# Solution to the Two Quarter Circles and a Semi-Circle II Puzzle +-- {.image} [[TwoQuarterCirclesandaSemiCircleII.png:pic]] > Two quarter circles and a semicircle. What’s the total shaded area? =-- ## Solution by [[Circle Area]], [[Angle in a Semi-Circle]], and [[Pythagoras' Theorem]] +-- {.image} [[TwoQuarterCirclesandaSemiCircleIILabelled.png:pic]] =-- With the diagram annotated as above, the left-hand quarter circle has area: $$ \frac{1}{4} \pi a^2 $$ and the right-hand one: $$ \frac{1}{4} \pi b^2 $$ The area of the semi-circle is: $$ \frac{1}{2} \pi 2^2 $$ The triangle formed by the segments is a [[right-angled triangle]] as the [[angle in a semi-circle]] is a right-angle. Therefore, by [[Pythagoras' theorem]]: $$ a^2 + b^2 = 4^2 $$ In terms of the labelled areas, the left-hand quarter circle is $A + B + C$, the right-hand quarter circle is $C + D + E$, and the semi-circle is $B + C + D$. The shaded area is $A + C + E$ which can be written as: $$ A + C + E = (A + B + C) + (C + D + E) - (B + C + D) $$ Putting these together, the shaded area is given by: $$ \begin{aligned} A + C + E &= \frac{1}{4} \pi a^2 + \frac{1}{4} \pi b^2 - \frac{1}{2} \pi 2^2 \\ &= \frac{1}{4} \pi (a^2 + b^2) - \frac{1}{2} \pi 2^2 \\ &= \frac{1}{4} \pi 4^2 - \frac{1}{2} \pi 2^2 \\ &= 4 \pi - 2 \pi \\ &= 2 \pi \end{aligned} $$