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GMATPrep Test 2 Q8

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Abhijit K Really wants to Beat The GMAT! Default Avatar
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GMATPrep Test 2 Q8 Post Sun Feb 15, 2015 8:03 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    The perimeter of a certain isosceles right triangle is 16+16root2. What is the length of the hypotenuse of the triangle?
    A.8
    B.16
    C.4root2
    D.8root2
    E.16root2

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    Post Sun Feb 15, 2015 8:10 am
    Quote:
    The perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse?
    a. 8
    b. 16
    c. 4√2
    d. 8√2
    e. 16√2
    The sides of an isosceles right triangle are in the following ratio: s : s : s√2.
    So if s=side and h=hypotenuse, then h = s√2 and s = h/√2.

    We can plug in the answers, which represent the length of the hypotenuse.

    Answer choice C: h = 4√2
    s = (4√2)/√2 = 4.
    p = 4 + 4 + 4√2 = 8 + 4√2.
    Eliminate C. The perimeter needs to be quite a bit larger.

    Answer choice B: h = 16
    s = 16/√2 = (16*√2)/(√2*√2) = (16√2)/2 = 8√2.
    p = 8√2 + 8√2 + 16 = 16 + 16√2. Success!

    The correct answer is B.

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    Post Sun Feb 15, 2015 8:25 am
    Quote:
    The Perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle?

    A) 8
    B) 16
    C) 4√2
    D) 8√2
    E) 16√2
    An IMPORTANT point to remember is that, in any isosceles right triangle, the sides have length x, x, and x√2 for some positive value of x.

    Note: x√2 is the length of the hypotenuse, so our goal is to find the value of x√2

    From here, we can see that the perimeter will be x + x + x√2

    In the question, the perimeter is 16 + 16√2, so we can create the following equation:
    x + x + x√2 = 16 + 16√2,
    Simplify: 2x + x√2 = 16 + 16√2
    IMPORTANT: Factor x√2 from the left side to get : x√2(√2 + 1) = 16 + 16√2
    Now factor 16 from the right side to get: x√2(√2 + 1) = 16(1 + √2)
    Divide both sides by (1 + √2) to get: x√2 = 16

    Answer = B

    Cheers,
    Brent

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    Post Sun Feb 15, 2015 6:17 pm
    We'll start with x + x + x√2 = 16 + 16√2.

    We know that ONE of these numbers (16 or 16√2) is the hypotenuse, and the other is the sum of the two legs. If 2x = 16, then x = 8, and x√2 = 8√2. That doesn't match our numbers, so it can't work.

    Hence the OTHER number, 16√2, must be the sum of the two legs. This gives us legs of 8√2, and a hypotenuse of 8√2(√2), or 16. Success!

    If we wanted to do this algebraically, we could. Let x = one of the legs, and x√2 = the hypotenuse.

    2x + √2x = 16 + 16√2
    x * (2 + √2) = 16 + 16√2
    x = (16 + 16√2)/(2 + √2)
    x = ((16 + 16√2) * (2 - √2)) / ((2+√2)(2-√2))
    x = (16√2)/2
    x = 8√2

    So our legs are each 8√2, and our hypotenuse is 16.

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