Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
3. Areas of Regular Polygons
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Exercise 19 Page 632

Find the perimeter and the apothem. Then substitute their values in the formula A= 12ap to find the area of the regular polygon.

93.5m^2

Practice makes perfect

The area of a regular polygon is half the product of the apothem and the perimeter. Note that we are only given the side length.

We will first need to find the perimeter and the apothem of the polygon. Then we will use the formula A= 12ap to find the area.

Finding the Perimeter

Since we have a regular hexagon whose side length is 6m, we can find the perimeter by multiplying 6 by 6. 6* 6= 36m

The perimeter of the given polygon is 36m.

Finding the Apothem

Let's now find the apothem. To do so, we will start by drawing the radii of the hexagon. Be aware that the radii divide a regular hexagon into six congruent isosceles triangles.

Since corresponding angles of congruent figures are congruent, the vertex angles of the isosceles triangles formed by the radii are congruent. Moreover, since a full turn measures 360^(∘), we can divide 360 by 6 to obtain their measures. 360/6=60^(∘) The vertex angles of the isosceles triangles measure 60^(∘) each.

The apothem bisects both the vertex angle of the isosceles triangle and the opposite side of the vertex, which is a side of the hexagon. As a result, a 30^(∘)-60^(∘)-90^(∘) triangle is created. The length of its shorter leg is 6÷ 2=3m.

In this type of special triangle the length of the longer leg, which is the apothem, is sqrt(3) times the length of the shorter leg. Longer Leg: sqrt(3)* 3=3sqrt(3)m Therefore, the length of the apothem is 3sqrt(3)m.

Finding the Area

To find the area of the given regular polygon, we will substitute a=3sqrt(3) and p= 36 in the formula A= 12ap.
A=1/2ap
A=1/2(3sqrt(3))( 36)
Evaluate right-hand side
A=1/2(108sqrt(3))
A=108sqrt(3)/2
A=93.53074
A≈ 93.5
The area of the given regular polygon is 93.5m^2, correct to the nearest tenth.