A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #9 # and #6 # 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 · 2018-07-04T08:56:34

#color(maroon)("Total Surface Area " A_T = A_B + A_L = 52.16 + 145.77 = 197.93#

#l = 9, b = 6, theta = (5pi)/12, h = 9#

#"To find the Total Surface Area T S A"#

#"Area of parallelogram base " A_B = l b sin theta#

#A_B = 9 * 6 * sin ((5pi)/12) = 52.16#

#S_1 = sqrt(h^2 + (b/2)^2) = sqrt(9^2 + 3^2) = 9.49#

#S_2 = sqrt(h^2 + (l/2)^2) = sqrt(9^2 + (9/2)^2) = 10.06#

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

#A_L = (cancel2 * cancel(1/2)) * (l * S_1 + b * S_2) = (9 * 9.49 + 6 * 10.06)#

#A_L = 145.77#

#color(maroon)("Total Surface Area " A_T = A_B + A_L = 52.16 + 145.77 = 197.93#