Start by finding the side length so that you can find the perimeter.
Then, find the apothem.
Finally, substitute the values of the perimeter and the apothem into the formula A= 12ap to find the area of the polygon.
We will first calculate the side length to find the perimeter.
Then, we will find the apothem.
Finally, since the area of a regular polygon is half the product of the apothem and the perimeter, we will use the obtained values to find the area. Let's do it!
Side Length and Perimeter
By drawing the 12 radii we can divide the polygon into 12 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 36012=30^(∘).
Now, let's consider just one of the isosceles triangles and the apothem a.
Recall that the apothem bisects the angle and the side of the regular polygon. Therefore, we obtain a right triangle with an acute angle that measures 15^(∘). Let x be the shorter leg of this right triangle.
This is a right triangle with a hypotenuse that measures 1mi. To find the length of the side that is opposite to the acute angle that measures 15^(∘), we will use the sine ratio.
sin θ =Opposite/Hypotenuse
Let's substitute the corresponding values into this ratio and solve for x.
The length of the shorter leg of the right triangle is sin 15^(∘) mi. As previously mentioned, the apothem bisects the side of the polygon. Therefore, the side length is twice the length of the shorter leg of the triangle.
Finally, since we have a 12-sided regular polygon we can find the perimeter by multiplying 12 by the side length.
The perimeter of the given regular polygon is 6.2mi, to the nearest tenth.
Apothem and Area
Once more, let's consider the right triangle whose longer leg is the apothem of the polygon.
Remember, this a right triangle with a hypotenuse that measures 1mi. To find the length of the side that is adjacent to the acute angle that measures 15^(∘), we will use the cosine ratio.
cos θ =Adjacent/Hypotenuse
Let's substitute the corresponding values into this ratio and solve for a — the length of the longer leg of the triangle — and the apothem of the polygon.
We found that the apothem is cos 15^(∘)mi. We already found that the perimeter is about 6.2mi. To find the area, we will substitute these values in the formula A= 12ap.
