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 #2 #. 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 · 2018-01-22T14:58:39

Total Surface Area (T S A )= #A_T = color(purple)(38.9734)#

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Area of base parallelogram

#A_p = l * b sin (theta) = 9 * 2 * sin ((5pi)/6) = 9#

Area of slant triangle with length as base,

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

#A_l = (1/2) * 9 * sqrt(2^2 + (2/2)^2) ~~10.0623#

Area of slant triangle with breadth as base,

#A_b = (1/2) * b * sqrt(h^2 + (l/2)^2)#

#A_b = (1/2) * 2 * sqrt(2^2 + 4.5^2) ~~ 4.9244#

Total Surface Area #A_T = A_p + (2 * A_l) + (2 * A_b)#

#A_T = 9 + (2*10.0623) + (2*4.9244) = color (purple)(38.9734)#