A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $7$ and $5$ and the pyramid's height is $8$ . If one of the base's corners has an angle of $(3pi)/8$ , what is the pyramid's surface area?

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Answer 1 · 2017-12-25T12:18:49

T S A = 83.50111

$CH = b1 = 5 * sin ((3pi)/8) = 4.6194$
Area of parallelogram base $= a * b1 = 7*4.6194 = color(red)(32.3358)$

$EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(8^2+ (7/2)^2)= 8.7321$
Area of $Delta AED = BEC = (1/2)*b*h_1 = (1/2)*5* 8.7321=$ color(red)(21.8303)#

$EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(8^2+(5/2)^2 )= 8.3815$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*8.3815 = color(red)( 29.3353)$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$=( 2 * 21.8303)+ (2* 29.3353) = color(red)(51.1653)$

Total surface area =Area of parallelogram base + Lateral surface area $= 32.3358 + 51.1653 = 83.5011$

Total Surface Area # T S A = **83.5011**#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 7 7 and 5 5 and the pyramid's height is 8 8 . If one of the base's corners has an angle of (3pi)/8(3pi)/8, what is the pyramid's surface area?