A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $2$ and $12$ 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 · 2018-01-07T10:31:22

Total Surface Area $= A_(base) + A_L = color(red)(68.3344)$

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Area of base $A_(base) = DC * AD * sin D = 12 * 2 * sin (pi/4)$

$A_(base) = 16.9706$

Area of $Delta ADE = (1/2) * AD * sqrt(EM^2 + ((DC)/2)^2)$

$A_(ADE) = (1/2) * 2 * sqrt(3^2 + ((12)/2)^2) = 6.7082$

$A_(ABE) = (1/2) * AB * sqrt(AB^2+ ((AD)/2)^2)$

$A_(ABE) = (1/2) * 12 * sqrt(3^2 + (2/2)^2) = 18.9737$

Lateral Surface Area $A_L = (2 * A_(ADE)) + (2 * A_(ABE))$

$A_L= (2*6.7082) + (2 * 18.9737) = 51.3638$

Total Surface Area
$A_T = A_(base) + A_L = 16.9706 + 51.3638 = color(red)68.3344$

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 2 2 and 12 12 and the pyramid's height is 3 3 . If one of the base's corners has an angle of pi/4pi/4, what is the pyramid's surface area?