11.164. The side of the base of a regular triangular prism is less than the lateral edge and is equal to $a$. A plane is drawn through the side of the upper base, which forms an angle of $45^{\circ}$ with the base plane and divides the prism into two parts. Determine the volume and the total surface area of the upper part of the prism.
## Solution.

In the right prism $A B C A_{1} B_{1} C_{1} \quad B_{1} C_{1}=a, B_{1} C_{1}<A A_{1}, E$ is the intersection point of the plane passing through the side $B_{1} C_{1}$ of the upper base and the line $A A_{1}$ (Fig. 11.58). Let $K$ be the midpoint of $B_{1} C_{1}$. Then $A_{1} K \perp B_{1} C_{1} ; A_{1} K=\frac{a \sqrt{3}}{2} \cdot A_{1} K$ is the projection of the segment $E K$ on the plane $A_{1} B_{1} C_{1}$. Then $E K \perp B_{1} C_{1}, \angle A_{1} K E$ is the angle between the intersecting plane $B_{1} E C_{1}$ and the plane $A_{1} B_{1} C_{1}, \angle A K E=45^{\circ}$.

Thus, $A_{1} E=A_{1} K=\frac{a \sqrt{3}}{2}$, and since $B_{1} C_{1}<A A_{1}$, then $A_{1} E<A A_{1}$, i.e., the point $E$ belongs to the segment $A A_{1}$. Therefore, the upper part of the prism, mentioned in the problem, is a pyramid with the base $A_{1} B_{1} C_{1}$ and height $E A_{1}$, the volume of which is $V=\frac{1}{3} S_{\triangle A_{1} B_{1} C_{1}} \cdot E A_{1}=\frac{1}{3} \frac{a^{2} \sqrt{3}}{4} \cdot \frac{a \sqrt{3}}{2}=\frac{a^{3}}{8}$, $S_{\triangle E A_{1} C_{1}}=S_{\Delta E A_{1} B_{1}}=\frac{1}{2} E A_{1} \cdot A_{1} C_{1}=\frac{a^{2} \sqrt{3}}{4} \cdot \Delta A_{1} B_{1} C_{1}-$ is the projection of $\Delta E B_{1} C_{1}$ on

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0869.jpg?height=462&width=503&top_left_y=217&top_left_x=96)

Fig. 11.59

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0869.jpg?height=603&width=624&top_left_y=78&top_left_x=637)

Fig. 11.60

the plane $A_{1} B_{1} C_{1}$. Then $S_{\triangle E B_{1} C_{1}}=\frac{S_{\triangle A_{1} B_{1} C_{1}}}{\cos \angle A_{1} K E}=\frac{a^{2} \sqrt{6}}{4}$. The total surface area of the pyramid $E A_{1} B_{1} C_{1}$:

$S=2 S_{\Delta E A_{1} C_{1}}+S_{\Delta E B_{1} C_{1}}+S_{\Delta A_{1} B_{1} C_{1}}=\frac{2 a^{2} \sqrt{3}}{4}+\frac{a^{2} \sqrt{6}}{4}+\frac{a^{2} \sqrt{3}}{4}=\frac{a^{2} \sqrt{3}(3+\sqrt{2})}{4}$. Answer: $\frac{a^{3}}{8} ; \frac{a^{2} \sqrt{3}(3+\sqrt{2})}{4}$.