How can you halve the area of a circle by an arch on the circumferenc of of the circle?
We have a circle,its origine O and its radius=r,A is point on the circumference of this circle,Find R,such that if the second circle whose radius = R,and origin A is drawn and cross the first circle at B and C,the area of ABOC is 1/2 of the area of the first circle.
Work backwards. You know that for ABOC to be half the circle, BC must be a diameter of circle O.
Now, note that AB = AC = R, so triangles ABO and ACO are similar and angle AOB is a right angle.
Finally, R² = r² + r², so R = √(2r²) = r√2.
Work backwards. You know that for ABOC to be half the circle, BC must be a diameter of circle O.
Now, note that AB = AC = R, so triangles ABO and ACO are similar and angle AOB is a right angle.
Finally, R² = r² + r², so R = √(2r²) = r√2.
References :
It usually helps to draw a picture:
http://www.flickr.com/photos/dwread/2647641726/
Note that ABOC is the area bounded by the arcs AB and AC and line segments BO and OC.
BO = OC = r
Since ABOC contains half the first circle, BOC must be a diameter.
Then AOB is a right triangle with legs BO = AO = r, and R is the hypotenuse.
By the Pythagorean Theorem,
R² = r² + r²
R = √(2r²) = r√2
References :