Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `9` and `6` and the pyramid's height is `9`. If one of the base's corners has an angle of `(5pi)/12`, what is the pyramid's surface area?

Answer 1

`color(maroon)("Total Surface Area " A_T = A_B + A_L = 52.16 + 145.77 = 197.93`

Explanation:

`l = 9, b = 6, theta = (5pi)/12, h = 9`

`"To find the Total Surface Area T S A"`

`"Area of parallelogram base " A_B = l b sin theta`

`A_B = 9 * 6 * sin ((5pi)/12) = 52.16`

`S_1 = sqrt(h^2 + (b/2)^2) = sqrt(9^2 + 3^2) = 9.49`

`S_2 = sqrt(h^2 + (l/2)^2) = sqrt(9^2 + (9/2)^2) = 10.06`

`"Lateral Surface Area " A_L = 2 * ((1/2) l * S_1 + (1/2) b * S_2`

`A_L = (cancel2 * cancel(1/2)) * (l * S_1 + b * S_2) = (9 * 9.49 + 6 * 10.06)`

`A_L = 145.77`

`color(maroon)("Total Surface Area " A_T = A_B + A_L = 52.16 + 145.77 = 197.93`