Which of the following is the formula that is used to calculate the area of a regular 𝑛-sided polygon with side length 𝑥? Is it option (A) 𝑛𝑥 squared divided by four times the cot of 𝜋 by 𝑛? Option (B) 𝑛𝑥 squared divided by four times the tan of 𝜋 by 𝑛. Is it option (C) 𝑛𝑥 squared divided by two tan of 𝜋 by 𝑛? Option (D) 𝑛𝑥 divided by four tan of 𝜋 by 𝑛. Or is it option (E) 𝑛𝑥 squared divided by four times the tan of 𝜋?
In this question, we’re asked to recall which of five given options is the formula for calculating the area of a regular 𝑛-sided polygon with side length 𝑥. And of course, we could just answer this question by recalling the formula for the area of a regular polygon. We recall the area of a regular 𝑛-sided polygon of side length 𝑥 with internal angle measured in radians is given by 𝑛𝑥 squared multiplied by the cot of 𝜋 by 𝑛.
And this is not exactly the same as any of the five given options. However, we can rewrite this by recalling multiplying by the cotangent of an angle is the same as dividing by the tangent of an angle. This then gives us 𝑛𝑥 squared divided by four times the tan of 𝜋 by 𝑛, which we can see is option (B). And we could stop here since it’s very useful to commit this formula to memory. However, it’s also useful to see where this formula comes from.
So let’s derive this formula by considering a regular pentagon with side length 𝑥. And before we start with this proof, it’s worth noting this proof generalizes to any regular 𝑛-sided polygon. However, we’re just going to consider a pentagon in this case. To find the area of this pentagon, we’re going to split the pentagon into five triangles. We’re going to do this by connecting each of the vertices of this pentagon to the center of the pentagon. And at this point, there’s many different ways of showing all five of these triangles are congruent.
One way is to note that each line from the vertex to the center of the pentagon is a perpendicular bisector of the interior angle of the pentagon. Therefore, all of these angles are equal. And of course, we know we have a regular pentagon. So all of the side lengths of the pentagon are side length 𝑥. And then by using the angle–side–angle criterion, we can show all five of these triangles are congruent. In particular, we can also show that all five of these triangles are isosceles because they have two equal angles. Since these five triangles make up the pentagon and they’re all congruent, the area of the pentagon is five times the area of one of these triangles.
And we can recall the area of a triangle is one-half the length of the base multiplied by its perpendicular height. By using one of the side lengths 𝑥 as the base of this triangle, the area of one of these triangles is one-half 𝑥 multiplied by ℎ. To find the value of ℎ, we’re going to need to find the angle at the center of this pentagon. We know that each of the five triangles is congruent, so each of the five angles at the center of this pentagon are equal. And since these five angles sum to give a full revolution, the sum of their measures is two 𝜋. This means that they must have measure two 𝜋 by five.
We can now use this to determine one of the angles in the right triangle involving ℎ. We need to use the fact that the perpendicular bisector in an isosceles triangle bisects the angle. Therefore, the measure of this angle is two 𝜋 by five divided by two, which is 𝜋 by five. We can sketch just the right triangle as shown. Remember, we want to use all of this to determine an expression for ℎ. We now have a right triangle where we know one of the angles of this right triangle and one of the side lengths. And we need to determine another side length. We can do this by using trigonometry.
And to apply right triangle trigonometry, we first need to label the sides of this right triangle relative to the angle of 𝜋 by five. We’ll start by labeling the side opposite the angle of 𝜋 by five. That’s the side of length 𝑥 over two. Next, the side opposite the right angle is the longest side in this right triangle. This means it’s the hypotenuse. Finally, the remaining side adjacent to the angle of 𝜋 by five is the adjacent side. Therefore, we want the trigonometric ratio involving the opposite side and the adjacent side in a right triangle. That’s the tangent function.
Now, by recalling the tan of 𝜃 is equal to the length of the opposite side divided by the length of the adjacent side in a right triangle, we can substitute 𝜃 is equal to 𝜋 by five. The length of the opposite side is 𝑥 over two, and the length of the adjacent side is equal to ℎ. We get the tan of 𝜋 by five is equal to 𝑥 over two divided by ℎ.
We now want to rearrange this to find an expression for ℎ. We multiply both sides of the equation through by ℎ, and we divide both sides of the equation by tan of 𝜋 by five. This gives us that ℎ is equal to 𝑥 divided by two times the tan of 𝜋 by five.
Now, we can substitute this expression for ℎ into our formula for the area. However, it’s quicker to multiply this area through by five because, remember, five times this area is the area of the regular pentagon. So multiplying this equation through by five and substituting the expression for ℎ, we get the area of a regular pentagon is five over two times 𝑥 multiplied by 𝑥 divided by two tan of 𝜋 by five. This gives us five 𝑥 squared divided by four tan of 𝜋 by five. And we can see this is exactly equal to the expression in option (B), where 𝑛 is equal to five.
And it’s worth pointing out that this proof can be generalized to any 𝑛-sided polygon. The only differences would’ve been we would have had 𝑛 congruent triangles, and our internal angle would’ve been two 𝜋 by 𝑛. Then when we found our right triangles, the angle we know would’ve been 𝜋 divided by 𝑛. Then when we found our expression for ℎ, we would’ve found that ℎ is equal to 𝑥 divided by two times the tan of 𝜋 by 𝑛. And finally, when we found the area of our regular 𝑛-sided polygon, we would’ve needed to multiply the expression by 𝑛. And applying all of these changes to our proof, we would’ve shown that the formula used to calculate the area of a regular 𝑛-sided polygon with side length 𝑥 is 𝑛𝑥 squared divided by four tan of 𝜋 by 𝑛, which is option (B).