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# M27-05

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Math Expert
Joined: 02 Sep 2009
Posts: 42305

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16 Sep 2014, 01:26
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Difficulty:

65% (hard)

Question Stats:

43% (01:12) correct 57% (01:21) wrong based on 21 sessions

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

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

(2) The area of the triangle is 50.
[Reveal] Spoiler: OA

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16 Sep 2014, 01:26
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Official Solution:

This is a 700+ question.

(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, $$a+b \gt c \gt 15$$, which means 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: $$Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50$$. Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.

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Senior Manager
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01 Aug 2016, 05:19
I think this is a high-quality question and I agree with explanation.

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Senior Manager
Status: You have to have the darkness for the dawn to come
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03 Mar 2017, 09:34
Bunuel wrote:
Official Solution:

This is a 700+ question.

(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, $$a+b \gt c \gt 15$$, which means 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: $$Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50$$. Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.

Could you please explain statement 2 in detail. I am unable to understand the solution for statement 2.
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04 Mar 2017, 03:12
daboo343 wrote:
Bunuel wrote:
Official Solution:

This is a 700+ question.

(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, $$a+b \gt c \gt 15$$, which means 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: $$Area_{equilateral}=s^2*\frac{\sqrt{3}}{4}=(\frac{30}{3})^2*\frac{\sqrt{3}}{4}=25*\sqrt{3} \lt 50$$. Since even an equilateral triangle with perimeter of 30 cannot produce the area of 50, then the perimeter must be more than 30. Sufficient.

Could you please explain statement 2 in detail. I am unable to understand the solution for statement 2.

From (2) we have that if even an equilateral triangle with perimeter of 30 cannot have the area of 50, then the perimeter must be more that 30.
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14 Jun 2017, 06:16
Hi Sir,

I believe it is the same if I assume this way right?

For a give area, equilateral triangle has the smallest perimeter.

S^2 (sq.rt3/4)=50 when simplified will show clearly that a side is >12.

Hence sufficient.

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16 Oct 2017, 11:04
Nice answer, in my opinion doing it in your way would be more useful for other problems!
Thanks!

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Re: M27-05   [#permalink] 16 Oct 2017, 11:04
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# M27-05

Moderators: Bunuel, chetan2u

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