Question

interior angles: non - regular polygons
the sum of the measures of the interior angles of any polygon is _.
find the sum of the measures of the interior angles of the polygon with ( n ) sides.

1. ( n = 8 ) 2) ( n = 15 ) 3) ( n = 20 )

find the measure of ( angle a ).

1. ( angle a=) _ 5) ( angle a=) _ 6) ( angle a=) _

find the measures of the missing angles (hint: first determine the sum of the interior angles of the polygon).
7.
8.
10.

Explanation:

Step1: Recall the formula for the sum of interior angles of a polygon

The formula for the sum of the interior angles of a polygon with \(n\) sides is \(S=(n - 2)\times180^{\circ}\).

Step2: Solve for \(n = 8\)

Substitute \(n = 8\) into the formula: \(S=(8 - 2)\times180^{\circ}=6\times180^{\circ}=1080^{\circ}\).

Step3: Solve for \(n = 15\)

Substitute \(n = 15\) into the formula: \(S=(15 - 2)\times180^{\circ}=13\times180^{\circ}=2340^{\circ}\).

Step4: Solve for \(n = 20\)

Substitute \(n = 20\) into the formula: \(S=(20 - 2)\times180^{\circ}=18\times180^{\circ}=3240^{\circ}\).

Step5: Solve for problem 4 (quadrilateral, \(n = 4\))

The sum of interior angles \(S=(4 - 2)\times180^{\circ}=360^{\circ}\). Let \(\angle A=x\), then \(x + 101^{\circ}+92^{\circ}+68^{\circ}=360^{\circ}\). So \(x=360^{\circ}-(101^{\circ}+92^{\circ}+68^{\circ})=360^{\circ}-261^{\circ}=99^{\circ}\).

Step6: Solve for problem 5 (pentagon, \(n = 5\))

The sum of interior angles \(S=(5 - 2)\times180^{\circ}=540^{\circ}\). Let \(\angle A=x\), then \(x + 113^{\circ}+80^{\circ}+130^{\circ}+90^{\circ}=540^{\circ}\). So \(x=540^{\circ}-(113^{\circ}+80^{\circ}+130^{\circ}+90^{\circ})=540^{\circ}-413^{\circ}=127^{\circ}\).

Step7: Solve for problem 6 (hexagon, \(n = 6\))

The sum of interior angles \(S=(6 - 2)\times180^{\circ}=720^{\circ}\). Let \(\angle A=x\), then \(x + 102^{\circ}+146^{\circ}+120^{\circ}+124^{\circ}+158^{\circ}=720^{\circ}\). So \(x=720^{\circ}-(102^{\circ}+146^{\circ}+120^{\circ}+124^{\circ}+158^{\circ})=720^{\circ}-650^{\circ}=70^{\circ}\).

Step8: Solve for problem 7 (quadrilateral, \(n = 4\))

The sum of interior angles \(S=(4 - 2)\times180^{\circ}=360^{\circ}\). Let \(n^{\circ}=x\), then \(x+37^{\circ}+152^{\circ}+44^{\circ}=360^{\circ}\). So \(x=360^{\circ}-(37^{\circ}+152^{\circ}+44^{\circ})=360^{\circ}-233^{\circ}=127^{\circ}\).

Step9: Solve for problem 8 (octagon, \(n = 8\))

The sum of interior angles \(S=(8 - 2)\times180^{\circ}=1080^{\circ}\). Let \(n^{\circ}=x\), then \(x + 147^{\circ}+145^{\circ}+122^{\circ}+139^{\circ}+147^{\circ}+141^{\circ}+99^{\circ}=1080^{\circ}\). So \(x=1080^{\circ}-(147^{\circ}+145^{\circ}+122^{\circ}+139^{\circ}+147^{\circ}+141^{\circ}+99^{\circ})=1080^{\circ}-940^{\circ}=140^{\circ}\).

Step10: Solve for problem 9 (quadrilateral, \(n = 4\))

The sum of interior angles \(S=(4 - 2)\times180^{\circ}=360^{\circ}\). Let \(\angle X=x\) and \(\angle Y = y\). Since \(\angle W = 90^{\circ}\) and \(\angle Z=120^{\circ}\), then \(x + y+90^{\circ}+120^{\circ}=360^{\circ}\), \(x + y=150^{\circ}\). If it is a parallelogram - like (assuming \(x=y\)), then \(x=y = 75^{\circ}\).

Step11: Solve for problem 10 (hexagon, \(n = 6\))

The sum of interior angles \(S=(6 - 2)\times180^{\circ}=720^{\circ}\). Let \(\angle Y=x\), \(\angle X = y\) (assuming \(y = 90^{\circ}\) as a right - angle). Then \(x+90^{\circ}+149^{\circ}+103^{\circ}+108^{\circ}+90^{\circ}=720^{\circ}\). So \(x=720^{\circ}-(90^{\circ}+149^{\circ}+103^{\circ}+108^{\circ}+90^{\circ})=720^{\circ}-540^{\circ}=180^{\circ}\).

Answer:

1. \(1080^{\circ}\)
2. \(2340^{\circ}\)
3. \(3240^{\circ}\)
4. \(99^{\circ}\)
5. \(127^{\circ}\)
6. \(70^{\circ}\)
7. \(127^{\circ}\)
8. \(140^{\circ}\)
9. (Assuming \(x=y\)) \(75^{\circ}\)
10. \(180^{\circ}\)