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superrrom: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Mon Sep 01, 2008 6:27 am; Thanked: 2 times

by superrrom » Sun Sep 28, 2008 4:58 pm

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Source: Beat The GMAT — Problem Solving | Sun Sep 28, 2008 4:58 pm

LSB: Master | Next Rank: 500 Posts; Posts: 229; Joined: Thu Nov 08, 2007 3:47 pm; Thanked: 14 times; GMAT Score:750

by LSB » Sun Sep 28, 2008 6:30 pm

The rectangle with the corners (0,0) (7,0) (0,4) and R has the area:

7*4=28

Now you need to deduct the area of the three triangles that are part of this rectangle and surround PQR

Triangle 1 (0,0),P,Q
(4*3)/2 = 6

Triangle 2: (7,0),R,P
(4*3)/2 = 6

Triangle 3:
(7*1)/2 = 3.5

28-6-6-3.5= 12.5 (Ans A)

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manulath: Senior | Next Rank: 100 Posts; Posts: 35; Joined: Tue Sep 09, 2008 10:12 pm; Location: Lko; Thanked: 3 times

by manulath » Sun Sep 28, 2008 10:36 pm

including the both axis, it forms a trapezoid

deduct the area of two triangle from the area of trapezoid.

METHOD II: CO-ORDINATE

by quick glance you can see that both triangles are 3-4-5 triangles

hence two sides are 5 each

also the lines are perpendicular to each other (can be verified by the slopes)

hence you have base and height of a rt. angled triangle

area = 12.5

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