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MGMAT Geometry - p. 119, Question #6

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bml1105 Master | Next Rank: 500 Posts
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MGMAT Geometry - p. 119, Question #6

Post Sat Jan 11, 2014 6:07 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    ttp://postimg.org/image/8fhg9g5s7/" target="_blank">

    Can someone explain how to get the answer? I don't understand the MGMAT explanation at all.

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    Post Sat Jan 11, 2014 9:32 pm
    ttp://postimg.org/image/vu799tqsh/" target="_blank">

    I added some letters to help guide the solution.

    Area of triangle = (1/2)(base)(height)
    IMPORTANT CONCEPT: we can use ANY of the three sides as our base.

    So, for example, if we want to find the area of triangle ABC, we can use side AB as the base, or we can use side AC as the base, or we can use side BC as the base.

    If we use side AB as the base, then the base has length 12 and the height is 3
    So, area of triangle ABC = (1/2)(12)(3)

    If we use side AC as the base, then the base has length 7 and the height is x
    So, area of triangle ABC = (1/2)(7)(x)

    IMPORTANT: If we use side AB as the base, the area of the triangle will be the same as the area we get if we use side AC as the base.

    So, (1/2)(12)(3) = (1/2)(7)(x) [solve for x]
    Divide both sides by 1/2 to get: (12)(3) = (7)(x)
    Divide both sides by 7 to get: 36/7 = x

    Cheers,
    Brent

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    Post Sun Jan 12, 2014 4:23 am
    As Brent mentioned, any side of a triangle can be deemed the base, and each base has a corresponding height.
    More practice with this concept:

    http://www.beatthegmat.com/geometry-helppppp-t117047.html
    http://www.beatthegmat.com/geometry-t271788.html
    http://www.beatthegmat.com/need-to-understand-the-concept-based-solution-t268314.html

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    bml1105 Master | Next Rank: 500 Posts
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    Post Sun Jan 12, 2014 11:08 am
    Thanks Brent & GMATGuru!

    I think I was getting mixed up because I was trying to find the area of the entire diagram (included the dotted lines) without realizing that I didn't actually need to do that to find x.

    stuffs88 Newbie | Next Rank: 10 Posts Default Avatar
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    Post Thu Apr 10, 2014 8:25 am
    Hi,

    Can you please explain the concept with similar triangles? Which 2 triangles are they comparing in the book?

    Thank You!

    Post Thu Apr 10, 2014 8:49 am
    stuffs88 wrote:
    Hi,

    Can you please explain the concept with similar triangles? Which 2 triangles are they comparing in the book?

    Thank You!
    ttp://postimg.org/image/vu799tqsh/" target="_blank">

    ∆ABE is similar to ∆ACD
    We know this because ∠BAE = ∠CAD, and because ∠AEB = ∠ADC (both are 90º)
    If two triangles share 2 equal angles, their 3rd angles must also be equal.
    If the triangles share 3 equal angles, they are SIMILAR TRIANGLES, which means the ratios of their respective sides must be equal.

    The hypotenuse of ∆ABE has length 12. The hypotenuse of ∆ACD has length 7.
    In ∆ABE, the side opposite ∠A has length x. In ∆ACD, the side opposite ∠A has length 3.

    So, we can write: 12/7 = x/3
    Cross multiply to get: (12)(3) = 7x
    Divide both sides by 7 to get: x = 36/7

    Cheers,
    Brent

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