A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $4$ and $5$ and the pyramid's height is $6$ . If one of the base's corners has an angle of $pi/3$ , what is the pyramid's surface area?

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Answer 1 · 2017-12-20T12:36:07

T S A = 74.9435.

$CH = 4 * sin (pi/3) = 3.4641$
Area of parallelogram base $= a * b1 = 5*3.4641 = color(red)(17.3205)$

$EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(6^2+ (5/2)^2)= 6.5$
Area of $Delta AED = BEC = (1/2)*b*h_1 = (1/2)*4* 6.5=$ color(red)(13)#

$EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(6^2+(4/2)^2 )= 6.3246$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*5*6.3246 = color(red)( 15.8115)$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$=( 2 * 13)+ (2* 15.8115) = color(red)(57.623)$

Total surface area =Area of parallelogram base + Lateral surface area $= 17.3205 + 57.623 = 74.9435$

Total Surface Area # T S A = **74.9435**#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 4 4 and 5 5 and the pyramid's height is 6 6 . If one of the base's corners has an angle of pi/3 pi/3 , what is the pyramid's surface area?