Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `1` and `8` and the pyramid's height is `2`. If one of the base's corners has an angle of `(5pi)/12`, what is the pyramid's surface area?

Answer 1

`28.527\ \text{unit}^2`

Explanation:

Area of parallelogram base with sides `1` & `8` & an interior angle `{5\pi}/12`

`=1\cdot 8\sin({5\pi}/12)=7.727`

The parallelogram shaped base of pyramid has its semi-diagonals

`1/2\sqrt{1^2+8^2-2\cdot 1\cdot 8\cos({5\pi}/12)}=3.9` &

`1/2\sqrt{1^2+8^2-2\cdot 1\cdot 8\cos({7\pi}/12)}=4.157`

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

`\sqrt{2^2+(3.9)^2}=4.383` &

`\sqrt{2^2+(4.157)^2}=4.613`

There are two pairs of opposite identical triangular lateral faces of pyramid. One pair of two opposite triangular faces has the sides `1, 4.383` & `4.613` and another pair of two opposite triangular faces has the sides `8, 4.383` & `4.613`

1) Area of each of two identical triangular lateral faces with sides `1, 4.383` & `4.613`

semi-perimeter of triangle, `s={1+4.383+4.613}/2=4.998`

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

`=\sqrt{4.998(4.998-1)(4.998-4.383)(4.998-4.613)}`

`=2.175`

2) Area of each of two identical triangular lateral faces with sides `8, 4.383` & `4.613`

semi-perimeter of triangle, `s={8+4.383+4.613}/2=8.498`

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

`=\sqrt{8.498(8.498-8)(8.498-4.383)( 8.498 -4.613)}`

`=8.225`

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(2.175)+2(8.225)+7.727`

`=28.527\ \text{unit}^2`