1. ## Chord problem

Hey
1.a chord of length 6 cm is drawn in a circle of radiue 7cm. Find:
a)the area of the smaller region inside the circle cut off by the chord.

My working out:
Formula i used is ½r^2 (theta-sintheta)
So r=7, theta=?
Find theta.
Sine theta/2 =3/7
=50.75 degrees conversion to radians is 0.89.
So plug out degree and radius into the equation
½ r^2(theta-sintheta)
½r^2(0.89-sin0.89)
=21.42 ??, the actual answer is 2.73 cm^2

my diagram is below

2. Using the chord formula as you done, we should get the angle theta as being

$\displaystyle 14sin(\frac{\theta}{2})=6$

$\displaystyle {\theta}=2sin^{-1}(\frac{3}{7})\approx .885822 \;\ \text{radians}$ or about 50.75 degrees.

Then, using the formula for the area of a circular segment gives:

$\displaystyle \frac{1}{2}(7)^{2}(2sin^{-1}(\frac{3}{7})-sin(2sin^{-1}(\frac{3}{7})))\approx 2.73$

3. Originally Posted by galactus
Using the chord formula as you done, we should get the angle theta as being

$\displaystyle 14sin(\frac{\theta}{2})=6$

$\displaystyle {\theta}=2sin^{-1}(\frac{3}{7})\approx .885822 \;\ \text{radians}$ or about 50.75 degrees.

Then, using the formula for the area of a circular segment gives:

$\displaystyle \frac{1}{2}(7)^{2}(2sin^{-1}(\frac{3}{7})-sin(2sin^{-1}(\frac{3}{7})))\approx 2.73$
hey thats the exact thing i did, but in radians, i plugged your formula into the calc and it says 1224.4 ???

4. Make sure your calculator is not set to degrees even though you're working in radians. Alas, something is wrong there somewhere.

That is your problem. I just checked. Your calculator is set to degrees.

5. YES! my calc is in degree mode .. thanks

6. Originally Posted by smmmc
Hey
1.a chord of length 6 cm is drawn in a circle of radiue 7cm. Find:
a)the area of the smaller region inside the circle cut off by the chord.

My working out:
Formula i used is ½r^2 (theta-sintheta)
So r=7, theta=?
Find theta.
Sine theta/2 =3/7
=50.75 degrees conversion to radians is 0.89.
So plug out degree and radius into the equation
½ r^2(theta-sintheta)
½r^2(0.89-sin0.89)
=21.42 ??, the actual answer is 2.73 cm^2

my diagram is below
Using the Cosine Rule, we find

$\displaystyle b^2 = a^2 + c^2 - 2ac \cos{\theta}$

$\displaystyle 6^2 = 7^2 + 7^2 - 2\cdot 7\cdot 7 \cdot \cos{\theta}$

$\displaystyle 36 = 49 + 49 - 98\cos{\theta}$

$\displaystyle 36 = 98 - 98\cos{\theta}$

$\displaystyle 98\cos{\theta} = 62$

$\displaystyle \cos{\theta} = \frac{31}{49}$

$\displaystyle \theta = \arccos{\frac{31}{49}}$.

As long as the angle is in degrees, the Area of the sector is given by

$\displaystyle A_{\textrm{sector}} = \frac{\theta}{360} \pi r^2$

$\displaystyle = \frac{1}{360}\cdot \arccos{\frac{31}{49}} \cdot \pi \cdot 7^2$

$\displaystyle = \frac{49\pi}{360} \cdot \arccos{\frac{31}{49}}$.

The Area of the triangle is found using Heron's Formula.

$\displaystyle s = \frac{a + b + c}{2}$

$\displaystyle = \frac{7 + 7 + 6}{2}$

$\displaystyle = \frac{20}{2}$

$\displaystyle = 10$.

$\displaystyle A_{\textrm{triangle}} = \sqrt{s(s - a)(s - b)(s - c)}$

$\displaystyle = \sqrt{10(10 - 7)(10 - 7)(10 - 6)}$

$\displaystyle = \sqrt{10\cdot 3 \cdot 3 \cdot 4}$

$\displaystyle = \sqrt{360}$

$\displaystyle = 6\sqrt{10}$.

So the Shaded Area is given by

$\displaystyle A = \frac{49\pi}{360} \cdot \arccos{\frac{31}{49}} - 6\sqrt{10} \,\textrm{units}^2$