Question 47625

Dec 23, 2017

Area of parallelogram $A B C D$ is $90$ Sq.unit.

Explanation:

$A \left(- 12 , 2\right) , B \left(6 , 2\right) , C \left(- 2 , - 3\right) \mathmr{and} D \left(- 20 , - 3\right)$

Let a digonal $B D$ is drawn . Then we get two triangles

$\Delta D A B \mathmr{and} \Delta D C B$ ; for $\Delta D A B$ ,

$D \left(- 20 , - 3\right) , A \left(- 12 , 2\right) , B \left(6 , 2\right)$ Area $\Delta D A B =$

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

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

$= | \frac{1}{2} \left(0 - 60 - 30\right) | = | \frac{1}{2} \cdot \left(- 90\right) | = | - 45 | = 45$

For $\Delta D C B$

$D \left(- 20 , - 3\right) , C \left(- 2 , - 3\right) , B \left(6 , 2\right)$ Area $\Delta D C B =$

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

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

$= | \frac{1}{2} \left(100 - 10 + 0\right) | = | \frac{1}{2} \cdot \left(90\right) | = | 45 | = 45$

Area of parallelogram $A B C D = {A}_{D A B} + {A}_{D C B}$

$= 45 + 45 = 90$ Sq.unit