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High School Geometry Help

  1. Jan 23, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the area of the shaded region. Leave your answer in terms of pi.
    The circle has a right triangle inside of it, with the leg length of 18ft.

    Area of a shaded region of a circle = Area of a sector of a circle minus the Area of the triangle.

    2. Relevant equations


    Area of a shaded region = the measure of the arc divided by 360 multiplied by the area of a circle.

    Area of a circle = the product of pi and the square of the radius.

    Area of a triangle = the product of the base and the height divided by half.
    3. The attempt at a solution

    Area of a shaded region= 90/360= multiplied by pi18squared.
    = .250 multiplied by pi324
    = 81pi

    Area of a triangle = half (18 multiplied by 18)
    = half (324)
    =162ft

    Some the end equation should be 81pi minus 162=?

    The problem is I don't know how to get the answer in terms of pi.

    So how would I write this in terms of pi.


    Thanks for the help
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 23, 2007 #2
    Its hard to tell without a picture, but I think you want the area of the whole circle minus the triangle
     
  4. Jan 23, 2007 #3
    It looks like you want the area of a shaded region like that shown here:
    [​IMG]

    Your solution is correct (just make sure you write ft2, "square feet" instead of just "feet" for area). Your answer IS in terms of pi. You're done! "In terms of pi" just means you should describe the answer with pi, which is what you've done. (For example, an expression in terms of x might be 5x - 3, or 6x2 - 12x - 9.)

    If they did not want the answer in terms of pi but instead wanted a decimal number, you would use 3.14... in place of pi and multiply everything out (81*3.14 - 162) to get somewhere around 92.5 which is the same answer in a different form. :)
     
    Last edited: Jan 23, 2007
  5. Jan 23, 2007 #4
    Oh! Well now I feel a little foolish. I apreciate the help, so thank you .
    Just have to label the problem and then I can sleep!!
     
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