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Find the length of largest square

  1. Sep 10, 2014 #1
    Find the length of largest square accomodated in a right angled triangle whose perpendicular length is 16 m and base's length is 8 m.

    My attempt
    I have tried a lot to find the length but every time it comes in fraction , i think that the figure i have made is wrong
    . The answer I am getting is 16/3,could any body tell me whether i am right or wrong.
     

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  2. jcsd
  3. Sep 10, 2014 #2

    Dick

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    Seems right. How you are getting the answer is maybe more interesting than just the diagram.
     
  4. Sep 10, 2014 #3
    I have only applied the concept of similarity in which triangle ADE Is similar to Triangle ABC and thus from the photo below u could find the value of x.
     

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  5. Sep 10, 2014 #4

    Dick

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    Sounds right. I think 16/3 is correct.
     
  6. Sep 10, 2014 #5
    I know from this way the answer is correct but if the squre in the triangle could be accomodated in any other way then,
     
  7. Sep 11, 2014 #6

    haruspex

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    You are right to consider there may be another way to fit a square in the triangle.
    First, suppose the square does not touch all sides of the triangle. Show that there must be a larger square.
    Now suppose no side of the square is parallel to a side of the triangle. Is there a small change that could be made to increase the size of the square?
    And so on, until you only have two cases to consider.
     
  8. Sep 11, 2014 #7
    Suppose the square is in this case then
     

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  9. Sep 11, 2014 #8

    haruspex

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    Quite so. It remains to compute the size of the square in that arrangement and see which is bigger.
     
  10. Sep 11, 2014 #9
    My mind is not working in finding the length in this way , can u suggest how to find the length
     
  11. Sep 12, 2014 #10

    Mentallic

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    Going by your diagram, EF is parallel to AC and since the gradient of AC = -2, then let F cut the y-axis at y=a. Now you can find where E cuts the x-axis in terms of a, then you can find the length of EF, and so you just need to find an expression in terms of a for the parallel distance between EF and AC. What gradient would a line perpendicular to AC have?
     
  12. Sep 12, 2014 #11

    ehild

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    Find similar triangles again. What is the ratio y/x comparing the small right triangle PQC with the big one, BAC? Also, x2+y2=a2. How are x and y related to a?

    The yellow triangle is also similar to the big one. You can write an equation among the sides...
     

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  13. Sep 12, 2014 #12
    Thanks sir ehild I got it , I was knowing about you from one of my friend #satvik pandey , you are great.Thanks man.
     
  14. Sep 12, 2014 #13

    ehild

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    You are welcome:biggrin:

    ehild
     
  15. Sep 13, 2014 #14
    I tried this.

    As ##\triangle ACB\sim \triangle QCP##


    So ##\frac { 8 }{ x } =\frac { 8\sqrt { 5 } }{ a } ##


    or ##\frac { 8 }{ x } =\frac { 8\sqrt { 5 } }{ { x }^{ 2 }+{ y }^{ 2 } } ##


    or ##x\sqrt { 5\quad } ={ x }^{ 2 }+{ y }^{ 2 }##

    But ##y=2x##

    So ##x=\frac { 8 }{ \sqrt { 5 } } ##

    but ##a=\sqrt { 5 } x##.

    So ##a=8##

    This is larger than 16/3.
    Is this the answer of this question?
     
  16. Sep 13, 2014 #15

    ehild

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    ##a=\sqrt{x^2+y^2}## and y=2x, so your equation transforms to ##\frac { 8 }{ x } =\frac { 8\sqrt { 5 } }{ \sqrt{ x ^ 2 + y ^ 2 } }=\frac { 8\sqrt { 5 } }{ \sqrt{5 x^2 }}=\frac{8}{x}##:tongue2:

    ehild
     
  17. Sep 13, 2014 #16
    Yes I did a silly mistake.Sorry:tongue2:
    As ##\triangle ABC\sim \triangle PFB##

    F is a point of intersection of red line passing through P and AB.

    So ##\frac { 16 }{ a } =\frac { 8\sqrt { 5 } }{ 8-x } ##

    or ##16-2x=\sqrt { 5 } a##

    but ##a=\sqrt { 5 } x##

    so ##x=\frac { 16 }{ 7 } ##

    So ##a=\frac { 16\sqrt { 5 } }{ 7 } ##
     
    Last edited: Sep 13, 2014
  18. Sep 13, 2014 #17

    ehild

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    It is correct now! :thumbs:

    ehild
     
  19. Sep 13, 2014 #18
    Thank you ehild for helping me.:smile:
     
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