Surface area and Volume

[vampirewitchreine]

I'm not really sure how to post this as a question, but here's the problem:

I've got a figure that's a triangular prism. The base of it is 9 (the base of the triangle) and the height of the figure is 4. The triangles are all equilateral, making the rectangles that form the rest of the figure 9 by 4. I've determined that the height of the triangle to be about 7.79(Some really long number, it's the square root of 60.75). I know that the volume of my figure is about 140.3 (I checked my answer in the back of the book).

What I'd like to know, is how can I find the surface area of my figure?


I'm not really sure how to do this with this type of figure, but I do understand how to do rectangular, hexagonal, octagonal, and cylindrical prisms.

 

[soroban]

Hello, vampirewitchreine!

I don't understand your difficulty.

I've got a a triangular prism.
The base is an equilateral triangle of side is 9; the height is 4.
Find the surface area of the prism.

I'm not really sure how to do this with this type of figure,
but I do understand how to do rectangular, hexagonal, octagonal, and cylindrical prisms.
This is the part I don't understand.
You can solve it if the base is a rectangle, hexagon, octagon, or a circle.
But you can't solve it when the base is a triangle ??

An equilateral triangle of side \(\displaystyle x\) has area: .\(\displaystyle A \:=\:\dfrac{\sqrt{3}}{4}\,x^2\) .**
. . and we have two such triangles.

We also have three \(\displaystyle 9\times4\) rectangles.

Go for it!


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** .Proof

              *
             /|\
            / | \
           /  |  \
        x /   |h  \
         /    |    \
        /     |     \
       /      |      \
      * - - - * - - - *
         x/2

We have a right triangle with sides \(\displaystyle h,\:\frac{x}{2},\,\text{ and }x.\)

Pythagorus says: .\(\displaystyle h^2 + \left(\dfrac{x}{2}\right)^2 \:=\:x^2\)

Then: .\(\displaystyle h^2 + \dfrac{x^2}{4} \:=\:x^2 \quad\Rightarrow\quad h^2 \:=\:\dfrac{3x^2}{4} \)

Hence: .\(\displaystyle h \:=\:\sqrt{\dfrac{3x^2}{4}} \:=\:\dfrac{\sqrt{3}}{2}x\)

Therefore: .\(\displaystyle \text{Area} \:=\:\frac{1}{2}\text{(base)(height)} \:=\:\frac{1}{2}(x)\left(\frac{\sqrt{3}}{2}x\right) \:=\:\dfrac{\sqrt{3}}{4}\,x^2\)