12.366. In a truncated cone, the diagonals of the axial section are mutually perpendicular and the length of each is $a$. The angle between the generatrix and the base plane is $\alpha$. Find the total surface area of the truncated cone.
## Solution.

Isosceles trapezoid $ABCD$ (Fig. 12.226) is the axial section of the given truncated cone, $AC \perp BD, AC=BD=a, \angle CDA=\alpha, O_{1}$ is the midpoint of $AD$ - the center of the lower base of the truncated cone, $O_{2}$ is the midpoint of $BC$ - the center of the upper base.

Let $O_{2}B=r, O_{1}A=R, CD=L$.

The lateral surface area of the truncated cone $S_{6}=\pi(R+r) L$, and the total surface area $S=\pi\left((R+r) L+R^{2}+r^{2}\right)$.
$\triangle AED$ and $\triangle BEC$ are isosceles and right triangles, $EO_{2}=r$, $EO_{1}=R$.

$CF$ is the height of the trapezoid, $CF=O_{1}O_{2}=R+r$. $\triangle AFC$ is a right triangle with an angle of $45^{\circ}$. Therefore, $AF=CF=R+r=\frac{a}{\sqrt{2}}$ and $FD=\frac{AD-BC}{2}=R-r$.

In $\triangle CFD\left(\angle CFD=90^{\circ}\right): R-r=FD=CF \operatorname{ctg} \angle CDA=\frac{a}{\sqrt{2}} \operatorname{ctg} \alpha$;

$$
\begin{aligned}
& L=CD=\frac{CF}{\sin \angle CDA}=\frac{R+r}{\sin \alpha}. R+r=\frac{a}{\sqrt{2}}, R-r=\frac{a}{\sqrt{2}} \operatorname{ctg} \alpha \Rightarrow \\
& R^{2}+2Rr+r^{2}=\frac{a^{2}}{2} ; R^{2}-2Rr+r^{2}=\frac{a^{2}}{2} \operatorname{ctg}^{2} \alpha . \text { Adding, we have } \\
& 2R^{2}+2r^{2}=\frac{a^{2}}{2}\left(1+\operatorname{ctg}^{2} \alpha\right) \Leftrightarrow R^{2}+r^{2}=\frac{a^{2}}{4 \sin ^{2} \alpha} .
\end{aligned}
$$

Then

$$
\begin{aligned}
& S=\pi\left((R+r) L+R^{2}+r^{2}\right)=\pi\left(\frac{(R+r)^{2}}{\sin \alpha}+R^{2}+r^{2}\right)= \\
& =\pi\left(\frac{a^{2}}{2 \sin \alpha}+\frac{a^{2}}{4 \sin ^{2} \alpha}\right)=\frac{\pi a^{2}}{2 \sin ^{2} \alpha}\left(\sin \alpha+\frac{1}{2}\right)= \\
& =\frac{\pi a^{2}}{2 \sin ^{2} \alpha}\left(\sin \alpha+\sin 30^{\circ}\right)=\frac{\pi a^{2} \sin \left(\frac{\alpha}{2}+15^{\circ}\right) \sin \left(\frac{\alpha}{2}-15^{\circ}\right)}{\sin ^{2} \alpha} . \\
& \text { Answer: } \frac{\pi a^{2} \sin \left(\frac{\alpha}{2}+15^{\circ}\right) \sin \left(\frac{\alpha}{2}-15^{\circ}\right)}{\sin ^{2} \alpha} .
\end{aligned}
$$