A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $3$ and $7$ and the pyramid's height is $4$ . 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 · 2017-12-07T05:42:05

T S A = 65.2506

$CH = 3 * sin ((3pi)/8) = 4 sin (60) = 2.7716$
Area of parallelogram base $= 7* b1 = 7*2.7716= color(red)(19.4012 )$

$EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(4^2+ (7/2)^2)= 5.3151$
Area of $Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3* 5.3151=$ color(red)(7.9727)#

$EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(3/2)^2 )= 4.272$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*4.272 = color(red)( 14.952)$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$=( 2 * 7.9727)+ (2* 14.952) = color(red)(45.8494)$

Total surface area =Area of parallelogram base + Lateral surface area $= 19.4012 + 45.8494 = 65.2506$

Total Surface Area # T S A = **65.2506**#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 3 3 and 7 7 and the pyramid's height is 4 4 . If one of the base's corners has an angle of (3pi)/8(3pi)/8, what is the pyramid's surface area?