A chord with a length of #4 # runs from #pi/12 # to #pi/3 # radians on a circle. What is the area of the circle?

Question

A chord with a length of #4 # runs from #pi/12    # to #pi/3   # radians on a circle. What is the area of the circle?

Noah Atkinson · Accepted Answer

Area of the circle #A_c = color (purple)(25.1327)##

Given : Chord length #d = 4#

#delta = pi/3 - pi/12 = pi/4#

#R = d / (2 * sin delta) = 4 / (2 * sin (pi/4)) = 2sqrt2#

Area of circle #A_c = pi R^2 = pi * (2sqrt2)^2 = 8pi = color (purple)(25.1327)#

Ella Armstrong · Answer

undefined The length of a chord in a circle and the angle it subtends at the center of the circle can be used to find the radius of the circle. Once the radius is known, the area of the circle can be calculated using the formula for the area of a circle.

Given the chord length $ L = 4 $ and the angle subtended by the chord $ \theta = \frac{\pi}{3} - \frac{\pi}{12} = \frac{\pi}{4} $, we can use the formula:

$$ L = 2r \sin\left(\frac{\theta}{2}\right) $$

Solve for $ r $, the radius of the circle:

$$ r = \frac{L}{2 \sin\left(\frac{\theta}{2}\right)} $$

Substitute the given values:

$$ r = \frac{4}{2 \sin\left(\frac{\pi/4}{2}\right)} $$

$$ r = \frac{4}{2 \sin\left(\frac{\pi}{8}\right)} $$

$$ r = \frac{4}{2 \sin\left(22.5^\circ\right)} $$

$$ r = \frac{4}{2 \sin\left(\frac{\pi}{8}\right)} $$

$$ r = \frac{4}{2 \cdot \frac{\sqrt{2 + \sqrt{2}}}{2}} $$

$$ r = \frac{4}{\sqrt{2 + \sqrt{2}}} $$

$$ r = \frac{4\sqrt{2 - \sqrt{2}}}{2} $$

$$ r = 2\sqrt{2 - \sqrt{2}} $$

Now, use the radius to find the area of the circle:

$$ A = \pi r^2 $$

$$ A = \pi (2\sqrt{2 - \sqrt{2}})^2 $$

$$ A = 4\pi(2 - \sqrt{2}) $$