Problem 705

Label the centre of the square $A$, the midpoint of the base $B$, and the bottom-right corner of the
square $C$. Then $\triangle ABC$ has a right angle at $B$ and $AB=CB$, as shown.

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The flower is made up of four "petals", each of which is made up of
two copies of the region formed by taking a sector of the circle of
angle $90\degree$ and removing the
right-isosceles triangle formed by the two radii.

The radius of each circle is $\dfrac{10}{2}=5$, so the area of each
sector is $\dfrac{\pi(5)^2}{4}=\dfrac{25\pi}{4}$. The
area of the triangle being removed is $\dfrac{1}{2}(5)^2=\dfrac{25}{2}$.

Therefore, the area of each petal is $\$2\times\left(\dfrac{25\pi}{4}-\dfrac{25}{2}\right) = \dfrac{25\pi-50}{2}\$.$

The area of the flower is $4$
times the area of each petal, which is $$\begin{align*} 4\times\left(\dfrac{25\pi-50}{2}\right) &= 2\times (25\pi - 50) \\ &= 50\pi-100 \\ &\approx 57.08\end{align*}$$ The integer closest to the area
of the flower is $57$.