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 #9 #. If one of the base's corners has an angle of #(5pi)/12#, what is the pyramid's surface area?

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Answer 1 · 2018-04-06T11:06:47

#"Pyramid's Total Surface Area " = 51.22 " sq units"#

#"Given : l = 3, b = 2, h = 9#

#S_1 = sqrt(h^2 + (b/2)^2) = sqrt(9^2 + 1^2) = 9.06#

#S_2 = sqrt(h^2 + (l/2)^2) = sqrt (9^2 + 1.5^2) = 9.12#

#"Area of Pyramid base " A_b = l * b * sin theta = 3 *2 * sin ((5pi)/12) = 5.8#

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

#=> cancel2 * cancel(1/2) * (3 * 9.06 + 2 * 9.12) = 45.42#

#"Total Surface Area " T S A = L S A + A_b = 45.42 + 5.8 = 51.22#