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M27-05

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16 Sep 2014, 01:26
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95% (hard)

Question Stats:

34% (01:16) correct 66% (01:29) wrong based on 187 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.

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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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01 Aug 2016, 05:19
I think this is a high-quality question and I agree with explanation.
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03 Mar 2017, 09:34
1
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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13 Sep 2018, 08:15
EDIT: Got it. Number 2 is sufficient that perimeter will always be >30.
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13 Sep 2018, 10:52
1
Statement 1 can be easily shown as sufficient as the difference between the two sides must be less than the third side. So the third side would be at least 15 and as given in statement the difference between the two sides is 15 so their sum would obviously be greater than 15. So perimeter will be greater than 30.

Statement 2: Area of a triangle is given to be 50. As it is a fact that for a given perimeter the equilateral triangle will have maximum area. So let us assume the side of a triangle is 10 then (root3/4)*(10)2 then we will get the area as 43.25 which less than 50. hence the perimeter has to be more than 30. Sufficient.

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13 Sep 2018, 23:34
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.

HI,

I understood Statement 1 is sufficient.

However, I have issues understanding statement 2.

How can we confirm that "For a given perimeter equilateral triangle has the largest area.".

Can it be proved by a theorem?

Thanks
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13 Sep 2018, 23:50
honneeey 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.

HI,

I understood Statement 1 is sufficient.

However, I have issues understanding statement 2.

How can we confirm that "For a given perimeter equilateral triangle has the largest area.".

Can it be proved by a theorem?

Thanks

Here is a proof of this theorem: https://www.mathalino.com/reviewer/diff ... -perimeter

Hope it helps.
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13 Sep 2018, 23:54
Bunuel wrote:
honneeey 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.

HI,

I understood Statement 1 is sufficient.

However, I have issues understanding statement 2.

How can we confirm that "For a given perimeter equilateral triangle has the largest area.".

Can it be proved by a theorem?

Thanks

Here is a proof of this theorem: https://www.mathalino.com/reviewer/diff ... -perimeter

Hope it helps.

Useful video:

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20 Sep 2018, 11:16
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.

This is a damn tricky one for sure
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