Question

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.

Answer 1

$a=8,~b=2\sqrt2,~r=2,$

1)

$S_{rec}=ab=8\cdot 2\sqrt2=16\sqrt2,$

2)

$S_{com}=2(\frac 12\cdot\frac b2 \cdot2\sqrt{r^2-(\frac b2)^2}+\frac{\pi r^2}{4})=2\cdot (\frac12\cdot \sqrt2 \cdot 2 \cdot \sqrt 2+{\pi}{})=4+2\pi.$