## Task B-3.7.

The base of a right pyramid is a triangle with side lengths $25 \mathrm{~cm}, 29 \mathrm{~cm}, 36 \mathrm{~cm}$. All lateral faces of the pyramid form an angle of $75^{\circ}$ with the base plane. The volume of the pyramid is $k+l \sqrt{m} \mathrm{~cm}^{3}$. Determine the natural numbers $k, l$ and $m$.
## Solution.

If all the lateral faces of the pyramid form the same angle with the base plane, the orthogonal projection of the pyramid's apex is the center of the inscribed circle of the base. (or the image)

![](https://cdn.mathpix.com/cropped/2024_05_30_a25c851e6461b5bf7f1cg-14.jpg?height=728&width=559&top_left_y=1104&top_left_x=677)

Let the sides of the triangle be $a=25 \mathrm{~cm}, b=29 \mathrm{~cm}, c=36 \mathrm{~cm}$.

Then the semi-perimeter is $s=\frac{a+b+c}{2}=\frac{25+29+36}{2}=45 \mathrm{~cm}$.

The area of the base is

$$
B=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{45 \cdot 20 \cdot 16 \cdot 9}=30 \cdot 4 \cdot 3=360 \mathrm{~cm}^{2}
$$

The radius of the inscribed circle of the base is

$$
r=\frac{B}{s}=8 \mathrm{~cm}
$$

From the shaded triangle in the image, it follows that

$$
\tan 75^{\circ}=\frac{h}{r}
$$

Thus,

$$
h=r \tan\left(30^{\circ}+45^{\circ}\right)=8 \cdot \frac{\tan 30^{\circ}+\tan 45^{\circ}}{1-\tan 30^{\circ} \cdot \tan 45^{\circ}}=8 \cdot(2+\sqrt{3})
$$

Then the volume is

$$
V=\frac{1}{3} \cdot 360 \cdot 8 \cdot(2+\sqrt{3})=960 \cdot(2+\sqrt{3})=1920+960 \cdot \sqrt{3} \mathrm{~cm}^{3}
$$

1 point

It follows that $k=1920, l=960, m=3$.

1 point

## SCHOOL/CITY COMPETITION IN MATHEMATICS

4th grade - high school - B variant
January 5, 2015.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE TASK, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.