A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $9$ $9$ and $6$ $6$ and the pyramid's height is $9$ $9$ . If one of the base's corners has an angle of $\frac{5 π}{12}$ $(5pi)/12$ , what is the pyramid's surface area?

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A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $9$ $9$ and $6$ $6$ and the pyramid's height is $9$ $9$ . If one of the base's corners has an angle of $\frac{5 π}{12}$ $(5pi)/12$ , what is the pyramid's surface area?

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Answer 1 · 2018-07-04T08:56:34

$Total Surface Area A_{T} = A_{B} + A_{L} = 52.16 + 145.77 = 197.93$ $color(maroon)("Total Surface Area " A_T = A_B + A_L = 52.16 + 145.77 = 197.93$

Explanation:

https://socratic.org/questions/a-pyramid-has-a-parallelogram-shaped-base-and-a-peak-directly-above-its-center-i-123

$l = 9, b = 6, θ = \frac{5 π}{12}, h = 9$ $l = 9, b = 6, theta = (5pi)/12, h = 9$

$To find the Total Surface Area T S A$ $"To find the Total Surface Area T S A"$

$Area of parallelogram base A_{B} = l b sin θ$ $"Area of parallelogram base " A_B = l b sin theta$

$A_{B} = 9 \cdot 6 \cdot sin (\frac{5 π}{12}) = 52.16$ $A_B = 9 * 6 * sin ((5pi)/12) = 52.16$

$S_{1} = \sqrt{h^{2} + {(\frac{b}{2})}^{2}} = \sqrt{9^{2} + 3^{2}} = 9.49$ $S_1 = sqrt(h^2 + (b/2)^2) = sqrt(9^2 + 3^2) = 9.49$

$S_{2} = \sqrt{h^{2} + {(\frac{l}{2})}^{2}} = \sqrt{9^{2} + {(\frac{9}{2})}^{2}} = 10.06$ $S_2 = sqrt(h^2 + (l/2)^2) = sqrt(9^2 + (9/2)^2) = 10.06$

$Lateral Surface Area A_{L} = 2 \cdot ((\frac{1}{2}) l \cdot S_{1} + (\frac{1}{2}) b \cdot S_{2}$ $"Lateral Surface Area " A_L = 2 * ((1/2) l * S_1 + (1/2) b * S_2$

$A_L = (cancel2 * cancel(1/2)) * (l * S_1 + b * S_2) = (9 * 9.49 + 6 * 10.06)$

$A_L = 145.77$

$color(maroon)("Total Surface Area " A_T = A_B + A_L = 52.16 + 145.77 = 197.93$

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A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $9$ $9$ and $6$ $6$ and the pyramid's height is $9$ $9$ . If one of the base's corners has an angle of $\frac{5 π}{12}$ $(5pi)/12$ , what is the pyramid's surface area?

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A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of $9$ $9$ and $6$ $6$ and the pyramid's height is $9$ $9$ . If one of the base's corners has an angle of $\frac{5 π}{12}$ $(5pi)/12$ , what is the pyramid's surface area?