geometry

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geometry

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