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# M09 Q7

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Joined: 13 May 2010
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M09 Q7 [#permalink]

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25 Jun 2012, 18:07
On the picture below, is the area of the triangle $$ABC$$ greater than 1?

m09-7.gif

$$\angle ABC \lt 90^\circ$$
Perimeter of the triangle $$ABC$$ is greater than $$\frac{4}{a}$$

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient

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

$$\text{area}=\frac{1}{2}*\text{base}*\text{height}=\frac{1}{2}*\frac{2}{a}*a^2=a$$ . So, the question basically ask whether $$a \gt 1$$ ?

(1) $$\angle ABC \lt 90^\circ$$ $$\right$$ 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 $$\text{leg}=BC=AB=\frac{2}{a \sqrt{2}}=\frac{\sqrt{2}}{a}$$ .

But $$BC=\frac{\sqrt{2}}{a}$$ also equals to $$\sqrt{(\frac{1}{a})^2+(a^2)^2}$$ , so we have $$\frac{\sqrt{2}}{a}=\sqrt{(\frac{1}{a})^2+(a^2)^2}$$ $$\right$$ $$2=1+a^6$$ $$\right$$ $$a^6=1$$ $$\right$$ $$a=1$$ (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 $$\frac{4}{a}$$ $$\right$$ $$P=BC+AB+AC=2BC+AC=2\sqrt{(\frac{1}{a})^2+(a^2)^2}+\frac{2}{a} \gt \frac{4}{a}$$ $$\right$$ $$\sqrt{\frac{1+a^6}{a^2}} \gt \frac{1}{a}$$ $$\right$$ $$a^6 \gt 0$$ $$\right$$ $$a \gt 0$$ . Hence we don't know whether $$a \gt 1$$ is true. Not sufficient.

I understand the above explanation. Is there some shorter and faster method of solving this question? Anyone has a good strategic guess for this one?
Kellogg MMM ThreadMaster
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Re: M09 Q7 [#permalink]

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25 Jun 2012, 20:58
Picture missing.

Regards,
Math Expert
Joined: 02 Sep 2009
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Re: M09 Q7 [#permalink]

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26 Jun 2012, 00:54
teal wrote:
On the picture below, is the area of the triangle $$ABC$$ greater than 1?

m09-7.gif

$$\angle ABC \lt 90^\circ$$
Perimeter of the triangle $$ABC$$ is greater than $$\frac{4}{a}$$

Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
EACH statement ALONE is sufficient
Statements (1) and (2) TOGETHER are NOT sufficient

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

$$\text{area}=\frac{1}{2}*\text{base}*\text{height}=\frac{1}{2}*\frac{2}{a}*a^2=a$$ . So, the question basically ask whether $$a \gt 1$$ ?

(1) $$\angle ABC \lt 90^\circ$$ $$\right$$ 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 $$\text{leg}=BC=AB=\frac{2}{a \sqrt{2}}=\frac{\sqrt{2}}{a}$$ .

But $$BC=\frac{\sqrt{2}}{a}$$ also equals to $$\sqrt{(\frac{1}{a})^2+(a^2)^2}$$ , so we have $$\frac{\sqrt{2}}{a}=\sqrt{(\frac{1}{a})^2+(a^2)^2}$$ $$\right$$ $$2=1+a^6$$ $$\right$$ $$a^6=1$$ $$\right$$ $$a=1$$ (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 $$\frac{4}{a}$$ $$\right$$ $$P=BC+AB+AC=2BC+AC=2\sqrt{(\frac{1}{a})^2+(a^2)^2}+\frac{2}{a} \gt \frac{4}{a}$$ $$\right$$ $$\sqrt{\frac{1+a^6}{a^2}} \gt \frac{1}{a}$$ $$\right$$ $$a^6 \gt 0$$ $$\right$$ $$a \gt 0$$ . Hence we don't know whether $$a \gt 1$$ is true. Not sufficient.

I understand the above explanation. Is there some shorter and faster method of solving this question? Anyone has a good strategic guess for this one?

This question is discussed here: ds-triangle-m09q07-72173.html
_________________
Re: M09 Q7   [#permalink] 26 Jun 2012, 00:54
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# M09 Q7

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