Question

. The co-ordinates of the consecutive vertices of a quadrilateral are (-2, -3), (6, -5). (18, 9) and (0, 12). Find the area of the quadrilateral.​

Answer 1

Answer:

244 Sq. units

Step-by-step explanation:

Here we are given with the vertices of a quadrilateral.

So let us name it as ABCD having vertices A(-2, -3) B(6,-5)

C(18,9) and D(0, 12)

To find the area of a quadrilateral let us divide it in two parts which forms a triangle.

On dividing it we have two triangle, namely ABD and BDC

NOW,

here x1= -2, x2 = 6, x3 =0

whereas y1 = -3, y2 = -5, y3= 9

Hence area of triangle ABD

=1/2[x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]

=1/2[-2(-5-9) + 6(9+3) + 0(-3+5)]

=1/2[-2(-14) + 6(12)]

=1/2[28 + 72]

=1/2(100)

=50 Sq. units

Similarly finding area of triangle BDC

area of ️BDC= 144 Sq. units (do it yourself)

Therefore area of quadrilateral ABCD= area of ABD+ BDC

=100 Sq. units + 144 Sq. units

=244 Sq. units