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If triangle AOB is equilateral, O is the midpoint of AC, and

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If triangle AOB is equilateral, O is the midpoint of AC, and [#permalink] New post 26 Sep 2006, 06:23
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If triangle AOB is equilateral, O is the midpoint of AC, and the radius of semicircle ABC is 6, what is the length of line segment BC?

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 [#permalink] New post 26 Sep 2006, 07:34
If triangle AOB is equilateral, O is the midpoint of AC, and the radius of semicircle ABC is 6, what is the length of line segment BC?
1. 3* (3^1/2)
2. 6*(2^1/2)
3. 6*(3^1/2)
4. 6*(5^1/2)
5. 12


Radius is 6 so AC = 12; AB = 6 (equilateral triangle). So BC = 12^2 - 6^2 = 108 ^ 1/2 = 6 * 1/3^2

Ans C
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 [#permalink] New post 26 Sep 2006, 23:31
Another way to solve this is to split the equilateratal triangle into two 30-60-90 triangles. This will give you the height of a right triangle with segment BC as the hypotnuse.

Use the pythagorean theorum to solve.

sqrt108 ---> 6sqrt3

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 [#permalink] New post 27 Sep 2006, 04:17
yes the answer is C

I solved it same way GMATT 73 did
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 [#permalink] New post 02 Feb 2007, 22:31
gk3.14 wrote:
If triangle AOB is equilateral, O is the midpoint of AC, and the radius of semicircle ABC is 6, what is the length of line segment BC?
1. 3* (3^1/2)
2. 6*(2^1/2)
3. 6*(3^1/2)
4. 6*(5^1/2)
5. 12


Radius is 6 so AC = 12; AB = 6 (equilateral triangle). So BC = 12^2 - 6^2 = 108 ^ 1/2 = 6 * 1/3^2

Ans C

why is this the equation for the hypotenuse?

BC = 12^2 - 6^2 = x
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 [#permalink] New post 03 Feb 2007, 04:44
BC is not the hypotenuse. AC^2=AB^2+BC^2, where AC is the hypo. Hence, BC^2=AC^2-AB^2
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 [#permalink] New post 03 Feb 2007, 10:19
This question can be solved using a special property of Pythagoras theorem called Appollonius Principle. You get the answer directly.

The expression in this case would be,

AB² + BC² = 2OB² + 2OA²

6² + BC² = 2(6)² + 2(6)²
BC² = 108
BC = 6√3
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 [#permalink] New post 03 Feb 2007, 10:31
Triangle is going to be right trangle as AC is the diameter and B is pt on semicircle
Thus bc^2 = ac^2-ab^2
= 12^2 -6^2
= 6sqrt(3)
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 [#permalink] New post 03 Feb 2007, 10:52
Answer is 6 * sq.rt(3)

Because we know that angle inscribed in a semi-cricle is a right angle triange there for angleABC is 90.

We know AC = 12
and becacuse triangle BAO is equilateral triangle AB = 6

hence AC^2 = AB^2 + BC^2

therefore, BC^2 = 12^2 - 6^2

BC^2 = 108
BC = 6 * sq.rt(3)

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  [#permalink] 03 Feb 2007, 10:52
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