Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `3` and `1` and the pyramid's height is `2`. If one of the base's corners has an angle of `(3pi)/8`, what is the pyramid's surface area?

Answer 1

`11.36\ \text{unit}^2`

Explanation:

Area of parallelogram base with sides `3` & `1` & an interior angle `{3\pi}/8`

`=3\cdot 1\sin({3\pi}/8)=2.772`

The parallelogram shaped base of pyramid has its semi-diagonals

`1/2\sqrt{3^2+1^2-2\cdot 3\cdot 1\cos({3\pi}/8)}=1.388` &

`1/2\sqrt{3^2+1^2-2\cdot 3\cdot 1\cos({5\pi}/8)}=1.753`

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

`\sqrt{2^2+(1.388)^2}=2.434` &

`\sqrt{2^2+(1.753)^2}=2.659`

There are two pairs of opposite identical triangular lateral faces of pyramid. One pair of two opposite triangular faces has the sides `3, 2.434` & `2.659` and another pair of two opposite triangular faces has the sides `1, 2.434` & `2.659`

1) Area of each of two identical triangular lateral faces with sides `3, 2.434` & `2.659`

semi-perimeter of triangle, `s={3+2.434+2.659}/2=4.0465`

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

`=\sqrt{4.0465(4.0465-3)(4.0465-2.434)(4.0465-2.659)}`

`=3.078`

2) Area of each of two identical triangular lateral faces with sides
`1, 2.434` & `2.659`

semi-perimeter of triangle, `s={1+2.434+2.659}/2=3.0465`

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

`=\sqrt{3.0465(3.0465-1)(3.0465-2.434)(3.0465-2.659)}`

`=1.216`

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(3.078)+2(1.216)+2.772`

`=11.36\ \text{unit}^2`