{"id":599660,"date":"2022-05-06T05:08:56","date_gmt":"2022-05-06T05:08:56","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=599660"},"modified":"2022-05-10T05:37:54","modified_gmt":"2022-05-10T05:37:54","slug":"selina-class-6-icse-solutions-mathematics-chapter-28-polygons","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/selina-class-6-icse-solutions-mathematics-chapter-28-polygons\/","title":{"rendered":"Selina Class 6 ICSE Solutions Mathematics : Chapter 28-\u00a0Polygons"},"content":{"rendered":"\n

Class 6: Maths Chapter 28 solutions. Complete Class 6 Maths Chapter 28 Notes.<\/p>\n\n\n\n

Selina Class 6 ICSE Solutions Mathematics : Chapter 28- Polygons<\/h2>\n\n\n\n

Selina 6th Maths Chapter 28, Class 6 Maths Chapter 28 solutions<\/p>\n\n\n\n

Exercise 28(A)<\/h3>\n\n\n\n

1. State, which of the following are polygons:<\/strong><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) The given figure is not closed.<\/p>\n\n\n\n

Hence, the figure is not a polygon.<\/p>\n\n\n\n

(ii) The given figure is closed.<\/p>\n\n\n\n

Hence, the figure is a polygon.<\/p>\n\n\n\n

(iii) The given figure is closed.<\/p>\n\n\n\n

Hence, the figure is a polygon.<\/p>\n\n\n\n

(iv) In the given figure, one of the sides is an arc.<\/p>\n\n\n\n

Hence, the figure is not polygon.<\/p>\n\n\n\n

(v) The side intersects each other in the given figure.<\/p>\n\n\n\n

Hence, the figure is not polygon.<\/p>\n\n\n\n

2. Find the sum of interior angles of a polygon with:<\/strong><\/p>\n\n\n\n

(i) 9 sides<\/strong><\/p>\n\n\n\n

(ii) 13 sides<\/strong><\/p>\n\n\n\n

(iii) 16 sides<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 9 sides<\/p>\n\n\n\n

Number of sides n = 9<\/p>\n\n\n\n

The Sum of interior angles of polygon = (2n \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

= (2 \u00d7 9 \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

= (18 \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

= 14 \u00d7 900<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 12600<\/sup><\/p>\n\n\n\n

(ii) 13 sides<\/p>\n\n\n\n

Number of sides n = 13<\/p>\n\n\n\n

The sum of interior angles of polygon = (2n -4) \u00d7 900<\/sup><\/p>\n\n\n\n

= (2 \u00d7 13 \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

= (26 \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

= 22 \u00d7 900<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 19800<\/sup><\/p>\n\n\n\n

(iii) 16 sides<\/p>\n\n\n\n

Number of sides n = 16<\/p>\n\n\n\n

The sum of interior angles of polygon = (2n \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

= (2 \u00d7 16 \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

= (32 \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

= 28 \u00d7 900<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 25200<\/sup><\/p>\n\n\n\n

3. Find the number of sides of a polygon, if the sum of its interior angles is:<\/strong><\/p>\n\n\n\n

(i) 14400<\/sup><\/strong><\/p>\n\n\n\n

(ii) 16200<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 14400<\/sup><\/p>\n\n\n\n

The sum of interior angles of polygon = 14400<\/sup><\/p>\n\n\n\n

Let the number of sides = n<\/p>\n\n\n\n

The sum of interior angle of polygon is (2n \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

The side of polygon can be calculated as,<\/p>\n\n\n\n

(2n \u2013 4) \u00d7 900<\/sup> = 14400<\/sup><\/p>\n\n\n\n

2n \u2013 4 = 14400<\/sup> \/ 900<\/sup><\/p>\n\n\n\n

2(n \u2013 2) = 14400<\/sup> \/ 900<\/sup><\/p>\n\n\n\n

n \u2013 2 = 14400<\/sup> \/ (2 \u00d7 900<\/sup>)<\/p>\n\n\n\n

On further calculation, we get<\/p>\n\n\n\n

n \u2013 2 = 160<\/sup> \/ 2<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

n \u2013 2 = 8<\/p>\n\n\n\n

n = 8 + 2<\/p>\n\n\n\n

n = 10<\/p>\n\n\n\n

Hence, the side of polygon = 10<\/p>\n\n\n\n

(ii) 16200<\/sup><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

The sum of interior angles of polygon = 16200<\/sup><\/p>\n\n\n\n

Let number of sides = n<\/p>\n\n\n\n

The sum of interior angles of polygon = (2n \u2013 4) \u00d7 <\/strong>900<\/sup><\/p>\n\n\n\n

The side of polygon can be calculated as,<\/p>\n\n\n\n

(2n \u2013 4) \u00d7 900<\/sup> = 16200<\/sup><\/p>\n\n\n\n

2(n \u2013 2) \u00d7 900<\/sup> = 16200<\/sup><\/p>\n\n\n\n

n \u2013 2 = 16200<\/sup> \/ (2 \u00d7 900<\/sup>)<\/p>\n\n\n\n

n \u2013 2 = 8100<\/sup> \/ 900<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

n \u2013 2 = 9<\/p>\n\n\n\n

n = 9 + 2<\/p>\n\n\n\n

n = 11<\/p>\n\n\n\n

Hence, the side of polygon = 11<\/p>\n\n\n\n

4. Is it possible to have a polygon, whose sum of interior angles is 10300<\/sup>.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

The sum of interior angles of polygon = 10300<\/sup><\/p>\n\n\n\n

Let us consider the number of sides = n<\/p>\n\n\n\n

The sum of interior angle of polygon = (2n \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

The side of polygon is calculated as,<\/p>\n\n\n\n

(2n \u2013 4) \u00d7 900<\/sup> = 10300<\/sup><\/p>\n\n\n\n

2(n \u2013 2)= 10300<\/sup> \/ 900<\/sup><\/p>\n\n\n\n

On further calculation, we get<\/p>\n\n\n\n

(n \u2013 2) = 10300<\/sup> \/ (2 \u00d7 <\/strong>900<\/sup>)<\/p>\n\n\n\n

(n \u2013 2) = 1030<\/sup> \/ 180<\/sup><\/p>\n\n\n\n

n = 5.72 + 2<\/p>\n\n\n\n

n = 7.72<\/p>\n\n\n\n

which is not a whole number.<\/p>\n\n\n\n

Therefore, it is not a polygon, whose sum of interior angles is 10300<\/sup><\/p>\n\n\n\n

5. (i) If all the angles of a hexagon arc equal, find the measure of each angle.<\/strong><\/p>\n\n\n\n

(ii) If all the angles of an octagon are equal, find the measure of each angle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Number of sides of polygon n = 6<\/p>\n\n\n\n

Let us consider each angle be = x0<\/sup><\/p>\n\n\n\n

We know,<\/p>\n\n\n\n

The sum of interior angles of hexagon = 6x0<\/sup><\/p>\n\n\n\n

The sum of interior angle of polygon = (2n \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

The sum of the interior angles of polygon can be calculated as,<\/p>\n\n\n\n

(2n \u2013 4) \u00d7 900<\/sup> = Sum of angles<\/p>\n\n\n\n

(2 \u00d7 6 \u2013 4) \u00d7 900<\/sup> = 6x0<\/sup><\/p>\n\n\n\n

(12 \u2013 4) \u00d7 900<\/sup> = 6x0<\/sup><\/p>\n\n\n\n

6x0<\/sup> = 8 \u00d7 900<\/sup><\/p>\n\n\n\n

x0<\/sup> = (8 \u00d7 900<\/sup>) \/ 6<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 1200<\/sup><\/p>\n\n\n\n

Therefore, each angle of hexagon = 1200<\/sup><\/p>\n\n\n\n

(ii) Number of sides of octagon n = 8<\/p>\n\n\n\n

Let us consider each angle be = x0<\/sup><\/p>\n\n\n\n

We know that,<\/p>\n\n\n\n

The sum of interior angles of octagon = 8x0<\/sup><\/p>\n\n\n\n

The sum of interior angles of polygon = (2n \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

The sum of interior angles of polygon can be calculated as,<\/p>\n\n\n\n

(2n \u2013 4) \u00d7 900<\/sup> = Sum of angles<\/p>\n\n\n\n

(2n \u2013 4) \u00d7 900<\/sup> = 8x0<\/sup><\/p>\n\n\n\n

(2 \u00d7 8 \u2013 4) \u00d7 900<\/sup> = 8x0<\/sup><\/p>\n\n\n\n

12 \u00d7 900<\/sup> = 8x0<\/sup><\/p>\n\n\n\n

8x0<\/sup> = 12 \u00d7 900<\/sup><\/p>\n\n\n\n

x0<\/sup> = (12 \u00d7 900<\/sup>) \/ 8<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x0<\/sup> = 1350<\/sup><\/p>\n\n\n\n

Therefore, each angle of octagon = 1350<\/sup><\/p>\n\n\n\n

6. One angle of a quadrilateral is 900<\/sup> and all other angles are equal; find each equal angle<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider all the three equal angle of a quadrilateral be x0<\/sup><\/p>\n\n\n\n

The sum of angles of a quadrilateral = 3600<\/sup><\/p>\n\n\n\n

x + x + x + 900<\/sup> = 3600<\/sup><\/p>\n\n\n\n

3x + 900<\/sup> = 3600<\/sup><\/p>\n\n\n\n

3x = 3600<\/sup> \u2013 900<\/sup><\/p>\n\n\n\n

3x = 2700<\/sup><\/p>\n\n\n\n

x = 2700<\/sup> \/ 3<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 900<\/sup><\/p>\n\n\n\n

The measure of each equal angle = 900<\/sup><\/p>\n\n\n\n

7. If angles of quadrilateral are in the ratio 4: 5: 3: 6; find each angle of the quadrilateral.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider the angles of quadrilateral be 4x, 5x, 3x and 6x<\/p>\n\n\n\n

We know,<\/p>\n\n\n\n

The sum of angles of quadrilateral = 3600<\/sup><\/p>\n\n\n\n

4x + 5x + 3x + 6x = 3600<\/sup><\/p>\n\n\n\n

18x = 3600<\/sup><\/p>\n\n\n\n

x = 3600<\/sup> \/ 18<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 200<\/sup><\/p>\n\n\n\n

Now, all the angles are,<\/p>\n\n\n\n

4x = 4 \u00d7 200<\/sup><\/p>\n\n\n\n

= 800<\/sup><\/p>\n\n\n\n

5x = 5 \u00d7 200<\/sup><\/p>\n\n\n\n

= 1000<\/sup><\/p>\n\n\n\n

3x = 3 \u00d7 200<\/sup><\/p>\n\n\n\n

= 600<\/sup><\/p>\n\n\n\n

6x = 6 \u00d7 200<\/sup><\/p>\n\n\n\n

= 1200<\/sup><\/p>\n\n\n\n

Therefore, the angles of the quadrilateral are 800<\/sup>, 1000<\/sup>, 600<\/sup> and 1200<\/sup><\/p>\n\n\n\n

8. If one angle of a pentagon is 1200<\/sup> and each of the remaining four angles is x0<\/sup>, find the magnitude of x.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given<\/p>\n\n\n\n

One angle of a pentagon = 1200<\/sup><\/p>\n\n\n\n

Number of sides of pentagon n = 5<\/p>\n\n\n\n

Let us consider all other equal angle of pentagon be x<\/p>\n\n\n\n

The sum of interior angle of polygon is (2n \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

The sum of the interior angle of pentagon can be calculated as,<\/p>\n\n\n\n

(2n \u2013 4) \u00d7 900<\/sup> = (2 \u00d7 5 \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

= 6 \u00d7 900<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 5400<\/sup><\/p>\n\n\n\n

Therefore, the sum of interior angles of pentagon is 5400<\/sup><\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

x + x + x + x + 1200<\/sup> = 5400<\/sup><\/p>\n\n\n\n

4x + 1200<\/sup> = 5400<\/sup><\/p>\n\n\n\n

4x = 5400<\/sup> \u2013 1200<\/sup><\/p>\n\n\n\n

4x = 4200<\/sup><\/p>\n\n\n\n

x = 4200<\/sup> \/ 4<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 1050<\/sup><\/p>\n\n\n\n

Hence, the value of x = 1050<\/sup><\/p>\n\n\n\n

9. The angles of a pentagon are in the ratio 5: 4: 5: 7: 6; find each angle of the pentagon.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider all the angle of pentagon as 5x, 4x, 5x, 7x and 6x<\/p>\n\n\n\n

The sum of the interior angle of polygon is (2n \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

The sum of the interior angle of pentagon can be calculated as,<\/p>\n\n\n\n

(2n \u2013 4) \u00d7 900<\/sup> = (2 \u00d7 5 \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

= 6 \u00d7 900<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 5400<\/sup><\/p>\n\n\n\n

The sum of interior angles of pentagon = 5400<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

5x + 4x + 5x + 7x + 6x = 5400<\/sup><\/p>\n\n\n\n

27x = 5400<\/sup><\/p>\n\n\n\n

x = 5400<\/sup> \/ 27<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 200<\/sup><\/p>\n\n\n\n

Thus, each angle,<\/p>\n\n\n\n

5x = 5 \u00d7 200<\/sup><\/p>\n\n\n\n

= 1000<\/sup><\/p>\n\n\n\n

4x = 4 \u00d7 200<\/sup><\/p>\n\n\n\n

= 800<\/sup><\/p>\n\n\n\n

5x = 5 \u00d7 200<\/sup><\/p>\n\n\n\n

= 1000<\/sup><\/p>\n\n\n\n

7x = 7 \u00d7 200<\/sup><\/p>\n\n\n\n

= 1400<\/sup><\/p>\n\n\n\n

6x = 6 \u00d7 200<\/sup><\/p>\n\n\n\n

= 1200<\/sup><\/p>\n\n\n\n

Therefore, all the angles of a pentagon are 1000<\/sup>, 800<\/sup>, 1000<\/sup>, 1400<\/sup> and 1200<\/sup><\/p>\n\n\n\n

10. Two angles of a hexagon are 900<\/sup> and 1100<\/sup>. If the remaining four angles arc equal, find each equal angle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let us consider all the angle of hexagon as x<\/p>\n\n\n\n

Number of sides in hexagon n = 6<\/p>\n\n\n\n

The sum of interior angle of polygon is (2n \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

The sum of interior angle of hexagon can be calculated as,<\/p>\n\n\n\n

(2n \u2013 4) \u00d7 900<\/sup> = (2 \u00d7 6 \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

= (12 \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

= 8 \u00d7 900<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

= 7200<\/sup><\/p>\n\n\n\n

The sum of interior angles of pentagon is 7200<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

900<\/sup> + 1100<\/sup> + x + x + x + x = 7200<\/sup><\/p>\n\n\n\n

2000<\/sup> + 4x = 7200<\/sup><\/p>\n\n\n\n

4x = 7200<\/sup> \u2013 2000<\/sup><\/p>\n\n\n\n

4x = 5200<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 1300<\/sup><\/p>\n\n\n\n

Hence, the measure of each equal angle = 1300<\/sup><\/p>\n\n\n\n

\n\n\n\n

Exercise 28(B)<\/strong><\/h3>\n\n\n\n

1. Fill in the blanks:<\/strong><\/p>\n\n\n\n

In case of regular polygon, with<\/strong><\/p>\n\n\n\n

Number of sides<\/strong><\/td>Each exterior angle<\/strong><\/td>Each interior angle<\/strong><\/td><\/tr>
(i) 6<\/strong><\/td>\u2026\u2026\u2026\u2026.<\/strong><\/td>\u2026\u2026\u2026\u2026..<\/strong><\/td><\/tr>
(ii) 8<\/strong><\/td>\u2026\u2026\u2026\u2026.<\/strong><\/td>\u2026\u2026\u2026\u2026..<\/strong><\/td><\/tr>
(iii) \u2026\u2026\u2026\u2026<\/strong><\/td>360<\/sup><\/strong><\/td>\u2026\u2026\u2026\u2026..<\/strong><\/td><\/tr>
(iv) \u2026\u2026\u2026\u2026<\/strong><\/td>200<\/sup><\/strong><\/td>\u2026\u2026\u2026\u2026..<\/strong><\/td><\/tr>
(v) \u2026\u2026\u2026\u2026<\/strong><\/td>\u2026\u2026\u2026\u2026\u2026<\/strong><\/td>1350<\/sup><\/strong><\/td><\/tr>
(vi) \u2026\u2026\u2026..<\/strong><\/td>\u2026\u2026\u2026\u2026\u2026<\/strong><\/td>1650<\/sup><\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Number of sides<\/td>Each exterior angle<\/td>Each interior angle<\/td><\/tr>
(i) 6<\/td>600<\/sup><\/td>1200<\/sup><\/td><\/tr>
(ii) 8<\/td>450<\/sup><\/td>1350<\/sup><\/td><\/tr>
(iii) 10<\/td>360<\/sup><\/td>1440<\/sup><\/td><\/tr>
(iv) 18<\/td>200<\/sup><\/td>1600<\/sup><\/td><\/tr>
(v) 8<\/td>450<\/sup><\/td>1350<\/sup><\/td><\/tr>
(vi) 24<\/td>150<\/sup><\/td>1650<\/sup><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

(i) Each exterior angle = 3600<\/sup> \/ 6<\/p>\n\n\n\n

= 600<\/sup><\/p>\n\n\n\n

Each interior angle = 1800<\/sup> \u2013 600<\/sup><\/p>\n\n\n\n

= 1200<\/sup><\/p>\n\n\n\n

(ii) Each exterior angle = 3600<\/sup> \/ 8<\/p>\n\n\n\n

= 450<\/sup><\/p>\n\n\n\n

Each interior angle = 1800<\/sup> \u2013 450<\/sup><\/p>\n\n\n\n

= 1350<\/sup><\/p>\n\n\n\n

(iii) Given that, each exterior angle = 360<\/sup><\/p>\n\n\n\n

So, number of sides = 3600<\/sup> \/ 360<\/sup><\/p>\n\n\n\n

= 10 sides<\/p>\n\n\n\n

Each interior angle = 1800<\/sup> \u2013 360<\/sup><\/p>\n\n\n\n

= 1440<\/sup><\/p>\n\n\n\n

(iv) Given that, each exterior angle = 200<\/sup><\/p>\n\n\n\n

Hence, number of sides = 3600<\/sup> \/ 200<\/sup><\/p>\n\n\n\n

= 18 sides<\/p>\n\n\n\n

Each interior angle = 1800<\/sup> \u2013 200<\/sup><\/p>\n\n\n\n

= 1600<\/sup><\/p>\n\n\n\n

(v) Given that, each interior angle = 1350<\/sup><\/p>\n\n\n\n

Hence, exterior angle = 1800<\/sup> \u2013 1350<\/sup><\/p>\n\n\n\n

= 450<\/sup><\/p>\n\n\n\n

Therefore, number of sides = 3600<\/sup> \/ 450<\/sup><\/p>\n\n\n\n

= 8 sides<\/p>\n\n\n\n

(vi) Given that, each interior angle = 1650<\/sup><\/p>\n\n\n\n

Hence, exterior angle = 1800<\/sup> \u2013 1650<\/sup><\/p>\n\n\n\n

= 150<\/sup><\/p>\n\n\n\n

Therefore, the number of sides = 3600<\/sup> \/ 150<\/sup><\/p>\n\n\n\n

= 24 sides<\/p>\n\n\n\n

2. Find the number of sides in a regular polygon, if its each interior angle is:<\/strong><\/p>\n\n\n\n

(i) 1600<\/sup><\/strong><\/p>\n\n\n\n

(ii) 1500<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 1600<\/sup><\/p>\n\n\n\n

Let the number of sides of a regular polygon = n<\/p>\n\n\n\n

Each interior angle = 600<\/sup><\/p>\n\n\n\n

The sum of interior angle of polygon can be calculated as,<\/p>\n\n\n\n

(2n \u2013 4) \u00d7 900<\/sup> = 1600<\/sup> \u00d7 n<\/p>\n\n\n\n

1800<\/sup>n \u2013 3600<\/sup> = 1600<\/sup>n<\/p>\n\n\n\n

1800<\/sup>n \u2013 1600<\/sup>n = 3600<\/sup><\/p>\n\n\n\n

200<\/sup>n = 3600<\/sup><\/p>\n\n\n\n

n = 3600<\/sup> \/ 200<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

n = 18<\/p>\n\n\n\n

Hence, the number of sides = 18<\/p>\n\n\n\n

(ii) 1500<\/sup><\/p>\n\n\n\n

Let us consider the number of sides of regular polygon be n<\/p>\n\n\n\n

The sum of the interior angle of polygon = (2n \u2013 4) \u00d7 <\/strong>900<\/sup><\/p>\n\n\n\n

Each interior angle = 1500<\/sup><\/p>\n\n\n\n

The sum of the interior angle of polygon can be calculated as,<\/p>\n\n\n\n

(2n \u2013 4) \u00d7 900<\/sup> = 1500<\/sup> \u00d7 n<\/p>\n\n\n\n

1800<\/sup>n \u2013 3600<\/sup> = 1500<\/sup>n<\/p>\n\n\n\n

1800<\/sup>n \u2013 1500<\/sup>n = 3600<\/sup><\/p>\n\n\n\n

300<\/sup>n = 3600<\/sup><\/p>\n\n\n\n

n = 3600<\/sup> \/ 300<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

n = 12<\/p>\n\n\n\n

Hence, the number of sides = 12<\/p>\n\n\n\n

3. Find number of sides in a regular polygon, if its each exterior angle is:<\/strong><\/p>\n\n\n\n

(i) 300<\/sup><\/strong><\/p>\n\n\n\n

(ii) 360<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 300<\/sup><\/p>\n\n\n\n

Let us assume the number of sides be n<\/p>\n\n\n\n

Each exterior angle = 300<\/sup><\/p>\n\n\n\n

Each exterior angle of polygon = 3600<\/sup> \/ n<\/p>\n\n\n\n

Now, we have<\/p>\n\n\n\n

3600<\/sup> \/ n = 300<\/sup><\/p>\n\n\n\n

n = 3600<\/sup> \/ 300<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

n = 12<\/p>\n\n\n\n

Hence, the number of sides = 12<\/p>\n\n\n\n

(ii) 360<\/sup><\/p>\n\n\n\n

Let us assume the number of sides be n<\/p>\n\n\n\n

Each exterior angle = 360<\/sup><\/p>\n\n\n\n

Each exterior angle of polygon = 3600<\/sup> \/ n<\/p>\n\n\n\n

Now, we have<\/p>\n\n\n\n

3600<\/sup> \/ n = 360<\/sup><\/p>\n\n\n\n

n = 3600<\/sup> \/ 360<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

n = 10<\/p>\n\n\n\n

Hence, the number of sides = 10<\/p>\n\n\n\n

4. Is it possible to have a regular polygon whose each interior angle is:<\/strong><\/p>\n\n\n\n

(i) 1350<\/sup><\/strong><\/p>\n\n\n\n

(ii) 1550<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 1350<\/sup><\/p>\n\n\n\n

Let the number of sides of regular polygon be n<\/p>\n\n\n\n

The sum of the interior angle of polygon = (2n \u2013 4) \u00d7 900<\/sup><\/p>\n\n\n\n

Each interior angle = 1350<\/sup><\/p>\n\n\n\n

The sum of interior angle of polygon can be calculated as,<\/p>\n\n\n\n

(2n \u2013 4) \u00d7 900<\/sup> = 1350<\/sup> \u00d7 n<\/p>\n\n\n\n

1800<\/sup>n \u2013 3600<\/sup> = 1350<\/sup>n<\/p>\n\n\n\n

1800<\/sup>n \u2013 1350<\/sup>n = 3600<\/sup><\/p>\n\n\n\n

450<\/sup>n = 3600<\/sup><\/p>\n\n\n\n

n = 3600<\/sup> \/ 450<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

n = 8<\/p>\n\n\n\n

Since, it is a whole number<\/p>\n\n\n\n

Therefore, it is possible to have a regular polygon whose interior angle is 1350<\/sup><\/p>\n\n\n\n

(ii) 1550<\/sup><\/p>\n\n\n\n

Let the number of sides of a regular polygon is n<\/p>\n\n\n\n

The sum of the interior angle of polygon is (2n \u2013 4) \u00d7 <\/strong>900<\/sup><\/p>\n\n\n\n

Each interior angle = 1550<\/sup><\/p>\n\n\n\n

The sum of the interior angle of polygon can be calculated as,<\/p>\n\n\n\n

(2n \u2013 4) \u00d7 900<\/sup> = 1550<\/sup> \u00d7 n<\/p>\n\n\n\n

1800<\/sup>n \u2013 3600<\/sup> = 1550<\/sup>n<\/p>\n\n\n\n

1800<\/sup>n \u2013 1550<\/sup>n = 3600<\/sup><\/p>\n\n\n\n

250<\/sup>n = 3600<\/sup><\/p>\n\n\n\n

n = 3600<\/sup> \/ 250<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

n = 72 \/ 5<\/p>\n\n\n\n

Since, it is not a whole number<\/p>\n\n\n\n

Therefore, it is not possible to form a regular polygon whose interior angle is 1550<\/sup><\/p>\n\n\n\n

5. Is it possible to have a regular polygon whose each exterior angle is:<\/strong><\/p>\n\n\n\n

(i) 1000<\/sup><\/strong><\/p>\n\n\n\n

(ii) 360<\/sup><\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) 1000<\/sup><\/p>\n\n\n\n

Let the number of sides be n<\/p>\n\n\n\n

Each exterior angle = 1000<\/sup><\/p>\n\n\n\n

Each exterior angle of a polygon is calculated as,<\/p>\n\n\n\n

3600<\/sup> \/ n<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

3600<\/sup> \/ n = 1000<\/sup><\/p>\n\n\n\n

n = 3600<\/sup> \/ 1000<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

n = 18 \/ 5<\/p>\n\n\n\n

Since, it is not a whole number<\/p>\n\n\n\n

Therefore, it is not possible to form a regular polygon<\/p>\n\n\n\n

(ii) 360<\/sup><\/p>\n\n\n\n

Let us consider the number of sides be n<\/p>\n\n\n\n

Each exterior angle = 360<\/sup><\/p>\n\n\n\n

Each exterior angle of polygon = 3600<\/sup> \/ n<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

3600<\/sup> \/ n = 360<\/sup><\/p>\n\n\n\n

n = 3600<\/sup> \/ 360<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

n = 10<\/p>\n\n\n\n

Since, it is a whole number<\/p>\n\n\n\n

Therefore, it is possible to form a regular polygon<\/p>\n\n\n\n

6. The ratio between the interior angle and the exterior angle of a regular polygon is 2: 1. Find:<\/strong><\/p>\n\n\n\n

(i) each exterior angle of this polygon.<\/strong><\/p>\n\n\n\n

(ii) number of sides in the polygon.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given<\/p>\n\n\n\n

Interior angle: exterior angle = 2: 1<\/p>\n\n\n\n

Let us assume the interior angle = 2x0<\/sup> and the exterior angle = x0<\/sup><\/p>\n\n\n\n

\"Selina<\/figure>\n\n\n\n

The sum of the interior angle and exterior angle is 1800<\/sup><\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

2x0<\/sup> + x0<\/sup> = 1800<\/sup><\/p>\n\n\n\n

3x = 1800<\/sup><\/p>\n\n\n\n

x = 1800<\/sup> \/ 3<\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

x = 600<\/sup><\/p>\n\n\n\n

Therefore, each exterior angle = 600<\/sup><\/p>\n\n\n\n

(ii) Let us assume the number of sides be n<\/p>\n\n\n\n

Each exterior angle = 600<\/sup><\/p>\n\n\n\n

Each exterior angle of polygon = 3600<\/sup> \/ n<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

360 \/ n = 600<\/sup><\/p>\n\n\n\n

n = 3600<\/sup> \/ 600<\/sup><\/p>\n\n\n\n

We get,<\/p>\n\n\n\n

n = 6<\/p>\n\n\n\n

Hence, the number of sides = 6<\/p>\n\n\n\n

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