If the area of triangle formed by the points $ (2 a, b),(a+b, 2 b+a) $ and $ (2 b, 2 a) $ is $ 2 $ sq units, then the area of the triangle whose vertices are $ (a+b, a-b),(3 b-a, b+3 a) $ and $ (3 a-b, 3 b-a) $ will be

Question

If the area of triangle formed by the points $ (2 a, b),(a+b, 2 b+a) $ and $ (2 b, 2 a) $ is $ 2 $ sq units, then the area of the triangle whose vertices are $ (a+b, a-b),(3 b-a, b+3 a) $ and $ (3 a-b, 3 b-a) $ will be,

MARKS App · Accepted Answer

Given A a + b , a - b , B - a + 3 b , 3 a + b & C 3 a - b , - a + 3 b Midpoint of A B = 2 b , 2 a = P Midpoint of B C = a + b , a + 2 b = Q Midpoint of C A = 2 a , b = R Given the area of triangle P Q R = 2 sq. units Since, triangle P Q R is formed by joining the mid points of the sides of triangle A B C , Therefore, area of triangle A B C = 4 × area of triangle P Q R = 8 sq. units

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