# Find the length of largest square

1. Sep 10, 2014

### Kaustubh sri

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. Sep 10, 2014

### Dick

Seems right. How you are getting the answer is maybe more interesting than just the diagram.

3. Sep 10, 2014

### Kaustubh sri

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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4. Sep 10, 2014

### Dick

Sounds right. I think 16/3 is correct.

5. Sep 10, 2014

### Kaustubh sri

I know from this way the answer is correct but if the squre in the triangle could be accomodated in any other way then,

6. Sep 11, 2014

### haruspex

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.

7. Sep 11, 2014

### Kaustubh sri

Suppose the square is in this case then

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8. Sep 11, 2014

### haruspex

Quite so. It remains to compute the size of the square in that arrangement and see which is bigger.

9. Sep 11, 2014

### Kaustubh sri

My mind is not working in finding the length in this way , can u suggest how to find the length

10. Sep 12, 2014

### Mentallic

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?

11. Sep 12, 2014

### ehild

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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12. Sep 12, 2014

### Kaustubh sri

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

13. Sep 12, 2014

### ehild

You are welcome

ehild

14. Sep 13, 2014

### Satvik Pandey

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?

15. Sep 13, 2014

### ehild

$a=\sqrt{x^2+y^2}$ 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

16. Sep 13, 2014

### Satvik Pandey

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
17. Sep 13, 2014

### ehild

It is correct now! :thumbs:

ehild

18. Sep 13, 2014

### Satvik Pandey

Thank you ehild for helping me.