# In a regular triangular pyramid, the side of the base is 15√3 cm, and the side edge is 17 cm

**In a regular triangular pyramid, the side of the base is 15√3 cm, and the side edge is 17 cm. Calculate the area of the section drawn through the side edge and the height of the pyramid.**

In a regular triangular pyramid, the side faces are isosceles triangles.

In this case, the base is an equilateral triangle.

All this means that the plane that will pass through the lateral edge and the height of the pyramid will pass through the height of the face (isosceles triangle), which is opposite to the lateral edge.

As a result of the intersection, a triangle will be obtained, the sides of which are equal: the length of the edge of the pyramid, the height of the equilateral triangle of the base, the height of the side face.

To calculate the cross-sectional area, we find the height of the base and the side face.

Since the base is an equilateral triangle, and the side face is isosceles, the height will also be the median (it will divide the base of the triangle in half).

Hence, using the Pythagorean theorem, we obtain:

1) Let’s denote the height of the base through a, then a ^ 2 = (15√3) ^ 2 – (15√3 / 2) ^ 2 = 225 * 3 – 225 * 3/4 = 675 – 168.75 = 506.25 = 22 , 5 ^ 2.

Hence a = 22.5.

2) We denote the height of the side face by b, then b ^ 2 = 17 ^ 2 – (15√3 / 2) ^ 2 = 289 – 168.75 = 120.25

Hence b = √120.25 = 5√4.81.

To calculate the cross-sectional area, we use Heron’s formula:

S = √p (p-a) (p-b) (p-c), where p = half the perimeter, a, b, c are the sides of the triangle.

p = (17 + 22.5 + 5√4.81) / 2 ~ (39.5 + 11) / 2 = 50.5.

Hence S = √50.5 (50.5 – 22.5) (50.5 – 5√4.81) (50.5 – 17) ~ √50.5 * 28 * 39.5 * 33.5 = √1871075.5 ~ 1367.873 cm ^ 2.

Answer: the cross-sectional area is approximately equal to 1367.873 cm ^ 2 (accurate to thousandths).