In the following right $\Delta ABC$, $AC = BC = 1$ and $DEF$ is an arc of a circle with center $A$. Suppose the shaded areas $BDE$ and $CEF$ are equal. Then $AD = \frac{x}{\sqrt{x}}.$ Find $x$ .
What I Tried: Here is a picture :-
No idea for this. I cannot understand how I can get the position of point $E$ given that the area of both the red sections are the same. I cannot think of any idea to get the areas either since they are formed from arc of the circle.
Can anyone give me a small idea of how to start this? Thank You.
Edit: So the hint from @Student1058 , I think I have actually solved the problem but getting stuck somewhere.
We see that the area of $\Delta ABC$ is same as the area of the sector $ADF$.
We have :- $$\rightarrow [\Delta ABC] = \frac{(1 * 1)}{2} = 0.5$$
Also we have the central angle of the sector to be $45^\circ$ as this is an isosceles right-angled triangle.
So area of a sector given its radius and central angle would be :-
$$\rightarrow \frac{(r^2 * 45)}{2} = 0.5$$
$$\rightarrow r^2 = \frac{1}{45}$$
$$\rightarrow r = AD = \frac{1}{\sqrt{45}} = \frac{1}{3\sqrt{5}}$$
But the question says the answer should be of the form of $\frac{x}{\sqrt{x}}$ , where did I go wrong?
Best Answer
$$[\triangle ABC]=(ADF)$$
$$\frac{1\times1}{2}=\frac{\pi (\sqrt{x})^2 }{8}$$
$$x=\frac{4}{\pi}$$
Area of a sector with $\alpha$ is $\frac{{\pi} r^2\times \alpha }{360}$