Question

find the area of quadrilateral ABCD having vertices at A(1,2) , B(1,0) , C(4,0) ,D(4,4)​

Answer 1

Answer:

In triangle ABD

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

=1/2[1(0-4)+1(4-2)+4(2-0)]

=1/2[-4+2+8]

=1/2×6

=3units

In triangle BDC

area ∆BDC=1/2[1(4-0)+4(0-0)+4(0-4)]

=1/2[4+0-16]

=1/2×-12

=6units

area ABCD= areaABD+areaBDC

=3+6

=9units

Answer 2

Given: A(1,2) , B(1,0) , C(4,0) ,D(4,4)​

To find: area of quadrilateral ABCD

Solution:

• To find area of the quadrilateral, we need to divide this in two triangles

• Let them be triangle ABD and triangle BCD.

• Now,

• In triangle BDC

• area of triangle BDC = 1/2 { 1(4-0) + 4(0-0) + 4(0-4) }

           =1/2[4+0-16]

           =1/2×-12

           =6 units

• In triangle ABD

• area of triangle ABD = 1/2 { x1(y2-y3) + x2(y3-y1) + x3(y1-y2)}

           =1/2{ 1(0-4) + 1(4-2) + 4(2-0) }

           =1/2[-4+2+8]

           =1/2×6

           =3 units

• Since, we have got the area of the two triangles, adding them we will get the area of the quadrilateral.

• Area of quadrilateral ABCD = area of triangle ABD + area of triangle BDC

             =3+6

             =9units

Answer:

            Area of quadrilateral ABCD is 9 units