Question

Find the area of the quadrilateral ABCD, whose vertices are A(−3, −1), B (−2, −4), C(4, − 1) and D (3, 4).

Answer 1

The given quadrilateral i.e., ABCD whose vertices are A (−3, −1), B (−2, −4), C (4, −1) and D (3, 4) can be drawn as follows: 

Here, B is joined with D. 

We know that the area of a triangle whose vertices are (x₁ , y₁ ), (x₂ , y₂ ) and (x₃ , y₃ ) is given by 

${\displaystyle =\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]}$

${\displaystyle =\frac{1}{2}[-3(-8)-2\left(5\right)+3\left(3\right)]}$

${\displaystyle =\frac{1}{2}[24-10+9]}$

${\displaystyle =\frac{23}{2}}$

${\displaystyle =11.5sq.\in its}$

∴ar(ΔABD)

${\displaystyle =\frac{1}{2}[-3(-4-4)+(-2)(4+1)+3(-1+4)]}$

∴ar (ΔCDB)

${\displaystyle =\frac{1}{2}[4(4+4)+3(-4+1)+(-2)(-1-4)]}$

${\displaystyle =\frac{1}{2}[(4\times 8)+(3x-3)-2\times (-5)]}$

${\displaystyle =\frac{1}{2}[32-9+10]}$

${\displaystyle =\frac{33}{2}}$

${\displaystyle =16.5sp.unit}$

Thus, ar (ABCD) = ar (ΔABD) + ar (ΔCDB) = (11.5 + 16.5) sq units = 28 sq units