Finding the Length of a Small Square

[dirk_mec1]

Homework Statement

2. Homework Equations [/B]

The Attempt at a Solution

I can find the length of the large square.
The small square is where the trouble starts.

If I look at the part of the circle where the small square is and put a center of a coordinate system at the bottom part of the circle I can setup an equation for the arc from the lowest point on the circle to the lower right corner of the large square.

$$f(x) = R-\sqrt{R^2-x^2}$$

I know that there is some x at which the height up until the large square is twice the x-value:

2x = H - f(x)

with H the length of the lowest point of the circle up until the large square.

$$H = R-Rcos(45)$$

Now if I solve 2x = H - f(x) I get:

$$5x^2 + 2\sqrt{2} Rx + 0.5R^2 = 0$$

Which does not lead to the correct answer. The answer of the length of the small square is
$$\frac{R \sqrt{2}}{5}$$

What am I doing wrong?

 

[Mark44]

Homework Statement

View attachment 209175

2. Homework Equations [/B]

The Attempt at a Solution

I can find the length of the large square.
The small square is where the trouble starts.

If I look at the part of the circle where the small square is and put a center of a coordinate system at the bottom part of the circle I can setup an equation for the arc from the lowest point on the circle to the lower right corner of the large square.

$$f(x) = R-\sqrt{R^2-x^2}$$

I know that there is some x at which the height up until the large square is twice the x-value:

2x = H - f(x)

with H the length of the lowest point of the circle up until the large square.

$$H = R-Rcos(45)$$

Now if I solve 2x = H - f(x) I get:

$$5x^2 + 2\sqrt{2} Rx + 0.5R^2 = 0$$

Which does not lead to the correct answer. The answer of the length of the small square is
$$\frac{R \sqrt{2}}{5}$$

What am I doing wrong?

I don't have time to check your work at the moment. What I would do is use some basic right triangle trig to get the dimensions of the two squares. For simplicity, I would assume that the radius of the circle is 1. Then find the length of a side of the large square. If you know the dimensions of the large square, the length of a side of the small square is 1 - (1/2)(length of a side of the large square). Knowing the length of a side of each square, you can get the ratio of the areas of the two squares.

 

[Dick]

the length of a side of the small square is 1 - (1/2)(length of a side of the large square

Mmm. Can't really agree. It's actually a bit less that that.

 

[dirk_mec1]

I know that there are other methods. I want to know where MY method is going wrong.

 

[mjc123]

Your quadratic equation ought to read 5x² + 2√2Rx - 0.5R² = 0

 

[Mark44]

Mmm. Can't really agree. It's actually a bit less that that.

Yes, you are right -- my mistake. I wasn't taking into account that the distance from the lower edge of the large square to the circle isn't equal to the length of a side of the smaller square.

 

[dirk_mec1]

Your quadratic equation ought to read 5x² + 2√2Rx - 0.5R² = 0

True this was a typo but I still do not get the right answer.

 

[SammyS]

True this was a typo but I still do not get the right answer.

When I solve the corrected quadratic equation:

$5x^2 + 2\sqrt{2} Rx - 0.5R^2 = 0$​

I get the correct result.

 

[dirk_mec1]

Yes I figured it out. When I solve it I need to multiply with 2 to get the length!