If 𝐴(–5, 7), 𝐵(–4, –5), 𝐶(–1, –6) 𝑎𝑛𝑑 𝐷(4, 5)are the vertices of a quadrilateral, find the area of the quadrilateral 𝐴𝐵𝐶𝐷

Given that, 𝐴(–5, 7), 𝐵(–4, –5), 𝐶(–1, –6) 𝑎𝑛𝑑 𝐷(4, 5)are the vertices of a quadrilateral 𝐴𝐵𝐶𝐷 Now we have to find the area of the quadrilateral 𝐴𝐵𝐶𝐷

Joining the vertices 𝐴 𝑎𝑛𝑑 𝐶 we get two triangles 𝐴𝐵𝐶 & 𝐴𝐷𝐶  

So, area of the quadrilateral   𝐴𝐵𝐶𝐷 is equal to the  sum of areas of the two triangles 𝐴𝐵𝐶 & 𝐴𝐷𝐶. 

 We know that,

Area of triangle$$= \frac{1}{2}\left|x₁\left(y₂-y₃\right)+x₂\left(y₃-y₁\right)+x₃\left(y₁-y₂\right)\right| sq units$$ Here, 𝐴(–5, 7), 𝐵(–4, –5), 𝐶(–1, –6)

$\begin{array}{r}\mathrm{area}\left(\mathrm{△}ABC\right)=\frac{1}{2}|-5(-5+6)-4(-6-7)-1(7+5\left)\right|\end{array}$

                            $\begin{array}{r} \\ =\frac{1}{2}|-5(1)-4(-13)-1(12\left)\right|\end{array}$

                            $=\frac{1}{2}|-5+52-12|$

                                $\begin{array}{r}=\frac{1}{2}\left|35\right|\\ =\frac{35}{2}\text{ sq. units }\end{array}$   

$$\Rightarrow area\left(𝛥ABC\right)= \frac{35}{2} sq. units$$ 

$here, A(-5,7), D(4,5), C(-1,-6)$

$\begin{array}{r}\mathrm{area} \left(\mathrm{\Delta }ADC\right)=\frac{1}{2}|-5(5+6)+4(-6-7)-1(7-5\left)\right|\end{array}$

$\begin{array}{r}=\frac{1}{2}|-5(11)+4(-13)-1(2\left)\right|\end{array}$

$\begin{array}{r}=\frac{1}{2}|-55-52-2|\end{array}$

$\begin{array}{r}=\frac{1}{2}|-109|\end{array}$

$\begin{array}{r}\Rightarrow \text{ area }\left(\mathrm{\Delta }ADC\right)=\frac{109}{2}\text{ sq. units }\\ \end{array}$

$$\begin{gathered} Now, Area of quadrilateral ABCD = area\left(\Delta ADC\right)+ area\left(\Delta ABC\right) \\ =\frac{35}{2}+\frac{109}{2} sq. units \\ =\frac{144}{2} \\ =72 sq units \end{gathered}$$

$\Rightarrow Area of quadrilateral 𝐴𝐵𝐶𝐷 = 72 𝑠𝑞. 𝑢𝑛𝑖𝑡𝑠$