A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $1$ and $5$ and the pyramid's height is $2$ . If one of the base's corners has an angle of $(5pi)/12$ , what is the pyramid's surface area?

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Answer 1 · 2017-10-21T22:49:40

Total Surface Area $T S A = **14.4755**$

$CH = 1*sin (5pi/12) = 0.9659$
Area of parallelogram base $= a* b1 = 5 * 0.9659$

$EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(2^2+ (5/2)^2)= 3.2016$
Area of $Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1*3.2016 = 1.6008$

$EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(2^2+(1/2)^2 ) = 2.0616$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*5*2.0616 = 5.154$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$=( 2 * 1.6008)+ (2*5.154 ) = 13.5096$

Total surface area =Area of parallelogram base + Lateral surface area $= 0.9659 + 13.5096 = 14.4755$

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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 1 1 and 5 5 and the pyramid's height is 2 2 . If one of the base's corners has an angle of (5pi)/12(5pi)/12, what is the pyramid's surface area?