## Task B-3.6.

A parallelogram with diagonals of length $20 \mathrm{~cm}$ and $30 \mathrm{~cm}$ that intersect at an angle of $60^{\circ}$ is the base of a right prism with a volume of $1500 \sqrt{3} \mathrm{~cm}^{3}$. What is the surface area of this prism?
Solution.

First, let's sketch the parallelogram (the base of the prism) and mark the given elements.

![](https://cdn.mathpix.com/cropped/2024_05_30_d455efeee432fadf0574g-16.jpg?height=311&width=609&top_left_y=381&top_left_x=655)

Let's calculate the area \( B \) of the parallelogram using the formula for the area of a quadrilateral given the lengths of the diagonals \( e \) and \( f \) and the measure of the angle \( \varphi \) between them.

\[ B = \frac{1}{2} ef \sin \varphi = \frac{1}{2} \cdot 20 \cdot 30 \cdot \sin 60^\circ \]

1 point

Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), the area of the parallelogram is \( B = 300 \cdot \frac{\sqrt{3}}{2} = 150 \sqrt{3} \, \text{cm}^2 \).

1 point

Using the formula for the volume of the prism \( V = B \cdot h \), we can determine its height \( h \):

\[ h = \frac{V}{B} = \frac{1500 \sqrt{3}}{150 \sqrt{3}} = 10 \, \text{cm} \].

1 point

To determine the surface area of the prism, we need to find the lengths of its base edges, that is, the lengths of the sides of the given parallelogram.

The intersection of the diagonals of the parallelogram divides both diagonals into two equal parts, so the lengths of the sides of the parallelogram can be calculated using the cosine rule.

\[ a^2 = 15^2 + 10^2 - 2 \cdot 10 \cdot 15 \cdot \cos 120^\circ = 225 + 100 + 150 = 475 \]

\[ a = \sqrt{475} = 5 \sqrt{19} \, \text{cm} \]

\[ b^2 = 15^2 + 10^2 - 2 \cdot 10 \cdot 15 \cdot \cos 60^\circ = 225 + 100 - 150 = 175 \]

\[ b = \sqrt{175} = 5 \sqrt{7} \, \text{cm} \]

It remains to calculate the surface area of the prism.

\[ O = 2B + P = 2B + 2ah + 2bh \]

1 point

\[ O = 2 \cdot 150 \sqrt{3} + 2 \cdot 5 \sqrt{19} \cdot 10 + 2 \cdot 5 \sqrt{7} \cdot 10 = (300 \sqrt{3} + 100 \sqrt{19} + 100 \sqrt{7}) \, \text{cm}^2 \]

Note: The student can achieve all points in the task without partial root extraction of the lengths of the sides of the parallelogram.