A chord with a length of #15 # runs from #pi/4 # to #pi/3 # radians on a circle. What is the area of the circle?
Area of the circle = 10,372.3476
Chord length = 15
Area of the circle =
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To find the area of the circle when a chord with a length of 15 runs from ( \frac{\pi}{4} ) to ( \frac{\pi}{3} ) radians, use the following steps:

Determine the radius of the circle using the chord length and the central angle: ( r = \frac{l}{2 \sin(\frac{\theta}{2})} ) ( r = \frac{15}{2 \sin(\frac{\pi/3  \pi/4}{2})} ) ( r = \frac{15}{2 \sin(\frac{\pi}{12})} ) ( r = \frac{15}{2 \times \frac{\sqrt{6}  \sqrt{2}}{4}} ) ( r = \frac{15 \times 4}{2(\sqrt{6}  \sqrt{2})} ) ( r = \frac{60}{\sqrt{6}  \sqrt{2}} ) ( r = \frac{60(\sqrt{6} + \sqrt{2})}{(\sqrt{6}  \sqrt{2})(\sqrt{6} + \sqrt{2})} ) ( r = \frac{60(\sqrt{6} + \sqrt{2})}{6  2} ) ( r = \frac{60(\sqrt{6} + \sqrt{2})}{4} ) ( r = 15(\sqrt{6} + \sqrt{2}) )

Use the radius to find the area of the circle: ( A = \pi r^2 ) ( A = \pi (15(\sqrt{6} + \sqrt{2}))^2 ) ( A = \pi (225(6 + 2\sqrt{12} + 2) ) ( A = \pi (225(8 + 2\sqrt{3})) ) ( A = \pi (1800 + 450\sqrt{3}) ) ( A ≈ 5647.83 ) square units.
Therefore, the area of the circle is approximately ( 5647.83 ) square units.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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