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Geometry

This topic has 2 expert replies and 0 member replies
juanierik Newbie | Next Rank: 10 Posts Default Avatar
Joined
29 Jul 2013
Posted:
3 messages

Geometry

Post Sat Sep 06, 2014 2:30 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    ttp://postimage.org/" target="_blank">

    O is the center of the circle above, OB=2, and angle AOB measures 120 degrees. What is the area of triangular region AOB?

    A 1
    B 2
    C (√3)/2
    D √ 3
    E 2 √ 3

    I assumed that OB and AO are radii of the circle and thus have the same length. With this assumption I came up with area 2 (answer B). This is not correct.

    Why can't I assume that OB and AO are radii?

    OA: D

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    Post Sat Sep 06, 2014 2:37 pm
    Hi juanierik,

    I'm going to give you a "nudge" so that you can retry this question.

    You ARE correct that OA and OB are both radii, so they both have a length of 2.

    1) Draw the triangle on your pad. Notice that it's an Isosceles triangle and fill in all 3 angles.

    2) Cut the triangle in half, right down the middle. You'll now have two identical triangles with right angles. What else do you know about those triangles? Can you calculate all 3 sides? (Think about the angles; what kind of triangles are they?).

    3) Now calculate the area of one of the smaller triangles….

    GMAT assassins aren't born, they're made,
    Rich

    _________________
    Contact Rich at Rich.C@empowergmat.com

    Post Sun Sep 07, 2014 9:08 am
    juanierik wrote:
    ttp://postimage.org/" target="_blank">

    O is the center of the circle above, OB=2, and angle AOB measures 120 degrees. What is the area of triangular region AOB?

    A) 1
    B) 2
    C) (√3)/2
    D) √3
    E) 2√3

    First, since OA and OB are radii, they both have the same length (length 2)
    This means ∆AOB is an ISOSCELES triangle, which means the other 2 angles are 30º each.
    ttp://postimg.org/image/xanlfb1v1/" target="_blank">

    Once I see the 30º angles, I start thinking of the "special" 30-60-90 right triangle, which we know a lot about. If we draw a line from point O so that it's PERPENDICULAR to AB, we can see that we have two 30-60-90 right triangles hiding within this diagram. PERFECT!
    ttp://postimg.org/image/9l2r4csvh/" target="_blank">

    I've added (in purple) a 30-60-90 right triangle with measurements.
    We can use this to determine the lengths of some important sides.
    ttp://postimg.org/image/xrdefhezx/" target="_blank">

    At this point, we can see that our original triangle has a base with length 2√3 and height 1
    ttp://postimg.org/image/6s9jkbsj1/" target="_blank">

    Area = (base)(height)/2
    So, the area of ∆AOB = (2√3)(1)/2
    = √3
    = D

    Cheers,
    Brent

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