A chord with a length of #3 # runs from #pi/12 # to #pi/2 # radians on a circle. What is the area of the circle?

Answer 1

Area of the circle is #19.07# sq.unit.

Chord length #L= 3 # unit

Angle subtended at center by the chord is

#c= (pi/2-pi/12)= 90-15=75^0#
We know chord length #L = 2r sin (c/2) ;r# is radius .
#:. 3 =2*r* sin (75/2) :. r = 3/ (2*sin37.5)~~2.464 # unit.
Area of the circle is #A_c= pi*r^2 = pi* (2.464)^2 ~~19.07# sq.unit

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Answer 2

To find the area of the circle, we need to first find its radius.

The chord length (3) is equal to twice the radius multiplied by the sine of half the angle between the endpoints of the chord.

Let ( r ) be the radius and ( \theta ) be the angle between the endpoints of the chord.

Given:

  • Chord length, ( c = 3 )
  • Angle between the endpoints of the chord, ( \theta = \frac{\pi}{2} - \frac{\pi}{12} = \frac{5\pi}{12} )

Using the chord formula: [ c = 2r \sin\left(\frac{\theta}{2}\right) ]

[ 3 = 2r \sin\left(\frac{5\pi}{24}\right) ]

Solving for ( r ): [ r = \frac{3}{2\sin\left(\frac{5\pi}{24}\right)} ]

Once we have the radius, we can find the area of the circle using the formula: [ A = \pi r^2 ]

[ A = \pi \left(\frac{3}{2\sin\left(\frac{5\pi}{24}\right)}\right)^2 ]

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Answer from HIX Tutor

When evaluating a one-sided 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 one-sided 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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When evaluating a one-sided 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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