Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `3` and `8` and the pyramid's height is `8`. 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 T. S. A. = 68.2216

Explanation:

`CH = 3 * sin (pi/4) = 3sin (45) = 3/sqrt2 = 2.1213`
Area of parallelogram base `= 8* b1 = 8*2.1213 = color(red)(16.9704 )`

`EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(8^2+ (8/2)^2)= 8.9443`
Area of # Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3* 8.9443 = #color(red)(13.4165)#

`EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(8^2+(3/2)^2 )= 8.1394`
Area of `Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*3*8.1394 = color(red)( 12.2091)`

Lateral surface area = `2* DeltaAED + 2*Delta CED`
`=( 2 * 13.4165)+ (2* 12.2091) = color(red)(51.2512)`

Total surface area =Area of parallelogram base + Lateral surface area `= 16.9704 + 51.2512 = 68.2216`