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ranjeet75 Really wants to Beat The GMAT! Default Avatar
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length of segment Post Sun Mar 11, 2012 10:49 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    What is the length of segment BC?

    (1) Angle ABC is 90 degrees. (2) The area of the triangle is 30.

    OA is [spoiler]A[/spoiler] but I think the answer should be D because if area and two sides are given then by the formula: Area = sqrt s(s-a)(s-b)(s-c) we can find the 3rd side.

    where i am wrong?
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    Anurag@Gurome GMAT Instructor
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    Post Sun Mar 11, 2012 11:17 pm
    ranjeet75 wrote:
    What is the length of segment BC?

    (1) Angle ABC is 90 degrees. (2) The area of the triangle is 30.

    OA is A but I think the answer should be D because if area and two sides are given then by the formula: Area = sqrt s(s-a)(s-b)(s-c) we can find the 3rd side.

    where i am wrong?
    (1) Angle ABC is 90 degrees.
    Since ABC is right angled at B, we can easily find the length of BC using Pythagoras Theorem; SUFFICIENT.

    (2) The area of the triangle is 30.
    Although Heron's formula is out of GMAT scope, but still let me clarify it here that why statement 2 is NOT sufficient.

    Let us assume that BC = x
    Then s = (5 + 13 + x)/2 = (18 + x)/2 = 9 + (x/2)
    Now area of triangle, A = √[s(s - a)(s - b)(s - c)] or A² = [s(s - a)(s - b)(s - c)]
    (30)² = [9 + (x/2)] * [9 + (x/2) - 5] * [9 + (x/2) - 13] * [9 + (x/2) - x]
    900 = [9 + (x/2)] * [4 + (x/2)] * [-4 + (x/2)] * [9 - (x/2)]
    900 = [81 - (x/2)²][(x/2)² - 16]
    Let (x/2)² = y
    900 = [81 - y][y - 16]
    900 = 81y - y² + 16y - 1296
    y² - 97y + 2196 = 0
    y² - 36y - 61y + 2196 = 0
    y(y - 36) - 61(y - 36) = 0
    (y - 61)(y - 36) = 0
    y = 36, 61
    (x/2)² = 36, (x/2)² = 61, which clearly implies that we are getting 2 values of x, which means 2 values of BC. So, statement 2 is NOT sufficient.

    The correct answer is A.

    I hope that helps.

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    vinaypat Just gettin' started! Default Avatar
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    Post Thu Mar 15, 2012 4:27 am
    What if we draw a perpendicular from point B to line AC. This perpendicular touches AC at D. Assume the length of this line is x

    from 2) 1/2 * x * 13 = 30
    i.e. x = 60/13

    in right angle triangle ABD (ADB is 90) we can find AD = sqrt ( (5*5) - (60/13 * 60/13)) == assume y

    So, DC = 13 - y

    Now in right angle triangle BDC. BC and be found by
    sqrt ( (60/13 * 60/13) + (13-y)(13-y)) == So we can find the solution

    I know this is complicated calculation but in DS we don't have to solve the equations

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    Post Thu Mar 15, 2012 6:58 am
    vinaypat wrote:
    What if we draw a perpendicular from point B to line AC. This perpendicular touches AC at D. Assume the length of this line is x

    from 2) 1/2 * x * 13 = 30
    i.e. x = 60/13

    in right angle triangle ABD (ADB is 90) we can find AD = sqrt ( (5*5) - (60/13 * 60/13)) == assume y
    AD = 25 - 21.6 ~= 4. Sqrt of 4 we get + or - 2

    so AD or y = + or - 2.

    Quote:
    So, DC = 13 - y

    Now in right angle triangle BDC. BC and be found by
    sqrt ( (60/13 * 60/13) + (13-y)(13-y)) == So we can find the solution

    I know this is complicated calculation but in DS we don't have to solve the equations
    Now BC will also have two values based on y. You cannot arrive at a single solution. Hope I am right and this helps !!

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    vinaypat Just gettin' started! Default Avatar
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    Post Thu Mar 15, 2012 4:29 pm
    There is no such thing of taking -ve value when calculating length of the Triangle. Only +ve values are valid as any side of a triangle can't be less then difference of other 2 sides

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    Post Fri Mar 16, 2012 12:55 am
    Anurag@Gurome


    well I do not think that the Heron's formula is out of scope in GMAT. I really do not think so...

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