Find the length of largest square

In summary, the conversation discusses finding the length of the largest square that can be accommodated in a right angled triangle with perpendicular length of 16 m and base's length of 8 m. The answer derived through the concept of similarity is 16/3, but it is acknowledged that there may be other ways to fit a square in the triangle. After further discussion and calculations, it is determined that the correct answer is 16√5/7.
  • #1
Kaustubh sri
22
0
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.
 

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  • #2
Kaustubh sri said:
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.
 
  • #3
Dick said:
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.
 

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  • #4
Kaustubh sri said:
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.
 
  • #5
Dick said:
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,
 
  • #6
Kaustubh sri said:
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.
 
  • #7
haruspex said:
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
 

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  • #8
Kaustubh sri said:
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.
 
  • #9
My mind is not working in finding the length in this way , can u suggest how to find the length
 
  • #10
Kaustubh sri said:
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?
 
  • #11
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
ehild said:
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.
 
  • #13
You are welcome:biggrin:

ehild
 
  • #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?
 
  • #15
Satvik Pandey said:
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}##:tongue2:

ehild
 
  • #16
ehild said:
##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
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:
  • #17
Satvik Pandey said:
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 } ##


It is correct now! :thumbs:

ehild
 
  • #18
ehild said:
It is correct now! :thumbs:

ehild
Thank you ehild for helping me.:smile:
 

1. What is the purpose of finding the length of the largest square?

The purpose of finding the length of the largest square is to determine the maximum possible size of a square that can be inscribed within a given shape or area.

2. How do you find the length of the largest square?

The length of the largest square can be found by using the Pythagorean theorem, which states that the square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.

3. What information do you need to find the length of the largest square?

In order to find the length of the largest square, you will need to know the dimensions of the given shape or area, such as its length and width. You may also need to know the coordinates of its vertices or the length of its diagonal.

4. Can the length of the largest square be found for any shape or area?

No, the length of the largest square can only be found for certain shapes and areas, such as rectangles, triangles, and circles. It is not possible to find the length of the largest square for irregular shapes or areas with curved boundaries.

5. Why is it important to find the length of the largest square?

Knowing the length of the largest square is important in various fields, including architecture, engineering, and mathematics. It can help determine the maximum amount of space that can be utilized within a given area, and can also be used in problem solving and optimization scenarios.

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