finding surface area of square pyramid, given height, volume

[ocgirl]

I have a square pyramid that has a height of 5ft. The volume of the pyramid is 60ft³. I need to find the surface area of it.

I know how to get the volume if I have the a side of the square base. I'm not sure how to get the the sides if I have the volume. I think I need the volume for this question right? I'm not really sure how to approach it. Can someone help me?

 

[stapel]

If you can find the volume, given the height and the length of a side of the base, then you must have some formula and/or process.

Just use that formula/process in reverse, working backwards from the volume and the height, to find the length of a side of the base.

Once you have the height and the length of a side of the base, plug these into whatever formulas they've given you to find the surface area.

If you get stuck, please reply showing what you have tried, and providing the formulas and/or processes you are expected to use.

Thank you.

Eliz.

___________
Edit: Ne'mind. The answer has been posted below.

 

[soroban]

Re: help help help

Hello, ocgirl!

You have to work backwards . . . step by step . . . carefully.

I have a square pyramid that has a height of 5 ft.
The volume of the pyramid is 60 ft³.
I need to find the surface area of it.

We know that the volume of the pyramid is: \(\displaystyle \,V\;=\;\frac{1}{3}Bh\)
\(\displaystyle \;\;\)where \(\displaystyle B\) is the area of the base, \(\displaystyle h\) os the height.

We know the height and the volume.
\(\displaystyle \;\;\)So we have: \(\displaystyle \:60\;=\;\frac{1}{3}(B)(5)\;\;\Rightarrow\;\;B\,=\,36\)
Hence, the base is a 6-by-6 square . . . right?

The sides are four isosceles triangles with base 6 and unknown height.

Look at a cross-section of the pyramid:

            *
           /:\
          / : \
         /  :  \h
        /   :5  \
       /    :    \
      * - - + - - *
               3

Note: This is not a triangular face of the pyramid.
\(\displaystyle \;\;\;\;\;\)It is a cross-section through the center of the pyramid.

The hypotenuse \(\displaystyle h\) will be the height of the triangular faces.
We have: \(\displaystyle \,h\:=\:\sqrt{3^2\,+\,5^2}\:=\:\sqrt{34}\)

Now visualize the triangular faces.
They have a base of \(\displaystyle 6\) and a height of \(\displaystyle \sqrt{34}.\)
They have an area of: \(\displaystyle \frac{1}{2}(6)(\sqrt{34})\:=\:3\sqrt{34}\)

The four triangular faces have an area of: \(\displaystyle 4\,\times\,3\sqrt{34}\:=\:12\sqrt{34}\) ft².

If the base-area is to be included, add \(\displaystyle \,6^2\:=\:36\) ft².