Hello, fleurdel!
The figure below is made up of a square of side 28 cm.
One semicircle and one quadrant are drawn inside the square.
Given that the total area of portions A and C is two times the area of B,
find the area of B. .$\displaystyle \left(\text{Take }\pi = \tfrac{22}{7}\right)$
Code:*-*-*-------------*-* | * A * | | * B | | * * | 14 | | | * * | | * * + | * C | | *| | D * | 14 | * * | * * *-----------------*-* 28
$\displaystyle \text{(Area of }A+B) \;=\;\text{(Area of the square)} - \text{(Area of the quadrant)}$
. . . . . . . $\displaystyle A + B \;=\;28^2 - \tfrac{1}{4}\pi(28^2) \;=\;784 - 196\pi$ .[1]
$\displaystyle \text{(Area of }B+C) \;=\;\text{(Area of semicircle)}$
. . . . . . .$\displaystyle B + C \;=\;\tfrac{1}{2}\pi(14^2) \;=\;98\pi$ .[2]
Add [1] and [2]:
. . $\displaystyle \underbrace{A + C}_{\text{This is }2B} + 2B \;=\;(784-196\pi) + 98\pi$
. . . . . . $\displaystyle 4B \;=\;784 - 98\pi $
$\displaystyle \text{Therefore: }\:B \:=\:196 - \tfrac{49}{2}\pi \:=\:196 - \tfrac{49}{2}\left(\tfrac{22}{7}\right) $
. . . . . . . . $\displaystyle B \;=\;119\text{ cm}^2$