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Find the area of quadrilateral ABCD whose vertices are A(1,1), B(7,-3), C(12,2) and D(7,21). |
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Answer» The coordinates of A,B,C & D are A(1,1),B(7,3),C(12,2),D(7,21).Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACDArea of triangle = 21\u200b ∣[x 1\u200b (y 2\u200b −y 3\u200b )+x 2\u200b (y 3\u200b −y 1\u200b )+x 3\u200b (y 1\u200b −y 2\u200b )]∣Area of triangle ABC = 21\u200b ∣[1(3−2)+7(2−1)+12(1−3)]∣= 21\u200b ∣[1+7−24]∣= 21\u200b (16)=8 sq. unitsArea of triangle ACD = 21\u200b ∣[1(2−21)+12(21−1)+7(1−2)]∣= 21\u200b ∣[−19+240−7]∣= 21\u200b (214)=107 sq. unitsHence, area of quadrilateral ABCD = 107+8=115 sq. units Rtfho |
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