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The coordinates of the triangle ABC are A(3,10), B (1,5) and C (9,3). AD bisects BC, then find the area of triangle ABD. |
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Answer» Coordinates of D are:- {(1 + 9) / 2}, {(5 + 3) / 2} = (5, 4) The vertices of ∆ABD are:- A(3,10), B(1, 5), D(5, 4) Ar(ABD) = \(\frac{ 1}{2}|(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))|\) = 1/2 ({3 × 1} + {1 × -6} + {5 × 5}) = 1/2 (3 - 6 + 25) = 22/2 = 11 sq. units Therefore, the area formed by the ∆ABD is 11 sq.units. |
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