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Math Help - Area of a triangle

  1. #1
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    Area of a triangle

    Use calculus to find the area of the triangle with the given vertices: (0,0) (10,1) (-1,2)


    This completely blindsided me. What's going on here?
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  2. #2
    Senior Member eumyang's Avatar
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    First, find the equations of the three lines that go through these points.
    L1 -> the line going through (0, 0) and (-1, 2)
    L2 -> the line going through (0, 0) and (10, 1)
    L3 -> the line going through (-1, 2) and (10, 1)
    Put all equations in slope-intercept form (y = mx + b). I'm going to call these three functions L_1(x), L_2(x), L_3(x) respectively.

    Draw the triangle. From x = -1 to x = 0, you can find the area of the part of the triangle to the left of the y-axis by finding the integral of the difference of the two functions L3(x) - L1(x):
    A = \displaystyle \int_{-1}^0 L_3(x) - L_1(x) \,dx \ldots

    Then, from x = 0 to x = 10, you find the area of the part of the triangle to the right of the y-axis by finding the integral of the difference of the two functions L3(x) - L2(x):
    \ldots +  \displaystyle \int_{0}^{10} L_3(x) - L_2(x) \,dx

    So the complete integral will be
    A = \displaystyle \int_{-1}^0 L_3(x) - L_1(x) \,dx +  \displaystyle \int_{0}^{10} L_3(x) - L_2(x) \,dx
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  3. #3
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    Do you have to use calculus?

    You could just use the area of a triangle \text{Area} =\frac{1}{2}ab\sin(C)
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  4. #4
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    I would break the problem up into two pieces: the area in the triangle between x = -1 and x = 0, and the area in the triangle between x = 0 and x = 10. For each of those triangles, you're going to have to compute the following:

    \int_{x_{\text{min}}}^{x_{\text{max}}}f_{\text{upp  er}}(x)\,dx-\int_{x_{\text{min}}}^{x_{\text{max}}}f_{\text{low  er}}(x)\,dx.

    Then you add the two together to find your total area. Make sense? I should say that this isn't the only way to solve this problem using calculus, but it's the way I would do it.
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  5. #5
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    So the area is 12?

    And yes, it has to be with calculus.
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  6. #6
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    Can you elaborate on that?
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  7. #7
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    I'm still not sure how to get it with this information...I've gotten closer but I don't know what the variables are in your work.
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  8. #8
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    Quote Originally Posted by bobsanchez View Post
    I'm still not sure how to get it with this information...I've gotten closer but I don't know what the variables are in your work.


    You need to find the line segments mentioned in post #2. Are you firmilar with finding y=mx+c given 2 points?
    Last edited by pickslides; August 11th 2010 at 09:14 PM.
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  9. #9
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    No...I mean, that reminds me of y = mx + b, but I don't know that specifically.
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  10. #10
    Senior Member eumyang's Avatar
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    pickslides is talking about the same thing.

    You'll need to find the slope between the two points for each line (this is m), and then you'll need to find the y-intercept (b). Although in two instances you are already given the y-intercept.
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  11. #11
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    I didn't recognize it when it was added together (what form was that?) but I understand y = mx + c. I was just trying to clarify that that was indeed the process to go with.
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  12. #12
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    Quote Originally Posted by bobsanchez View Post
    I didn't recognize it when it was added together (what form was that?) but I understand y = mx + c.
    Hi Bob, that was just a typo, all fixed now. You need to find the line for each of these from post #2

    Quote Originally Posted by eumyang View Post
    First, find the equations of the three lines that go through these points.
    L1 -> the line going through (0, 0) and (-1, 2)
    L2 -> the line going through (0, 0) and (10, 1)
    L3 -> the line going through (-1, 2) and (10, 1)
    Put all equations in slope-intercept form (y = mx + b). I'm going to call these three functions L_1(x), L_2(x), L_3(x) respectively.

    Draw the triangle. From x = -1 to x = 0, you can find the area of the part of the triangle to the left of the y-axis by finding the integral of the difference of the two functions L3(x) - L1(x):
    A = \displaystyle \int_{-1}^0 L_3(x) - L_1(x) \,dx \ldots

    Then, from x = 0 to x = 10, you find the area of the part of the triangle to the right of the y-axis by finding the integral of the difference of the two functions L3(x) - L2(x):
    \ldots +  \displaystyle \int_{0}^{10} L_3(x) - L_2(x) \,dx

    So the complete integral will be
    A = \displaystyle \int_{-1}^0 L_3(x) - L_1(x) \,dx +  \displaystyle \int_{0}^{10} L_3(x) - L_2(x) \,dx
    After that, integrate as explained in post #2.
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