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?

1 Answer

73.788\ \text{unit}^2

Explanation:

Area of rombus base with each side 5 & an interior angle \pi/6

=5\cdot 5\sin(\pi/6)=12.5

The rhombus base of pyramid has its semi-diagonals 5\cos(\pi/12) & 5\sin(\pi/12)

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

\sqrt{6^2+(5\cos(\pi/12))^2}=7.702 &

\sqrt{6^2+(5\sin(\pi/12))^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={5+7.702+6.138}/2=9.42

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

=\sqrt{9.42(9.42-5)(9.42-7.702)(9.42-6.138)}

=15.322

Hence, the total surface area of pyramid

=4(\text{area of lateral triangular face})+\text{area of rhombus base}

=4(15.322)+12.5

=73.788\ \text{unit}^2