In the diagram, $A B C$ is a quarter-circle centred at $B$. Each of square $P Q R S$, square $S R T B$ and square $R U V T$ has side length 10. Points $P$ and $S$ are on $A B$, points $T$ and $V$ are on $B C$, and points $Q$ and $U$ are on the quartercircle. Line segment $A C$ is drawn. Three triangular regions are shaded, as shown. What is the integer closest to the total area of the shaded regions?

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-005.jpg?height=429&width=434&top_left_y=1222&top_left_x=1301)
Since $A B$ and $B C$ are both radii of the circle, then $A B=B C$.

Since $A B C$ is a quarter-circle centred at $B$, then $\angle A B C=90^{\circ}$.

Thus, $\triangle A B C$ is isosceles and right-angled, which means that $\angle B A C=\angle B C A=45^{\circ}$.

We add some additional labels to the diagram:

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-008.jpg?height=429&width=425&top_left_y=409&top_left_x=904)

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

Since $\angle P A D=\angle B A C=45^{\circ}$, this means that

$$
\angle P A D=\angle A D P=\angle Q D E=\angle D E Q=\angle R E F=\angle E F R=\angle U F G=\angle U G F=45^{\circ}
$$

This in turn means that each of $\triangle A P D, \triangle D Q E, \triangle E R F$, and $\triangle F U G$ is right-angled and isosceles.

Since the side length of each square is 10 , then $B T=10$ and $T Q=T R+R Q=20$.

Since $\angle B T Q=90^{\circ}$, then by the Pythagorean Theorem,

$$
B Q=\sqrt{B T^{2}+T Q^{2}}=\sqrt{10^{2}+20^{2}}=\sqrt{500}
$$

We note that $\sqrt{500}=\sqrt{100 \times 5}=\sqrt{100} \times \sqrt{5}=10 \sqrt{5}$.

Since $B Q$ is a radius of the circle, then $B Q=B A=B C=10 \sqrt{5}$.

Since $B P=T Q=20$, then $A P=B A-B P=10 \sqrt{5}-20$.

Thus, $P D=A P=10 \sqrt{5}-20$.

Since $P Q=10$, then $D Q=P Q-P D=10-(10 \sqrt{5}-20)=30-10 \sqrt{5}$.

Thus, $Q E=D Q=30-10 \sqrt{5}$.

Since $P Q=Q R$ and $D Q=Q E$, then $P D=E R=10 \sqrt{5}-20$.

Using similar reasoning, $E R=R F=10 \sqrt{5}-20$ and $U F=U G=30-10 \sqrt{5}$.

The total area, $\mathcal{A}$, of the shaded regions equals the sum of the areas of $\triangle D Q E, \triangle E R F$ and $\triangle F U G$.

Therefore,

$$
\begin{aligned}
\mathcal{A} & =\frac{1}{2} \times D Q \times Q E+\frac{1}{2} \times E R \times R F+\frac{1}{2} U F \times U G \\
& =\frac{1}{2}(30-10 \sqrt{5})^{2}+\frac{1}{2}(10 \sqrt{5}-20)^{2}+\frac{1}{2}(30-10 \sqrt{5})^{2} \\
& =(30-10 \sqrt{5})^{2}+\frac{1}{2}(10 \sqrt{5}-20)^{2} \\
& =\left(30^{2}-2 \times 30 \times 10 \sqrt{5}+(10 \sqrt{5})^{2}\right)+\frac{1}{2}\left((10 \sqrt{5})^{2}-2 \times 10 \sqrt{5} \times 20+20^{2}\right) \\
& =(900-600 \sqrt{5}+500)+\frac{1}{2}(500-400 \sqrt{5}+400) \\
& =1400-600 \sqrt{5}+450-200 \sqrt{5} \\
& =1850-800 \sqrt{5} \\
& \approx 61.14
\end{aligned}
$$

and so the integer closest to $\mathcal{A}$ is 61 .