Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `8` and `7` 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

Total Surface Area `T S A = **70.4414**`

Explanation:

`CH = 7 * sin (pi/3)= (7sqrt3)/2`
Area of parallelogram base `= 8 * b1 = (8 * 7sqrt3)/2 = color(red)(48.4974)`

`EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(4^2+ (8/2)^2)= 5.6569`
Area of `Delta AED = BEC = (1/2)*b*h_1 = (1/2)*7*5.6569 = color (red)(5.6569)`

`EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(7/2)^2 )= 5.3151`
Area of `Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*5.3151 = color (red)(5.3151)`

Lateral surface area = `2* DeltaAED + 2*Delta CED`
`=( 2 * 5.6569)+ (2* 5.3151)= color(blue)21.944`

Total surface area =Area of parallelogram base + Lateral surface area `= 48.4974 + 21.944= color (green)(70.4414)`