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Question
Answer the following question:
Find the area of quadrilateral whose vertices are A(0, −4), B(4, 0), C(−4, 0), D(0, 4)
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Solution
A(0, −4), B(4, 0), C(−4, 0), D(0, 4)
A`(square"ABDC")` = A(ΔABC) + A(ΔBDC)
Area of a triangle = `1/2|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|`
∴ A(ΔABC) = `1/2|(0, -4, 1),(4, 0, 1),(-4, 0, 1)|`
= `1/2[0 - (-4)(4 + 4) + 1(0 - 0)]`
= `1/2(32)`
= 16 sq. units
A(ΔBDC) = `1/2|(4, 0, 1),(0, 4, 1),(-4, 0, 1)|`
= `1/2[4(4 - 0) - 0 + 1(0 + 16)]`
= `1/2[4(4) + 1(16)]`
= `1/2(16 + 16)`
= `1/2(32)`
= 16 sq. units
∴ A`(square"ABDC")` = A(ΔABC) + A(ΔBDC)
= 16 + 16
= 32 sq. units
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