A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #2 # and #6 # 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 · 2018-01-31T12:03:37

#T S A = color(green)(35.9443)#

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Area of parallelogram base

#A_b = l * b * sin theta = 6 * 2 * sin (pi/4) = color(brown)(8.4853)#

Lateral Surface Area of the pyramid

#A_l = 2 * Delta ADE + 2 * Delta CDE#

#A_l = 2 * (1/2) * (w) * sqrt((l/2)^2 + h^2) + 2 * ( 1/2) * (l) * sqrt((w/2)^2 + h^2)#

#A_l = (2 * (1/2) * 2 * sqrt(3^2 + 3^2)) = (2 * (1/2) * 6 * sqrt(1^2 + 3^2))#

#A_l = 8.4853 + 18.9737 = color(brown)(27.459)#

Total Surface Area # T S A = A_b + A_l #

#T S A = 8.4853 + 27.459 = color(green)(35.9443)#