To find the shaded area ithts all it syas need hlep

You have not supplied an adequate amount of details to even begin to help you. Write in detail the instructions contained in the problem.
 
Your picture contains NO measurements....don't you think that things like the radius (or diameter) of the circle might be important? Or, the angle formed by the two radii?

We can't possibly help you without knowing ALL information you're given for the problem (but, having the diagram is nice).
 
Hello, arthurhernandez123!

Find the shaded area.
Code: 
            *   *
        *:::::::::::
    A * *::::::::::::
      | \    *:::::::*
      |   \     *:::::
    x |     \    *:::*
      |       \   :::
      |         \ *:*
      + - - - - - *
    O      x      B
Click to expand...

I assume we have a quarter-circle with radius \(\displaystyle x\),
. . and a semicircle with diameter \(\displaystyle AB\).

Since \(\displaystyle AB\,=\,\sqrt{2}x\), the radius of the semicircle is is: \(\displaystyle \frac{x}{\sqrt{2}}\)
. . and the area of the semicircle is: \(\displaystyle \,A1 \:=\:\frac{1}{2}\pi\left(\frac{x}{\sqrt{2}}\right)^2 \:=\:\frac{1}{4}\pi x^2\)


The area of the quarter-circle is: \(\displaystyle \:\frac{1}{4}\pi x^2\)
The area of \(\displaystyle \Delta AOB\) is: \(\displaystyle \:\frac{1}{2}x^2\)
. . Hence, the area of the segment is: \(\displaystyle \,A_2\;=\;\frac{1}{4}\pi x^2\,-\,\frac{1}{2}x^2\)


Therefore, the area of the shaded region is:
. . \(\displaystyle A_1\,-\,A_2\;=\;\frac{1}{4}\pi x^2\,-\,\left(\frac{1}{4}\pi x^2\,-\,\frac{1}{2}x^2\right) \;=\;\L\frac{1}{2}x^2\)


Amazing! . . . The area of the "lune" equals the area of the right triangle.

 
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