Question

If A(–7, 5), B(–6, –7), C(–3, –8) and D(2, 3) are the vertices of a quadrilateral ABCD then find the area of the quadrilateral.

Answer 1

Consider the figure.
Construction: Produce AC by joining points A to C to form two triangles, $∆\mathrm{ABC}\mathrm{and}∆\mathrm{ADC}$.
In $∆\mathrm{ABC},$
$x_1=-7,x_2=-6\mathrm{and}{x}_3=-3;y_1=5,y_2=-7\mathrm{and}{y}_3=-8$
We know that,
$$\begin{gathered} ar(∆\mathrm{ABC})=\left|\frac{1}{2}\left\{\left(x_1\right)(y_2-y_3)+\left(x_2\right)(y_3-y_1)+\left(x_3\right)(y_1-y_2)\right\}\right| \\ \Rightarrow ar(∆\mathrm{ABC})=\left|\frac{1}{2}\left\{(-7)(-7+8)+(-6)(-8-5)+(-3)(5+7)\right\}\right| \\ \Rightarrow ar(∆\mathrm{ABC})=\left|\frac{1}{2}\left\{(-7)\left(1\right)+(-6)(-13)+(-3)\left(12\right)\right\}\right| \\ \Rightarrow ar(∆\mathrm{ABC})=\left|\frac{1}{2}\left\{(-7)+78+(-36)\right\}\right| \\ \Rightarrow ar(∆\mathrm{ABC})=\left|\frac{1}{2}\left\{35\right\}\right| \\ \Rightarrow ar(∆\mathrm{ABC})=\frac{35}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$
Similarly, in $∆\mathrm{ADC},$
$x_1=-7,x_2=2,x_3=-3,y_1=5,y_2=3\mathrm{and}{y}_3=-8$
$$\begin{gathered} \Rightarrow ar(∆\mathrm{ADC})=\left|\frac{1}{2}\left\{(-7)(3+8)+\left(2\right)(-8-5)+(-3)(5-3)\right\}\right| \\ \Rightarrow ar(∆\mathrm{ADC})=\left|\frac{1}{2}\left\{(-7)\left(11\right)+\left(2\right)(-13)+(-3)\left(2\right)\right\}\right| \\ \Rightarrow ar(∆\mathrm{ADC})=\left|\frac{1}{2}\left\{(-77)-26-6\right\}\right| \\ \Rightarrow ar(∆\mathrm{ADC})=\left|\frac{1}{2}\left\{-109\right\}\right| \\ \Rightarrow ar(∆\mathrm{ADC})=\frac{109}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$
Now, ar(quad. ABCD) = $ar(∆\mathrm{ABC})+ar(∆\mathrm{ADC})$

ar(quad. ABCD) = $\frac{35}{2}+\frac{109}{2}=\frac{144}{2}=72$
Therefore, area of quadrilateral ABCD is 72 sq. units

Disclaimer: The answer thus calculated does not match with the answer given in the book.
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