Square inscribed in circle...

**zoso** · July 19th 2006, 02:21 PM

What is the method for determining the perimeter of a square inscribed in a circle, given the diameter of the circle (4 cm)?

**Quick** · July 19th 2006, 03:02 PM

Originally Posted by zoso

What is the method for determining the perimeter of a square inscribed in a circle, given the diameter of the circle (4 cm)?

The diameter of a circle with perimeter 4 is $\frac{4}{\pi}$ The diameter of the circle is the length from one corner of the square to the opposite corner, in other words it is the hypotenuse of a triangle formed by two sides of the square. The perimeter of the square is the length of a side times 4.. . let's write this out...

$P_{square}=4s$

$d^2=s^2+s^2$

$d=\frac{c}{\pi}$

therefore...

$d^2=2s^2\Longrightarrow\sqrt{\frac{d^2}{2}=s$

substitute: $\sqrt{\left(\frac{c}{\pi}\right)^2\div2}}=s$

$\sqrt{\frac{c^2}{\pi^2}\times \frac{1}{2}}=s$

$\sqrt{\frac{c^2}{2\pi^2}}=s$

$\frac{c}{\pi\sqrt{2}}=s$

therefore

$\frac{4c}{\pi\sqrt{2}}=4s=P_{square}$

rationalize the denominator: $\frac{4c\sqrt{2}}{2\pi}=P_{square}$

Hello, you've discovered my hidden text! hurray for you!
~ $Q\!u\!i\!c\!k$

**malaygoel** · July 19th 2006, 06:24 PM

The diameter of the circle is the length from one corner of the square to the opposite corner,

Just for better understanding of the problem(for Zoso),
Why the above condition has to be true?
How many squares of different area can be inscribed in the circle? Why?
How many squares of different perimeter can be inscribed in the circle? Why?

Keep Smiling
Malay

**Quick** · July 19th 2006, 06:35 PM

Originally Posted by malaygoel

Just for better understanding of the problem(for Zoso),
Why the above condition has to be true?

Right, for Zoso

, the above condition is only true for the largest possible square you can inscribe in a circle. If that doesn't help I'll show Zoso a diagram.

How many squares of different area can be inscribed in the circle? Why?
How many squares of different perimeter can be inscribed in the circle? Why?

Keep Smiling
Malay

It's possible to have practically infinite squares inscribed in a circle, maybe if you set up a size restriction I could help you out more?

**malaygoel** · July 19th 2006, 06:42 PM

It's possible to have practically infinite squares inscribed in a circle, maybe if you set up a size restriction I could help you out more?

Yes, it is possible have infinite squares, but all of them will have the same area.
I think are misinterpreting the word 'inscribe'. I means that the vertices of the square lie on the circumference of the circle.

Right, for Zoso

, the above condition is only true for the largest possible square you can inscribe in a circle. If that doesn't help I'll show Zoso a diagram.

There is no largest square, all of the inscibed squares have the same area.
The above condition is true because the vertex angle of square is 90 degrees.

Keep Smiling
Malay

**Quick** · July 19th 2006, 06:51 PM

Originally Posted by malaygoel

Yes, it is possible have infinite squares, but all of them will have the same area.
I think are misinterpreting the word 'inscribe'. I means that the vertices of the square lie on the circumference of the circle.

I said practically infinite.

There is no largest square, all of the inscibed squares have the same area.
The above condition is true because the vertex angle of square is 90 degrees.

Keep Smiling
Malay

There is but one possible size of a square per circle that has corners that touch the outside of the circle. I've drawn one below (pay no attention to the fact it's a graph program and looks oval)

as you can clearly see in the diagram, the diameter of the circle is the distance between two opposite corners of the square.

**ThePerfectHacker** · July 19th 2006, 07:36 PM

The diameter of the circle is the diagnol of the square.

**zoso** · July 20th 2006, 11:35 AM

Originally Posted by Quick

The diameter of a circle with perimeter 4 is $\frac{4}{\pi}$ The diameter of the circle is the length from one corner of the square to the opposite corner, in other words it is the hypotenuse of a triangle formed by two sides of the square. The perimeter of the square is the length of a side times 4.. . let's write this out...

$P_{square}=4s$

$d^2=s^2+s^2$

$d=\frac{c}{\pi}$

therefore...

$d^2=2s^2\Longrightarrow\sqrt{\frac{d^2}{2}=s$

substitute: $\sqrt{\left(\frac{c}{\pi}\right)^2\div2}}=s$

$\sqrt{\frac{c^2}{\pi^2}\times \frac{1}{2}}=s$

$\sqrt{\frac{c^2}{2\pi^2}}=s$

$\frac{c}{\pi\sqrt{2}}=s$

therefore

$\frac{4c}{\pi\sqrt{2}}=4s=P_{square}$

rationalize the denominator: $\frac{4c\sqrt{2}}{2\pi}=P_{square}$

Hello, you've discovered my hidden text! hurray for you!
~ $Q\!u\!i\!c\!k$

so if the diameter of the circle IS 4 cm, the result would be $16\sqrt {2} cm$ or something near 22 cms for the perimeter?

**Quick** · July 20th 2006, 11:42 AM

Originally Posted by zoso

so if the diameter of the circle IS 4 cm, the result would be $16\sqrt {2} cm$ or something near 22 cms for the perimeter?

It would be...

$\frac{4c\sqrt{2}}{2\pi}=\frac{4(4)\sqrt{2}}{2\pi}= \frac{16\sqrt{2}}{2\pi}=\frac{8\sqrt{2}}{\pi}<br /> \approx<br /> 3.601265265cm$

**zoso** · July 20th 2006, 12:13 PM

k, I see what I did wrong.

**thedarktemplar** · July 21st 2006, 06:25 PM

The diameter is 4cm...not the circumference. So, using pythagoras' theorem, 2 squard+ 2 sqard=8
root 8= 2.828
2.828 x 4= ~11.3

**Quick** · July 21st 2006, 06:32 PM

Originally Posted by thedarktemplar

The diameter is 4cm...not the circumference.

It's a good thing you noticed. This actually makes the equation easier...

Originally Posted by Quick

The diameter of a circle with perimeter 4 is \frac{4}{\pi} The diameter of the circle is the length from one corner of the square to the opposite corner, in other words it is the hypotenuse of a triangle formed by two sides of the square. The perimeter of the square is the length of a side times 4.. . let's write this out...

$P_{square}=4s$

$d^2=s^2+s^2$

$d=\frac{c}{\pi}$

therefore...

$d^2=2s^2\Longrightarrow\sqrt{\frac{d^2}{2}}=s$

expand on that and you'll get...
$\frac{4d}{\sqrt{2}}=4s=P_{square}$

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