Question

A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of `7` and `5` and the pyramid's height is `4`. If one of the base's corners has an angle of `pi/3`, what is the pyramid's surface area?

Answer 1

`A=40sin(pi/3)+7sqrt(16+25/4sin^2(pi/3))+5sqrt(28.25)~=93.05`

Explanation:

Let's first try and visualize the pyramid. It would look something like this from an angled top-down perspective:

Let's start with the area of the parallelogram (the base). It looks like this: To be able to solve for the area of this parallelogram, we need to solve for it's height, `h`. Using a bit of trigonometry, we can express the following equation:
`sin(pi/3)=h/5`
If we solve for h, we get that it is equal to:
`h=5sin(pi/3)~=4.33`

The area of a parallelogram is equal to `base*height`, so if we plug in `7` and `5sin(pi/3)` we get:
`A_(parall\e\l\ogram)=7*5sin(pi/3)=40sin(pi/3)~=34.64`

Now that we have the base finished, let's have a look at the triangles in the upper part of the pyramid. We will start with the ones marked in blue:
To do this, we will need to solve for the height of the triangle, `h`, in the picture above. To do this we will consider a triangle on the inside of the pyramid:
The bottom left corner is the center of the base parallelogram in the pyramid, so the side with length `4` is the height of the pyramid, the side with length `5/2sin(pi/3)`, is half the width we calculated from before (since the distance from the edge to the center point in the parallelogram is half the width of it), and the hypotenuse with length `h`, is the same `h` in the blue triangle.

Using the pythagorean theorem, `a^2+b^2=c^2`, we can express `h` in the following way:
`4^2+(5/2sin(pi/3))^2=h^2`
Solving for `h`, we get:
`h=sqrt(4^2+(5/2sin(pi/3))^2)=sqrt(16+25/4sin^2(pi/3))~=4.55`

Now we can calculate the area of one of the blue triangles. It will be equal to the height of the blue triangle multiplied by `7`, all over `2`:
`A_(blue)=7/2sqrt(16+25/4sin^2(pi/3))~=15.92`
Let's then multiply this by two to get the area of both of the blue triangles:
`2A_(blue)=7sqrt(16+25/4sin^2(pi/3))~=31.84`

Now, we can go through a similar process to solve the red triangles. This time we will consider a triangle like this:
Here, `h` is equal to the height of the red triangle, `4` is once again equal to the height of the pyramid, and `3.5` is the distance from the center point to the edge (the side to side distance was `7`, so this length is half that, `3.5`). Using the same process as the blue triangle, we get this expression for `h`:
`h=sqrt(4^2+3.5^2)=sqrt(28.25)~=5.32`

We can once again just multiply this `h` by `5` and then divide by `2` to get the area of a red triangle:
`A_(red)=5/2sqrt(28.25)~=13.29`

Then we need to multiply by two because we have `2` red triangles:
`2A_(red)=5sqrt(28.25)~=26.58`

Now all we need to to is add all of the areas together:
`A=A_(parall\e\l\ogram)+2A_(blue)+2A_(red)`

`A=40sin(pi/3)+7sqrt(16+25/4sin^2(pi/3))+5sqrt(28.25)~=93.05`