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