The area of a regular polygon is half the product of the apothem and the perimeter. We will first find the apothem and then the side length to obtain the perimeter. Finally, we will use this information to find the area.
Let's do it!
Apothem
By drawing the three radii, we can divide the equilateral triangle into three isosceles triangles. Since the triangles are congruent and a full turn measures 360^(∘), the central angles of the isosceles triangles formed by the radii measure 3603=120^(∘).
Now, let's consider just one of these isosceles triangles. We will also draw the apothem a of the regular polygon, which is perpendicular to the side.
Remember that an apothem bisects the central angle and the side of the regular polygon. Therefore, we obtain a right triangle with an acute angle that measures 60^(∘).
By the Triangle Angle Sum Theorem we know that the sum of the three interior angles of a triangle add to 180^(∘). With this information we can find the measure of the unknown acute angle.
180- 90- 60=30^(∘)
The third angle of the right triangle measures 30^(∘). Therefore, we have a 30^(∘)-60^(∘)-90^(∘) triangle.
In this type of special triangle, the shorter leg is 12 times the hypotenuse. Keep in mind that the shorter leg of this triangle is the apothem of the polygon.
a=1/2 * 2= 1
The apothem of the equilateral triangle is 1m.
Perimeter
Consider the 30^(∘)-60^(∘)-90^(∘) triangle one more time.
In this type of special triangle, the longer leg is sqrt(3)2 times the hypotenuse.
Longer Leg:=sqrt(3)/2 * 2= sqrt(3)
As previously mentioned, the apothem bisects the side of the equilateral triangle. Therefore, the length of the side of the given polygon is twice the length of the longer leg of the above triangle.
Consequently, the side length of the equilateral triangle is 2sqrt(3)m. Since this polygon has three congruent sides, we will multiply the side length by 3 to find its perimeter.
Perimeter: 3* 2sqrt(3)=6sqrt(3)m
Area
Now that we know that the apothem of the figure is 1m and that the perimeter is 6sqrt(3)m we are ready to find its area. To do this, we will substitute these values in the formula A= 12ap. Let's do it!
