- #1
John O' Meara
- 325
- 0
Show that there is a square with a diagonal that is between r and 2r and an area that is half the area of a circle of radius r.
The Theorem: If f is continuous on a closed interval [a,b] and k is any number between f(a) and f(b), inclusive, then there is at least one number x in the interval [a,b] such that f(x)=k.
Let l be a length of a side of the square, then the diagonal's length is \sqrt{2}l \mbox{ and } r < \sqrt{2}l < 2r \mbox{ that implies } f(a)=f(r) = r^2 \mbox{ and } f(b)=f(2r)= 4r^2. Can I find k = \frac{\pi}{2}r^2? I just cannot visualize k on the graph of l^2 and in between f(a) and f(b), because it is impossible. So where am I going wrong? I'm doing this for my own interest. Please help. Thank you.
The Theorem: If f is continuous on a closed interval [a,b] and k is any number between f(a) and f(b), inclusive, then there is at least one number x in the interval [a,b] such that f(x)=k.
Let l be a length of a side of the square, then the diagonal's length is \sqrt{2}l \mbox{ and } r < \sqrt{2}l < 2r \mbox{ that implies } f(a)=f(r) = r^2 \mbox{ and } f(b)=f(2r)= 4r^2. Can I find k = \frac{\pi}{2}r^2? I just cannot visualize k on the graph of l^2 and in between f(a) and f(b), because it is impossible. So where am I going wrong? I'm doing this for my own interest. Please help. Thank you.
