Answer
A. 28.53 units²
Explanation
Finding the area of irregular quadrilateral ABCD, we divide the given figure into shapes (two triangles) as shown below:
Then, we find the area of the two triangles.
Triangle ABD:
[tex]\begin{gathered} Area=\sqrt{s(s-a)(s-b)(s-c)} \\ \\ s=\frac{a+b+c}{2}=\frac{2.89+8.59+8.6}{2}=\frac{20.08}{2}=10.04 \\ \\ Area\text{ }of\text{ }triangle\text{ }ABD=\sqrt{10.04(10.04-2.89)(10.04-8.59)(10.04-8.6)} \\ \\ Area\text{ }of\text{ }triangle\text{ }ABD=\sqrt{10.04(7.15)(1.45)(1.44)} \\ \\ Area\text{ }of\text{ }triangle\text{ }ABD=\sqrt{149.889} \\ \\ Area\text{ }of\text{ }triangle\text{ }ABD=12.24\text{ }unit^2 \end{gathered}[/tex]
Triangle ADC:
[tex]\begin{gathered} Area=\sqrt{s(s-a)(s-b)(s-c)} \\ \\ s=\frac{a+b+c}{2}=\frac{4.3+7.58+8.6}{2}=\frac{20.48}{2}=10.24 \\ \\ Area\text{ }of\text{ }triangle\text{ }ADC=\sqrt{10.24(10.24-4.3)(10.24-7.58)(10.24-8.6)} \\ \\ Area\text{ }of\text{ }triangle\text{ }ADC=\sqrt{10.24(5.94)(2.66)(1.64)} \\ \\ Area\text{ }of\text{ }triangle\text{ }ADC=\sqrt{265.346} \\ \\ Area\text{ }of\text{ }triangle\text{ }ADC=16.29\text{ }unit^2 \end{gathered}[/tex]
Therefore, the area of the quadrilateral ABCD = the Sum of the two triangles
The area of the quadrilateral ABCD = 12.24 units² + 16.29 units² = 28.53 units²
