Question

The circumference of the circle D is and IJ bisects the sides GH and GF is the square HGFD. Which is the value of IJ?

Answer 1

Given the circumgerence of the circle = c= \[2pir\] then \[r = c/2\pi\]
So JG = GI = r/2
By Pythagoras in triangle JGI, \[IJ^2=JG^2+GI^2=2r^2/4=r^2/2\]

Answer 2

so whats the answer?

Answer 3

What is the circumference of the circle? The answer is \[\sqrt{[}(c/2\pi)^2/2]=c/\pi2\sqrt{2}\]

By the way, what program did you use to draw the diagram so neatly?

Answer 4

i use Microsoft pain

Answer 5

*paint*

Answer 6

Thanks - are you happy with my answer? Were you given the circumference of the circle?