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In the figure above, showing circle with center O and points

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In the figure above, showing circle with center O and points

by gmattesttaker2 » Sun Jun 15, 2014 9:44 pm
Hello,

For the following:

In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)= 12 sq. root (3) , what is the radius of the circle?

OA: 12


I am not sure how to calculate the value of angle AOB here. Given that minor arc (AB) = 1/3 circumference, does it mean that circumference is 360 degrees?

Thanks for your help.

Regards,
Sri
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In the figure above, showing circle with center O and points

by GMATGuruNY » Mon Jun 16, 2014 3:53 am
gmattesttaker2 wrote:Hello,

For the following:

In the figure above, showing circle with center O and points A, B, and C on the circle, given that minor arc (AB) is one-third of the total circumference and length (AB)= 12 sq. root (3) , what is the radius of the circle?
Image

Since AC is the diameter, arc ABC is 1/2 the circumference.
Since arc AB is 1/3 of the circumference, we get:
Arc BC = 1/2 - 1/3 = 1/6 of the circumference.
Thus, the degree measurement of arc BC = (1/6)(360) = 60º.

An inscribed angle is formed by 2 chords.
Thus, ∠BAC is an inscribed angle.
The two chords of an inscribed angle intercept an arc on the circle.
Here, inscribed ∠BAC intercepts arc BC.
The degree measurement of an inscribed angle is equal to 1/2 the degree measurement of the intercepted arc.
Thus:
∠BAC = (1/2)(60) = 30º.

An inscribed angle that intercepts the diameter is a right angle.
Thus, inscribed ∠ABC is a right angle, implying that ∆ABC is a 30-60-90 triangle.
The sides of a 30-60-90 triangle are in the following ratio:
x : x√3 : 2x.
Since AB = 12√3, the sides of ∆ABC are as follows:
BC = 12, AB = 12√3, AC = 24.

Since diameter AC=24, r=12.
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In the figure above, showing circle with center O and points

by GMATinsight » Mon Jun 16, 2014 6:30 am
Please find attached the solution



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by GMAT Mantra » Mon Jun 25, 2018 3:11 am
[url]https://youtu.be/RlaqljEHe5s
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