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# C2 radians question (difficult) watch

1. Ok so I am completely stuck, any idea on how I would approach this question?

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2. If you can take away the area of two quarter circles from the entire square, do you see how you'll be left with the area of two of the white parts? So via that you could get the area of 4 white parts, and take that away from the entire square to get the shaded area.

Now how do you get the area of the two quarter circles? Well you know the formula for the area of a circle. You just need the radius.

Hint for that: Look at the diagonal of the square.. What does that represent?
3. (Original post by Student403)
If you can take away the area of two quarter circles from the entire square, do you see how you'll be left with the area of two of the white parts? So via that you could get the area of 4 white parts, and take that away from the entire square to get the shaded area.

Now how do you get the area of the two quarter circles? Well you know the formula for the area of a circle. You just need the radius.

Hint for that: Look at the diagonal of the square.. What does that represent?
Can you draw it out for me? I am bad at visualising, also you use Pythagoras to get the radius, but what about area of triangle? Also confused on "white parts" what does that mean?
4. (Original post by TSRforum)
Can you draw it out for me? I am bad at visualising, also you use Pythagoras to get the radius, but what about area of triangle? Also confused on "white parts" what does that mean?
Sorry it's a bit hard to draw. White parts are the 4 unshaded bits on the corners.

No need to calculate any triangle areas?

Try re-reading it now that you know what "white parts" I'm referring to
5. Perhaps split the problem into smaller parts. This can be done by making forming a sector of 45 degrees and drawing a triangle inside of it. I'll leave you to find the radius but this shaded area is 1/8th of the problem. hope this helps. The side lengths of the trangle are 4 as the circles meet in the centre.

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6. You are right to use Pythagoras for the radius.

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