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Geometry: Perimeter of a triangle vs that of a square

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topspin360 Rising GMAT Star Default Avatar
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Geometry: Perimeter of a triangle vs that of a square Post Thu Jan 03, 2013 7:32 pm
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    Can someone please explain how you'd go about tackling this question?

    Question: Is the perimeter of a triangle T greater than the perimeter of square S?

    1) The length of the longest side of T is twice the length of a side S.

    2) T is isoceles.

    I assume you start by picking a variable for either the side of S or T. How do you quickly deduce from there?

    the answer is

    A

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    Post Thu Jan 03, 2013 8:13 pm
    I did this way,

    We know that " the sum of two sides of a triangle are always greater than the third side."

    So, let the length of first, second and third sides of triangle be x, y and z respectively where z is the longest side.

    So, we know that x + y >= z ... (i)

    And for perimeter ... x + y + z >= 2 z ( from i )

    OK Now, lets go to options..

    St. 1 -- It says that the longest side is 2*a ( where a is the side of square )
    So.. from above analysis... we can say that permeter of triangle will be >= 2 ( 2 *a) >= 4*a

    and we know that perimeter of square is 4*a...so this statement is sufficient.

    St2 -- It does not say anything about square, so insufficient.

    Hence, answer is A. Hope this clarifies your concern

    topspin360 wrote:
    Can someone please explain how you'd go about tackling this question?

    Question: Is the perimeter of a triangle T greater than the perimeter of square S?

    1) The length of the longest side of T is twice the length of a side S.

    2) T is isoceles.

    I assume you start by picking a variable for either the side of S or T. How do you quickly deduce from there?

    the answer is

    A

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    Whitney Garner GMAT Instructor
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    Post Thu Jan 03, 2013 8:15 pm
    Hi topspin360!

    This is a tricky problem to do algebraically because of the various configurations for the triangle (is it right, isosceles, equilateral, nothing??). So maybe we just start making up some numbers that work. We are only asked about perimeters, so as long as we make the sides of the triangle "legal" (no side is longer than the sum of the other 2 sides), then we're good.

    (1) The length of the longest side of T is twice the length of a side S.
    To be lazy here, I picked a well known right triangle: 3-4-5, perimeter=12. This means the side of the square is 1/2 of 5, or 2.5, perimeter=10. In this case, the perimeter of the triangle is LARGER than the perimeter of the square. Now I have to test something where the perimeter of the triangle would be LESS than the perimeter of the square....maybe if I make the sides of the triangle really small, and if I make the "longest" side as short as possible, or the same length as the other sides? What about an equilateral with all sides 1, perimeter=3. The square would have sides .5, perimeter 2...still smaller!! So let's think theory... the smallest "long" side for a triangle would be one where it was the same length as the other sides (equilateral), so if the triangle has side lengths t, perimeter 3t, the square would have sides .5T, perimeter 2t...always smaller than 3t!! So it looks like we were able to prove that under this constraint, the perimeter of the triangle is ALWAYS larger than that of the square! SUFFICIENT

    (2) T is isoceles.
    So here, we don't have ANY relationship between the triangle and the square so we can make up whatever we want. Let's make the triangle 1-1-1 (because ALL equilateral triangles are also isosceles), and the square 2-2-2-2... Here the square has a LARGER perimeter than the triangle. But swap them, make the triangle 2-2-2 and the square 1-1-1-1... Now the square is only 4 while the triangle is 6, so the triangle has a LARGER perimeter. NOT SUFFICIENT

    Therefore, the answer is A

    Hope this helps!
    Smile
    Whit

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    Post Thu Jan 03, 2013 11:55 pm
    topspin360 wrote:
    Can someone please explain how you'd go about tackling this question?

    Question: Is the perimeter of a triangle T greater than the perimeter of square S?

    1) The length of the longest side of T is twice the length of a side S.

    2) T is isoceles.

    I assume you start by picking a variable for either the side of S or T. How do you quickly deduce from there?

    the answer is

    A
    Statement 1: The length of the longest side of T is twice the length of a side pf S.
    Square S:
    Let each side = s.
    Then the perimeter = 4s.

    Triangle T:
    Let triangle T be ∆ABC, whose longest side is AC.
    Then AC = 2s.
    Since the third side of a triangle must be LESS than the sum of the lengths of the other 2 sides, AC < AB+BC.
    Thus, AB+BC > AC, implying that AB+BC > 2s.
    Since AC = 2s, and AB+BC > 2s, AB+BC+AC > 4s.
    In other words, the perimeter of triangle T is GREATER than 4s.

    Since the perimeter of triangle T > 4s, and the perimeter of square S = 4s, the perimeter of triangle T is greater than the perimeter of square S.
    SUFFICIENT.

    Statement 2: T is isosceles.
    No information about square S.
    INSUFFICIENT.

    The correct answer is A.

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