Calc triangles

L

Lizzie

Full Member
Joined
Sep 8, 2005
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317
Problem:
Use calculus to find the area of the triangle with vertices at:
(0,0), (2,1), and (3,0)

This is probably very simple...what I did was draw out the triangle and measure everything. I got A=3/2.

I am pretty sure there is a more calculus-ish way to do this. Any help would be appreciated. I'm sure this is just something I overlooked or wasn't thinking about.
 
Hello, Lizzie!

Use calculus to find the area of the triangle with vertices at: A(0,0), B(2,1), and C(3,0)
Click to expand...
Code: 
               B    
        |     (2,1)
        |      *
        |     /:\
        |   / 1: \
        | /    :  \
    ----*------+----*--
      (0,0)       (3,0)
        A           C
We already know the answer: \(\displaystyle \;A\;=\;\frac{1}{2}(3)(1)\;=\:\frac{3}{2}\)

But they want to see if we can set up the Calculus problem . . .


The line through \(\displaystyle A(0,0)\) and \(\displaystyle B(2,1)\) is: \(\displaystyle \;y\:=\:\frac{1}{2}x\)
. . The area under that line segment is: \(\displaystyle \L\;A_1\;=\;\int^{\;\;\;\;2}_0\frac{1}{2}x\,dx \;= \;1\)

The line through \(\displaystyle B(2,1)\) and \(\displaystyle C(3,0)\) is:\(\displaystyle \;y\;=\;-x\,+\,3\)
. . The area under that line segment is: \(\displaystyle \L\;A_2\;=\;\int^{\;\;\;\;3}_2(-x\,+\,3)\,dx\;=\;\frac{1}{2}\)

Therefore, the total area is: \(\displaystyle \L\;A\;=\;A_1\,+\,A_2\;=\;1\,+\,\frac{1}{2}\;=\;\frac{3}{2}\)
 
I wasn't asking for the answer...I just needed help getting started...sorry.
 
Lizzie said:
I wasn't asking for the answer...I just needed help getting started...sorry.
Click to expand...
Don't apologize -- you didn't ask for the fully-worked solution, and when you've had some idea how to start, you've shown that work.

But next time (since I think you want to figure out how to do these things for yourself, which is a good thing), it might be helpful if you specified that you're just needing help getting started.

Eliz.
 
Lizzie said:
I wasn't asking for the answer...I just needed help getting started...sorry.
Click to expand...
No worries. Really, I'm just curious what part was most beneficial. I'm always a little curious when the problem is presented with things like, "This is probably really easy, but..." Of course it's easy if you know how to do it and there is a simple solution. :wink:
 
Well, I knew the solution...I just didn't have a clue how to arrive at it using calculus. That was what I needed help with. I posted that I knew the solution was 3/2 already. I got there by sketching it and taking the area, but I knew that wasn't what was being asked of me.
 
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