Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `2` and `9` and the pyramid's height is `7`. If one of the base's corners has an angle of `(5pi)/6`, what is the pyramid's surface area?

Answer 1

T S A `color(blue)(A_T = A_b + A_L = 9 + 80.1033 = 89.1033`

Explanation:

Area of base `A_b = l * b * sin theta = 2 * 9 * sin ((5pi)/6) = 9`

`s_1 = sqrt ((b/2)^2 + h^2) = sqrt ((9/2)^2 + 7^2) = 8.3217`

`s_2 = sqrt ((l/2)^2 + h^2) = sqrt((2/2)^2 + 7^2) = 7.0711`

Lateral Surface Area

`A_L = 2 ( (1/2) l * s_1 + (1/2) b * s_2)`

`A_L = (2 * 8.3217 + 9 * 7.0711) = 80.1033`

T S A `A_T = A_b + A_L = 9 + 80.1033 = 89.1033`