Find the length of largest square

[Kaustubh sri]

Find the length of largest square accommodated 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 anybody tell me whether i am right or wrong.

 

[Dick]

Find the length of largest square accommodated 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 anybody tell me whether i am right or wrong.

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

 

[Kaustubh sri]

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

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.

 

[Dick]

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.

Sounds right. I think 16/3 is correct.

 

[Kaustubh sri]

Sounds right. I think 16/3 is correct.

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

 

[haruspex]

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

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.

 

[Kaustubh sri]

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.

Suppose the square is in this case then

 

[haruspex]

Suppose the square is in this case then

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

 

[Kaustubh sri]

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

 

[Mentallic]

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

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?

 

[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...

 

[Kaustubh sri]

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...

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

 

[ehild]

You are welcome

ehild

 

[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?

 

[ehild]

I tried this.

As $\triangle ACB\sim \triangle QCP$So $\frac { 8 }{ x } =\frac { 8\sqrt { 5 } }{ a }$or $\frac { 8 }{ x } =\frac { 8\sqrt { 5 } }{ { x }^{ 2 }+{ y }^{ 2 } }$

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

ehild

 

[Satvik Pandey]

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

ehild

Yes I did a silly mistake.Sorry
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 $a=\frac { 16\sqrt { 5 } }{ 7 }$

 

[ehild]

Yes I did a silly mistake.Sorry
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 $a=\frac { 16\sqrt { 5 } }{ 7 }$

It is correct now! :thumbs:

ehild

 

[Satvik Pandey]

It is correct now! :thumbs:

ehild

Thank you ehild for helping me.