A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $6$ and $3$ and the pyramid's height is $3$ . 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 · 2018-01-05T03:20:54

T S A = 45.5801

$CH = 3 * sin ((pi)/4) = 2.1213$
Area of parallelogram base $= a * b1 = 6* 2.1213 = color(red)(12.7279)$

$EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(3^2+ (6/2)^2)= 4.2426$
Area of $Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3* 4.2426=$ color(red)(6.3638)#

$EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(3^2+(3/2)^2 )= 3.3541$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*6*3.3541 = color(red)( 10.0623)$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$=( 2 * 6.3638)+ (2* 10.0623) = color(red)(32.8522)$

Total surface area =Area of parallelogram base + Lateral surface area $= 12.7279 + 32.8522 = 45.5801$

Total Surface Area # T S A = **45.5801**#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 6 6 and 3 3 and the pyramid's height is 3 3 . If one of the base's corners has an angle of pi/4pi/4, what is the pyramid's surface area?