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Math Help - Homework Help - Area of Trapezium

  1. #1
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    Homework Help - Area of Trapezium

    Struggling with this question and so are my mates.

    Attached is the Graph.

    The line PQ has equation y=2x+4

    (a) Find, without using calculus, the area of the shaded trapezium shown in the diagram.

    (b) Express the area of this trapezium as a definite integral.

    (c) Evaluate this integral

    I just dont know where to go about tackling this question at all, I feel its the english putting me off. If anyone can help me to find the solutions to this question it will help me and assure I know what I am doing when I face similar questions in future.

    Many Thanks

    PS. The Q should be right at the end of the line with P..... the digit close to the P is a 0 not a Q, sorry about that
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  2. #2
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    The area of a trapezoid is:
    A_{\text{trapezoid}} = \frac{h (b_1 +b_2)}{2}

    Where h is the height and b_1 and b_2 are the upper and lower bases.

    You might need to change your perspective. Remember that the bases have to be parallel sides!

    Now for the next question....you have to know that the integral is the area under the curve (or line). Here, you're trying to find the area under the curve between 2 and 5.

    As for question (c): I guess you have to integrate f(x) = 2x + 4
    Last edited by Chop Suey; September 3rd 2008 at 01:19 PM.
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  3. #3
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    Quote Originally Posted by Chop Suey View Post
    The area of a trapezoid is:
    A_{\text{trapezoid}} = \frac{h (b_1 +b_2)}{2}

    Where h is the height and b_1 and b_2 are the upper and lower bases.

    You might need to change your perspective. Remember that the bases have to be parallel sides!
    Or, you can think of the trapezium as a combination of rectangle and a triangle.

    Recall that:
    A_{Triangle} = \frac{1}{2}bh

    A_{Rectangle} = b \cdot h
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  4. #4
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    Quote Originally Posted by Chop Suey View Post
    Or, you can think of the trapezium as a combination of rectangle and a triangle.

    Recall that:
    A_{Triangle} = \frac{1}{2}bh

    A_{Rectangle} = b \cdot h
    So the base is 3. But I dont know the height for the rectangle or the triangle. So unless I'm missing something (which I probably am due to tiredness) I cant manage to work out the area of the triangle nor the rectangle?

    I know what Im doing with and (b) and (c) now though Thanks!
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  5. #5
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    The height of the rectangle is given by f(x) = 2x+4. The height of the triangle is given by f(5) - f(2). Do you understand why?
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  6. #6
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    Quote Originally Posted by Chop Suey View Post
    The height of the rectangle is given by f(x) = 2x+4. The height of the triangle is given by f(5) - f(2). Do you understand why?
    y = 2x+4 is for the line PQ so I dont understand how that becomes the height of the rectangle.
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  7. #7
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    The endpoints of the two segments lie on the line y = 2x + 4. To find the length of these two lines, we find f(5) and f(2).
    Last edited by Chop Suey; September 3rd 2008 at 01:34 PM.
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  8. #8
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    Quote Originally Posted by ajnokia View Post
    y = 2x+4 is for the line PQ so I dont understand how that becomes the height of the rectangle.

    your base goes from 2 to 5

    just find y then x=5
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  9. #9
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    Quote Originally Posted by Chop Suey View Post
    The endpoint of the two lines lie on the line y = 2x + 4. To find the length of these two lines, we find f(5) and f(2).
    gah, ya beat me
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