Question

# For â–³ABC whose vertices are given by A(3,6),B(âˆ’4,7) and C(10,âˆ’7), if A1,B1,C1 represent the mid points of sides opposite to vertices A,B,C respectively, of â–³ABC and A2,B2,C2 represent the mid points of sides opposite to vertices A1,B1,C1 respectively, of â–³A1B1C1 and so on, then which of the following is/are correct?

A
Area (A2B2C2)Area (ABC)=116
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B
Area (A3B3C3)Area (ABC)=116
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C
The centroid of A3B3C3 is (3,2)
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D
The centroid of A2B2C2 is (3,2)
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Solution

## The correct option is D The centroid of â–³A2B2C2 is (3,2)Given vertices of â–³ABC are A(3,6),B(âˆ’4,7) and C(10,âˆ’7) We know that, Centroid of triangle = Centroid of triangle formed by midpoints. So, centroid of triangles â–³ABC=â–³A1B1C1=â–³A2B2C2â€¦=(3âˆ’4+103,6+7âˆ’73)=(3,2) Area of triangle formed by midpoints =14Ã—Area of original triangle. So, Area (â–³A2B2C2)Area (â–³A1B1C1)=14 and Area (â–³A1B1C1)Area (â–³ABC)=14 âˆ´Area (â–³A2B2C2)Area (â–³ABC)=116 Similarly, Area (â–³A3B3C3)Area (â–³ABC)=164

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