A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #2 # and #9 # and the pyramid's height is #7 #. If one of the base's corners has an angle of #(5pi)/6#, what is the pyramid's surface area?

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Answer 1 · 2018-02-20T05:20:46

T S A #color(blue)(A_T = A_b + A_L = 9 + 80.1033 = 89.1033#

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Area of base #A_b = l * b * sin theta = 2 * 9 * sin ((5pi)/6) = 9#

#s_1 = sqrt ((b/2)^2 + h^2) = sqrt ((9/2)^2 + 7^2) = 8.3217#

#s_2 = sqrt ((l/2)^2 + h^2) = sqrt((2/2)^2 + 7^2) = 7.0711#

Lateral Surface Area

#A_L = 2 ( (1/2) l * s_1 + (1/2) b * s_2)#

#A_L = (2 * 8.3217 + 9 * 7.0711) = 80.1033#

T S A #A_T = A_b + A_L = 9 + 80.1033 = 89.1033#