So the curved lines in both parts are quadrants with radius being the length of one side of the square.
You're given the area of the square, so you can calculate the length of one of its sides (1 unit).
For part b:
Let \(b\) square units, \(p\) square units, and \(y\) square units be the area of a blue region, a pink region, and the yellow region respectively. (You may have to argue about how the four blue regions have the same area as each other by symmetry (same for the pink regions)? I'm not sure.)
\[
2b + 3p + y = \frac{1}{4} \pi(1^2) = \frac{\pi}{4}
\quad \text{(Area of one quadrant.)}
\]
From part a, you know \(2p+y=\frac{\pi}{2}-1\). We will substitute \(y=\frac{\pi}{2}-1-2p\) into our equation above.
\[
\begin{align*}
2b + 3p + \bigg(\frac{\pi}{2} - 1 - 2p\bigg) &= \frac{\pi}{4} \\
2b + p &= 1 - \frac{\pi}{4}
\end{align*}
\]
We're going to try and find the area bound by 2 adjacent quadrants \((b+2p+y)\). If we call the bottom vertex of the yellow region \(C\) and the top 2 vertices of the square \(A\) and \(B\) then
\[
AB = AC = BC = 1
\quad \text{(Radii of a circle are have the same length.)}
\]
\(ABC\) is an equilateral triangle, so \(\angle{ABC} = \angle{BAC} = \frac{\pi}{3}\). The area we want is therefore:
\[
\begin{align}
&\hphantom{{}={}}
\text{(Area of the sector centred at \(A\) with arc \(BC\))} + \text{(Area of the sector centred at \(B\) with arc \(AC\))} - \text{(Area of triangle \(ABC\))} \\
&= \frac{1}{6}\pi(1^2) + \frac{1}{6}\pi(1^2) - \frac{1}{2}(1^2)\sin\bigg(\frac{\pi}{3}\bigg) \\
&= \frac{\pi}{3} - \frac{\sqrt{3}}{4}
\end{align}
\]
Now, this is also \(b + 2p + y = b + 2p + \bigg(\frac{\pi}{2} - 1 - 2p\bigg) = b + \frac{\pi}{2} - 1\). To obtain \(y\), solve \(b + \frac{\pi}{2} - 1 = \frac{\pi}{3} - \frac{\sqrt{3}}{4}\) and \(2b + p = 1 - \frac{\pi}{4}\), and then substitute the obtained value of \(p\) into the equation \(y = \frac{\pi}{2} - 1 - 2p\).
And the final answer is (if I haven't done anything wrong) \(y = 1 - \sqrt{3} + \frac{\pi}{3}\).
Aaaaaaaaaand units.
The area of the yellow region is \(1 - \sqrt{3} + \frac{\pi}{3}\) square units.
EDIT: I cannot LaTeX.
Graveyard EDIT: After 5+ years and a masters degree, I can now \(\LaTeX\).
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