A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 9 9 and 3 3 and the pyramid's height is 9 9. If one of the base's corners has an angle of (5pi)/125π12, what is the pyramid's surface area?

1 Answer
sankarankalyanam
Nov 23, 2017

Total surface area of the pyramid = color(blue)(138.36)138.36

Explanation:

CH = 3 * sin ((5pi)/12)= 3 sin (75) = 2.9CH=3⋅sin(5π12)=3sin(75)=2.9
Area of parallelogram base = 9* b1 = 9*2.9=color(red)(26.1)=9⋅b1=9⋅2.9=26.1

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(9^2+ (9/2)^2)= 10.06EF=h1=√h2+(a2)2=√92+(92)2=10.06
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3*10.06=color(red)(15.09)

EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(9^2+(3/2)^2 )= 9.12
Area of Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*9*9.12= color(red)(41.04)

Lateral surface area = 2* DeltaAED + 2*Delta CED
=( 2 * 15.09)+ (2* 41.04)= color(red)(112.26)

Total surface area =Area of parallelogram base + Lateral surface area = 26.1 + 112.26 = 138.36

Total Surface Area # T S A = **33.2695**#enter image source here

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