By varying the arbitrary inner radius an equally varied range of areas can be encompassed by the shaded area.
A simple exercise with pencil and compass confirms this.
If the ends of the 10cm line are A & B , its mid point is D and the centre of the circles is C ;
a generalised formula for the shaded area might read : ( pi times the square of the inner radius CD times the angle ACD over 360) minus 5CD square cm.

If R and r are the larger and smaller circle radii, they form a right triangle with half of the 10 cm line. R^2 = r^2 + 5^2
So the orange area is (R^2 – r^2) pi which is (5^2) pi or 25 pi cm^2

25 pi cm

yup

The area is surely in cm^2.

Since the radius of the inner circle is arbitrary, make it zero, so the 10cm measurement is the orange circle’s diameter.

Now that IS a smart approach!

Wouldn’t that make it half of a 10cm diameter circle’s area, which is to say 25/2 pi cm^2?

In the other answers, it’s stated as 25pi, am I missing something?

That would make it half the area of a 10cm diameter circle, which would be 25/2 pi cm^2, but you state the answer as 25 pi. Am I missing something?

Nevermind, I WAS missing something.

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Is the answer pi(10)^2?

25pi

25*pi

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By varying the arbitrary inner radius an equally varied range of areas can be encompassed by the shaded area.

A simple exercise with pencil and compass confirms this.

If the ends of the 10cm line are A & B , its mid point is D and the centre of the circles is C ;

a generalised formula for the shaded area might read : ( pi times the square of the inner radius CD times the angle ACD over 360) minus 5CD square cm.

Addendum to the above.

The angle ACD is found by Tan ACD = 5/AD

Sorry , I misled myself by seeking to find only the area of the arc above the 10cm line . I’m an idiot.

If R and r are the larger and smaller circle radii, they form a right triangle with half of the 10 cm line. R^2 = r^2 + 5^2

So the orange area is (R^2 – r^2) pi which is (5^2) pi or 25 pi cm^2