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# The sides of a triangle is a, b, and c. Are the three angles

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The sides of a triangle is a, b, and c. Are the three angles [#permalink]

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27 Oct 2008, 08:36
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The sides of a triangle is a, b, and c. Are the three angles all less than 90 measure degrees?
1). The areas of the semi-circles with the radius a, b, c are 4, 5, 6, respectively.
2). c<a+b<c+2

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rampuria

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27 Oct 2008, 10:12
It must be A.

1). The areas of the semi-circles with the radius a, b, c are 4, 5, 6, respectively.

This gives a^2,b^2 and c^2 values to be 8/Pie , 10/Pie and 12/Pie. None of them add up to form the third side square. Thus all of the angles are below 90.

2). c<a+b<c+2

This only shows a+b is between C and C+2. This is not enough to prove.

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27 Oct 2008, 10:43
rampuria wrote:
The sides of a triangle is a, b, and c. Are the three angles all less than 90 measure degrees?
1). The areas of the semi-circles with the radius a, b, c are 4, 5, 6, respectively.
2). c<a+b<c+2

I'm also getting A.

Rationale:
1) By reversing the area calculation, you now know the lengths of a,b,c.
It doesn't matter what the lengths are, but with this data the size of all the angles can be calculated using sine/cosine functions (which I forgot how to use ). Anyway, regardless of what the size of the angles are, the data in (1) is sufficient to find the angles.

2) Insufficient. A large range of sizes could fit into that with varying answers.
Eg. Contrast
c=200 a=180 b=21 (all angles less than 90 ... i tihnk )
c=0.3 a=0.4 b=0.5 (right-angled triangle)

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17 Nov 2008, 08:11
very good question. It would be great to get some detailed explanations.

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Re: DS   [#permalink] 17 Nov 2008, 08:11
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