Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `6` and `4` and the pyramid's height is `3`. If one of the base's corners has an angle of `pi/4`, what is the pyramid's surface area?

Answer 1

`51.565\ \text{unit}^2`

Explanation:

Area of parallelogram base with sides `6` & `4` & an interior angle `{\pi}/4`

`=6\cdot 4\sin({\pi}/4)=16.971`

The parallelogram shaped base of pyramid has its semi-diagonals

`1/2\sqrt{6^2+4^2-2\cdot 6\cdot 4\cos({\pi}/4)}=2.125` &

`1/2\sqrt{6^2+4^2-2\cdot 6\cdot 4\cos({3\pi}/4)}=4.635.`

Now, the sides of triangular lateral face of pyramid are given as

`\sqrt{3^2+(2.125)^2}=3.676` &

`\sqrt{3^2+(4.635)^2}=5.521`

There are two pairs of opposite identical triangular lateral faces of pyramid. One pair of two opposite triangular faces has the sides `6, 3.676` & `5.521` and another pair of two opposite triangular faces has the sides `4, 3.676` & `5.521`

1) Area of each of two identical triangular lateral faces with sides `6, 3.676` & `5.521`

semi-perimeter of triangle, `s={6+ 3.676+5.521 }/2=7.5985`

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

`=\sqrt{7.5985(7.5985-6)(7.5985-3.676)(7.5985-5.521)}`

`=9.949`

2) Area of each of two identical triangular lateral faces with sides `4, 3.676` & `5.521`

semi-perimeter of triangle, `s={4+ 3.676+5.521 }/2=6.5985`

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

`=\sqrt{6.5985(6.5985-4)(6.5985-3.676)(6.5985-5.521)}`

`=7.348`

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(9.949)+2(7.348)+16.971`

`=51.565\ \text{unit}^2`