## Task B-1.6.

A circle is circumscribed around rectangle $A B C D$. Point $P$ is the midpoint of side $\overline{A D}$. Line segment $\overline{B P}$ intersects diagonal $\overline{A C}$ at point $T$. If $|B T|=10 \mathrm{~cm}$ and $|A P|=9 \mathrm{~cm}$, what is the area of the shaded part in the figure?

![](https://cdn.mathpix.com/cropped/2024_05_30_9b9f9f32e16538559f3ag-05.jpg?height=294&width=299&top_left_y=1224&top_left_x=1461)
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_9b9f9f32e16538559f3ag-05.jpg?height=368&width=357&top_left_y=1695&top_left_x=787)

Point $P$ is the midpoint of segment $\overline{A D}$, so $|A D| = |B C| = 2|A P| = 18 \text{ cm}$. 1 point

Point $S$ is the intersection of the diagonals $\overline{B D}$ and $\overline{A C}$ of rectangle $A B C D$. This means that point $S$ is the midpoint of segment $\overline{B D}$, so we conclude that segments $\overline{B P}$ and $\overline{A S}$ are medians of triangle $A B D$. 1 point Therefore, point $T$ is the centroid of triangle $A B D$. 1 point

The centroid of a triangle divides the median in the ratio 2:1, starting from the vertex of the triangle. Therefore,

$|B T| = \frac{2}{3}|B P|$, that is, $10 = \frac{2}{3}|B P| \text{ cm}$, so $|B P| = 15 \text{ cm}$.

By the Pythagorean theorem in triangle $A B P$:

$|B P|^2 = |A B|^2 + |A P|^2$, so $|A B| = 12 \text{ cm}$.

1 point

By the Pythagorean theorem in triangle $A B D$ (or $A B C$):

$|B D|^2 = |A D|^2 + |A B|^2$, so $|B D| = |A C| = \sqrt{468} = 6 \sqrt{13} \text{ cm}$.

1 point

The center of the circle circumscribed around the rectangle is at the intersection of the diagonals $S$ and $2 r = |B D|$, so $r = 3 \sqrt{13} \text{ cm}$.

1 point

The area of the circle circumscribed around the rectangle is $P_{K} = r^2 \pi = 117 \pi \text{ cm}^2$.

1 point

The area of the rectangle is $P_{P} = |A B| \cdot |A D| = 216 \text{ cm}^2$.

1 point

The area of the shaded part is $P = \frac{1}{2}(P_{K} - P_{P}) = \frac{1}{2}(117 \pi - 216) \text{ cm}^2$.

1 point