## 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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