If the area of the triangle formed by the points 2a,b a+b, 2b+a, and 2b,2a is 2 sq. units, then the area of the triangle whose vertices and a+b,a−b,3b−a, b+3a and 3a−b,3b−a will be

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answer is 8.

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Detailed Solution

We know that the area of the triangle formed by joining the midpoints of any triangle is one-fourth of that triangle. Therefore,Required area =8

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