Question

Question #47625

Answer 1

Area of parallelogram `ABCD` is `90` Sq.unit.

Explanation:

`A(-12,2) , B(6,2),C(-2,-3) and D(-20,-3)`

Let a digonal `BD` is drawn . Then we get two triangles

`Delta DAB and Delta DCB` ; for `Delta DAB` ,

`D(-20,-3),A(-12,2) , B(6,2)` Area `Delta DAB =`

`A_(DAB) = |1/2(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))|`

`=|1/2(-20(2-2)+(-12(2+3)+6(-3-2)|`

`=|1/2(0-60-30)| =|1/2 *(-90)| = | -45| =45`

For `Delta DCB`

`D(-20,-3),C(-2,-3) , B(6,2)` Area `Delta DCB =`

`A_(DCB) = |1/2(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))|`

`=|1/2(-20(-3-2)+(-2(2+3)+6(-3+3)|`

`=|1/2(100-10+0)| =|1/2 *(90)| = | 45| =45`

Area of parallelogram `ABCD = A_(DAB)+A_(DCB)`

`=45+45 =90` Sq.unit