Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `3` and `2` and the pyramid's height is `8`. If one of the base's corners has an angle of `(3pi)/8`, what is the pyramid's surface area?

Answer 1

`45.939\ \text{unit}^2`

Explanation:

Area of parallelogram base with sides `3` & `2` & an interior angle `{3\pi}/8`

`=3\cdot 2\sin({3\pi}/8)=5.543`

The parallelogram shaped base of pyramid has its semi-diagonals

`1/2\sqrt{3^2+2^2-2\cdot 3\cdot 2\cos({3\pi}/8)}=1.45` &

`1/2\sqrt{3^2+2^2-2\cdot 3\cdot 2\cos({5\pi}/8)}=2.097`

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

`\sqrt{8^2+(1.45)^2}=8.13` &

`\sqrt{8^2+(2.097)^2}=8.27`

There are two pairs of opposite identical triangular lateral faces of pyramid. One pair of two opposite triangular faces has the sides `3, 8.13` & `8.27` and another pair of two opposite triangular faces has the sides `2, 8.13` & `8.27`

1) Area of each of two identical triangular lateral faces with sides `3, 8.13` & `8.27`

semi-perimeter of triangle, `s={3+8.13+8.27}/2=9.7`

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

`=\sqrt{9.7(9.7-3)(9.7-8.13)(9.7-8.27)}`

`=12.079`

2) Area of each of two identical triangular lateral faces with sides
`2, 8.13` & `8.27`

semi-perimeter of triangle, `s={2+8.13+8.27}/2=9.2`

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

`=\sqrt{9.2(9.2-2)(9.2-8.13)(9.2-8.27)}`

`=8.119`

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(12.079)+2(8.119)+5.543`

`=45.939\ \text{unit}^2`