A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $5$ and $3$ and the pyramid's height is $4$ . 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 · 2017-12-06T13:55:02

T S A = 48.5015

$CH = 3 * sin (pi/3) = 3sin (60)= 2.5981$
Area of parallelogram base $= 5* b1 = 5*2.5981 = color(red)(12.9905 )$

$EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(4^2+ (5/2)^2)= 4.717$
Area of $Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3* 4.717 =$ color(red)(7.0755)#

$EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(4^2+(3/2)^2 )= 4.272$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*5*4.272 = color(red)( 10.68)$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$=( 2 * 7.0755)+ (2* 10.68) = color(red)(35.511)$

Total surface area =Area of parallelogram base + Lateral surface area $= 12.9905 + 35.511 = 48.5015$

Total Surface Area # T S A = **48.5015**#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 5 5 and 3 3 and the pyramid's height is 4 4 . If one of the base's corners has an angle of pi/3pi/3, what is the pyramid's surface area?