Solution:
We are given an irregular pentagon. To calculate the area, we will have to break the pentagon into component shapes
We can break the pentagon into a trapezium and a triangle
The dimensions of the trapezium are :
1st base = 3 units
2nd base = 4 units
Height = 3 units
[tex]\begin{gathered} Area\text{ of the trapezium = }\frac{1}{2}(a+b)h \\ =\frac{1}{2}(3+4)(3) \\ =10.5\text{ square unit} \end{gathered}[/tex]The dimensions of the triangle are:
Base= 4 units
Height=2 units
[tex]\begin{gathered} Area\text{ of the triangle = }\frac{1}{2}\text{ x base x height} \\ =\frac{1}{2}\text{ x 4 x 2} \\ =4\text{ square units} \end{gathered}[/tex]Area of the polygon = area of trapezium + area of triangle
= 10.5 square unit + 4 square unit = 14.5 square unit
The answer is 14.5 square units