Q29 of 50 Page 1

If A ( -3, 5), B (-2, -7), C (1, -8) and D (6, 3) are the vertices of a quadrilateral ABCD, find its area.

Given vertices of quadrilateral ABCD,


A (-3, 5), B (-2, -7), C (1, -8) and D (6, 3)


We know that area of quadrilateral ABCD = area of Δ ABC + area of Δ ACD and area of Δ ABC = 1/2 |x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) |


Consider Δ ABC


⇒ Area of Δ ABC = 1/2 | (-3) (-7 – (-8)) + (-2) (-8 – 5) + (1) (5 – (-7)) |


= 1/2 | (-3) (1) + (-2) (-13) + 1 (12) |


= 1/2 | -3 – 26 + 12|


= 1/2 |-17|


= 1/2 (17)


⇒ Area of Δ ABC = 17/2 square units


Now in Δ ACD,


⇒ Area of Δ ACD = 1/2 | (-3) (-8 – 3) + (1) (3 – 5) + (6) (5 – (-8)) |


= 1/2 | (-3) (-11) + (1) (-2) + 6 (13) |


= 1/2 | -33 – 2 + 78|


= 1/2 |43|


⇒ Area of Δ ACD = 43/2


∴ Area of quadrilateral = 17/2 + 43/2 = 60/2 = 30 square units


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