# A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of 8  and 2  and the pyramid's height is 9 . If one of the base's corners has an angle of (5pi)/12, what is the pyramid's surface area?

May 3, 2018

color(maroon)("Total Surface Area " = color(purple)(A_T = A_B + A_L = color(crimson)(15.45 + 94.12 = 109.57

#### Explanation:

$l = 8 , b = 2 , \theta = \frac{5 \pi}{12} , h = 9$

$\text{To find the Total Surface Area T S A}$

$\text{Area of parallelogram base } {A}_{B} = l b \sin \theta$

${A}_{B} = 8 \cdot 2 \cdot \sin \left(\frac{5 \pi}{12}\right) = 15.45$

${S}_{1} = \sqrt{{h}^{2} + {\left(\frac{b}{2}\right)}^{2}} = \sqrt{{9}^{2} + {1}^{2}} = 9.06$

${S}_{2} = \sqrt{{h}^{2} + {\left(\frac{l}{2}\right)}^{2}} = \sqrt{{9}^{2} + {4}^{2}} = 10.82$

"Lateral Surface Area " A_L = 2 * ((1/2) l * S_1 + (1/2) b * S_2

${A}_{L} = \left(\cancel{2} \cdot \cancel{\frac{1}{2}}\right) \cdot \left(l \cdot {S}_{1} + b \cdot {S}_{2}\right) = \left(8 \cdot 9.06 + 2 \cdot 10.82\right)$

${A}_{L} = 94.12$

$\text{Total Surface Area } {A}_{T} = {A}_{B} + {A}_{L} = 15.45 + 94.12 = 109.57$