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Given isosceles triangle ABC with \(base=AC=\frac{2}{a}\) and \(height=a^2\).

Question: is \(area=\frac{1}{2}*base*height=\frac{1}{2}*\frac{2}{a}*a^2=a>1\). So we see that basically the question aska: is \(a>1\) true?

(1) \(\angle ABC < 90^\circ\) --> assume \(\angle ABC=90^\circ\) then hypotenuse is \(AC=\frac{2}{a}\), as ABC becomes isosceles right triangle (\(45-45-90=1-1-\sqrt{2}\)) then the \(leg=BC=AB=\frac{2}{a\sqrt{2}}=\frac{\sqrt{2}}{a}\).

But \(BC=\frac{\sqrt{2}}{a}\) also equals to \(\sqrt{(\frac{1}{a})^2+(a^2)^2}\), so we have \(\frac{\sqrt{2}}{a}=\sqrt{(\frac{1}{a})^2+(a^2)^2}\) --> \(2=1+a^6\) --> \(a^6=1\) --> \(a=1\) (as \(a\) per diagram is positive). Now, if we increase \(a\) then the base \(\frac{2}{a}\) will decrease and the height \(a^2\) will increase thus making angle ABC smaller than 90 and if we decrease \(a\) then the base \(\frac{2}{a}\) will increase and the height \(a^2\) will decrease thus making the angle ABC greater than 90. So, as angle ABC is less than 90, \(a\) must be more than 1. Sufficient.

(2) Perimeter of the triangle \(ABC\) is greater than \(\frac{4}{a}\) --> \(P=BC+AB+AC=2BC+AC=2\sqrt{(\frac{1}{a})^2+(a^2)^2}+\frac{2}{a}>\frac{4}{a}\) --> \(\sqrt{\frac{1+a^6}{a^2}}>\frac{1}{a}\) --> \(a^6>0\) --> \(a>0\). Hence we don't know whether \(a>1\) is true. Not sufficient.

Re: DS : TRIANGLE (m09q07) [#permalink]
01 Mar 2013, 20:56

5

This post received KUDOS

Area of the triangle : \(\frac{1}{2}*AC*OB\) = \(\frac{1}{2}*\frac{2}{a}*a^2\) = a

So we to find that a > 1 or not ??

Statement 1:

angle ABC < 90 => angle ABO < 45. mean angle BAO is greater than angle ABO that further implies that side OB > side OA \(a^2 >\frac{1}{a}\) => \(a^3> 1\) => \(a > 1\) since length of side can not be negative

So A is sufficient

Statement 2:

Perimeter of the triangle is greater than 4/a

So fine perimeter, first we have to find AB

\(AB^2 = OB^2 + OA ^2\)

\(AB^2 = (a^2)^2 + \frac{1}{a^2}\)

\(AB^2 = \frac{(a^6 +1)}{a ^2}\)

\(AB =\sqrt{(a^6 +1)}/a\)

Perimeter of triangle = AB + BC +CA since AB = BC,

Re: DS : TRIANGLE (m09q07) [#permalink]
02 Dec 2010, 11:56

2

This post received KUDOS

Expert's post

MathMind wrote:

Bunuel,

Great explanation - I am not sure about your conclusion "Hence a>1 is not true as when we decrease angle ABC, a will decrease as well and will become less than 1"

I think this may lead to "As the angle ABC decreases, since ht=a^2 when base = 2a (ht/base ratio = a/2), the base will become broader => a will increase

=> This triangle will have area = 1 for ABC = 90*

=> This triangle will have area > 1 for ABC < 90*

=> This triangle will have area < 1 for ABC > 90* "

Thx, JS

First of all base=2/a (and not 2a) and height/base=a^3/2.

Though I did have a typo there. It should be: if we increase \(a\) then the base \(\frac{2}{a}\) will decrease and the height \(a^2\) will increase thus making angle ABC smaller than 90 and if we decrease \(a\) then the base \(\frac{2}{a}\) will increase and the height \(a^2\) will decrease thus making the angle ABC greater than 90. So, as angle ABC is less than 90, \(a\) must be more than 1. _________________

Re: DS : TRIANGLE (m09q07) [#permalink]
18 Feb 2014, 00:49

1

This post received KUDOS

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.

Re: DS : TRIANGLE (m09q07) [#permalink]
05 Oct 2009, 17:08

Hey guys..not clear..could somebody explain it in more detail? I understood that area of the triangle=a so we have to see whther a>1.. Got stuck in stmt 1:How does it imply AB^2 + BC^2 > AC^2 ?? and then 2*(a^4 + 1/a^2) > 4/a^2?????? _________________

Re: DS : TRIANGLE (m09q07) [#permalink]
22 Feb 2010, 05:24

tejal777 wrote:

Hey guys..not clear..could somebody explain it in more detail? I understood that area of the triangle=a so we have to see whther a>1.. Got stuck in stmt 1:How does it imply AB^2 + BC^2 > AC^2 ?? and then 2*(a^4 + 1/a^2) > 4/a^2??????

got the same question: How does it imply AB^2 + BC^2 > AC^2 ?? _________________

Re: DS : TRIANGLE (m09q07) [#permalink]
22 Feb 2010, 06:45

flyingbunny wrote:

tejal777 wrote:

Hey guys..not clear..could somebody explain it in more detail? I understood that area of the triangle=a so we have to see whther a>1.. Got stuck in stmt 1:How does it imply AB^2 + BC^2 > AC^2 ?? and then 2*(a^4 + 1/a^2) > 4/a^2??????

got the same question: How does it imply AB^2 + BC^2 > AC^2 ??

Although I'm still confused with the problem (actually the significance of angleABC < 90-degree), still "trying to force" a logic for the above algebric inequality. Since the AD (y-axis) is perpendicular to BC (x-axis), AB^2 = AD^2 + BD^2. Thus, AB^2 = (a^6 + 1)/ a^2 = AC^2 ------- (i) Now for triangle ABC, length of 2-sides must be greater than that of the 3rd side. So, AB + AC > BC. => 2 AB > BC => sqrt (2(a^6 + 1)/ a^2) > 2/a ------- (ii) i.e. 2(a^6 + 1)/ a^2 > 4/a^2 => a^6 + 1 > 2 => a^6 > 1. Since "a" (measure of side of a triangle) cannot be negative, "a^6 > 1" implies a > 1