A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 8 8 and 1 1 and the pyramid's height is 7 7. If one of the base's corners has an angle of (5pi)/65π6, what is the pyramid's surface area?

1 Answer
sankarankalyanam
Dec 6, 2017

T S A = 68.2048

Explanation:

CH = 8 * sin ((5pi)/6) = 4CH=8⋅sin(5π6)=4
Area of parallelogram base = 1* b1 = 1*4 = color(red)(4 )=1⋅b1=1⋅4=4

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(7^2+ (8/2)^2)= 8.0623EF=h1=√h2+(a2)2=√72+(82)2=8.0623
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1* 8.0623 = color(red)(4.0312)#

EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(7^2+(1/2)^2 )= 7.0178
Area of Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*7.0178 = color(red)( 28.0712)

Lateral surface area = 2* DeltaAED + 2*Delta CED
=( 2 * 4.0312)+ (2* 28.0712) = color(red)(64.2048)

Total surface area =Area of parallelogram base + Lateral surface area = 4 + 64.2048 = 68.2048

Total Surface Area # T S A = **68.2048**#enter image source here

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