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Math Help - Equations with Geometry

  1. #1
    Member Rimas's Avatar
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    Equations with Geometry

    The square with vertices (-1,-1)(1,-1)(-1,1)(1,1) is ciut by the line
    y=x/2+1
    Inro a triangle and a pentagon. What is the number of square units in the area of the pentagon? express your answer as a decimal to th nearest hundreth
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Rimas View Post
    The square with vertices (-1,-1)(1,-1)(-1,1)(1,1) is ciut by the line
    y=x/2+1
    Inro a triangle and a pentagon. What is the number of square units in the area of the pentagon? express your answer as a decimal to th nearest hundreth
    Look at the figure below.

    The trick is that the area of the pentagon is the area of the square minus the area of the triangle.

    So the question is, can you find the area of the triangle? or do you need help...
    Attached Thumbnails Attached Thumbnails Equations with Geometry-square.jpg  
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  3. #3
    Member Rimas's Avatar
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    Quote Originally Posted by Quick View Post
    Look at the figure below.

    The trick is that the area of the pentagon is the area of the square minus the area of the triangle.

    So the question is, can you find the area of the triangle? or do you need help...
    I need help with the area of the triangle to
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  4. #4
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by Rimas View Post
    I need help with the area of the triangle to
    Alright, notice that it's a right triangle, so its area is 1/2 base times height.

    That's good, but what's the base length?

    (by the way, for this problem, I'm calling the base the vertical side)

    To do that, you need to figure out for what value of y does x equal -1?

    So: y=\frac{x}{2}+1

    Substitute: y=\frac{-1}{2}+1

    Then: y=\frac{1}{2}

    Now how long is the base? Well it's 1-\frac{1}{2}=\frac{1}{2} because that is the distance from the vertex to the point of intersection.

    Can you find height?
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