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Square of maximum area

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A square of maximum area is inscribed in a semi-circle as shown.
    What percent (rounded to the nearest tenth) of the circle is outside the square.

    Circle is centered at C.


    2. Relevant equations

    None

    3. The attempt at a solution

    I dont know how to begin solving this since i cant get the radius of the circle or the side lengths of the square. A square of maximum area mean that it is the biggest it can possibly be inside the semi-circle without going out, right?
     
  2. jcsd
  3. Nov 8, 2009 #2

    jgens

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    I don't think that you need the actual radius of the circle to solve this problem: Just let the radius have length [itex]r[/itex]. Now, looking at your diagram, extend two line segments of length [itex]r[/itex] from [itex]C[/itex] to the far corners of the square. What do you notice about these new triangles that might help you solve for the side length of the square?
     
  4. Nov 8, 2009 #3
    the hypotenuse of each triangle is r, but the center triangle has two length r's. Is the center equilateral? If it then the length of the bottom sides of each side triangle would be 1/2 r. But how do i prove that the center one is equilateral?

    Also, if the middle one is equilateral, the bottom sides are 1/2 r, making both sides a 30-60-90 triangle making the last side of the two triangles on the side (r *sqrt(3))
     
  5. Nov 8, 2009 #4
    i just looked up properties of squares and it said that if two sides are equal the last side is r√2.

    that wouild mean the bottom sides of both side riangles are (r√2)/2. Which complies with the 30 60 90 triangle but would mean the last side is (r√2*√3) which is r√6.

    But if all that is true, how do i solve for the percentage covered.
     
  6. Nov 8, 2009 #5

    jgens

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    I can't make sense of what you posted, sorry! Alright, from the modified diagram you drew, you should have found that one of the triangles had hypotenuse [itex]r[/itex] and side lengths [itex]s[/itex] and [itex]\frac{s}{2}[/itex] if [itex]s[/itex] is the side length of the square. Can you think of a formula that relates the sides of a right triangle?

    To find the percent covered you need only divide the area of the square and the area of the circle. However, notice that the problem asks to find the percentage that is uncovered.
     
  7. Nov 8, 2009 #6
    [tex]\frac{s}{2}^2 + s^2 = r^2[/tex]

    but i dont see how that helps find the area.
     
  8. Nov 8, 2009 #7

    jgens

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    First, your equation should read:

    [tex]\frac{s^2}{4} + s^2 = s^2(1 + \frac{1}{4}) = r^2[/tex]

    Now, since [itex]s[/itex] is the side length of the square, then [itex]s^2 = A[/itex]. Does that help?
     
  9. Nov 8, 2009 #8
    so the area of each side triangle is [tex]\frac{r^2}{1+\frac{1}{4}}[/tex]

    what about the middle triangle?

    EDIT: wait since i know the middle triangle has a base of S and a height of S cant i just say it is [tex]1/2s^2[/tex]

    but i still dont understand how to use these areas to find the percent covered
     
    Last edited: Nov 8, 2009
  10. Nov 8, 2009 #9
    wait, how do i solve:

    [tex](\pi~r^2) - (2(\frac{r^2}{1+\frac{1}{4}}) + \frac{1}{2}s^2)[/tex]
     
  11. Nov 8, 2009 #10
    augh, i have no idea where im going!!!! please help someone!!!
     
  12. Nov 8, 2009 #11

    jgens

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    Alright, I'm back now! So from my previous hint you should know that [itex]A_{square} = s^2[/itex] and [itex]5s^2 = 4r^2[/itex], where [itex]A_{square}[/itex] is the area of the square, [itex]r[/itex] is the radius length, and [itex]s[/itex] is the side length of the square.

    Now, the area of a circle is given by [itex]A_{circle} = \pi r^2[/itex]. The percentage of the area covered by the square is then [itex]A_{square}/A_{circle}[/itex]. Can you take it from there?
     
  13. Nov 8, 2009 #12
    yea, thanks for all your help.
     
    Last edited: Nov 8, 2009
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