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A triangle is inscribed in the semicircle. If the length of

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A triangle is inscribed in the semicircle. If the length of  [#permalink]

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New post 09 Nov 2008, 18:37
A triangle is inscribed in the semicircle. If the length of line segment AB is 8 and the length of line segment BC is 6, what is the length of arc ABC?

A. 15
B. 12
C. 10
D. 7
E. 5

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Re: semicircle  [#permalink]

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New post 09 Nov 2008, 19:25
is there any kind of graph to go with it?
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Re: semicircle  [#permalink]

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New post 09 Nov 2008, 19:38
I don't think this question can be answered without some diagram. Nothing in the questions states where the vertices of the triangle are located. We need to know visually the relationshipbetween Segment AB and Segment BC.
haichao wrote:
a triangle is inscribed in the semicircle. If the length of line segment AB is 8 and the length of line segment BC is 6, what is the length of arc ABC?

A. 15
B. 12
C. 10
D. 7
E. 5

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Re: semicircle  [#permalink]

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New post 10 Nov 2008, 03:27
I think, the question is complete. Since, the question asks for length of arc ABC, A, B and C must be on the semi circle.

Moreover, since, AB is 8 and BC is 6, in this case, AC has to be the diameter of the semi circle.
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Re: semicircle  [#permalink]

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New post 10 Nov 2008, 03:45
C

ABC is inscribed in the semicircle => ABC is a right triangle

AC = \(sqrt(AB^2 + BC^2)\) = 10
or

AC = \(sqrt(AB^2 - BC^2)\) (no answer choice for this)
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Re: semicircle  [#permalink]

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New post 10 Nov 2008, 03:56
scthakur is right, qs can be complete. But i was looking for a " approximate length of arc ABC " in the qs rather than "the length" since the exact number is missing from the answer choices.

Is the QA A ?
triangle ABC is right angled , since its incribed in a semicircle. Hence the third side AC is 10 and radius is 5. So half circumference is 5 * 3.14 which is slightly greater than 15. But since 15 is the highest choice in answers , it has to be A.
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Re: semicircle  [#permalink]

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New post 10 Nov 2008, 04:39
assuming that AC is diameter, so ac=10cm
radius =5 so length of arc=xr=3.14*5=15.7...A
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New post 10 Nov 2008, 04:57
prasun84 wrote:
assuming that AC is diameter, so ac=10cm
radius =5 so length of arc=xr=3.14*5=15.7...A


so the formula for arcABC = xr??, i'm not sure what is ABC here
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Re: semicircle  [#permalink]

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New post 10 Nov 2008, 07:04
Here is the problem I have with this question. A triangle inscribed in a semi-circle will be a right triangle. So we know that 6 & 8 are the bases, which means the 3rd side is 10, a.k.a. the diameter of the semicircle. The formula for finding the perimeter of a semi-circle is \(\frac{1}{2}*d*pi\). The answer to this is 5pi, but that's not an option. 15 is close, but not exactly it either 5*pi(~3.14) = 15.7.

haichao wrote:
a triangle is inscribed in the semicircle. If the length of line segment AB is 8 and the length of line segment BC is 6, what is the length of arc ABC?

A. 15
B. 12
C. 10
D. 7
E. 5


--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

_________________

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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: semicircle &nbs [#permalink] 10 Nov 2008, 07:04
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