7. Consider the pyramids $T A B C D$ with the following properties: the base of the pyramid is a convex quadrilateral $A B C D$, where $A B=B C=2, C D=D A=3$, and each of the planes of the lateral faces $T A B, T B C, T C D, T D A$ forms an angle of $30^{\circ}$ with the plane of the base.

a) Find the volume of such a pyramid if its height, dropped from the vertex $T$, is 2.

b) For what height is the volume of the considered pyramids maximal, and what is this volume?
Answer. a) $\frac{4}{\sqrt{3}}$; b) $4 \sqrt{3}$.

Let $TH$ be the height of the pyramid, $(TH=h)$, and $P$ be the projection of $T$ onto the line $AB$. Then $THP$ is a right triangle with angle $\angle TPH=30^{\circ}$, from which $HP=h \cdot \operatorname{ctg} 30^{\circ}=h \sqrt{3}$. Similarly, it can be proven that the point $H$ is at a distance $r=h \sqrt{3}$ from each of the lines $BC, CD, DA$ (in other words, the circle with radius $r$ and center $H$ is tangent to the lines $AB, BC, CD, DA$).

The triangles $BAD$ and $BCD$ are equal by three sides, so the quadrilateral $ABCD$ is symmetric with respect to the diagonal $BD$. Its area $S$ is equal to $2 S_{BAD}=AB \cdot AD \sin \angle BAD$, so $S \leqslant AB \cdot AD=6$. Equality is achieved when $\angle BAD=90^{\circ}$, so $S_{\max }=6$.

The point $H$ lies on the internal or external bisector of each of the angles of the quadrilateral $ABCD$. $BD$ is the internal bisector of angles $B$ and $D$. The external bisectors of angles $B$ and $D$ are parallel, so $H$ must lie on $BD$. Let $I$ and $J$ be the points of intersection of the internal and external bisectors of angle $A$ with the line $BD$. Then $I$ is the center of the inscribed circle of the quadrilateral $ABCD$ (let its radius be $r_{1}$); $J$ is the center of the circle tangent to the extensions of the sides of the quadrilateral $ABCD$ (the exscribed circle, let its radius be $r_{2}$). The area of a quadrilateral with an inscribed circle can be given by the formula $S=\frac{(AB+BC+CD+DA) r_{1}}{2}$, from which $r_{1}=\frac{S}{5}$. Also, $S=S_{ADJ}+S_{CDJ}-S_{ABJ}-S_{BCJ}=\frac{(AD+CD-AB-BC) r_{2}}{2}$, from which $r_{2}=S$.

The pyramid satisfies the condition of the problem if and only if (1) the height passes through the center of the inscribed circle in the base (i.e., $H=I$) and its length is $h=r_{1}=\frac{S}{5}$ or (2) the height passes through the center of the exscribed circle (i.e., $H=J$) and $h=r_{2}=S$.

a) When $h=2$, the first case is impossible $\left(S=5 r_{1}=5 h \sqrt{3}=10 \sqrt{3}>6\right)$. Therefore, only the second case remains, and then $S=r_{2}=h \sqrt{3}=2 \sqrt{3}$. The volume is $V=\frac{S h}{3}=\frac{4}{\sqrt{3}}$.

b) The volume in the first and second cases is $V_{1}=\frac{S h}{3}=\frac{S^{2}}{15 \sqrt{3}}$ and $V_{2}=\frac{S h}{3}=\frac{S^{2}}{3 \sqrt{3}}$. The maximum volume $V_{\max }=\frac{S_{\max }^{2}}{3 \sqrt{3}}=4 \sqrt{3}$.