Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `9` and `3` 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

Total surface area of the pyramid = `color(blue)(138.36)`

Explanation:

`CH = 3 * sin ((5pi)/12)= 3 sin (75) = 2.9`
Area of parallelogram base `= 9* b1 = 9*2.9=color(red)(26.1)`

`EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(9^2+ (9/2)^2)= 10.06`
Area of `Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3*10.06=color(red)(15.09)`

`EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(9^2+(3/2)^2 )= 9.12`
Area of `Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*9*9.12= color(red)(41.04)`

Lateral surface area = `2* DeltaAED + 2*Delta CED`
`=( 2 * 15.09)+ (2* 41.04)= color(red)(112.26)`

Total surface area =Area of parallelogram base + Lateral surface area `= 26.1 + 112.26 = 138.36`