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Is the perimeter of triangle T greater than the perimeter of square S ? (1) T is an isoceles right triangle. (2) The length of the longest side of T is equal to the length of a diagonal of S.

(1) T is an isoceles right triangle --> no info about the square. Not sufficient.

(2) The length of the longest side of T is equal to the length of a diagonal of S --> let the side of the square be \(s\) --> the longest side of the triangle will be \(\sqrt{2}s\) and \(P_{square}=4s\). Now the max perimeter of the triangle T will be if this triangle is equilateral, then \(P_{triangle}=3\sqrt{2}s>4s=P_{square}\), but if triangle T is half of the square S (isosceles right triangle), then \(P_{triangle}=\sqrt{2}s+s+s=s(\sqrt{2}+2)<4s=P_{square}\). Two different answers. Not sufficient.

(1)+(2) Statement (1) says that we have the second case from statement (2), hence \(P_{triangle}=\sqrt{2}s+s+s=s(\sqrt{2}+2)<4s=P_{square}\). Sufficient.

Re: Triangles and Squares - comparing perimeter [#permalink]

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02 Jul 2010, 22:07

I get B

i). Not sufficient

ii). a - longest side of triangle x - side of square Given, a = sqrt(2)*x, so perimeter of square, 4x = 2*sqrt(2)*a Perimeter of triangle (of sides a,b,c) = a+b+c is < 2a (sum of the two sides (b+c) < a)

Re: Triangles and Squares - comparing perimeter [#permalink]

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24 May 2011, 08:38

2

This post was BOOKMARKED

Is the perimeter of triangle T greater than the perimeter of square S ? (1) T is an isosceles right triangle. (2) The length of the longest side of T is equal to the length of a diagonal of S.

Solution: Bunnuel is right the ans is C. (1) No information about S is given, so insufficient (2) Let the one side of s = x, so Diagonal of s = x√2 Perimeter of S = 4x Thus, the longest side of T = x√2 No other information is given about other sides of T, So insufficient.

Considering C the sides of T = x, x, x√2 [Sides of triangle with 90 degree, 45 degree and 45 degree is s, s, s√2] Perimeter of T = x + x + x√2 which is Less than 4x. Ans. C
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Re: perimeter of triangle T greater than the perimeter of square [#permalink]

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09 Feb 2012, 21:09

subhajeet wrote:

Is the perimeter of triangle T greater than the perimeter of square S ?

(1) T is an isoceles right triangle.

(2) The length of the longest side of T is equal to the length of a diagonal of S.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

(D) EACH statement ALONE is sufficient.

(E) Statements (1) and (2) TOGETHER are NOT sufficient.

Can anyone please provide a solution for this.

Let a, b, and c denote the three sides of triangle T, with c as the largest side, and let s denote a side of square S. The question then becomes: is \((a+b+c)>4s\)?

(1) Does not give any information about the square S. Thus, Not Sufficient.

(2) \(c = s\sqrt{2}\). From the triangle property, we know that sum of lengths of any two sides of a triangle is always greater than the length of third side. Thus, for triangle T, we have: \(a+b>c\) \(a+b+c>2c\) \(a+b+c> 2\sqrt{2}s\) This still is insufficient data, as we cannot prove weather \(4s>(a+b+c)>2\sqrt{2}s\) OR \((a+b+c)>4s\) (true in case of equilateral triangles)

(1)+(2) a = b, and \(c = a\sqrt{2} = s\sqrt{2}\) Thus, we get a = s. \(a+b+c = a(2+\sqrt{2}) < 4s\). Thus, we can have our answer to the question: is (a+b+c)>4s? No!

So correct answer is (C): (1) and (2) are sufficient together, but not alone..

Re: Is the perimeter of triangle T greater than the perimeter of [#permalink]

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30 Jul 2014, 15:01

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Is the perimeter of triangle T greater than the perimeter of [#permalink]

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08 Nov 2015, 09:50

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is the perimeter of triangle T greater than the perimeter of square S?

(1) T is an isoceles right triangle. (2) The length of the longest side of T is equal to the length of a diagonal of S.

There are 3 variables (3 sides) in a triangle and 1 (one side) in a square. We need 4 equations when only 2 equations are given by the conditions, so there is high chance (E) will be our answer. Looking at the conditions together, perimeter of T < perimeter of S, so the answer is 'no' and this is sufficient. The answer becomes (C).

For cases where we need 3 more equation, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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