Exercise 43 Page 42

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=1/2bh 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 ⇔ A=5( 1/2bh) Let's simplify the expression.

A=5(1/2bh)

RemovePar

Remove parentheses

A=5* 1/2bh

MoveLeftFacToNumOne

a* 1/b= a/b

A=5/2bh

We want to solve this formula for the height h, which means we need to isolate it on one side.

A=5/2bh

MultEqn

LHS * 2=RHS* 2

2A=2*5/2bh

DenomMultFracToNumber

2 * a/2= a

2A=5bh

DivEqn

.LHS /5b.=.RHS /5b.

2A/5b=h

RearrangeEqn

Rearrange equation

h=2A/5b