A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #3 # and #2 # 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 · 2018-06-08T04:33:47

#color(indigo)(T S A = L S A + A_(base) = 19.98 + 4.24 = 24.22# sq units

#"Total Surface Area of Pyramid " T S A = L S A + A_(base)#

#l = 3, b = 2, theta = pi/4, h = 3#

#S_1 = sqrt (h^2 + (b/2)^2) = sqrt(3^2 + 1^2) = sqrt 10 = 3.16#

#S_2 = sqrt (h^2 + (l/2)^2) = sqrt(3^2 + (1.5)^2) = sqrt (11.25) = 3.5#

#L S A = 2((1/2) l * S_1 + (1/2) b * S_2) = l*S_1 + b*S_2#

#L S A = 3 * 3.16 + 3 + 3.5 = 19.98#

#A_(base) = l b sin theta = 3 * 2 * (1/sqrt2) = 3 sqrt2 = 4.24#

#color(indigo)(T S A = L S A + A_(base) = 19.98 + 4.24 = 24.22# sq units