Question

Find the area of quadrilateral ABCD whose vertices are: A(1,3), B(-3,-5), C(4,1) and D(-2,4).

Answer 1

Given,

• A = (1,3)
• B = (-3,-5)
• C = (4,1)
• D = (-2,4)

Let AC be the diagonal of quadrilateral ABCD.

Therefore,

Area of quadrilateral ABCD= Area of ΔABC + Area of ΔADC

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Area of ΔABC = 1/2[ x₁(y₂ - y₃) - x₂(y₃ - y₁) - x₃(y₁ - y₂) ]

Area = 1/2[ (1)(-5-3) + (-3)(1-3) +(4)(3+5) ]

Area =1/2[-8+6+32]

Area of ΔABC = 15 sq units

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Area of ΔADC = = 1/2[ x₁(y₂ - y₃) - x₂(y₃ - y₁) - x₃(y₁ - y₂) ]

Area = 1/2[ (1)(4-1) + (-2)(1-3) + (4)(3-4) ]

Area = 1/2[3+4-4]

Area of ΔADC = 3/2 sq units

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Hence,

Area of quadrilateral ABCD = 15 + (3/2)

Area of quadrilateral ABCD = 33\2 square units.