Question

how can you use what you know about the interior angle measures of triangles to find the sum of the interiors angle measures polygons? (hint: the first one has been done for you). pick 1 vertex to make your triangles.
a.
b.
c.
d.
e.
f.
number of sides (n) 3 4 5 6 7 8 9 10
number of triangles
interior angle sum (s)

Explanation:

Step1: Recall the formula for the number of triangles in a polygon

The number of triangles formed by connecting one vertex to other non - adjacent vertices in an \(n\) - sided polygon is \(n - 2\).

Step2: Recall the formula for the sum of interior angles of a polygon

Since the sum of interior angles of a triangle is \(180^{\circ}\), and if a polygon can be divided into \(n - 2\) triangles, then the sum of interior angles \(S=(n - 2)\times180^{\circ}\)

For \(n = 3\):

• Number of triangles: \(n-2=3 - 2=1\)
• Interior angle sum: \(S=(3 - 2)\times180^{\circ}=180^{\circ}\)

For \(n = 4\):

• Number of triangles: \(n - 2=4-2 = 2\)
• Interior angle sum: \(S=(4 - 2)\times180^{\circ}=360^{\circ}\)

For \(n = 5\):

• Number of triangles: \(n - 2=5 - 2=3\)
• Interior angle sum: \(S=(5 - 2)\times180^{\circ}=540^{\circ}\)

For \(n = 6\):

• Number of triangles: \(n - 2=6 - 2=4\)
• Interior angle sum: \(S=(6 - 2)\times180^{\circ}=720^{\circ}\)

For \(n = 7\):

• Number of triangles: \(n - 2=7 - 2=5\)
• Interior angle sum: \(S=(7 - 2)\times180^{\circ}=900^{\circ}\)

For \(n = 8\):

• Number of triangles: \(n - 2=8 - 2=6\)
• Interior angle sum: \(S=(8 - 2)\times180^{\circ}=1080^{\circ}\)

For \(n = 9\):

• Number of triangles: \(n - 2=9 - 2=7\)
• Interior angle sum: \(S=(9 - 2)\times180^{\circ}=1260^{\circ}\)

For \(n = 10\):

• Number of triangles: \(n - 2=10 - 2=8\)
• Interior angle sum: \(S=(10 - 2)\times180^{\circ}=1440^{\circ}\)

Answer:

Number of Sides (\(n\))	3	4	5	6	7	8	9	10
Interior Angle Sum (\(S\))	\(180^{\circ}\)	\(360^{\circ}\)	\(540^{\circ}\)	\(720^{\circ}\)	\(900^{\circ}\)	\(1080^{\circ}\)	\(1260^{\circ}\)	\(1440^{\circ}\)