A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 5 5 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
Dec 20, 2017

T S A = 88.132

Explanation:

CH = 3 * sin ((5pi)/12) = 2.8978CH=3⋅sin(5π12)=2.8978
Area of parallelogram base = a * b1 = 5*2.8978 = color(red)(14.489)=a⋅b1=5⋅2.8978=14.489

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(9^2+ (5/2)^2)= 9.3408EF=h1=√h2+(a2)2=√92+(52)2=9.3408
Area of Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3* 9.3408= color(red)(14.0112)#

EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(9^2+(3/2)^2 )= 9.1241
Area of Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*5*9.1241 = color(red)( 22.8103)

Lateral surface area = 2* DeltaAED + 2*Delta CED
=( 2 * 14.0112)+ (2* 22.8103) = color(red)(73.643)

Total surface area =Area of parallelogram base + Lateral surface area = 14.489 + 73.643 = 88.132

Total Surface Area # T S A = **88.132**#enter image source hereTotal Surface Area T S A = **88.132**enter image source here

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