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 $7$ . 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 · 2018-02-20T05:20:46

T S A $color(blue)(A_T = A_b + A_L = 9 + 80.1033 = 89.1033$

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Area of base $A_b = l * b * sin theta = 2 * 9 * sin ((5pi)/6) = 9$

$s_1 = sqrt ((b/2)^2 + h^2) = sqrt ((9/2)^2 + 7^2) = 8.3217$

$s_2 = sqrt ((l/2)^2 + h^2) = sqrt((2/2)^2 + 7^2) = 7.0711$

Lateral Surface Area

$A_L = 2 ( (1/2) l * s_1 + (1/2) b * s_2)$

$A_L = (2 * 8.3217 + 9 * 7.0711) = 80.1033$

T S A $A_T = A_b + A_L = 9 + 80.1033 = 89.1033$

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)/6(5pi)/6, what is the pyramid's surface area?