9. The base of the pyramid $\mathrm{TABCD}$ is an isosceles trapezoid $\mathrm{ABCD}$, the length of the larger base $A D$ of which is $12 \sqrt{3}$. 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 TABCD are inclined to the base plane at an angle of $30^{\circ}$. Find the volume of the pyramid SAKND, where points $\mathrm{K}$ and $\mathrm{N}$ are the midpoints of edges ТВ and $\mathrm{TC}$ respectively, and point $\mathrm{S}$ lies on edge $\mathrm{TD}$, such that $\mathrm{TS}: \mathrm{SD}=1: 2$.
# 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, $A D=a=12 \sqrt{3}, B C=b$.

According to the problem, we have

$S_{\text {MBCP }}=5 x, S_{\text {AMPD }}=7 x, \quad \frac{5}{7}=\frac{b+(a+b) / 2}{a+(a+b) / 2}=\frac{3 b+12 \sqrt{3}}{b+36 \sqrt{3}}, \quad b=6 \sqrt{3}$. Since a circle can be inscribed in trapezoid ABCD, then $A B+C D=a+b, \quad A B=C D=19 \sqrt{3}$. We calculate the height of the trapezoid $h=\sqrt{A B^{2}-(a-b)^{2} / 4}=6 \sqrt{6}$. Draw a line through point O perpendicular to the bases of the trapezoid, intersecting these bases at points $\mathrm{Q}$ and $\mathrm{R}, \mathrm{OR}=\mathrm{h}$. Since the lateral faces are inclined to the plane

![](https://cdn.mathpix.com/cropped/2024_05_06_9e353629715b7c8c09e0g-11.jpg?height=583&width=404&top_left_y=682&top_left_x=1594)

of the base at an angle of $30^{\circ}$, the height of the pyramid $T O=\frac{1}{2} Q R \operatorname{tg} 30^{\circ}=3 \sqrt{2}$.

Let TF be the height of the pyramid TAKND, TF is the perpendicular dropped from point $\mathrm{T}$ to line $\mathrm{QL}$, where $\mathrm{L}$ is the midpoint of $\mathrm{KN}$. We calculate the volume of the pyramid

SAKND:

$$
V_{S A K N D}=\frac{1}{3} \cdot \frac{A D+K N}{2} \cdot Q S \cdot \frac{2}{3} T F
$$

we can

$S_{\mathrm{TQS}}=\frac{Q R \cdot T O}{4}, S_{\mathrm{TQS}}=\frac{Q S \cdot T F}{2}, \quad Q S \cdot T F=\frac{Q R \cdot T O}{2}$,

$V_{\text {and }} V_{S K N D}=\frac{1}{6} \cdot \frac{A D+K N}{2} \cdot \frac{2}{3} Q R \cdot T O$.

From this, we get $V_{\text {SAKND }}=90$.

Answer: 90.