Ratio of area of two circles

**shane99** · December 11th 2009, 03:46 PM

A circle with center O intersects another circle with center P at points C and D. if angle COD = 30 and angle CPD = 60, what is the ratio of the area of the circle with center O to the area of the circle with center P?
Answer= 1: (2-√3)
Can someone paint me a picture of this because I don't completely understand this question.

**11rdc11** · December 11th 2009, 04:05 PM

Originally Posted by shane99

A circle with center O intersects another circle with center P at points C and D. if angle COD = 30 and angle CPD = 60, what is the ratio of the area of the circle with center O to the area of the circle with center P?
Answer= 1: (2-√3)
Can someone paint me a picture of this because I don't completely understand this question.

Are you sure it is not?

$2+\sqrt{3}$

**shane99** · December 11th 2009, 04:31 PM

Yes, I am 100% sure. (lol the answer is actually in the back of the book)

**11rdc11** · December 11th 2009, 04:44 PM

Originally Posted by shane99

Yes, I am 100% sure. (lol the answer is actually in the back of the book)

O then I'm not sure but I can tell you how I approached the problem. First draw out the problem.

Then I gave line CD value of 1.

now using this info and that angle cpd is 60 degrees I figure out the radius for the circle with center is 1. Then I used the equation to find area of a circle and got the area to be 1pi.

Now I use the same idea for the other circle. To find the radius of the other circle is a bit harder so I used the law of sines

$\frac{\sin{30}}{1} = \frac{\sin{75}}{r}$

Can you finish up from here

**skeeter** · December 11th 2009, 04:56 PM

Originally Posted by shane99

A circle with center O intersects another circle with center P at points C and D. if angle COD = 30 and angle CPD = 60, what is the ratio of the area of the circle with center O to the area of the circle with center P?
Answer= 1: (2-√3)
Can someone paint me a picture of this because I don't completely understand this question.

connect chord $CD$

$OC = OD = R$

$PC = PD = CD = r$

cosine law ...

$r^2 = R^2 + R^2 - 2R^2\cos(30)$

$r^2 = 2R^2[ 1 - \cos(30)]$

$r^2 = 2R^2\left(\frac{2-\sqrt{3}}{2}\right)$

$r^2 = R^2(2-\sqrt{3})$

$\frac{1}{2-\sqrt{3}} = \frac{R^2}{r^2}$

$\frac{1}{2-\sqrt{3}} = \frac{\pi R^2}{\pi r^2}$

**Gusbob** · December 11th 2009, 05:09 PM

Originally Posted by 11rdc11

Are you sure it is not?

$2+\sqrt{3}$

11rdc11:

$2+\sqrt{3} : 1 = 1: 2- \sqrt{3}$

Your method is also correct, just different ways of expressing the ratio.

**Soroban** · December 11th 2009, 05:45 PM

Hello, shane99!

A circle with center $O$ intersects another circle with center $P$ at points $C$ and $D.$
If $\angle COD = 30^o$ and $\angle CPD = 60^o,$
what is the ratio of the area of the circle $O$ to the area of the circle $P$ ?

Answer: . $1: (2-\sqrt{3})$

Here's the diagram ... without the circles.

Code:

                      C
                      *
             R    *   | *
              *       |   * r
          *           |     *
    O *  30°         d|   60° * P
          *           |     *
              *       |   * r
             R    *   | *
                      *
                      D

The radius of circle $O$ is $R.$
The radius of circle $P$ is $r.$

The ratio of the areas is: . $\text{Ratio} \:=\:\frac{\pi R^2}{\pi r^2} \;=\;\frac{R^2}{r^2}$

Apply the Law of Cosines to $\Delta OCD$ and $\Delta PCD:$

. . $R^2 + R^2 - 2R^2\cos30^o \:=\:d^2 \:=\:r^2 + r^2 - 2r^2\cos60^o$

. . $2R^2(1-\cos30^o) \;=\;2r^2(1-\cos60^o) \quad\Rightarrow\quad \frac{R^2}{r^2} \;=\;\frac{1-\cos60^o}{1-\cos30^o}$

Therefore: . $\text{Ratio} \;=\;\frac{R^2}{r^2} \;=\;\frac{1 - \frac{1}{2}}{1-\frac{\sqrt{3}}{2}} \;=\;\frac{2}{2}\cdot\frac{1-\frac{1}{2}}{1-\frac{\sqrt{3}}{2}} \;=\; \frac{1}{2-\sqrt{3}}$

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