Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `6` and `4` and the pyramid's height is `7`. If one of the base's corners has an angle of `(5pi)/6`, what is the pyramid's surface area?

Answer 1

T S A = 116.4108

Explanation:

`CH = 4 * sin ((5pi)/6) = 8`
Area of parallelogram base `= a * b1 = 6*8 = color(red)(48)`

`EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(7^2+ (6/2)^2)= 7.6158`
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4* 7.6158= #color(red)(15.2316)#

`EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(4/2)^2 )= 6.3246`
Area of `Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*6*6.3246 = color(red)( 18.9738)`

Lateral surface area = `2* DeltaAED + 2*Delta CED`
`=( 2 * 15.2316)+ (2* 18.9738) = color(red)(68.4108)`

Total surface area =Area of parallelogram base + Lateral surface area `= 48 + 68.4108 = 116.4108`