The base of the pyramid is a rhombus with diagonals of 6 cm and 8 cm. The height of the pyramid is lowered

The base of the pyramid is a rhombus with diagonals of 6 cm and 8 cm. The height of the pyramid is lowered to the point of intersection of its diagonals. The smaller side edges of the pyramid are 5 cm. Find the volume of the pyramid.

To solve this problem, remember that the volume of the pyramid is equal to the product of 1/3 by the area of ​​the base and the height. Let’s calculate the area of ​​a rhombus along its diagonals. The area of ​​a rhombus is half the product of its diagonals. The diagonals are 6 cm and 8 cm.
S = 6 + 8/2 = 7 cm ^ 2.
Now we need to calculate the height. Recall that the diagonals of the rhombus intersect at right angles and are halved at the point of intersection. Thus, half of the diagonal, side edge and height make up a right-angled triangle, in which one leg is equal to:
6/2 = 3 cm.
The second leg is the height we are looking for, and the hypotenuse is the edge of the pyramid – 5 cm. By the tower of Pythagoras, we find the height.
H ^ 2 = 5 * 5-3 * 3 = 16
H = √16 = 4 cm.
Substitute the values ​​into the formula and calculate the volume.
V = 1/3 * 7 * 4 = 28/3 = 9 1/3 cm ^ 3.
Answer: 9 whole 1/3 cm ^ 3.