A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 2 2 and 9 9 and the pyramid's height is 7 7. If one of the base's corners has an angle of (5pi)/65π6, what is the pyramid's surface area?

1 Answer
sankarankalyanam
Feb 20, 2018

T S A color(blue)(A_T = A_b + A_L = 9 + 80.1033 = 89.1033AT=Ab+AL=9+80.1033=89.1033

Explanation:

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Area of base A_b = l * b * sin theta = 2 * 9 * sin ((5pi)/6) = 9Ab=l⋅b⋅sinθ=2⋅9⋅sin(5π6)=9

s_1 = sqrt ((b/2)^2 + h^2) = sqrt ((9/2)^2 + 7^2) = 8.3217s1=√(b2)2+h2=√(92)2+72=8.3217

s_2 = sqrt ((l/2)^2 + h^2) = sqrt((2/2)^2 + 7^2) = 7.0711s2=√(l2)2+h2=√(22)2+72=7.0711

Lateral Surface Area

A_L = 2 ( (1/2) l * s_1 + (1/2) b * s_2)AL=2((12)l⋅s1+(12)b⋅s2)

A_L = (2 * 8.3217 + 9 * 7.0711) = 80.1033AL=(2⋅8.3217+9⋅7.0711)=80.1033

T S A A_T = A_b + A_L = 9 + 80.1033 = 89.1033AT=Ab+AL=9+80.1033=89.1033

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