A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 4 4 and 9 9 and the pyramid's height is 4 4. If one of the base's corners has an angle of (3pi)/83π8, what is the pyramid's surface area?

1 Answer
sankarankalyanam
Dec 7, 2017

T S A = 97.5919

Explanation:

CH = 4 * sin ((3pi)/8) = 3.6955CH=4⋅sin(3π8)=3.6955
Area of parallelogram base = 9* b1 = 9*3.6955 = color(red)(33.2595 )=9⋅b1=9⋅3.6955=33.2595

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(4^2+ (9/2)^2)= 6.0208EF=h1=√h2+(a2)2=√42+(92)2=6.0208
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4* 6.0208= color(red)(12.0416)#

EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(4/2)^2 )= 4.4721
Area of Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*9*4.4721 = color(red)( 20.1246)

Lateral surface area = 2* DeltaAED + 2*Delta CED
=( 2 * 12.0416)+ (2* 20.1246) = color(red)(64.3324)

Total surface area =Area of parallelogram base + Lateral surface area = 33.2595 + 64.3324 = 97.5919

Total Surface Area # T S A = **97.5919**#enter image source here

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