A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $7$ and $1$ and the pyramid's height is $5$ . 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 · 2017-12-23T12:35:04

T S A = 46.2275

$CH = 1 * sin ((pi)/4) = 0.7071$
Area of parallelogram base $= a * b1 = 7*0.7071 = color(red)(4.9497)$

$EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(5^2+ (7/2)^2)= 6.1033$
Area of $Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1* 6.1033=$ color(red)(3.0517)#

$EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(5^2+(1/2)^2 )= 5.0249$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*7*5.0249 = color(red)( 17.5872)$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$=( 2 * 3.0517)+ (2* 17.5872) = color(red)(41.2778)$

Total surface area =Area of parallelogram base + Lateral surface area $= 4.9497 + 41.2778 = 46.2275$

Total Surface Area # T S A = **46.2275**#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 1 1 and the pyramid's height is 5 5 . If one of the base's corners has an angle of pi/4pi/4, what is the pyramid's surface area?