Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `2` and `12` and the pyramid's height is `3`. If one of the base's corners has an angle of `pi/4`, what is the pyramid's surface area?

Answer 1

Total Surface Area `= A_(base) + A_L = color(red)(68.3344)`

Explanation:

Area of base `A_(base) = DC * AD * sin D = 12 * 2 * sin (pi/4)`

`A_(base) = 16.9706`

Area of `Delta ADE = (1/2) * AD * sqrt(EM^2 + ((DC)/2)^2)`

`A_(ADE) = (1/2) * 2 * sqrt(3^2 + ((12)/2)^2) = 6.7082`

`A_(ABE) = (1/2) * AB * sqrt(AB^2+ ((AD)/2)^2)`

`A_(ABE) = (1/2) * 12 * sqrt(3^2 + (2/2)^2) = 18.9737`

Lateral Surface Area `A_L = (2 * A_(ADE)) + (2 * A_(ABE))`

`A_L= (2*6.7082) + (2 * 18.9737) = 51.3638`

Total Surface Area
`A_T = A_(base) + A_L = 16.9706 + 51.3638 = color(red)68.3344`