# 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 6 , its base's sides have lengths of 5 , and its base has a corner with an angle of ( pi)/6 . What is the pyramid's surface area?

$73.788 \setminus \setminus {\textrm{u n i t}}^{2}$

#### Explanation:

Area of rombus base with each side $5$ & an interior angle $\setminus \frac{\pi}{6}$

$= 5 \setminus \cdot 5 \setminus \sin \left(\setminus \frac{\pi}{6}\right) = 12.5$

The rhombus base of pyramid has its semi-diagonals $5 \setminus \cos \left(\setminus \frac{\pi}{12}\right)$ & $5 \setminus \sin \left(\setminus \frac{\pi}{12}\right)$

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

$\setminus \sqrt{{6}^{2} + {\left(5 \setminus \cos \left(\setminus \frac{\pi}{12}\right)\right)}^{2}} = 7.702$ &

$\setminus \sqrt{{6}^{2} + {\left(5 \setminus \sin \left(\setminus \frac{\pi}{12}\right)\right)}^{2}} = 6.138$

Each of 4 identical lateral triangular faces of pyramid has the sides 5, 7.702 \ & \ 6.138

semi-perimeter of triangle, $s = \frac{5 + 7.702 + 6.138}{2} = 9.42$

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

$= \setminus \sqrt{9.42 \left(9.42 - 5\right) \left(9.42 - 7.702\right) \left(9.42 - 6.138\right)}$

$= 15.322$

Hence, the total surface area of pyramid

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

$= 4 \left(15.322\right) + 12.5$

$= 73.788 \setminus \setminus {\textrm{u n i t}}^{2}$