One regular polygon has twice as many sides as another. If the ratio of the interior angles of the first to that of second is 5:4, find the number of sides in each polygon?

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To solve this problem, we need to set up an equation using the given information.

Let's say the first polygon has "x" sides and the second polygon has "y" sides.

From the problem statement, we know that the first polygon has twice as many sides as the second polygon. Therefore, we can write the equation:

x = 2y ----(Equation 1)

We are also given that the ratio of the interior angles of the first polygon to that of the second is 5:4. Each interior angle of a regular polygon can be found using the formula:

Interior angle = (n-2) * 180 / n

where n represents the number of sides of the polygon.

Using this formula, we can write the equation for the first polygon as:

(x-2) * 180 / x

And the equation for the second polygon as:

(y-2) * 180 / y

Given that the ratio of the interior angles of the first to the second polygon is 5:4, we can write the equation:

((x-2) * 180 / x) / ((y-2) * 180 / y) = 5/4

Simplifying this equation, we have:

(4(x-2) * y) / (x * (y-2)) = 5/4

Cross multiplying, we get:

4(x-2)y = 5x(y-2)

Expanding and simplifying further, we have:

4xy - 8y = 5xy - 10x

Rearranging the terms and canceling out common factors, we get:

xy - 2x = 2y

Substituting equation 1 (x = 2y), we have:

(2y)y - 2(2y) = 2y

Simplifying, we get:

2y^2 - 4y - 2y = 0

2y^2 - 6y = 0

Factorizing, we get:

2y(y - 3) = 0

Since y cannot be equal to 0, we have:

y - 3 = 0

y = 3

Substituting the value of y into equation 1, we get:

x = 2y
x = 2 * 3
x = 6

Therefore, the first polygon has 6 sides, and the second polygon has 3 sides.

the interior angles of an n-gon are

(n-2)(180)/n

So, you have

((2n-2)/(2n)) / ((n-2)/n) = 5/4

Now just solve for n.

(2n-2)/2n *180