Question

A chord with a length of `5` runs from `pi/4` to `pi/2` radians on a circle. What is the area of the circle?

Answer 1

134.066

Explanation:

As shown in the figure ,the length of the chord AB is 5, which runs from `pi/4` to `pi/2`. This would lie in the 1st Quadrant. Now divide the circumference of the circle in eight equal parts , from 0 to`pi/4`, `pi/4` to `pi/2`, `pi/2` to `(3pi)/4`, `(3pi)/4` to `pi`, `pi` to `(5pi)/4`, `(5pi)/4` to `(3pi)/2`, `(3pi)/2` to `(7pi)/4` and `(7pi)/4` to `2pi`. The chord lengths joining these points on the circumference would all be equal to 5. If all these points are joined the resultant figure would be a regular octagon with side 5, as shown in the figure along side.

Now, let the radius of the circle be 'r'. In triangle OAB, angle O is `360/8` or `45^o`. The base angles A and B would thus be each `67 1/2` degrees.

Draw perpendicular OD from O to side AB. This would bisect AB because OAB is an isosceles triangle. This means AD=BD= 2.5. Since ODB is a rt triangle `r cos67 1/2 = 2.5` . This would give `r= 2.5 sec 67 1/2`

= 6.53

Area of circle would be `pi r^2= 3.14 (6.53)^2`= 134.066