help finding surface area of triangular prism (tent)

[ocgirl]

hi,

I need a little help finding the surface area of a tent shaped like a triangular prism. the floor of the tent would be a rectangle with the demensions 4ft by 6ft and the 2 triangular ends would be isosceles triandgles with base lengths of 4ft. The volume is 60ft cubed. I need a formula and a little help starting it. If anyone could help it would be greatly appreciated.

Thank you

 

[skeeter]

volume = (Base area)(Height)

since you know the volume and the Height ...

Base area = volume/Height

Base area = 60/6 = 10 ft²

the "front" of the tent is the Base, and it is also a triangle ...

Base area = (1/2)(base length)(height)

2(Base area)/(base length) = height

2(10)/4 = height

so ... the height of the "front" of the tent is 5 ft

draw in the altitude to the base from the top of the tent to the base ... two congruent right triangles are formed.

use the pythagorean theorem to find the "slant" height of both sides of the tent.
using that value, you can find the area of one slanted side (which is a rectangle 6 ft in length)

finish up ...

total surface area = (front + back areas) + 2(area of a single slanting side) + (area of bottom)

 

[soroban]

Re: help finding surface area

Hello, ocgirl!

There is no one formula for this problem.
You must work your way through it step by step,
$\displaystyle \;\;$using various formulas along the way.

Find the surface area of a tent shaped like a triangular prism.
The floor of the tent would be a rectangle with the demensions 4 ft by 6 ft.
The two triangular ends would be isosceles triangles with base lengths of 4 ft.
The volume is 60 ft³.

I assume you made a sketch . . .

                      *
                  *    \
              *         \x
          *              \
         /:\              *
       x/ : \x        *
       /  :h \    * 6
      * - + - *
          4

The volume is: $\displaystyle \,\text{(area of triangle) }\times\,6 \;=\;60$

Hence: $\displaystyle \,\text{area of triangle } =\:10$ ft²

We know the areas of the two triangular ends;
$\displaystyle \;\;$we want the areas of the two rectangular sides.

The sides are $\displaystyle 6$ by $\displaystyle x$ ft.


The triangular end looks like this:

            *
           /:\
          / : \
         /  :  \x
        /   :h  \
       /    :    \
      * - - + - - *
         2     2

We know that its area is 10 ft²

We know that: $\displaystyle \,A\;=\;\frac{1}{2}bh$

So we have: $\displaystyle \,\frac{1}{2}(4)h\;=\;10\;\;\Rightarrow\;\;h\,=\,5$ ft

Using Pythagorus: $\displaystyle \,x^2\;=\;h^2\,+\,2^2\,=\,5^2\,+\,2^2\,=\,29\;\;\Rightarrow\;\;x\,=\,\sqrt{29}$

Hence, a side rectangle has area: $\displaystyle \,\sqrt{29}\,\times\,6\:=\:6\sqrt{29}$


We have:
$\displaystyle \;\;$two triangles with area: $\displaystyle \,2\,\times\,10\:=\:20$ ft²
$\displaystyle \;\;$two rectangles with area: $\displaystyle \,2\,\times\,6\sqrt{29}\:=\:12\sqrt{29}$ ft²
$\displaystyle \;\;$a base with area: $\displaystyle \,4\,\times\,6\:=\:24$ ft² (if we include the floor)

Total surface area: $\displaystyle \,20\,+\,12\sqrt{29}\,+\,24\:=\:44\,+\,12\sqrt{29}$ ft²


Edit: skeeter's too fast for me . . . and a nice job, too!