A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $3$ and $8$ and the pyramid's height is $2$ . If one of the base's corners has an angle of $(5pi)/6$ , what is the pyramid's surface area?

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Answer 1 · 2017-12-12T10:21:45

T S A is 45.4164

$CH = b1 = 3 * sin (pi/6) = 1.5$
Area of parallelogram base $= a * b1 = 8*1.5 = color(red)(12 )$

$EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(2^2+ (8/2)^2)= 4.4721$
Area of $Delta AED = BEC = (1/2)*b*h_1 = (1/2)*3* 4.4721=$ color(red)(6.7082)#

$EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(2^2+(3/2)^2 )= 2.5$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*8*2.5 = color(red)( 10)$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$=( 2 * 6.7082)+ (2* 10) = color(red)(33.4164)$

Total surface area =Area of parallelogram base + Lateral surface area $= 12 + 33.4164 = 45.4164$

Total Surface Area # T S A = **45.4164**#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 3 3 and 8 8 and the pyramid's height is 2 2 . If one of the base's corners has an angle of (5pi)/6(5pi)/6, what is the pyramid's surface area?