Question 1 Draw a quadrilateral in the Cartesian plane, whose vertices are (–4, 5), (0, 7), (5, –5) and (–4, –2). Also, find its area.
Class X1 - Maths -Straight Lines Page 211
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see attachment, joining all vertices a quadrilateral ABCD is formed .
we see that ,
area of quadrilateral ABCD = area of traingle ∆ABC + area of traingle ∆ADC
now use the formula,
area of ∆ = 1/2{x₁(y₂ - y₃ ) + x₂(y₃ - y₁) + x₃(y₁ - y₂)}
area of ∆ADC = 1/2 [ (-4) (-2 + 5) +(-4)( -5 - 5) + 5(5 + 2) ]
= 1/2 [ -4 × 3 + (-4) × (-10) + 5 × 7 ]
= 1/2 [ -12 + 40 + 35 ]
= 1/2 [ 28 + 35 ]
= 63/2 sq unit
again, area of ∆ABC = 1/2 [ -4 ( 7 + 5) + 0( -5 - 5) + 5 ( 5 - 7)]
= 1/2 [ -4 × 12 + 0 + 5 × -2 ]
= 1/2| [ -48 - 10]|
= 29 sq unit
now,
area of quadrilateral ABCD = 63/2 + 29
= 121/2 sq unit
we see that ,
area of quadrilateral ABCD = area of traingle ∆ABC + area of traingle ∆ADC
now use the formula,
area of ∆ = 1/2{x₁(y₂ - y₃ ) + x₂(y₃ - y₁) + x₃(y₁ - y₂)}
area of ∆ADC = 1/2 [ (-4) (-2 + 5) +(-4)( -5 - 5) + 5(5 + 2) ]
= 1/2 [ -4 × 3 + (-4) × (-10) + 5 × 7 ]
= 1/2 [ -12 + 40 + 35 ]
= 1/2 [ 28 + 35 ]
= 63/2 sq unit
again, area of ∆ABC = 1/2 [ -4 ( 7 + 5) + 0( -5 - 5) + 5 ( 5 - 7)]
= 1/2 [ -4 × 12 + 0 + 5 × -2 ]
= 1/2| [ -48 - 10]|
= 29 sq unit
now,
area of quadrilateral ABCD = 63/2 + 29
= 121/2 sq unit
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