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in Straight Lines by (56.3k points)

Find the area of the quadrilaterals, the coordinates of whose vertices are

(-3, 2), (5, 4), (7, -6) and (-5, – 4)

1 Answer

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by (30.5k points)
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Best answer

Let A(-3, 2), B(5, 4), C(7,- 6) and D ( -5, – 4) be the given points.

Area of ∆ABC is given by

= \(\frac{1}{2}\)[-3(4 + 6) + 5(- 6 – 2) + 7(2 – 4)]

= \(\frac{1}{2}\)[-3×1 + 5×(-8) + 7(-2)]

= \(\frac{1}{2}\)[- 30 – 40 -14]

= – 42

As the area cannot be negative,

The area of ∆ADC = 42 square units

Now, area of ∆ADC is given by

= \(\frac{1}{2}\)[-3( – 6 + 4) + 7(- 4 – 2) + (- 5)(2 + 6)]

= \(\frac{1}{2}\)[- 3( – 2) + 7(- 6) – 5 × 8]

= \(\frac{1}{2}\)[6 – 42 – 40]

= \(\frac{1}{2}\) × – 76

= – 38

But, as the area cannot be negative,

The area of ∆ADC = 38 square units

Thus, the area of quadrilateral ABCD = Ar. of ABC+ Ar. of ADC

= (42 + 38)

= 80 sq. units

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