A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $2$ and $9$ and the pyramid's height is $6$ . If one of the base's corners has an angle of $pi/4$ , what is the pyramid's surface area?

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Answer 1 · 2017-12-14T12:59:38

T S A = 82.4708

$CH = 2 * sin (pi/4) = 1.414$
Area of parallelogram base $= a * b1 = 9*1.414 = color(red)(12.726)$

$EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(6^2+ (9/2)^2)= 7.5$
Area of $Delta AED = BEC = (1/2)*b*h_1 = (1/2)*2* 7.5=$ color(red)(7.5)#

$EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(2/2)^2 )= 6.0828$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*9*6.0828 = color(red)( 27.3724)$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$=( 2 * 7.5)+ (2* 27.3724) = color(red)(69.7448)$

Total surface area =Area of parallelogram base + Lateral surface area $= 12.726 + 69.7448 = 82.4708$

Total Surface Area # T S A = **82.4708**#enter image source here Total Surface Area $T S A = **82.4708**$ 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 2 2 and 9 9 and the pyramid's height is 6 6 . If one of the base's corners has an angle of pi/4 pi/4 , what is the pyramid's surface area?