9. The base of the pyramid $T A B C D$ is an isosceles trapezoid $A B C D$, the length of the smaller base $B C$ of which is $\sqrt{3}$. The ratio of the areas of the parts of the trapezoid $A B C D$, into which it is divided by the midline, is $5: 7$. All lateral faces of the pyramid $T A B C D$ are inclined to the base plane at an angle of $30^{\circ}$. The plane $A K N$, where points $K$ and $N$ are the midpoints of edges $T B$ and $T C$ respectively, divides the pyramid into two parts. Find the volume of the larger part.

$(16$ points)
# Solution.

Let $TO$ be the height of the pyramid. Since all lateral faces are inclined to the base at the same angle, $O$ is the center of the circle inscribed in the base. Let $MP$ be the midline of the trapezoid, $AD=a, BC=b=\sqrt{3}$. According to the problem, we have $S_{MB CP}=5x, S_{\text{AMPD}}=7x, \quad \frac{5}{7}=\frac{b+(a+b) / 2}{a+(a+b) / 2}=\frac{a+\sqrt{3}}{3a+\sqrt{3}}$, $a=2\sqrt{3}$. Since a circle can be inscribed in the trapezoid $ABCD$, then $AB+CD=a+b, \quad AB=CD=1.5\sqrt{3}$. We calculate the height of the trapezoid $h=\sqrt{AB^2-(a-b)^2 / 4}=\sqrt{6}$. Draw a line through point $O$ perpendicular to the bases of the trapezoid, intersecting these bases at points $Q$ and $R, OR=h$. Since the lateral faces are inclined to the base plane at an angle of $30^\circ$, the height of the pyramid $TO=\frac{1}{2} QR \tan 30^\circ=\frac{\sqrt{2}}{2}$.

Let $TF$ be the height of the pyramid TAKND, $TF$ is the perpendicular dropped from point $T$ to the line $QS$, where $S$ is the midpoint of $KN$. We calculate the volume of the pyramid TAKND:

$V_{\text{TAKND}}=\frac{1}{3} \cdot \frac{AD+KN}{2} \cdot QS \cdot TF \cdot$ The area of triangle $TQS$ can be calculated in two ways:

$S_{TQS}=\frac{QR \cdot TO}{4}, S_{TQS}=\frac{QS \cdot TF}{2}, \quad QS \cdot TF=\frac{QR \cdot TO}{2}$, and

$V_{\text{TAKND}}=\frac{1}{6} \cdot \frac{AD+KN}{2} \cdot QR \cdot TO$. From this, we get $V_{\text{TAKND}}=5/8$.

We calculate the volume of the pyramid TABCD:

![](https://cdn.mathpix.com/cropped/2024_05_06_3205747dae2ad15e28c2g-11.jpg?height=414&width=508&top_left_y=1049&top_left_x=1439)

$V_{\text{TABCD}}=\frac{1}{3} \cdot \frac{AD+BC}{2} \cdot QR \cdot TO=\frac{3}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3205747dae2ad15e28c2g-11.jpg?height=439&width=466&top_left_y=1471&top_left_x=1389)

$V_{\text{TABCD}}-V_{\text{TAKND}}=\frac{3}{2}-\frac{5}{8}=\frac{7}{8}=0.875$.

Answer: 0.875.