Task B-3.7. The length of the side of an equilateral triangle is $6 \mathrm{~cm}$. A line parallel to one of the sides has been drawn through its centroid. Calculate the volume of a right prism whose base is the resulting trapezoid, and the length of the prism's height is equal to the height of the trapezoid. Determine the cosine of the angle that the spatial diagonal of the prism forms with the base plane.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_06_03_ade005c4283eff439485g-17.jpg?height=542&width=1328&top_left_y=924&top_left_x=366)

(If the angle between the diagonal and the base is clearly marked)

(2 points)

Since the centroid divides the median in the ratio $2: 1$, and the median coincides with the height of the equilateral triangle, the height of the resulting trapezoid is

$$
h=\frac{1}{3} \cdot \frac{6 \sqrt{3}}{2}=\sqrt{3} \text{ cm}
$$

(2 points)

Similarly, we have

$$
\begin{aligned}
& \frac{c}{a}=\frac{2}{3} \\
& c=\frac{2}{3} \cdot 6=4 \text{ cm}
\end{aligned}
$$

The area of the trapezoid is

$$
B=\frac{a+c}{2} \cdot h=5 \sqrt{3} \text{ cm}^{2}
$$

The volume of the prism is

$$
V=B \cdot h=5 \sqrt{3} \cdot \sqrt{3}=15 \text{ cm}^{3}
$$

To calculate the desired cosine, let's calculate the length of the diagonal of the base and the length of the space diagonal.

$$
\begin{aligned}
d & =\sqrt{5^{2}+(\sqrt{3})^{2}}=\sqrt{28} \text{ cm} \\
D & =\sqrt{28+(\sqrt{3})^{2}}=\sqrt{31} \text{ cm}
\end{aligned}
$$

Now we have

$$
\cos \varphi=\frac{d}{D}=\sqrt{\frac{28}{31}}
$$

(1 point)

## SCHOOL/CITY COMPETITION IN MATHEMATICS

## 4th grade - high school - B variant

January 17, 2013.

IF A STUDENT HAS A DIFFERENT METHOD OF SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT METHOD APPROPRIATELY.