A chord with a length of $18$ $18$ runs from $\frac{π}{4}$ $pi/4$ to $\frac{π}{2}$ $pi/2$ radians on a circle. What is the area of the circle?

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Answer 1 · 2017-05-07T03:27:00

$Area = 162 π (2 + \sqrt{2})$ $"Area" = 162pi(2+sqrt2)$

The chord and radii drawn to each end of the chord form an isosceles triangle. The angle, $θ$ $theta$ , between the two radii is:

$θ = \frac{π}{2} - \frac{π}{4}$ $theta = pi/2-pi/4$

$θ = \frac{π}{4}$ $theta = pi/4$

We can use the Law of Cosines:

$c^{2} = a^{2} + b^{2} - 2 (a) (b) cos (θ)$ $c^2 = a^2 + b^2 - 2(a)(b)cos(theta)$

to find the value of $r^{2}$ $r^2$

Let $c = 18$ $c = 18$ , $a = r$ $a = r$ , and $b = r$ $b = r$

$18^{2} = r^{2} + r^{2} - 2 (r) (r) cos (\frac{π}{4})$ $18^2 = r^2 + r^2 - 2(r)(r)cos(pi/4)$

$18^{2} = 2 r^{2} - 2 r^{2} cos (\frac{π}{4})$ $18^2 = 2r^2 - 2r^2cos(pi/4)$

$r^{2} = \frac{18^{2}}{2 - 2 cos (\frac{π}{4})}$ $r^2 = 18^2/(2-2cos(pi/4)$

$r^{2} = \frac{18^{2}}{2 - \sqrt{2}}$ $r^2 = 18^2/(2-sqrt2)$

$r^{2} = 18^{2} \frac{2 + \sqrt{2}}{2}$ $r^2 = 18^2(2+sqrt2)/2$

$r^{2} = 162 (2 + \sqrt{2})$ $r^2 = 162(2+sqrt2)$

To find the area of a circle, multiply by $π$ $pi$ :

$Area = 162 π (2 + \sqrt{2})$ $"Area" = 162pi(2+sqrt2)$

A chord with a length of 1818 runs from π4pi/4 to π2pi/2 radians on a circle. What is the area of the circle?