A chord with a length of #5 # runs from #pi/12 # to #pi/6 # radians on a circle. What is the area of the circle?

Question

A chord with a length of #5  # runs from #pi/12    # to #pi/6  # radians on a circle. What is the area of the circle?

Maverick Smith · Accepted Answer

The area of the circle is #=1152.5u^2#

The angle subtended at the centre of the circle is

#hat(AOB)=theta=pi/6-pi/12=2/12pi-1/12pi=1/12pi#

The length of the chord is

#AB=5#

#AC=5/2=2.5#

#sin(theta/2)=(AC)/r#

The radius of the circle is

#r=(AC)/sin(theta/2)=2.5/sin(1/2*1/12pi)=2.5/sin(1/24pi)=19.2u#

The area of the circle is

#area=pir^2=pi*19.2^2=1152.5u^2#

Caroline Aucoin · Answer

undefined To find the area of the circle, we need to first determine the radius of the circle. The chord length (5 units) is equal to the straight-line distance between two points on the circumference of the circle. This chord forms an isosceles triangle with the radius of the circle, where the central angle is $ \frac{\pi}{6} - \frac{\pi}{12} = \frac{\pi}{12} $ radians.

Using trigonometric properties, we can find the radius (r) of the circle using the formula for the radius of an isosceles triangle:

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

where $ c $ is the length of the chord and $ \theta $ is the central angle.

Substituting the given values:

$$ r = \frac{5}{2 \sin\left(\frac{\pi/12}{2}\right)} $$

$$ r = \frac{5}{2 \sin\left(\frac{\pi}{24}\right)} $$

$$ r = \frac{5}{2 \cdot \sin\left(\frac{\pi}{24}\right)} $$

$$ r = \frac{5}{2 \cdot \left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)} $$

$$ r = \frac{20}{\sqrt{6} - \sqrt{2}} $$

Now, we can use the formula for the area (A) of a circle:

$$ A = \pi r^2 $$

Substituting the value of $ r $:

$$ A = \pi \left(\frac{20}{\sqrt{6} - \sqrt{2}}\right)^2 $$

$$ A = \pi \left(\frac{20}{\sqrt{6} - \sqrt{2}}\right) \cdot \left(\frac{20}{\sqrt{6} - \sqrt{2}}\right) $$

$$ A = \pi \cdot \frac{400}{6 - 2\sqrt{3} - 2\sqrt{3} + 2} $$

$$ A = \pi \cdot \frac{400}{8 - 4\sqrt{3}} $$

$$ A = \pi \cdot \frac{100}{2 - \sqrt{3}} $$

$$ A = \frac{100\pi}{2 - \sqrt{3}} $$