A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #3 # and #8 # 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 · 2018-01-24T12:44:58

Total Surface Area of the pyramid #color(brown)(A_T = 125.722)#

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Given : a = 8, b = 3, theta = (5pi)/12#

Area of parallelogram base #A_p = 3 * 8 * sin ((5pi)/12) = color(green)(23.1822)# sq. units

Area of slant triangle with b as base #A_b = (1/2) * b * sqrt (h^2 + (a/2)^2)#

#A_b = (1/2) * 3 * sqrt(9^2 + (8/2)^2) = color(green)(14.7733)#

Similarly,
#A_a = (1/2) * 8 * sqrt (9^2 + (3/2)^2) = color(green)(36.4966)#

Total Area of the parallelogram based pyramid

#A_T = A_p +( 2 * A_b )+ (2 * A_a )#

#A_T = 23.1822 + (2*14.7733) + (2*36.4966) = color(brown)(125.722)#