RS Aggarwal Class 8 Solutions Chapter 14 Polygons Ex 14A

RS Aggarwal Class 8 Solutions Chapter 14 Polygons Ex 14A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 14 Polygons Ex 14A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 14 Polygons Ex 14A
• RS Aggarwal Solutions Class 8 Chapter 14 Polygons Ex 14B

Question 1.
Solution:
We know that sum of exterior angles of a polygon is 360°
Then,
(i) Pentagon’s exterior angle = \(\frac { { 360 }^{ o } }{ 5 } \)
= 72°

Question 2.
Solution:
Each exterior angle a n sided polygon = 50°
No of sides = \(\frac { { 360 }^{ o } }{ 50 } \)
= \(7\frac { 1 }{ 5 } \)
Which is not possible to have \(7\frac { 1 }{ 5 } \) sides
Which is not a whole number

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Question 3.
Solution:
We know that each interior angle of a regular polygon of n sides = \(\\ \frac { 2n-4 }{ n } \) right angle
(i) Polygon having 10 sides, each interior

Question 4.
Solution:
Let interior angle of a polygon having n sides = 100°
\(\\ \frac { 2n-4 }{ n } \) x 90° = 100°
=>\(\\ \frac { 2n-4 }{ n } \)
= \(\\ \frac { 100 }{ 90 } \)
= \(\\ \frac { 10 }{ 9 } \)

Question 5.
Solution:
We know that sum of all interior angles = 2n – 4 right angles
(i) Pentagon
Sum of its angles = (2 x 5 – 4) x 90°
= 6 x 90° = 540°

Question 6.
Solution:
We know that number of diagonal of polygon having n sides = \(\frac { n\left( n-3 \right) }{ 2 } \)
(i) In heptagon, no of diagonals = \(\frac { 7\left( 7-3 \right) }{ 2 } \)

Question 7.
Solution:
We know that each exterior angle
= \( \frac { { 360 }^{ o } }{ n } \)
Where n sides are of polygon
(i) Each exterior angle = 40°

Question 8.
Solution:
We know that sum of all exterior angle of a polygon – 360°
Exterior ∠A + ∠B + ∠C + ∠D = 360°
=> 115° + x + 90° + 50° = 360°
=> 255° + x + 360°
=> x = 360° – 255°
=> x = 105°

Question 9.
Solution:
In the given figure polygon is of 5 sides and each interior angle is x
\(x=\frac { 2x-4 }{ n } \times { 90 }^{ o }=\frac { 2\times 5-4 }{ 5 } \times { 90 }^{ o } \)
= \(=\frac { 6 }{ 5 } \times { 90 }^{ o } \)
= 108°

 

Hope given RS Aggarwal Solutions Class 8 Chapter 14 Polygons Ex 14A are helpful to complete your math homework.

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