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The fastest method to do this is on a graph paper.
The sketch pad provided in the exam can be used as a coordinate pad. If you draw the figure to scale on the pad, the area can be calculated by counting the number of complete squares that are covered in the figure of the triangle. It will take less than a minute to figure out that the area will be greater than 24 but less than 36. Hence 30 will be the answer.

You can google- virtual graph paper-to practice more of this technique.
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If solving by subtracting the 3 right triangles that result from drawing a rectangle/square around the original triangle (saves from memorizing a formula):

    1. Draw rough picture of triangle and rectangle around it for reference. Coordinates of rectangle are (-4,3), (4,3), (4,-5), and P (-4,-5) based on the max and min x- and y-coordinates of P, Q, R
    2. Distance between the coordinates shows that this rectangle is a square with sides 8 units long
    3. Area of square = \(side^2\) = \(8^2\) = 64
    4. Distance between the coordinates of square corners and the points forming the triangle should give you three right triangles, where two have base/height of 8 and 2 and the third has a base/height of 6 each
    5. Areas of these triangles based on A= \(\frac{1}{2}\)*b*h are \(\frac{1}{2}\)*2*8 = 8, \(\frac{1}{2}\)*2*8 = 8, and \(\frac{1}{2}\)*6*6 = 18
    6. Area of triangle of interest = 64 - 8 - 8 - 18 = 30, which is D
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Area=1/2(∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣)


=1/2[-4(6)-2(2)+4(-8)]

= ½ |(-24-4-32)|
=60/2 =30


Option D
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