11.154. The distance between any two lateral edges of an oblique triangular prism is $a$. The lateral edge is equal to $l$ and is inclined to the base plane at an angle of $60^{\circ}$. Determine the total surface area of the prism.
Solution.

Let $A_{1} O$ be the height of the oblique prism $A B C A_{1} B_{1} C_{1}$ (Fig. 11.50).

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0861.jpg?height=464&width=537&top_left_y=113&top_left_x=95)

Fig. 11.50

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0861.jpg?height=510&width=628&top_left_y=72&top_left_x=639)

Fig. 11.51

$A O$ is the projection of $A A_{1}$ onto the base plane, $\angle A_{1} A O$ is the angle of inclination of the lateral edge $A A_{1}$ to the base plane, and by the condition $\angle A_{1} A O=60^{\circ}$. In $\triangle A O A_{1}\left(\angle A O A_{1}\right)=90^{\circ} ; A_{1} O=A A_{1} \sin \angle A_{1} A O=\frac{l \sqrt{3}}{2}$.

Consider $\triangle M N K$ - the perpendicular section of the prism. Then the length of its sides is the distance between the lateral edges of the prism. Since by the condition these distances are equal to $a$, the perimeter of the perpendicular section $P=3 a$, its area $S_{1}=\frac{a^{2} \sqrt{3}}{4}$. The lateral surface of the prism $S_{6}=P \cdot A A_{1}=3 a l$. The volume of the prism $V=S_{1} \cdot A A_{1}=S_{\triangle A B C} \cdot A_{1} O$.

From this, $\frac{a^{2} \sqrt{3}}{4} \cdot l=S_{\triangle A B C} \frac{l \sqrt{3}}{2} ; S_{\triangle A B C}=\frac{a^{2}}{2}$. The total surface area of the prism $S=S_{6}+2 S_{\triangle A B C}=3 a l+a^{2}$.

Answer: $3 a l+a^{2}$.

11.155. The base of an oblique prism is an equilateral triangle with a side length of $a$. The length of the lateral edge is $b$, and one of the lateral edges forms angles of $45^{\circ}$ with the adjacent sides of the base. Determine the lateral surface area of the prism.

Solution.

Let the equilateral triangle $A B C$ be the base of the oblique prism $A B C A_{1} B_{1} C_{1}, \angle B_{1} B A=\angle B_{1} B C=45^{\circ}$ (Fig. 11.51). Drop the perpendicular $A M$ to $B B_{1} \cdot \triangle B M A=\triangle B M C$ (by two sides and the angle between them). Then $\angle B M C=\angle B M A=90^{\circ}$ and $A M=C M=\frac{A B \sqrt{2}}{2}=\frac{a \sqrt{2}}{2}$. From this, it follows that $\triangle A M C$ is the perpendicular section of the given prism and its perimeter $P=a \sqrt{2}+a=a(\sqrt{2}+1)$. The lateral surface area of the prism $S_{6}=P \cdot B B_{1}=a b(\sqrt{2}+1)$.

Answer: $a b(\sqrt{2}+1)$.