Try our Worksheet on Interior Angles of a Polygon to solve a variety of problems such as finding the sum of interior angles of a polygon, finding the missing angle in a polygon, etc. These engaging activities in Interior Angles of a Polygon Worksheet PDFs with regular and irregular polygons are all you need to spruce things up for your learning!
Practice the questions in the Interior Angles of a Polygon Worksheet and enhance your math proficiency in the concept. Printable Interior Angles of a Polygon Worksheet is accessible for free thus making your learning much effective.
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Interior Angles of a Polygon Worksheet PDF
Example 1.
Find the sum of interior angles of each polygon
Solution:
i)In the first diagram, no. of sides is 7.
we know that Sum of interior angles of a polygon is (n-2) × 180.
Here n=7
=(7-2) × 180
=5 × 180
=9000
Therefore, Sum of interior angles of a polygon is 9000.
ii) In the second diagram, no. of sides is
As we know that Sum of interior angles of a polygon is (n-2) × 180.
Here n=4
=(4-2)× 180
=2 × 180
=3600
Therefore, Sum of interior angles of a polygon is 3600.
iii) In the third diagram, no. of sides is 5.
we know that Sum of interior angles of a polygon is (n-2) × 180.
Here n=5
=(5-2)× 180
=3× 180
=540
Therefore, Sum of interior angles of a polygon is 5400.
Example 2.
Find the interior angles of a regular heptagon and a regular octagon.
Solution:
The number of sides of a heptagon is 7.
The formula for finding the interior angles of a heptagon is, ((n-2)180°)/n.
Here, ‘n’ or the number of sides is 7.
Interior angles of a heptagon = ((7-2)180°)/8.
=5 ×180°/8
= 900/8
=112.5°
Therefore, the interior angles of a regular heptagon = 112.5° each.
The number of sides of an octagon is 8.
The formula for finding the interior angles of a polygon is, ((n-2)180°)/n.
Here, ‘n’ or the number of sides is 8.
Interior angles of a octagon = ((8-2)180°)/8.
= 6×180°/8
= 1080/8
= 135°
Therefore, the interior angles of a regular octagon = 135° each.
Example 3.
The four interior angles of a pentagon are 1050,1180,750,1300. Find the other angle?
Solution:
Given that,
The four interior angles of a pentagon are 1050,1180,750,1300.
We know that sum of angles of the pentagon is 540.
Let the other angle be x.
105+118+75+130+x=540
428+x=540
x=540-428
=112
Therefore, the other angle of a polygon is 1120.
Example 4.
Find the number of sides in a polygon if the sum of its interior angles is 10800.
Solution:
Given that,
The sum of its interior angles is 10800
Let no. of sides =n
Sum of angles of polygon =1080
(n−2)×180=1080
n−2=1080/180
n−2=6
n=6+2
n=8
Therefore, no. of sides of the polygon is 8.
Example 5.
The interior angle of a polygon is 60 degrees. The other interior angles are all equal to 120 degrees. How many sides does a polygon have?
Solution:
Given that,
The interior angle of a polygon is 60 degrees
The other interior angles are all equal to 110 degrees.
Let the number of sides be x.
There are x-1 no. of angles,120 degrees.
we know that total interior angles of a polygon=180(sides-2)
180(x-2)=60+120(x-1)
180x-360=60+120x-120
180x-360=-60+120x
60x=-60+360
60x=300
x=300/60=5
Therefore, the polygon has 5 sides.
Example 6.
If each interior angle is equal to 120°, then how many sides does a regular polygon have?
Solution:
Given that, each interior angle = 120°
We know that, Interior angle + Exterior angle = 180°
Exterior angle = 180°-120°
Therefore, the exterior angle is 60°
The formula for finding the number of sides of a regular polygon is =360° / Magnitude of each exterior angle
Therefore, the number of sides = 360° / 60° = 6 sides
Hence, the polygon has 6 sides.
Example 7.
The three angles of a polygon are 120°, 100°, 140°, and the other two angles are equal. Find the measure of these equal angles?
Solution:
Given that,
The three angles of a polygon are 120°, 100°, 140°.
Here the polygon has six angles. Therefore, it is a hexagon.
The sum of interior angles of a polygon=(n-2)180,n is no. of sides
=(6-2)×180
=4×180
=720
Let x be the measure of three angles that have the same measure.
3x+120+100+140=720
3x+360=720
3x=720-360
3x=360
x=360/3=1200
Therefore, the three angles with the same measure have a measurement of 1200.
Example 8.
The angles of a hexagon are (x – 5)°, (x – 8)°, (2x – 9)°, x,(2x-6)0, and (x – 4)°. Find the value of x and the measure of all angles.
Solution:
Given that the angle of a hexagon is (x – 5)°, (x – 8)°, (2x – 9)°, x, (2x – 6)°, and (x – 4)°.
We know that sum of angles of a hexagon is 7200.
Therefore, (x – 5)°+ (x – 8)°+ (2x – 9)°+ (2x – 6)°+ x+ (x – 4)°=7200
8x-32=7200
8x=752
x=752/8=940
Therefore, x=940
x-5=94-5=890
x-8=94-8=860
2x-9=2(94)-9=1790
2x-6=2(94)-6=1820
x-4=94-4=900.
Hence, the angles of a hexagon are 940,890,860,1790,1820,900.
Example 9.
The angle of a heptagon is in the ratio 1: 2 : 3: 4: 5: 6: 8. Find the measure of the smallest and the biggest angles.
Solution:
Given that the angles of a heptagon are in the ratio 1: 2 : 3: 4: 5: 6: 8.
Let each angle be of x then,
1=x,
2=2x
3=3x
4=4x
5=5x
6=6x
8=8x
We know that sum of angles of the heptagon is 9000.
1x+2x+3x+4x+5x+6x+8x=9000
29x=9000
x=9000/29=310
Here the smallest angle is x=310.
The largest angle is 8x=8(31)=2480.
Therefore, the smallest angle and the largest angle are 310,2470.
Example 10.
The four angles of the pentagon are equal and the fifth angle measure 120°. Find the measure of four equal angles?
Solution:
Given that,
The four angles of the pentagon are equal.
The fifth angle measure= 120°
Let x be the measure of four angles that have the same measure.
We know that the sum of the angles of the pentagon is 3600.
4x+120=3600
4x=360-120
4x=240
x=240/4=600.
Therefore, the measure of four equal angles is 600.
Example 11.
The seven angles of an octagon are 120° each. Find the measure of the eighth angle?
Solution:
Given that,
The seven angles of an octagon are 120° each.
An octagon has 8 sides.
The Sum of angles of an octagon is (n-2)180=(8-2)180
=6(180)
=1080
Let the other angle be x.
x+7(120)=1080
x+840=1080
x=1080-840
x=240
Hence, the measure of eight angles is 240°.
Example 12.
If the ratio of the number of sides of two regular polygons is 4:5 and the ratio of the sum of their interior angles is 11:14. Find the number of sides in both the polygons?
Solution:
Given that,
The ratio of the number of sides of two regular polygons = 4:5
Let them be 4X and 5X
The number of sides in the first polygon = 4X
The number of sides in the second polygon = 5X
We know that
The sum of the interior angles in a regular polygon of n sides is (n-2)×180°
The sum of the interior angles in the first polygon of 4X sides = (4X-2)×180°
The sum of the interior angles in the second polygon of 5X sides = (5X-2)×180°
The ratio of the sum of the interior angles of the two polygons = [(4X-2)×180°]:[(5X-2)×180°]
(4X-2) : (5X-2)
According to the given problem
The ratio of the sum of their interior angles = 11:14
=> (4X-2) : (5X-2) = 11:14
=> (4X-2) / (5X-2) = 11 / 14
On applying cross multiplication then
=>14(4X-2) = 11(5X-2)
=> 56X – 28 = 55X – 22
=> 56X – 55X = 28- 22
=> X = 6
So,
The number of sides in the first polygon
= 4(6) = 24.
The number of sides in the second polygon= 5(6) = 30.
Hence, the number of sides of polygons is 24 and 30.