## RS Aggarwal Class 8 Solutions Chapter 14 Polygons Ex 14B

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

**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:**

In a pentagon, no. of diagonals

\(=\frac { n\left( n-3 \right) }{ 2 }\)

\(=\frac { 5\left( 5-3 \right) }{ 2 }\)

\( =\frac { 5\times 2 }{ 2 } \)

= 5 (a)

**Question 2.**

**Solution:**

In a hexagon, no. of diagonals

\(=\frac { n\left( n-3 \right) }{ 2 }\)

\(=\frac { 6\left( 6-3 \right) }{ 2 }\)

\( =\frac { 6\times 3 }{ 2 } \)

= 9 (c)

**Question 3.**

**Solution:**

In an octagon, no. of diagonals

\(=\frac { n\left( n-3 \right) }{ 2 }\)

\(=\frac { 8\left( 8-3 \right) }{ 2 }\)

\( =\frac { 8\times 5 }{ 2 } \)

= 20 (d)

**Question 4.**

**Solution:**

In a polygon of 12 sides, no. of diagonals

\(=\frac { n\left( n-3 \right) }{ 2 }\)

\(=\frac { 12\left( 12-3 \right) }{ 2 }\)

\( =\frac { 12\times 9 }{ 2 } \)

= 54 (c)

**Question 5.**

**Solution:**

A polygon has 27 diagonal

Either n – 9 = 0, then n = 9

or n + 6 = 0, then n = – 6 but it is not possible being negative

No. of sides = 9 (c)

**Question 6.**

**Solution:**

Angles of a pentagon are x°, (x + 20)°, (x + 40)°, (x + 60°) and (x + 80)°

But sum of angle of a pentagon

**Question 7.**

**Solution:**

Measure of each exterior angle = 40°

No. of sides = \(\frac { { 360 }^{ o } }{ 40 }\)9 sides (b)

**Question 8.**

**Solution:**

Each interior angle of a polygon = 108°

**Question 9.**

**Solution:**

Each interior angle = 135°

**Question 10.**

**Solution:**

Let each exterior angle = x, then

Each interior angles = 3n

But sum of angle = 180°

x + 3x = 180°

=>4x = 180°

=> x = 45°

No. of sides = \(\frac { { 360 }^{ o } }{ 45 } \)

= 8 sides (b)

**Question 11.**

**Solution:**

Each interior angles of decagon

**Question 12.**

**Solution:**

Sum of all interior angles of a hexagon

= (2n – 4) x right angle

= (2 x 6 – 4) right angle

= 8 right angles (b)

**Question 13.**

**Solution:**

Sum of all interior angles of polygon = 1080°

Let n be the number of sides, then

(2n – 4) x 90°= 1080°

**Question 14.**

**Solution:**

Difference between each interior and exterior angle = 108°

Then each interior angle = x + 108°

x + x + 108°= 180°

(Sum of both angles = 180°)

=> 2x = 180° – 108° = 72°

x = \(\\ \frac { 72 }{ 2 } \)

= 36°

No. of sides = \( \frac { { 360 }^{ o } }{ { 36 }^{ o } } \)

= 10 (d)

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

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