**Find the number of sides of a regular polygon whose interior angle is twice the ...**

### Question

Find the number of sides of a regular polygon whose interior angle is twice the exterior angle.### Options

The correct answer is B.

### Explanation:

Let's begin by understanding the terms in the question. A *regular polygon* is a polygon with all sides and angles equal. An *interior angle* is an angle formed inside the polygon, while an *exterior angle* is an angle formed outside the polygon by extending one side of it.

Now, we are given that the interior angle is twice the exterior angle. Let's denote the exterior angle as \(x\), then the interior angle will be \(2x\).

Recall that the sum of the measures of the interior angles of a polygon with \(n\) sides is given by the formula \((n - 2) \times 180°\). Since the polygon is regular, all interior angles are equal, so each interior angle measures \(\frac{(n - 2) \times 180°}{n}\).

We also know that the sum of the measures of the exterior angles of any polygon is always \(360°\). For a regular polygon, all exterior angles are equal, so each exterior angle measures \(\frac{360°}{n}\).

Now, we can set up an equation using the given information that the interior angle is twice the exterior angle:

\[\frac{(n - 2) \times 180°}{n} = 2 \times \frac{360°}{n}\]Now, we solve the equation for \(n\):

\[(n - 2) \times 180° = 2 \times 360°\]Divide both sides by 180°:

\[n - 2 = 4\]Add 2 to both sides:

\[n = 6\]So, the regular polygon has **6 sides**, which is Option B.

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e+I=180

Where I is twice of e

e+2e=180

3e=180

Divide both sides by 3

e=60

360/60=6