6. Answer: 1) $S_{1}=\frac{3 \sqrt{3}}{2}$ 2) $S_{2}=6 \sqrt{6}$

Option 0

In a cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ with edge $a$, a section is made by a plane parallel to the plane $B D A^{\prime}$ and at a distance $b$ from it. Find the area of the section.
Answer: 1) $S_{1}=\frac{\sqrt{3}}{2}(a-b \sqrt{3})^{2}$ for $a>b \sqrt{3}$ 2) $S_{2}=\frac{\sqrt{3}}{2}\left(a^{2}+2 a b \sqrt{6}-12 b^{2}\right)$ for $a>b \sqrt{6}$ Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_706be2aaaec187922a3ag-10.jpg?height=571&width=714&top_left_y=1759&top_left_x=177)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_706be2aaaec187922a3ag-10.jpg?height=371&width=574&top_left_y=1962&top_left_x=889)

Fig. 2

Case 1. The section passes through point $P$ on edge $A A^{\prime}, A P=n a, 0<b \sqrt{3}$. Triangle $P Q R$ of the section is similar to triangle $A^{\prime} D B$ with a similarity coefficient of $n$, so $S_{P Q R}=n^{2} \cdot S_{A^{\prime} D B}=\frac{(a-b \sqrt{3})^{2}}{a^{2}} \cdot \frac{\sqrt{3}}{4} 2 a^{2}=\frac{\sqrt{3}}{2}(a-b \sqrt{3})^{2}$.

In the case where $1 a=2 \sqrt{3}, b=1$, the area of the section $S_{1}$ in Case 1 is $S_{1}=\frac{3 \sqrt{3}}{2}$

Case 2. The section passes through point $G$ on diagonal $A^{\prime} C^{\prime}$, at a distance $b$ from the plane $B D A^{\prime}$.

The perpendicular $G U$ (Fig. 2), dropped from point $G$ to the plane $B D A^{\prime}$, has a length of $b$:

$A^{\prime} G=\frac{b}{\cos \alpha}=b \sqrt{3}$. Through point $G$, we draw a line $K L$ (see Fig. 1) parallel to the diagonal $B^{\prime} D^{\prime}$. $A^{\prime} K=\sqrt{2} \cdot A^{\prime} G=b \sqrt{6}$. Through point $K$, we draw a line parallel to the diagonal $A^{\prime} D$ of the face until it intersects the extension of edge $A D$ at point $M$. Similar actions are performed with line $L N$ (Fig. 1). We connect points $M$ and $N$, which intersects the edges of the base of the cube at points $S$ and $T$.

Calculations:

$$
D M=A^{\prime} K=b \sqrt{6}, M S=D M \sqrt{2}=2 b \sqrt{3}
$$

$A M=a+D M=a+b \sqrt{6}, M N=\sqrt{2} \cdot A M=a \sqrt{2}+2 b \sqrt{3}$.

$S T=M N-2 \cdot M S=a \sqrt{2}+2 b \sqrt{3}-4 b \sqrt{3}=a \sqrt{2}-2 b \sqrt{3}>0 \rightarrow a>b \sqrt{6}$.

The area $S_{2}^{*}$ of the projection of the section in Case 2 is $S_{2}^{*}=a^{2}-S_{A^{\prime} K L}-S_{T S C}=a^{2}-3 b^{2}-\frac{1}{4} T S^{2}=a^{2}-3 b^{2}-\frac{(a \sqrt{2}-2 b \sqrt{3})^{2}}{4}=\frac{a^{2}+2 a b \sqrt{6}-12 b^{2}}{2}$.

The area of the section $S_{2}=\frac{S_{2}^{*}}{\cos \alpha}=\frac{\sqrt{3}}{2}\left(a^{2}+2 a b \sqrt{6}-12 b^{2}\right)$.

In the case where $1 a=2 \sqrt{3}, b=1$, i.e., $S_{2}=6 \sqrt{6}$.