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Question
If A(–5, 7), B(–4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD
Solution
Joining B to D, we get two triangles ABD and BCD.
Area of ΔABD = `1/2 [-5(-5-5) - 4(5-7)+4(7+5)]`
`= 1/2 (50 + 8 + 48)`
`= 106/2`
= 53 sq units
Area of ΔBCD = `1/2 [-4(-6-5) - 1(5 + 5) + 4(-5 + 6)]`
`= 1/2 (44 - 10 + 4)`
`= 38/2`
= 19 sq units.
∴ Area of quad. ABCD = Area of ∆ABD + Area of ∆BCD = 53 + 19 = 72 sq units.
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