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Source: — Problem Solving |

by Rahul@gurome » Sun Sep 05, 2010 5:39 am
Area of the triangle when coordinates of the vertices are (x1, y1), (x2, y2), (x3, y3)
= {|x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)|}/ 2

So, Area of PQR = {|4 (3 - 4) + 0 (4 - 0) + 7 (0 - 3)|}/ 2 = |-4 - 21|/2 = 25/2 = 12.5 sq units

The correct answer is [spoiler](A)[/spoiler].
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by GMATGuruNY » Sun Sep 05, 2010 8:03 am
Estimating is the quickest way to solve this problem. See the attached .pdf. The area of the rectangle drawn around triangle PQR is 28. PQR takes up less than half the rectangle, so PQR < 14. Only answer choice A works: 12.5 < 14.
Attachments
BTG_area PQR.pdf
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by kanaka » Sun Sep 05, 2010 9:17 am
I found a slightly different approach.
From the rectangle area subtract the three triangles (T1, T2 and T3) areas, please see the pic.
Area of the rectangle = 7*4 = 28
Area of T1 = 1/2*4*3 = 6
Area of T2 = 1/2 * 3 * 4 = 6
Area of T3 = 1/2 *1 * 7 = 7/2
Area of PQR = 28 - 6 - 6 - 7/2 = 16 - 7/2 = 25/2 = 12.5.
Thanks.
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by Ankur Barasia » Sun Sep 05, 2010 11:53 am
I also have a different approach...
first calculate the value of QR = (50)^0.5
then calculate mid point of QR, suppose S = (7/2,7/2)
now calculate PS= (50/4)^0.5
Area of triangle will be = QR.PS/2 = 12.5 units
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by kanaka » Sun Sep 05, 2010 12:25 pm
It worked for this question because PQ = PR but it in not guaranteed to be the mid point.
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