Let's divide the regular polygon into $5$ congruent triangles, where each has a base $b$ and a height $h .$

The area of the pentagon is equal to the sum of the areas of the

5

identical triangles. We can calculate the area

A_{T}

of each of them using the formula for the area of a triangle. In the formula

b

stands for the base and

h

stands for the corresponding height of the triangle.

A_{T} = \frac{1}{2} b h

If we multiply the above expression by

5,

we get the the total area of the

5

triangles making up the polygon. Therefore, we get the formula for the area

A

of the regular polygon.

A = 5 A_{T} \Leftrightarrow A = 5 (\frac{1}{2} b h)

Let's simplify the expression.

A = 5 (\frac{1}{2} b h)

Remove parentheses

A = 5 \cdot \frac{1}{2} b h

$a \cdot \frac{1}{b} = \frac{a}{b}$

A = \frac{5}{2} b h

We want to solve this formula for the height

h,

which means we need to isolate it on one side.

A = \frac{5}{2} b h

$LHS \cdot 2 = RHS \cdot 2$

2 A = 2 \cdot \frac{5}{2} b h

$2 \cdot \frac{a}{2} = a$

2 A = 5 b h

$LHS / 5 b = RHS / 5 b$

\frac{2 A}{5 b} = h

Rearrange equation

h = \frac{2 A}{5 b}

Exercise 43 Page 42

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