Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `1` and `5` and the pyramid's height is `2`. 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 `T S A = **14.4755**`

Explanation:

`CH = 1*sin (5pi/12) = 0.9659`
Area of parallelogram base `= a* b1 = 5 * 0.9659`

`EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(2^2+ (5/2)^2)= 3.2016`
Area of `Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1*3.2016 = 1.6008`

`EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(2^2+(1/2)^2 ) = 2.0616`
Area of `Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*5*2.0616 = 5.154`

Lateral surface area = `2* DeltaAED + 2*Delta CED`
`=( 2 * 1.6008)+ (2*5.154 ) = 13.5096`

Total surface area =Area of parallelogram base + Lateral surface area `= 0.9659 + 13.5096 = 14.4755`