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DS : TRIANGLE (m09q07)

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Joined: 28 Jan 2011
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Re: DS : TRIANGLE (m09q07) [#permalink]

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New post 18 Feb 2014, 01:49
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Bunuel's explanation is excellent, as usual.
Here's a different way that does not involve lots of algebra...

(1) Let's assume that a = 1, the simplest thing to try. Also, let's call the origin (0, 0) point D (got to name it so we can talk about it).

This gives us a triangle with height 1 (BD) and base 2 (AC), so the area is clearly 1.
It should be intuitively obvious, if you look at this triangle in your mind, that it's an isosceles right triangle (angle ABC = 90).
But if you're not sure, consider that if a = 1, BD = DC = 1, so triangle BDC is an isosceles right triangle.
This means that angle BCD is 45 degrees, and since angle BDC is 90 degrees, angle DBC is 45 degrees.
In the same way, we know that angle DBA is 45 degrees, and that therefore angle ABC is 90 degrees.

If we make a bigger, the triangle gets taller and narrower; try a = 2. Height 4, base 1, area 2.
Ah-hah! looks like as a gets bigger, area gets bigger.

One more quick check to validate: try \(a = \frac{1}{2}\). Height \(\frac{1}{4}\), base 4, area \(\frac{1}{2}\).

So it's clear, without doing any algebra: as the triangle gets taller and narrower, its area increases.
As the triangle gets taller and narrower (a > 1), angle ABC gets smaller, i.e. less than 90 degrees. That's what Statement (1) says. Sufficient.

(2) Again, let's try and avoid algebra with some quick numbers. If a = 1, BD = DC = 1, triangle BDC is an isosceles right triangle.
It must have sides in the ratio \(1:1:\sqrt{2}\); therefore BC = \(\sqrt{2}\).
AB will also = \(\sqrt{2}\), so the perimeter is \(2 + 2\sqrt{2}\), which is clearly > 4.
(If a = 1, \(\frac{4}{a}=4\)).

So when a = 1, the perimeter is > \(\frac{4}{a}\), but the area of triangle ABC is = 1 (see Statement (1) above) and therefore not > 1.
When a = 1, the answer to the overall question, whether the area > 1, is False.

Now if a becomes a larger number (a > 1), the perimeter will increase.

(Try a = 10. Now \(a^2\) = BD = 100. Clearly BC must be still larger.
Therefore the perimeter is much, much larger than \(\frac{4}{a}\), which is \(\frac{4}{10}\), and much larger than 4 as well.)

We already know from Statement (1) above that when a > 1, the answer to the question is True (area ABC > 1).
Therefore, without doing any more work, both True and False answers are possible and Statement (2) is Insufficient.
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Re: DS : TRIANGLE (m09q07) [#permalink]

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New post 21 Apr 2014, 09:35
The triangle's area = a^2*(2/a)*1/2=a
1) Angle ABC <90 degree -> angle ABO <45 and angle BAO>45 -> BAO>ABO ->a^2>1/a -> a^3>1 -> a>1: sufficient
2) Perimeter > 4/a -> 2sqrt(a^4+1/a^2)+2/a > 4/a -> a^4+1/a^2>1/a^2->a^4>0: insufficient
Choose A
Re: DS : TRIANGLE (m09q07)   [#permalink] 21 Apr 2014, 09:35

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DS : TRIANGLE (m09q07)

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