Question

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

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

(ii) (1, 2), (6, 2), (5, 3) and (3, 4)

(iii) (-4, -2), (-3, - 5), (3, - 2), (2, 3)

Answer 1

(i)

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

Area of ΔABC

But area cannot be negative

Therefore, Area of ΔADC = 42 square units

Area of ΔADC

But area cannot be negative

Therefore, Area of ADC=38 square units

Now, area of quadrilateral ABCD

Ar. of ABC + Ar. of ADC

=(42+38)

=80 square units

(ii)

Let A(1, 2), B(6, 2), C(5, 3) and D(3, 4) be the given points

Area of ΔABC

Now, Area of quadrilateral ABCD

=Area of ABC + Area of ADC

(iii)

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

Area of ΔABC

Area of ΔACD

But area can’t be negative, hence area of ΔACD=35/2

Now, area of quadrilateral (ABCD) = ar(ΔABC)+ar(ΔACD)

Area (quadrilateral ABCD)=21/2+35/2

Area (quadrilateral ABCD)=56/2

Area (quadrilateral ABCD) = 28 square Units