A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $2$ and $1$ and the pyramid's height is $3$ . 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 · 2017-12-12T11:34:31

T S A = 89.3341

$CH = 1 * sin (pi/4) = 0.707$
Area of parallelogram base $= a * b1 = 2*0.707 = color(red)(1.414 )$

$EF = h_1 = sqrt(h^2 + (a/2)^2) = sqrt(3^2+ (2/2)^2)= 3.1623$
Area of $Delta AED = BEC = (1/2)*b*h_1 = (1/2)*1* 3.1623=$ color(red)(1.5812)#

$EG = h_2 = sqrt(h^2+(b/2)^2 ) = sqrt(3^2+(1/2)^2 )= 3.0414$
Area of $Delta = CED = AEC = (1/2)*a*h_2 = (1/2)*9*6.1847 = color(red)( 23.8712)$

Lateral surface area = $2* DeltaAED + 2*Delta CED$
$=( 2 * 11.25)+ (2* 23.8712) = color(red)(70.2424)$

Total surface area =Area of parallelogram base + Lateral surface area $= 19.0917 + 70.2424 = 89.3341$

Total Surface Area # T S A = **89.3341**#enter image source here Total Surface Area $T S A = **89.3341**$ 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 2 2 and 1 1 and the pyramid's height is 3 3 . If one of the base's corners has an angle of pi/4 pi/4 , what is the pyramid's surface area?