Is it possible to adjust the sizes of square and circle inside a unit circle so the combined area is half of that unit circle. A square and a circle fits together inside an unit circle symmetrically. Is it possible to adjust the sizes of circle and the square so that their combined area is half of the area of that unit circle ?
I tried it like this : As the circle is of unit radius so radius is equal to 1 and so its area is irrational. Now let the length of the square be $a$ and that radius of circle be $r$  and now the question becomes is it possible such that $a^2$+$πr^2$ = $π/2$ since the combined area is half of the unit circle. There I got stuck as RHS is irrational . Any help would be appreciated.
 A: Trying to Proving the Existence ( Answering the Question in the title ) :
Where-ever we take the Centre $X$ of the Inner Circle , we can get the Combined total Area by Drawing the Square beside it.
We can move the Point $X$ from right Extreme to left Extreme.
In the Image , we can Draw the largest Square ( Blue ) inside the Circle & then Draw the Small Circle ( Blue ) beside it. Here $X$ is right Extreme.
Area of outer Circle is $\pi \approx 3.14$
Here , the Combined Blue total Area is more than half the outer Circle.
Diagonal of Square is $2$ , Area is $2$ , Combined Area will be a little more than $2$.
Now , the Purple Circle can be almost the whole Outer Circle, with a very small Purple Square. Here $X$ is left Extreme.
Hence , Combined Purple total Area is more than half the outer Circle.

Combined Area varies continuously from more than half to some minimum to more than half. Is there a Point $X$ where Combined Area is Exactly half ?
If YES , Where is that Point $X$ ?
We can figure that out by making a Equation.
Let the Inner Circle have Diameter $x$ , & the Square have Side $y$.
Combined Area is $ y^2 + \pi x^2 / 4 = \pi / 2 $ [[ EQ1 ]]

The left most Part $ z = 1 - \sqrt {1-y^2/4} $
Hence the Outer Circle Diameter is $ 2 = x+y+z $
$ 2 = x + y + 1 - \sqrt {1-y^2/4} $
$ x+y-1 = \sqrt {1-y^2/4} $
$ (x+y-1)^2 = 1-y^2/4 $ [[ EQ2 ]]
Solving  [[ EQ1 ]] &  [[ EQ2 ]] will give the necessary Point $X$.
Value will be cumbersome , involving square root , $\pi$ , square , Etc.
$ y^2 + \pi x^2 / 4 = \pi / 2 $
$ (x+y-1)^2 = 1-y^2/4 $
$ y^2 = \pi / 2 - \pi x^2 / 4 $
$ y = \sqrt { \pi / 2 - \pi x^2 / 4 } $
$ (x+\sqrt { \pi / 2 - \pi x^2 / 4 }-1)^2 = 1-\sqrt { ( \pi / 2 - \pi x^2 / 4 ) }^2/4 $
$ (x+\sqrt { \pi / 2 - \pi x^2 / 4 }-1)^2 = 1- ( \pi / 2 - \pi x^2 / 4 )/4 $
$ (x-1)^2+ ( \pi / 2 - \pi x^2 / 4 ) + 2(x-1)\sqrt { \pi / 2 - \pi x^2 / 4 } = 1- ( \pi / 2 - \pi x^2 / 4 )/4 $
$ 2(x-1)\sqrt { \pi / 2 - \pi x^2 / 4 } =  -(x-1)^2 - ( \pi / 2 - \pi x^2 / 4 ) + 1- ( \pi / 2 - \pi x^2 / 4 )/4 $
$ 4(x-1)^2 ( \pi / 2 - \pi x^2 / 4 ) = [ -(x-1)^2 - \pi / 2 + \pi x^2 / 4 + 1 - \pi / 8 + \pi x^2 / 16 ] ^2 $
We can use Wolfram to get the answer $x ≈ 0.995126$ & hence $y ≈ 0.890525$ . We can then verify this :
Combined Area is $ 0.890525^2 + \pi 0.995126 / 4  = 1.57460...$ which is quite close to half the Outer Circle Area.
