Question

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 `8`, its base has sides of length `9`, and its base has a corner with an angle of `(2 pi)/3`. What is the pyramid's surface area?

Answer 1

`230.328\ \text{unit}^2`

Explanation:

Area of rhombus base with each side `9` & an interior angle `{2\pi}/3`

`=9\cdot 9\sin({2\pi}/3)=70.148`

The rhombus shaped base of pyramid has its semi-diagonals

`9\cos({\pi}/6)=4.5` &

`9\sin({\pi}/6)=7.794`

Now, two unequal lateral edges of each triangular lateral face of pyramid are given as

`\sqrt{8^2+(4.5)^2}=9.179` &

`\sqrt{8^2+(7.794)^2}=11.169`

There are four identical triangular lateral faces of pyramid each has sides `9, 9.179` & `11.169`

Area of each of four identical triangular lateral faces with sides `9, 9.179` & `11.169`
semi-perimeter of triangle, `s={9+9.179+11.169}/2=14.674`

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

`=\sqrt{14.674(14.674-9)(14.674-9.179)(14.674-11.169)}`

`=40.045`

Hence, the total surface area of pyramid (including area of base)

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

`=4(40.045)+70.148`

`=230.328\ \text{unit}^2`

Note:
Height of pyramid is taken as vertical height of pyramid