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Find the area of the irregular polygon to the right show all work

Find The Area Of The Irregular Polygon To The Right Show All Work class=

Sagot :

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

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