4. The base edge of a right regular quadrilateral pyramid $A B C D T$ ($T$ is the apex of the pyramid) is $\sqrt{2}$, and the lateral edge is 2. A plane passes through vertex $A$ and is perpendicular to edge $T C$ ($A$ and $C$ are diagonally opposite on the base) and intersects the edges $T B, T C$, and $T D$ at points $P, Q$, and $R$ respectively.

Determine the volume of the intersected pyramid $A B C D P Q R$.
Solution. Let $M$ be the intersection of the diagonals of the base of the pyramid, and $O$ be the intersection of the diagonals of the quadrilateral $A P Q R$.

For the diagonal of the base, we have:

$$
\overline{A C}^{2}=\overline{A B}^{2}+\overline{B C}^{2}=2(\sqrt{2})^{2}=4
$$

i.e., $\overline{A C}=2$. Thus, $A C T$ and $B D T$ are two congruent equilateral triangles with side length 2. Since $T C \perp A Q$, it is clear that $A Q$ is the height of $\triangle A C T$. Similarly, the height of the pyramid $T M$ is the height of $\triangle A C T$. We have

$$
\overline{A Q}=\overline{T M}=\frac{2 \sqrt{3}}{2}=\sqrt{3}
$$

The point $O$ is the orthocenter of $\triangle A T C$ and

![](https://cdn.mathpix.com/cropped/2024_06_05_69e79dfbdd5b161398a0g-3.jpg?height=538&width=598&top_left_y=267&top_left_x=881)
${ }_{\triangle} B D T$, so, $\overline{R P}: \overline{B D}=2: 3$, from which we get $\overline{R P}=\frac{4}{3}$. The quadrilateral $A P Q R$ is a kite, so for its area $P$ we have

$$
P=\frac{\overline{A Q} \cdot \overline{R P}}{2}=\frac{2 \sqrt{3}}{3}
$$

Since $T Q$ is the height of the pyramid, we have

$$
V_{A P Q R T}=\frac{1}{3} \frac{2 \sqrt{3}}{3} \frac{2}{2}=\frac{2 \sqrt{3}}{9}
$$

The volume of the given pyramid is $V=\frac{1}{3}(\sqrt{2})^{2} \sqrt{3}=\frac{2 \sqrt{3}}{3}$. Finally, the desired volume is

$$
V_{A B C D P R Q}=\frac{2 \sqrt{3}}{3}-\frac{2 \sqrt{3}}{9}=\frac{4 \sqrt{3}}{9} .
$$

## II year