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Re: MGMAT - Circle [#permalink]
20 Feb 2008, 19:50

bmwhype2 wrote:

In the figure to the right, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region?

We have 10pi as the circumfrence. so 60*/360* --> 1/6*10pi --> 5/3pi but we must not forget we had an insribed angle so we must multiply this result by 2.

10/3pi

Now just draw a line from A to C. The hypotenous of this new triangle is 10. Thus the small side is 5 and the larger side 5sqrt3

Re: MGMAT - Circle [#permalink]
21 Feb 2008, 01:17

1

This post received KUDOS

Applying parallel lines cut by traversal rule: X = 30 Now applying the central angle to Inscribed angle rule Central angle = 2 * Inscribed angle Central Angle = 2 * 60 = 120

CAE perimeter = (120 / 360) * 2*pi*5 = 10pi/3 Sides BC and BE are equal and also angle B is equal to 60, so other two angles are equal to 60. Now apply 30-60-90 rule: BC = 5 sqrt3 Total perimeter = CAE + BC + BE = 10pi/3 + 5 sqrt3 + 5 sqrt3 = 10 pi/3 + 10 sqrt3

Re: In the figure to the right, circle O has center O, diameter [#permalink]
20 Nov 2013, 03:56

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Re: In the figure, circle O has center O, diameter AB and a radi [#permalink]
20 Nov 2013, 05:49

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Expert's post

In the figure, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region?

A. (5/3)pi + 5√3 B. (5/3) pi + 10√3 C. (10/3) pi + 5√3 D. (10/3) pi + 10√3 E. (10/3) pi + 20√3

Since CD is parallel to AB then <CBA=<BCD=30 --> <CBE=60 --> <COE=2*60=120 (according to the central angle theorem) --> length of minor arc CE is \frac{120}{360}*2\pi{r}=\frac{10}{3}*\pi;

Now, we should find the lengths of BC and BE (notice that they are equal). Since <CBA=30 then triangle ACB is 30-60-90 right triangle (AB=diameter means that <C=90), thus \frac{BC}{AB}=\frac{\sqrt{3}}{2}, (BC is opposite to 60 degrees so corresponds to \sqrt{3}) --> \frac{BC}{10}=\frac{\sqrt{3}}{2} (AB=diameter=2r=10) --> BC=5\sqrt{3};

Thus the perimeter of the shaded region is (minor arc CE)+BC+BE: \frac{10}{3}*\pi+5\sqrt{3}+5\sqrt{3}=\frac{10}{3}*\pi+10\sqrt{3}.