, 2023

Since $AB$ and $BC$ are both radii of the circle, then
$AB = BC$.

Since $ABC$ is a quarter-circle
centred at $B$, then $\angle ABC = 90\degree$.

Thus, $\triangle ABC$ is isosceles
and right-angled, which means that $\$\angle$
BAC = $\angle$ BCA = $45\degree\$.$

We add some additional labels to the diagram:

[[IMAGE0]]

We note that the angles between the straight lines at $P$, $Q$, $R$, and $U$ are all right angles.

Since $\$\angle$ PAD = $\angle$ BAC =
$45\degree$$, this means that$$\$\angle$
PAD = $\angle$ ADP = $\angle$ QDE = $\angle$ DEQ = $\angle$ REF = $\angle$ EFR =
$\angle$ UFG = $\angle$ UGF = $45\degree$$$$$ This in turn means that
each of $\triangle APD$, $\triangle DQE$, $\triangle ERF$, and $\triangle FUG$ is right-angled and
isosceles.

Since the side length of each square is 10, then $BT = 10$ and $TQ = TR + RQ = 20$.

Since $\angle BTQ = 90\degree$, then
by the Pythagorean Theorem, $$BQ = \sqrt{BT^2 + TQ^2} = \sqrt{10^2 + 20^2} = \sqrt{500}$$ We note that $\$\sqrt{500} = \sqrt{100 \times}$ 5} = $\sqrt{100} \times \sqrt{5} = 10\sqrt{5}\$.$

Since $BQ$ is a radius of the circle,
then $BQ = BA = BC =
$10\sqrt{5}\$.$

Since $BP = TQ = 20$, then $AP = BA - BP = 10\sqrt{5} - 20$.

Thus, $PD = AP = $10\sqrt{5}$ -
20$.

Since $PQ = 10$, then $DQ = PQ - PD = 10 - $(10\sqrt{5}$ - 20) = 30 -
$10\sqrt{5}\$.$

Thus, $QE = DQ = 30 -
$10\sqrt{5}\$.$

Since $PQ = QR$ and $DQ = QE$, then $PD = ER = 10\sqrt{5} - 20$.

Using similar reasoning, $ER = RF =
$10\sqrt{5}$ - 20$and$UF = UG =
$30-10\sqrt{5}\$.$

The total area, $\mathcal{A}$, of the
shaded regions equals the sum of the areas of $\triangle DQE$, $\triangle ERF$ and $\triangle FUG$.

Therefore, $$\begin{align*} \mathcal{A} & = \tfrac{1}{2} \times DQ \times QE + \tfrac{1}{2} \times ER \times RF + \tfrac{1}{2} UF \times UG \\ & = \tfrac{1}{2}(30-10\sqrt{5})^2 + \tfrac{1}{2} (10\sqrt{5}-20)^2 + \tfrac{1}{2}(30-10\sqrt{5})^2 \\ & = (30-10\sqrt{5})^2 + \tfrac{1}{2} (10\sqrt{5}-20)^2 \\ & = (30^2 - 2 \times 30 \times 10\sqrt{5} + (10\sqrt{5})^2) + \tfrac{1}{2}((10\sqrt{5})^2 - 2 \times 10\sqrt{5} \times 20 + 20^2) \\ & = (900 - 600\sqrt{5} + 500) + \tfrac{1}{2}(500 - 400\sqrt{5} + 400) \\ & = 1400 - 600\sqrt{5} + 450 - 200\sqrt{5} \\ & = 1850 - 800\sqrt{5} \\ & \approx 61.14\end{align*}$$ and so the integer closest to
$\mathcal{A}$ is 61.