# 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 5 , its base has sides of length 8 , and its base has a corner with an angle of (2 pi)/3 . What is the pyramid's surface area?

$152.75 \setminus \setminus {\textrm{u n i t}}^{2}$

#### Explanation:

Area of rombus base with each side $8$ & an interior angle $\frac{2 \setminus \pi}{3}$

$= 8 \setminus \cdot 8 \setminus \sin \left(\frac{2 \setminus \pi}{3}\right) = 55.426$

The rhombus base of pyramid has its semi-diagonals $8 \setminus \cos \left(\setminus \frac{\pi}{6}\right)$ & $8 \setminus \sin \left(\setminus \frac{\pi}{6}\right)$ i.e. $4 \setminus \sqrt{3}$ & $4$

Now, the sides of triangular lateral face of pyramid as given as

$\setminus \sqrt{{5}^{2} + {\left(4 \setminus \sqrt{3}\right)}^{2}} = \setminus \sqrt{73} = 8.544$ &

$\setminus \sqrt{{5}^{2} + {\left(4\right)}^{2}} = \setminus \sqrt{41} = 6.403$

Each of 4 identical lateral triangular faces of pyramid has the sides $8 , 8.544$ & $6.403$

semi-perimeter of triangle, $s = \frac{8 + 8.544 + 6.403}{2} = 11.4735$

Now, using heron's formula the area of lateral triangular face of pyramid

$= \setminus \sqrt{11.4735 \left(11.4735 - 8\right) \left(11.4735 - 8.544\right) \left(11.4735 - 6.403\right)}$

$= 24.331$

Hence, the total surface area of pyramid

$= 4 \left(\setminus \textrm{a r e a o f l a t e r a l t r i a n g \underline{a} r f a c e}\right) + \setminus \textrm{a r e a o f \rho m b u s b a s e}$

$= 4 \left(24.331\right) + 55.426$

$= 152.75 \setminus \setminus {\textrm{u n i t}}^{2}$