A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 3 3 and 8 8 and the pyramid's height is 9 9. If one of the base's corners has an angle of (5pi)/125π12, what is the pyramid's surface area?

1 Answer
sankarankalyanam
Jan 24, 2018

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

Explanation:

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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)Ap=3⋅8⋅sin(5π12)=23.1822 sq. units

Area of slant triangle with b as base A_b = (1/2) * b * sqrt (h^2 + (a/2)^2)Ab=(12)⋅b⋅√h2+(a2)2

A_b = (1/2) * 3 * sqrt(9^2 + (8/2)^2) = color(green)(14.7733)Ab=(12)⋅3⋅√92+(82)2=14.7733

Similarly,
A_a = (1/2) * 8 * sqrt (9^2 + (3/2)^2) = color(green)(36.4966)Aa=(12)⋅8⋅√92+(32)2=36.4966

Total Area of the parallelogram based pyramid

A_T = A_p +( 2 * A_b )+ (2 * A_a )AT=Ap+(2⋅Ab)+(2⋅Aa)

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

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