
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Expert's answer

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## Comments

You won't get anything done hoeing like that12.04.21, 21:06Oh my God just finish it. Solution: r=2 a=8 b=2√2 h=2-((2√2)/2)=2-2√2 α(rad)=90°π/180°=π/2 A(area of segment)=(2^2(π/2-sin(90°)))/2=4(π/2-sin(90°))/2=2(π/2-1)=π-2 A Δ=A1+A2 A Δ=(π-2)+(π-2)=2π-4 A(area of the common part)=4π-2π-4=2π-4 Answer: A(area of the common part)=2π-4

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