Concentric Circle & Square

Our aim is to find a side length of the square in terms of the radius of the circle so we can use the (given) fact that $\text{Area of square=Area of circle}$ .

First off let the radius of the circle be $r$ and let the altitude from $O$ to \ $EF$ meet $EF$ at $X$ . For brevity let a side length of the square be $s$ . Since $\triangle{OEF}$ is isosceles it follows that $EX=FX$ . Applying the Pythagorean theorem to $\triangle{OEX}$ we have:

$s=2\sqrt{r^2-\left(\frac{\sqrt{1600-400\pi}}{2}\right)^2}$

Since we are interested in the area of the square we can square this equation and set it equal to the area of the circle to obtain:

$s^2=4\left[r^2-\left(\frac{\sqrt{1600-400\pi}}{2}\right)^2\right]=4r^2-1600+400\pi=\pi{r}^2$

$\Leftrightarrow{r^2(4-\pi)=400(4-\pi)}$ .

Moreover $(4-\pi)$ is positive so we can divide it out leaving:

$r^2=400$ and thus, $r=\boxed{20}$ . (here we obviously took the positive value of $r$ since it is a length.)

Let the radius be $r$ . Then $\pi \times r^2 = AB^2$ , or $AB = r \sqrt{\pi}$ . EOF is an isosceles triangle with the height from $O = \frac {AB} {2} = \frac {r \sqrt{\pi}} {2}$ . Using Pythagoreans Theorem, we get $(EF/2)^2 + (AB/2)^2 = r^2$ . Solving this, we get $r = 15$

We extend line FO to meet line CD and let the intersection point of FO and CD be G. We also know $\pi r^2=|ABCD|=(BC)^2 \Rightarrow EG=BC=r\sqrt{\pi}$ , where r is the radius of the circle. We know that the triangle EFG is a right-angle triangle with $\angle FEG=90^\circ$ . So, by Pythagoras Theorem, we have $(EF)^2+(EG)^2-(FG)^2$ $=(\sqrt{1600-400\pi})^2+(r\sqrt{\pi})^2-(2r)^2$ $=1600-400\pi+\pi r^2-4r^2$ . $\Rightarrow r^2=400 \Rightarrow r=20$

Draw the figure..

Drop a perpendicular from O to AB to intersect at P. Let OA intersect circle $\tau$ at Q. Now, let AB=BC=CD=AD= $s$ (side of square)

AE = b,

AQ = a,

AC=BD = d (diagonal of square)= $s$ $\sqrt{2}$

OQ = r,

EF = $\sqrt{1600-400π}$ = 20 $\sqrt{4-π}$

Now, since ABCD and $\tau$ have the same area,

$s^2$ = $πr^2$

$\implies$ s = r $\sqrt{π}$

d=s $\sqrt{2}$ = $r \sqrt{2π}$

Let, c = 1/2 EF = 10 $\sqrt{4-π}$

b+c = $\frac{s}{2}$ = $\frac{r}{2}$ $\sqrt{π}$

$\implies$ b= [ $\frac{r}{2}$ $\sqrt{π}$ ] - [10 $\sqrt{4-π}$ ]

a+r = $\frac{d}{2}$ = $\frac{s}{√2}$

$\implies$ a = r $\sqrt{\frac{π}{2}}$ - r

since we have two secants from A, $b(s-b)$ = $a(d-a)$

[ $\frac{r}{2} \sqrt{π} -10\sqrt{4-π}$ ]*[ $\frac{r}{2} \sqrt{π} +10\sqrt{4-π}$ = [ $r \sqrt{\frac{π}{2}} - r$ ] * [ $r \sqrt{2π} -r\sqrt{\frac{π}{2}}+r$

Solving above equation, $r = \pm 20$

As radius is always positive, $r=20$

given that; EF=(1600-400 pi)^0.5; OE=OF ; area of square=(AB)^2 ; area of circle=pi (OE)^2 ; now (AB)^2=pi (OE)^2 ; hence AB=((pi)^0.5) (OE) ; if we connect the point E and F to the point O we will get a isosceles triangle ,(since OE=OF=radius of the circle.).

now if we draw a normal from point O to the line E F at point K it will bisect the line EF (since triangle is isosceles) and (KF)=(KE)=EF/2 and OK=AB/2; now in triangle OKF ; (OK)^2+(KF)^2=(OF)^2 ; (AB/2)^2+(EF/2)^2=(OF)^2 ; after solving this equation ; we will get ; OF=radius of the circle=20 .

Let the co-ordinates of A [Cartesian co-ordinate system] be (-l, l) Let the co-ordinates of B be (-l, -l). Let the co-ordinates of C be (l, -l). Let the co-ordinates of D be (l ,l).

Then the center of the square is (0, 0). Then the center of the circle is also (0, 0).

Let the radius of the circle be r units, so that the equation of the circle is x^2 + y^2= r^2.

Now the equation of AB is y= l.

So, at the points where the circle touches AB...

x^2 + l^2= r^2.

From here we get x= +-sqrt(r^2 - l^2).

So, the co-ordinates of E and F are (sqrt(r^2 - l^2), l) and (-sqrt(r^2 - l^2), l) in some order.

So, EF= 2sqrt(r^2 - l^2) units

By condition...

2sqrt(r^2 - l^2)= sqrt(900-225pi)

Or, r^2 - l^2= (900-225pi)/4= 225(1-pi/4).....(1)

But the length of the square is 2l units, and thus its area 4l^2 square units.

So, 4l^2= pi r^2 Or, l^2= pi r^2/4

Putting this value in (1)...

r^2(1-pi/4)= 225(1-pi/4) Or, r= 15 [since r must be positive]

So the radius of the circle is 15 units.

In this solution, I assume that the given length of EF is the length of the straight line EF, not of the arc EF.

Let center of circle be O

Now join O to E (OE will be radius of circle)

As Area of circle= Area of Square

        Pi. (OE^2)  =  (AB^2)

Join F to a point G on CD such that FG and BC are parallel ( FG = BC also).

Extend OE to meet it at point G.

Now EFG is right angle triangle with

EG=Hypotenuse of rt. angle triangle = diameter of circle.

now using pythagoras' theorem in rt. angle triangle EFG

(EG^2) = (EF^2)+(FG^2)

(EG^2) = (EF^2)+(BC^2)

(2OE)^2 = 900-225Pi+Pi(OE^2)

4(OE^2) =225(4-Pi) + Pi(OE^2)

4(OE^2) - Pi(OE^2) =225(4-Pi)

(4-Pi)(OE^2) =225(4-Pi)

OE^2=225

OE=15 is required measurement.

EF is a chord of the circle. Let G be the perpendicular from the radius of the circle O to EF. Then OG $\perp$ EF [A $\perp$ from the center always bisects the chord]

$\Rightarrow \bigtriangleup OGF$ is a right triangle. $\Rightarrow OG^2 + GF^2 = OF^2$

Let us assume the side of the square to be $a$ , the radius of the circle to be $r$ and the length EF (known) to be $x$ for simplicity and more readable math.

The the above equation can be reframed as

$(\frac{x}{2})^2+(\frac{a}{2})^2 = r^2$ $\Rightarrow x^2 + a^2 = 4r^2$

Now since $a^2 = \pi r^2$ ,

$\Rightarrow x^2 + \pi r^2 = 4r^2$ $\Rightarrow x^2 = r^2(4 - \pi)$ $\Rightarrow 225(4 - \pi) = r^2(4 - \pi)$ $\Rightarrow r^2 = 225$ $\Rightarrow r = 15$

Call the midpoint of $EF$ $M$ , the radius of $\Gamma$ $r$ , and the sidelength of $ABCD$ $s$ . We see that $OM = \frac {s}{2}$ , so $EM^2 = r^2 - OM^2 = r^2 - \frac {s^2}{4}$ . Because $M$ is the midpoint of $EF$ , $EF^2 = (2 EM)^2 = 4 (r^2 - \frac {s^2}{4}) = 1600 - 400 \pi$ . So $r^2 (4 - \frac {s^2}{r^2}) = 1600 - 400 \pi$ . We know that $\pi r^2 = s^2$ , so $\frac {s^2}{r^2} = \pi$ . This yields $r^2 (4 - \pi) = 1600 - 400 \pi$ or $r = 20$ .

Draw the diagram of a square and a circle intersecting it ( symmetry is important). Assume the center of the square/circle to be a point O. The line joining O to E and F will be of length R ( Radius of the circle ). The perpendicular from the center to the line segment EF will be of length A/2 [ where A is the side length if the square] . It is obvious that the length EF/2 = \sqrt{ R^2 - (A/2) ^2} . Substitute the value of A^2 in the given equation with pi* R^2 . Equate the value of EF in the given equation with the value of EF given in the question to get the value of R.

Consider triangle OAE. We have $OA=r\sqrt{\frac{\pi}{2}}$ , $OE=r$ , $AE=\frac{AB-EF}{2}=\frac{1}{2}\left({r\sqrt{\pi}-\sqrt{1600-400\pi}}\right)$ . With cosine rule, $OE^2=OA^2+AE^2-2.OA.AE.\cos{45^\circ}$ . After simplification we have $r^2=400-100\pi+\frac{\pi r^2}{4}$ . Thus, $r^2=400 \rightarrow r=20$

given area of circle or radius 'r' = area of square of side 'a' keep it as equation (1) As per the given data draw the figure and join center of circle O with E and F which are nothing but the radii of circle. Now draw a perpendicular from O to EF, of length equals to half of side of square. Now a right triangle is formed with first leg of length equals to half of side of square and the second leg of length equals to half of the length of EF and the hypotenuse is the radius of the circle. Now substitute all the three side lengths in this relation and convert everything in terms of radius 'r' using the given data and then simplify to get radius r = 15 units

Let $r$ be the radius of $\Gamma$ ,which has area $\pi r^2$ . Since the area of the square is equal to the area of the circle, therefore the length of the square is $\sqrt{\pi r^2} = r \sqrt{\pi}$ .

Let $G$ be the midpoint of $AB$ . By symmetry, $G$ is also the midpoint of $EF$ . Since $EF$ is a chord in $\Gamma$ , $OG$ must be perpendicular to $EF$ . Since $O$ is the center of the square, $OG$ is half the length of one of the side lengths of $ABCD$ , thus $OG = \frac{r \sqrt{\pi}}{2}$ .

Looking at triangle $OEG$ we have $OE = r$ , $EG = \frac{\sqrt{1600-400\pi}}{2}$ , $OG = \frac{r \sqrt{\pi}}{2}$ , and $\angle OGE = 90^\circ$ . Therefore by the Pythagorean theorem, we have $OE^2 = OG^2 + GE^2$ $\Rightarrow r^2 = \frac{\pi r^2}{4} + \frac{1600 - 400 \pi}{4}$ $\Rightarrow r^2 = \frac{1600 - 400 \pi}{4 - \pi} = \frac{400(4-\pi)}{4-\pi} = 400$ . Since $r >0$ we take the positive root and get that $r = 20$ .

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