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rectangular coordinate system

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by shibal » Sun Jul 12, 2009 1:27 pm
no idea how to do it....
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by nitya34 » Sun Jul 12, 2009 10:36 pm
solve it this way


[Area of Trapezium - Area of the left Triangle - Area of the right triangle]

=49/2 - 12/2 - 12/2
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by rah_pandey » Mon Jul 13, 2009 2:37 am
Its a isoceles rt angled triangle.
1/2*5*5.

Since coords are given, it would have been a good strategy to check what is the length of the sides. since nothing else was working
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by Mayur Sand » Mon Jul 13, 2009 9:32 am
Plz explain how did you get Area of trapezium as 49/2 ?
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by shibal » Mon Jul 13, 2009 4:29 pm
i didn't notice that there were two 3,4,5 triangles... thanks guys
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by ymach3 » Tue Jul 14, 2009 2:03 am
could someone explain what was considerd as Base and Height of Triangle PQR?

Was it 1/2*PQ*RS ( S being the Mid pt of PQ).???
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by kaulnikhil » Tue Jul 14, 2009 3:06 pm
use the determinat formulae ; smallest possible way
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by ymach3 » Tue Jul 14, 2009 3:58 pm
May i have the formula please?
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by kaulnikhil » Wed Jul 15, 2009 3:02 am
check out the file attached for formulae!!
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by cata1yst » Wed Jul 15, 2009 6:17 am
I have an easy way to solve this problem...

Make a rectangle out of the triangle and then subtract the area of the three small triangles.

We get:

Area of rectangle = 7 * 4 = 28

Area of bottom right triangle = (3 * 4) / 2 = 6

Area of bottom left triangle = (3 * 4) / 2 = 6

Area of top triangle = (7 * 1) / 2 = 3.5

28 - 6 - 6 - 3.5 = 12.5

A
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