Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `8` and `1` and the pyramid's height is `4`. If one of the base's corners has an angle of `pi/3`, what is the pyramid's surface area?

Answer 1

`color(magenta)("Total Surface Area " A_T = A_B + A_L = 6.93+ 37.9 = 44.83`

Explanation:

`l = 8, b = 1, theta = pi/3, h = 4`

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

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

`A_B = 8 * 1 * sin (pi/3) = 6.93`

`S_1 = sqrt(h^2 + (b/2)^2) = sqrt(4^2 + 0.5^2) = 4.03`

`S_2 = sqrt(h^2 + (l/2)^2) = sqrt(4^2 + (8/2)^2) = 5.66`

`"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) = (8 * 4.03 + 1 * 5.66)`

`A_L = 37.9`

`color(magenta)("Total Surface Area " A_T = A_B + A_L = 6.93+ 37.9 = 44.83`