A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $2$ and $7$ 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 · 2017-12-06T14:14:34

T S A = 66.37

$CH = 2 * sin (pi/4) = 2 sin (45) = 1.414$
Area of parallelogram base $= 7* b1 = 7*1.414 = color(red)(9.898 )$

$EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(6^2+ (7/2)^2)= 6.9462$
Area of $Delta AED = BEC = (1/2)*b*h_1 = (1/2)*2* 6.9462 =$ color(red)(6.9462)#

$EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(2/2)^2 )= 6.0828$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*6.0828 = color(red)( 21.2898)$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$=( 2 * 6.9462)+ (2* 21.2898) = color(red)(56.472)$

Total surface area =Area of parallelogram base + Lateral surface area $= 9.898 + 56.472 = 66.37$

Total Surface Area # T S A = **66.37**#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 7 7 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?