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M25-04

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New post 16 Sep 2014, 01:22
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New post 16 Sep 2014, 01:22
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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.


Answer: D
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New post Updated on: 02 Dec 2014, 02:39
1
Bunuel wrote:
Official Solution:


(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 can not produce the area of 50, then the perimeter must be more that 30. Sufficient.


Answer: D


I got my answer. Sorry to bother you. Now I know the reason why we used the equilateral triangle.

But anyways, I tried using both equilateral (the triangle is supposed to have the smallest perimeter for a given area) as well as isosceles triangle. See below:

1/2 ab= 50
or ab=100

a and b can be any of the following combinations or many more

1X 100 Now the third side "c" can be anything between 99-101. Hence, the smallest perimeter solution would be 1+100+100
2X 50 48-52. Hence, the smallest perimeter solution would be 2+49+50
4X 25 21-29. Hence, the smallest perimeter solution would be 4+22+25
5X 20 15-25. Hence, the smallest perimeter solution would be 5+16+20=41

Thus, in every case, the answer would be more than 30.

Similarly, in equilateral triangle with area 50, each side will come out to be 10.84

Originally posted by Rupalia on 18 Nov 2014, 09:21.
Last edited by Rupalia on 02 Dec 2014, 02:39, edited 1 time in total.
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New post 18 Nov 2014, 09:40
Rupalia wrote:
Bunuel wrote:
Official Solution:


(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 can not produce the area of 50, then the perimeter must be more that 30. Sufficient.


Answer: D


But does the question say that the area has to be the largest ( or equilateral triangle).
What if the triangle is a right angle triangle, then

1/2 ab= 50
or ab=100

a and b can be any of the following combinations or many more

1X 100 Now the third side "c" can be anything between 99-101. Hence, the smallest perimeter solution would be 1+100+100
2X 50 48-52. Hence, the smallest perimeter solution would be 2+49+50
4X 25 21-29. Hence, the smallest perimeter solution would be 4+22+25
5X 20 15-25. Hence, the smallest perimeter solution would be 5+16+20=41

Thus, in every case, the answer would be more than 30.

Similarly, in equilateral triangle with area 50, each side will come out to be 10.84


Sorry, but I don't follow you... In each case you got that the perimeter must be more than 30, so the same answer as in the solution.
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New post 16 Dec 2014, 07:25
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?
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New post 16 Dec 2014, 09:03
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.
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New post 16 Dec 2014, 09:19
Ah..thanks for that...wasn't aware of the property.
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New post 11 Apr 2015, 00:40
I had a problem with understanding how this statement, "So, a+b>c>15, from which it follows that a+b+c>30. Sufficient", came about. Took me quite a while but now I understand. So I'm posting this for future reference to those who didn't understand at first.

if 15 < c < a+b, it means that c is a value more than 15. Let's represent it as 15*
From the given statement, a - b = 15, we can conclude that a = 15 + b

So adding up all the sides to get the perimeter, a + b + c, and substituting a=15+b and c to 15*, we get
perimeter = (15 + b) + b + 15* = b + 30* (or a value more than 30)
therefore, the perimeter is definitely more than 30.

Please correct me if my logic is incorrect!
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New post 23 Sep 2015, 12:58
I think this is a poor-quality question.
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New post 23 Sep 2015, 22:10
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New post 25 Oct 2015, 16:40
I don't agree with the explanation.
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New post 26 Oct 2015, 17:52
please help understand why a+b+c>15 follows that a+b+c>30? Is the above explanation correct?
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New post 26 Oct 2015, 21:18
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New post 10 Dec 2015, 18:26
I didnt understand the explanation to fact (ii). Is it a formula for area of equilateral triangle - side^2 * \sqrt{3}/4?

Thanks!
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New post 12 Dec 2015, 08:40
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New post 13 Dec 2015, 06:35
I think this is a high-quality question and I agree with explanation. This is an awesome question - It just hits you on the basics and that is the most wonderful part of it
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New post 26 Dec 2015, 05:10
HK17 wrote:
please help understand why a+b+c>15 follows that a+b+c>30? Is the above explanation correct?

a+b>c
and c>15

so,
a+b+c>c+c
a+b+c>2c

2c>2*15

a+b+c>30
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New post 19 Dec 2016, 10:04
I don't agree with the explanation. for right angle triangle with two perpendicular sides of length 10 each, area would be 50 and perimeter would be 10+10+10 (root 2) > 30

So correct answer is A.
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New post 19 Dec 2016, 10:13
kDaenerys wrote:
I don't agree with the explanation. for right angle triangle with two perpendicular sides of length 10 each, area would be 50 and perimeter would be 10+10+10 (root 2) > 30

So correct answer is A.


You did no understand the explanation. Each statement gives an YES answer to the question. Please re-read the discussion above.
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Re: M25-04  [#permalink]

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New post 05 Jun 2017, 04:20
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
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