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Help!

This topic has 2 expert replies and 1 member reply
RiyaR Senior | Next Rank: 100 Posts Default Avatar
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Help!

Post Mon Oct 27, 2014 7:30 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    What is the area of a triangle created by the intersections of the lines x=4, y=5 and y=-{3/4}x+20?

    (A) 42

    (B) 54

    (C) 66

    (D) 72

    (E) 96

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    Post Mon Oct 27, 2014 7:33 pm
    Quote:
    What is the area of a triangle created by the intersections of the lines x=4, y=5 and y=-3/4x+20?

    A)42

    B)54

    C)66

    D)72

    E)96
    DRAW the figure:
    ttp://postimg.org/image/r6ppg5itp/" target="_blank">

    Vertex A is the intersection of x=4 and y=5:
    (4, 5).

    Vertex B is the intersection of x=4 and y=(-3/4)x + 20.
    Plugging x=4 into y=(-3/4)x + 20, we get:
    y = (-3/4)4 + 20
    y = 17.
    Thus, the coordinates of vertex B are (4, 17).

    Vertex C is the intersection of y=5 and y=(-3/4)x + 20.
    Plugging y=5 into y=(-3/4)x + 20, we get:
    5 = (-3/4)x + 20
    -15 = (-3/4)x
    -60 = -3x
    x = 20.
    Thus, the coordinates of vertex C are (20, 5).

    In triangle ABC, AC=16 and AB=12.
    Thus, the area of triangle ABC = (1/2)(16)(12) = 96.

    The correct answer is E.

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    Post Mon Oct 27, 2014 7:41 pm
    Quote:
    What is the area of a triangle created by the intersections of the lines x = 4, y = 5, and y= -3/4x + 20?

    A)42

    B)54

    C)66

    D)72

    E)96

    Let's first sketch the lines x = 4 and y = 5
    ttp://postimg.org/image/vtahxxyln/" target="_blank">

    To find the point where y = (-3/4)x + 20 intersects the line x = 4, replace x with 4 to get: y = (-3/4)4 + 20 = 17
    So the point of intersection is (4, 17)

    To find the point where y = (-3/4)x + 20 intersects the line y = 5, replace y with 5 to get: 5 = (-3/4)x + 20
    When we solve for x, we get x = 20
    So the point of intersection is (20, 5)

    Add this information to our sketch:
    ttp://postimg.org/image/dezytymaz/" target="_blank">

    From here, we can determine the length of the right triangle's base and height:
    ttp://postimg.org/image/qx6v68ygb/" target="_blank">

    Area = (1/2)(base)(height)
    = (1/2)(16)(12)
    = 96
    = E

    Cheers,
    Brent

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    Post Mon Oct 27, 2014 9:52 pm
    Hi RiyaR,

    As Mitch's and Brent's solutions have shown, it helps to be able to graph lines and equations. If a given line is horizontal (Y = a number) or vertical (X = a number), then those lines can be immediately placed into a graph. For any other line though, the ability to TEST VALUES and plot co-ordinates is usually helpful. Are you comfortable doing this? If so, then you'll find that most graphing questions are actually pretty straight-forward. Since you're not going to see that many graphing questions on Test Day (normally just 1-2), that category is not a big "point gainer", so you have to spend your study time accordingly. If you're close to your goal score, then nitpicking some of the easier, smaller categories (such as graphing) makes sense; if you're far from your goal score, then an emphasis on the bigger categories is warranted.

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

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