Coordinate Plane - triangle

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szDave Senior | Next Rank: 100 Posts Default Avatar
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Coordinate Plane - triangle

Post Mon Jan 21, 2013 10:52 am
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    I calculated the rectangulars area and subtracted the area of 3 perpendicular triangle, and got 8, but this is the wrong answer.


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    Post Mon Jan 21, 2013 11:13 am
    Sure, let's draw a rectangle around the triangle (as shown below) and then subtract from the rectangle's area (28) the areas of the 3 right triangles that surround the triangle in question.

    We get the following:


    So, the area of PQR = 28 - (3.5 + 6 + 6) = 12.5

    Answer = A

    Cheers,
    Brent

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    Post Mon Jan 21, 2013 7:36 pm
    szDave wrote:
    I calculated the rectangulars area and subtracted the area of 3 perpendicular triangle, and got 8, but this is the wrong answer.


    Here's one more approach, other than the one already explained by Brent.

    Formula for finding area of a triangle in a coordinate system = (1/2){(x1 - x2).(y2 - y3) - (y1 - y2).(x2 - x3)}

    In the given question, area of triangle = 1/2 {(-3)(1) - (-4).(7)} = 12.5 sq units

    The correct answer is A.

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    Post Tue Jan 22, 2013 7:25 am
    Here's an easier way to approach the problem: generally speaking, when the GMAT asks you to find areas of triangles in a coordinate plane, it's going to be an easy triangle to calculate. This almost always means a RIGHT TRIANGLE (or perhaps an equilateral that we can split into right triangles).

    Can we figure out if this is a right triangle? It looks like it. To know for sure, though, we'd have to prove that two of the sides are perpendicular. To do that, we need to calculate the slopes (vertical change/horizontal change, or rise/run)

    The side from Q to P has a slope of -3/4 (a rise of -3, a run of 4).
    The side from P to R has a slope of 4/3 (a rise of 4, a run of 3)

    The slopes are negative reciprocals, so the lines are perpendicular. It's a RIGHT TRIANGLE! Now all we have to do is figure out the length of sides QP and PR. It's probably not going to be hard to calculate these lengths - they are almost always "special" right triangles.

    To find QP, imagine that it's the hypotenuse of the triangle formed by the x and y axes. One side has a length of 3, the other is 4, so the hypotenuse - QP - must be 5.

    To find PR, draw an imaginary line down to the x axis, forming another triangle. It's a 3-4-5 triangle again!

    Both QP and PR have a length of 5, so we take (1/2)(base * height):

    (1/2)(5*5) = 12.5

    The answer is A.

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    Post Tue Jan 22, 2013 8:05 am
    Quote:
    In the rectangular coordinate system below, the are of triangular region PQR is

    12.5
    14
    10√2
    16
    25


    The area of the rectangle drawn around triangle PQR = 7*4 = 28.
    Since triangle PQR takes up less than half the rectangle, PQR < 14.

    The correct answer is A.

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    szDave Senior | Next Rank: 100 Posts Default Avatar
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    Post Wed Jan 23, 2013 3:06 am
    I am impressed by you! 4 ways to solve it! Smile

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