These are quarters of a circle cut out of the rectangle, so you need to calculate the surface area of these and deduct them.
The curved corners are the inverse of that, so you need the difference between area of a square and of a quarter of a circle, so then you get the area cut out to make a curved corner and you can deduct that from the rectangle.
As for R, I'm not familiar with this notation, but probably radius of these circles.
You can simplify (π5^2/4)-(π10^2/4) to just simply (π5^2)(1.25) cuz the 5 can be thought of as r and the 10 can be thought of as 2r, so if we have π(2r)^2, you'd get π4r^2. This becomes π4r^2/4 + πr^2/4, the 4/4 cancels each other out giving you a πr^2 + πr^2/4 factored to (πr^2)(1+1/4).
These are quarters of a circle cut out of the rectangle, so you need to calculate the surface area of these and deduct them.
The curved corners are the inverse of that, so you need the difference between area of a square and of a quarter of a circle, so then you get the area cut out to make a curved corner and you can deduct that from the rectangle.
As for R, I'm not familiar with this notation, but probably radius of these circles.
Yes, R = Radius seems likely.
Radius 5 is 5 units.
Solution to 1970 is:
30*20-(π5^2/4)-(π10^2/4)-2(2^2-π2^2)
You can simplify (π5^2/4)-(π10^2/4) to just simply (π5^2)(1.25) cuz the 5 can be thought of as r and the 10 can be thought of as 2r, so if we have π(2r)^2, you'd get π4r^2. This becomes π4r^2/4 + πr^2/4, the 4/4 cancels each other out giving you a πr^2 + πr^2/4 factored to (πr^2)(1+1/4).
And I thought I was over simplify writing "2(2^2-π2^2)" instead of "(2^2-π2^2) - (2^2-π2^2)"
lol
yep, too much thinking to solve this one. LOL