7. A chord AB of a circle makes an angle theta with the diameter passing through A. If the area of the minor segment is one-quarter the area of the circle, show that sin2theta=pi/2 - 2theta. Sovle this equation graphically.
8. A chord AB of a circle subtends an angle theta at the centre of the circl. If the perimeter of the minor segment is one-half the circumference of the circle, show that 2*sin(theta/2)=pi-theta. Solve this equation graphically.
First, make sure to draw out the relevant diagram, and label anything that's easy to figure out. Labeling the origin as 'O', and joining a line between O and B, we find that
As they are both the radius of the circle. As such,
As the base angles in an isosceles triangles are equal. Therefore, since the angle sum of a triangle is pi radians,
Using the area of a minor segment formula, we know that (for subtending angle alpha)
Subbing in the angle we found above, the area of the minor segment is going to be
As the area of the segment is a quarter of the area of the circle, we note that
Play around with this last line, and you'll quickly get the desired result. Then, plot the graph of sin(2theta) and pi/2 - 2theta. Find out where they intersect; it should be around 0.42.
I'm working on the second question now.
The second question is probably much easier; we know that the length of an arc will be
As we need the perimeter of the minor segment, so we also need the length of the chord AB. We can imagine that half the length is like a right angle triangle, with angle theta/2 and hypotenuse r. Therefore, the total length of the chord AB will be
The total perimeter is therefore
We know that the perimeter is half the circumference of the circle. Therefore,
Again, if you play around with this line, you'll get the desired result. Draw it graphically, you should get a number around 1.7