two regular polygons of the same number of sides have sides 5 ft. and 12 ft. in length, respectively. what is the length of the side of another regular polygon of the same number of sides, if its area is equal to the sum of the other two.
The areas are 25k and 144k, for some constant k.
The total area is 169k, so the side length is 13.
To find the length of the side of another regular polygon with the same number of sides, we can set up an equation based on the information given.
Let's call the number of sides in the polygons "n". We know that one polygon has sides of 5 ft in length, and the other has sides of 12 ft in length. The area of a regular polygon is given by the formula:
Area = (1/4) * n * s^2 * cot(π/n)
Where "s" is the length of a side and "cot" represents the cotangent function.
We can calculate the area of each polygon using this formula:
Area1 = (1/4) * n * (5^2) * cot(π/n)
Area2 = (1/4) * n * (12^2) * cot(π/n)
And according to the question, the area of the third polygon is equal to the sum of the areas of the first two polygons:
Area3 = Area1 + Area2
Now we can set up the equation:
(1/4) * n * (s^2) * cot(π/n) = (1/4) * n * (5^2) * cot(π/n) + (1/4) * n * (12^2) * cot(π/n)
We can simplify this equation by canceling out the common factors:
s^2 * cot(π/n) = 5^2 * cot(π/n) + 12^2 * cot(π/n)
Dividing both sides by cot(π/n):
s^2 = 5^2 + 12^2
Simplifying:
s^2 = 25 + 144
s^2 = 169
Taking the square root of both sides:
s = √169
s = 13 ft
Therefore, the length of the side of the third regular polygon is 13 ft.