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InterviewSolution
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If A (-3, 5), B (-2,-7), C (1,-8) and D (6, 3) are the vertices of a quadrilateral ABCD, find its area. |
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Answer» Given vertices of a quadrilateral ABCD are A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) Area of the quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3) = \(\frac{1}2\) |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)| Area of ∆ABC = \(\frac{1}2\) | − 3[ − 7 − ( − 8 )] + ( − 2) ( − 8 – 5 ) + 1 [ 5 − (− 7) ] | = \(\frac{1}2\) | − 3 + 26 + 12 | = \(\frac{35}2\) sq. units Area of ∆ACD = \(\frac{1}2\) |− 3( − 8 – 3 ) + 1( 3 – 5 ) + 6[ 5 − ( − 8 ) ] | = \(\frac{1}2\) | 33 – 2 + 78 | = \(\frac{109}2\) sq. units Area of the quadrilateral ABCD = \(\frac{35}2\) + \(\frac{109}2\) = 72 sq. units ∴Hence, the area of the quadrilateral is 72 sq. units. |
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