A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $4$ and $8$ and the pyramid's height is $7$ . If one of the base's corners has an angle of $(5pi)/6$ , what is the pyramid's surface area?

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Answer 1 · 2017-10-18T19:33:10

Total surface area $= 74.2409$

$CH = 4*sin ((5pi)/6) = 2$
Area of parallelogram base $= a* b1 = 8*2=16$

$EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(7^2+4^2)= sqrt65$
Area of $Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4*sqrt65=2sqrt65$

$EG = h_2 = sqrt(h^2+(b/2)^2 = sqrt(7^2+2^2) = sqrt53$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*sqrt53=4sqrt53$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$= 2*2sqrt65 + 2*4sqrt53= 32.249 + 58.2409$

Total surface area =Area of parallelogram base + Lateral surface area $= 16 + 58.2409 = 74.2409$

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A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 4 4 and 8 8 and the pyramid's height is 7 7 . If one of the base's corners has an angle of (5pi)/6(5pi)/6, what is the pyramid's surface area?