# A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is 9 , its base's sides have lengths of 4 , and its base has a corner with an angle of (5 pi)/6 . What is the pyramid's surface area?

##### 1 Answer
Oct 18, 2017

Total Surface Area $= \ast 85.7944 \ast$

#### Explanation:

AB = BC = CD = DA = a = 4
Height OE = h = 9
OF = a/2 = 2
$E F = \sqrt{E {O}^{2} + O {F}^{2}} = \sqrt{{h}^{2} + {\left(\frac{a}{2}\right)}^{2}} = \sqrt{{9}^{2} + {2}^{2}} = \sqrt{85}$

Area of $D C E = \left(\frac{1}{2}\right) \cdot a \cdot E F = \left(\frac{1}{2}\right) \cdot 4 \cdot \sqrt{95} = 19.4936$
Lateral surface area $= 4 \cdot \Delta D C E = 4 \cdot 19.4936 = 77.9744$

$\angle C = \frac{5 \pi}{6} , \angle \frac{C}{2} = \frac{5 \pi}{12}$
diagonal $A C = {d}_{1}$ & diagonal $B D = {d}_{2}$
$O B = {d}_{2} / 2 = B C \cdot \sin \left(\frac{C}{2}\right) = 4 \cdot \sin \left(\frac{5 \pi}{12}\right) = 3.8637$
$O C = {d}_{1} / 2 = B C \cos \left(\frac{C}{2}\right) = 4 \cdot \cos \left(\frac{5 \pi}{12}\right) = 1.0353$

Area of base ABCD $= \left(\frac{1}{2}\right) \cdot {d}_{1} \cdot {d}_{2} = \left(\frac{1}{2}\right) \left(2 \cdot 3.8637\right) \left(2 \cdot 1.0353\right) = 8$

Total Surface Area = Lateral surface area + Base area. T S A  77.7944 + 8 = 85.7944# 