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# M25-04

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16 Sep 2014, 01:22
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35% (01:23) correct 65% (01:28) wrong based on 514 sessions

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

(1) $$a-b=15$$

(2) The area of the triangle is 50

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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.

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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.

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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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.

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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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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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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16 Dec 2014, 09:19
Ah..thanks for that...wasn't aware of the property.
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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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23 Sep 2015, 12:58
I think this is a poor-quality question.
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23 Sep 2015, 22:10
achilies wrote:
I think this is a poor-quality question.

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25 Oct 2015, 16:40
I don't agree with the explanation.
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26 Oct 2015, 17:52
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26 Oct 2015, 21:18
HarleenKhanuja17 wrote:

Please re-read the solution: it's a + b > c > 15, from which it follows that a + b + c > 30.
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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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12 Dec 2015, 08:40
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roopz2scorpio wrote:
I didnt understand the explanation to fact (ii). Is it a formula for area of equilateral triangle - side^2 * \sqrt{3}/4?

Thanks!

Please go through the following post: math-triangles-87197.html

Hope it helps.
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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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26 Dec 2015, 05:10
HK17 wrote:

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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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

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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

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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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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