The base of the pyramid is a rhombus, the diagonals of which are equal to 6 in 8 in. The height of the pyramid passes

univerkov | January 6, 2021 | education

The base of the pyramid is a rhombus, the diagonals of which are equal to 6 in 8 in. The height of the pyramid passes through the intersection of the diagonals of the rhombus. Find the side surface of the pyramid if the apothem is 8 dm.

Let ABCD be a rhombus at the base of the pyramid, diagonal AC = 8, diagonal BD = 6, O be the point of intersection of the diagonals.
Then AO = 4, BO = 3, by the Pythagorean theorem AB = 5.
The area of the side face S resting on the edge AB is equal to AB * h / 2, where h is the apothem.
S = 5 * 8/2 = 20.
The area of the side surface of the pyramid F is equal to the sum of all its four faces.
F = 4 * S = 80 sq. Dm.
Answer: 80 sq. Dm.

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