1. ## Square and Circle

I'm trying to figure out the length of a side of a square:
A square is inscribed in a circle of radius r. Express the length of the side of the square in terms of r.
I tried to work it out this way:

$x^2+x^2=r^2$ since the the lines insribed on the insides of a square are right angles right?

$2x^2=2r^2$

$x^2=r^2$

$x=r$

but the answer says r doesn't = x so what am I doing wrong here and how can I get the right answer?

2. Originally Posted by Kataangel
I'm trying to figure out the length of a side of a square:

I tried to work it out this way:

$x^2+x^2=r^2$ since the the lines insribed on the insides of a square are right angles right?

$2x^2= 2r^2$

$x^2=r^2$

$x=r$

but the answer says r doesn't = x so what am I doing wrong here and how can I get the right answer?
You are close:

$x^2 + x^2 = r^2$
is equal to this: $2x^2 = r^2$

which can be changed to
$x^2 = \dfrac{r^2}{2}$
&
$x = \sqrt{ \dfrac{r^2}{2} }$

3. Originally Posted by Kataangel
I'm trying to figure out the length of a side of a square:
A square is inscribed in a circle of radius r. Express the length of the side of the square in terms of r.
Take notice that the diagonals of the square have length $2r$, a diameter.
So $2s^2 = 4r^2 \, \Rightarrow \,s = r\sqrt 2$

4. Originally Posted by Kataangel
I'm trying to figure out the length of a side of a square:

I tried to work it out this way:

$x^2+x^2=r^2$ since the the lines insribed on the insides of a square are right angles right?

$2x^2=2r^2$

$x^2=r^2$

$x=r$

but the answer says r doesn't = x so what am I doing wrong here and how can I get the right answer?
Hi Kataangel,

I interpreted it this way.

A square is inscribed in a circle of radius r. Express the length of the side x of the square in terms of r.

Draw r from the center of the circle to a vertex on the square.

Drop a perpendicular from the center of the circle to the side of the square.

You have formed a 45-45-90 triangle with r as the hypotenuse.

If x represents a side of the square, then x/2 will represent the sides of the isosceles triangle.

Use the Pythagorean Theorem to represent r in terms of x.

$r^2=\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\r ight)^2$

$r^2=\frac{x^2}{4}+\frac{x^2}{4}$

$r^2=\frac{2x^2}{4}$

$r^2=\frac{x^2}{2}$

$r=\frac{x}{\sqrt{2}}$

$x=r\sqrt{2}$

5. Originally Posted by Plato
Take notice that the diagonals of the square have length $2r$, a diameter.
So $2s^2 = 4r^2 \, \Rightarrow \,s = r\sqrt 2$
Yes, Plato, much easier your way.

6. Ahh... I love this forum! =)

I've been drooling over this prolbem for over 30 minutes, and couldn't have solved it without help from you guys. Thanks a lot!