Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `8` and `9` and the pyramid's height is `7`. If one of the base's corners has an angle of `(5pi)/6`, what is the pyramid's surface area?

Answer 1

`160.334\ \text{unit}^2`

Explanation:

Area of parallelogram base with sides `8` & `9` & an interior angle `{5\pi}/6`

`=8\cdot 9\sin({5\pi}/6)=36`

The parallelogram shaped base of pyramid has its semi-diagonals

`1/2\sqrt{8^2+9^2-2\cdot 8\cdot 9\cos({5\pi}/6)}=8.211` &

`1/2\sqrt{8^2+9^2-2\cdot 8\cdot 9\cos({\pi}/6)}=2.252`

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

`\sqrt{7^2+(8.211)^2}=10.79` &

`\sqrt{7^2+(2.252)^2}=7.353`

There are two pairs of opposite identical triangular lateral faces of pyramid. One pair of two opposite triangular faces has the sides `8, 10.79` & `7.353` and another pair of two opposite triangular faces has the sides `9, 10.79` & `7.353`

1) Area of each of two identical triangular lateral faces with sides `8, 10.79` & `7.353`

semi-perimeter of triangle, `s={8+10.79+7.353}/2=13.0715`

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

`=\sqrt{13.0715(13.0715-8)(13.0715-10.79)(13.0715-7.353)}`

`=29.409`

2) Area of each of two identical triangular lateral faces with sides
`9, 10.79` & `7.353`

semi-perimeter of triangle, `s={9+10.79+7.353}/2=13.5715`

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

`=\sqrt{13.5715(13.5715-9)(13.5715-10.79)(13.5715-7.353)}`

`=32.758`

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

`=2(\text{area of lateral triangular face of type-1})+2(\text{area of lateral triangular face of type-2})+\text{area of parallelogram base}`

`=2(29.409)+2(32.758)+36`

`=160.334\ \text{unit}^2`