A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $7$ and $2$ and the pyramid's height is $8$ . If one of the base's corners has an angle of $(3pi)/8$ , what is the pyramid's surface area?

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Answer 1 · 2017-12-23T12:42:02

T S A = 86.835

$CH = 2 * sin ((3pi)/8) = 1.8478$
Area of parallelogram base $= a * b1 = 7*1.8478 = color(red)(12.9346)$

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

$EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(8^2+(2/2)^2 )= 8.0623$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*8.0623 = color(red)( 28.2181)$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$=( 2 * 8.7321)+ (2* 28.2181) = color(red)(73.643)$

Total surface area =Area of parallelogram base + Lateral surface area $= 12.9346 + 73.9004 = 86.835$

Total Surface Area # T S A = **86.835**#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 7 7 and 2 2 and the pyramid's height is 8 8 . If one of the base's corners has an angle of (3pi)/8(3pi)/8, what is the pyramid's surface area?