A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $4$ and $7$ and the pyramid's height is $8$ . If one of the base's corners has an angle of $pi/3$ , what is the pyramid's surface area?

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Answer 1 · 2017-12-06T15:00:02

T S A = 116.9005

$CH = 4 * sin (pi/3) = 4 sin (60) = 3.4641$
Area of parallelogram base $= 7* b1 = 7*3.46 = color(red)(24.2487 )$

$EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(8^2+ (7/2)^2)= 8.7321$
Area of $Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4* 8.7321=$ color(red)(17.4642)#

$EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(8^2+(4/2)^2 )= 8.2642$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*8.2462 = color(red)( 28.8617)$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$=( 2 * 17.4642)+ (2* 28.8617) = color(red)(92.6518)$

Total surface area =Area of parallelogram base + Lateral surface area $= 24.2487 + 92.6518 = 116.9005$

Total Surface Area # T S A = **116.9005**#enter image source here

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 4 4 and 7 7 and the pyramid's height is 8 8 . If one of the base's corners has an angle of pi/3pi/3, what is the pyramid's surface area?