A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 3 3 and 8 8 and the pyramid's height is 8 8. If one of the base's corners has an angle of pi/4π4, what is the pyramid's surface area?

1 Answer
sankarankalyanam
Dec 4, 2017

Total Surface Area T. S. A. = 68.2216

Explanation:

CH = 3 * sin (pi/4) = 3sin (45) = 3/sqrt2 = 2.1213CH=3⋅sin(π4)=3sin(45)=3√2=2.1213
Area of parallelogram base = 8* b1 = 8*2.1213 = color(red)(16.9704 )=8⋅b1=8⋅2.1213=16.9704

EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(8^2+ (8/2)^2)= 8.9443EF=h1=√h2+(a2)2=√82+(82)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

Total Surface Area # T S A = **68.2216**#enter image source here

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