A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #5 # and #5 # and the pyramid's height is #8 #. If one of the base's corners has an angle of #pi/4#, what is the pyramid's surface area?

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Answer 1 · 2018-06-10T08:55:29

#color(indigo)("Total Surface Area of pyramid " A_T = 101.5 " sq units"#

#"Total Surface Area of Pyramid " (A_T) = "Base Area of parallelogram " (A_p) + "Lateral Surface Area of Pyramid "( A_L)#

#"Since base = 5, it's a rhombus"#

#A_p = b^2 sin theta = 5^2 * sin (pi/4) = 17.68#

#L S A " " A_L = 4 * (1/2) b * sqrt((b/2)^2 + h^2)#

#A_L = 2 * 5 * sqrt((5/2)^2 + 8^2) = 83.82#

#color(indigo)("Total Surface Area " A_T = A_p + A_L = 17.68 + 83.82 = 101.5 " sq units"#