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# M25-04

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Math Expert
Joined: 02 Sep 2009
Posts: 51229

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05 Jun 2017, 06:06
Mns1203 wrote:
Bunuel wrote:
rvc27 wrote:
Hi

I am not sure I understand why we use an equilateral triangle to test. Does it imply that for a given area, an equilateral triangle has the largest perimeter?

Because if even an equilateral triangle with perimeter of 30 cannot have the area of 50, then the perimeter must be more that 30.

As for the properties:
For a given area equilateral triangle has the minimum possible perimeter.
For a given perimeter equilateral triangle has the maximum possible area.

Hi Bunuel
I wonder whether these properties are considered just in case of triangle because i have learn that for a given area, circle has a minimum possible perimeter.
https://gmatclub.com/forum/gmat-diagnos ... 79366.html

Yes, we are talking only about the triangles there.
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23 Jan 2018, 01:32
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Bunuel
Hi. Question M25-04 and M27-05 are repeats.
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Joined: 02 Sep 2009
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23 Jan 2018, 01:42
Zksgmat wrote:
Bunuel
Hi. Question M25-04 and M27-05 are repeats.

Thank you for noticing. Removed M27-05 from database.
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09 Jun 2018, 08:38
Bunuel wrote:
Official Solution:

Is the perimeter of triangle with the sides $$a$$, $$b$$ and $$c$$ greater than 30?

(1) $$a-b=15$$. Must know for the GMAT: the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. So, $$a+b \gt c \gt 15$$, from which it follows that $$a+b+c \gt 30$$. Sufficient.

(2) The area of the triangle is 50. For a given perimeter equilateral triangle has the largest area. Now, if the perimeter were equal to 30 then it would have the largest area if it were equilateral.

Let's find what this area would be: $$Area=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50$$. Since even equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more that 30. Sufficient.

The way #2 is written is confusing.

I would rephrase as follows:
The equilateral triangle maximizes area while minimizing perimeter. If you can remember this, then you can safely assume that if you can prove an equilateral triangle has a larger perimeter than 30 while having an area of 50, then ALL other triangles MUST also have larger perimeters.

Area of any triangle $$= \frac{1}{2}*b*h$$

An equilateral triangle can be split into two $$30:60:90$$ triangles, whose sides are expressed as $$x:x\sqrt{3}:2x$$

Thus, half of the equilateral triangle $$= 5:5\sqrt{3}:10$$

Thus, area of our entire equilateral triangle: $$= \frac{1}{2}*b*h = \frac{1}{2}*10*5\sqrt{3} = 25\sqrt{3}$$

We just proved that a triangle with sides = 10 (and thus perimeter = 30) has an area of $$25\sqrt{3}$$, or approximately 35. So if the area of the triangle in statement #2 is 50, the perimeter MUST be more than 30.

I also personally solved this by a+b > c > a-b, the same rules with which you would solve statement #1; however, I feel there is some danger with this method because this would assume that $$b*h$$ in the area formula $$\frac{1}{2} * b * h$$ is the same as sides a and b. I can't prove this wrong, but thinking aloud maybe there is something wrong with that line of reasoning.
Re: M25-04 &nbs [#permalink] 09 Jun 2018, 08:38

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# M25-04

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