Question

The interior angle of a regular polygon exceeds its exterior angle by 108 °. How many sides does the polygon have?​

Answer 1

Answer:

number of sides = 10

Step-by-step explanation:

Let the exterior angle be x

we are given that the interior angle of a regular polygon exceeds its exterior angle by 108

So, interior angle = x+108

Since we know that the sum of interior angle and exterior angle is 180°

So

⇒x+108+x =  180

⇒2x+108= 180

⇒2x = 180-108

⇒2x = 72

⇒x = 36

Formula for number of sides = 180 × n-2/n = interior angle

⇒180 × n-2/n = 36+108

⇒180n- 360 = 144n

⇒180n - 144n = 360

⇒36n  = 360

⇒n = 10

Answer 2

Answer:

The polygon have 10 sides.

Step-by-step explanation:

Let's assume the regular polygon is of n sides.  Since any two adjacent vertices form a side.  To be able to draw any diagonal, we need to have vertices other than 3 vertices: the vertex itself from which any diagonal can be drawn and the it's two adjacent vertices.  So, number of diagonals from any vertex, will be 3 less than the number of vertices.

∴ From any vertex of the regular polygon, there can be (n - 3) digonals drawn.

These (n - 3) diagonals will make (n - 2) triangles.  Sum of the interior angles will measure (n - 2)x180°

Since we are dealing with a regular Polygon, each of the interior angles will measure $\frac{(n - 2)180}{n}$.

∴ The exterior angle will measure $180(1 - \frac{n - 2}{n} ) = \frac{180 times 2}{n}$

According to the problem:

$\frac{(n - 2)180}{n} - \frac{180 times 2}{n} = 108$

$=> (n -2) - 2 = \frac{108n}{180}$

$=> n(1 - \frac{108}{180} ) = 4$

$=> n = \frac{4 times 180}{180 - 108} = 10$

∴ The polygon have 10 sides.