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Exercise 28(A)

**1. State, which of the following are polygons:**

**Solution:**

(i) The given figure is not closed.

Hence, the figure is not a polygon.

(ii) The given figure is closed.

Hence, the figure is a polygon.

(iii) The given figure is closed.

Hence, the figure is a polygon.

(iv) In the given figure, one of the sides is an arc.

Hence, the figure is not polygon.

(v) The side intersects each other in the given figure.

Hence, the figure is not polygon.

**2. Find the sum of interior angles of a polygon with:**

**(i) 9 sides**

**(ii) 13 sides**

**(iii) 16 sides**

**Solution:**

(i) 9 sides

Number of sides n = 9

The Sum of interior angles of polygon = (2n â€“ 4) Ã— 90^{0}

= (2 Ã— 9 â€“ 4) Ã— 90^{0}

= (18 â€“ 4) Ã— 90^{0}

= 14 Ã— 90^{0}

We get,

= 1260^{0}

(ii) 13 sides

Number of sides n = 13

The sum of interior angles of polygon = (2n -4) Ã— 90^{0}

= (2 Ã— 13 â€“ 4) Ã— 90^{0}

= (26 â€“ 4) Ã— 90^{0}

= 22 Ã— 90^{0}

We get,

= 1980^{0}

(iii) 16 sides

Number of sides n = 16

The sum of interior angles of polygon = (2n â€“ 4) Ã— 90^{0}

= (2 Ã— 16 â€“ 4) Ã— 90^{0}

= (32 â€“ 4) Ã— 90^{0}

= 28 Ã— 90^{0}

We get,

= 2520^{0}

**3. Find the number of sides of a polygon, if the sum of its interior angles is:**

**(i) 1440 ^{0}**

**(ii) 1620 ^{0}**

**Solution:**

(i) 1440^{0}

The sum of interior angles of polygon = 1440^{0}

Let the number of sides = n

The sum of interior angle of polygon is (2n â€“ 4) Ã— 90^{0}

The side of polygon can be calculated as,

(2n â€“ 4) Ã— 90^{0} = 1440^{0}

2n â€“ 4 = 1440^{0} / 90^{0}

2(n â€“ 2) = 1440^{0} / 90^{0}

n â€“ 2 = 1440^{0} / (2 Ã— 90^{0})

On further calculation, we get

n â€“ 2 = 16^{0} / 2

We get,

n â€“ 2 = 8

n = 8 + 2

n = 10

Hence, the side of polygon = 10

(ii) 1620^{0}

Given

The sum of interior angles of polygon = 1620^{0}

Let number of sides = n

The sum of interior angles of polygon = (2n â€“ 4) **Ã— **90^{0}

The side of polygon can be calculated as,

(2n â€“ 4) Ã— 90^{0} = 1620^{0}

2(n â€“ 2) Ã— 90^{0} = 1620^{0}

n â€“ 2 = 1620^{0} / (2 Ã— 90^{0})

n â€“ 2 = 810^{0} / 90^{0}

We get,

n â€“ 2 = 9

n = 9 + 2

n = 11

Hence, the side of polygon = 11

**4. Is it possible to have a polygon, whose sum of interior angles is 1030 ^{0}.**

**Solution:**

Given

The sum of interior angles of polygon = 1030^{0}

Let us consider the number of sides = n

The sum of interior angle of polygon = (2n â€“ 4) Ã— 90^{0}

The side of polygon is calculated as,

(2n â€“ 4) Ã— 90^{0} = 1030^{0}

2(n â€“ 2)= 1030^{0} / 90^{0}

On further calculation, we get

(n â€“ 2) = 1030^{0} / (2 **Ã— **90^{0})

(n â€“ 2) = 103^{0} / 18^{0}

n = 5.72 + 2

n = 7.72

which is not a whole number.

Therefore, it is not a polygon, whose sum of interior angles is 1030^{0}

**5. (i) If all the angles of a hexagon arc equal, find the measure of each angle.**

**(ii) If all the angles of an octagon are equal, find the measure of each angle.**

**Solution:**

(i) Number of sides of polygon n = 6

Let us consider each angle be = x^{0}

We know,

The sum of interior angles of hexagon = 6x^{0}

The sum of interior angle of polygon = (2n â€“ 4) Ã— 90^{0}

The sum of the interior angles of polygon can be calculated as,

(2n â€“ 4) Ã— 90^{0} = Sum of angles

(2 Ã— 6 â€“ 4) Ã— 90^{0} = 6x^{0}

(12 â€“ 4) Ã— 90^{0} = 6x^{0}

6x^{0} = 8 Ã— 90^{0}

x^{0} = (8 Ã— 90^{0}) / 6

We get,

x = 120^{0}

Therefore, each angle of hexagon = 120^{0}

(ii) Number of sides of octagon n = 8

Let us consider each angle be = x^{0}

We know that,

The sum of interior angles of octagon = 8x^{0}

The sum of interior angles of polygon = (2n â€“ 4) Ã— 90^{0}

The sum of interior angles of polygon can be calculated as,

(2n â€“ 4) Ã— 90^{0} = Sum of angles

(2n â€“ 4) Ã— 90^{0} = 8x^{0}

(2 Ã— 8 â€“ 4) Ã— 90^{0} = 8x^{0}

12 Ã— 90^{0} = 8x^{0}

8x^{0} = 12 Ã— 90^{0}

x^{0} = (12 Ã— 90^{0}) / 8

We get,

x^{0} = 135^{0}

Therefore, each angle of octagon = 135^{0}

**6. One angle of a quadrilateral is 90 ^{0} and all other angles are equal; find each equal angle**

**Solution:**

Let us consider all the three equal angle of a quadrilateral be x^{0}

The sum of angles of a quadrilateral = 360^{0}

x + x + x + 90^{0} = 360^{0}

3x + 90^{0} = 360^{0}

3x = 360^{0} â€“ 90^{0}

3x = 270^{0}

x = 270^{0} / 3

We get,

x = 90^{0}

The measure of each equal angle = 90^{0}

**7. If angles of quadrilateral are in the ratio 4: 5: 3: 6; find each angle of the quadrilateral.**

**Solution:**

Let us consider the angles of quadrilateral be 4x, 5x, 3x and 6x

We know,

The sum of angles of quadrilateral = 360^{0}

4x + 5x + 3x + 6x = 360^{0}

18x = 360^{0}

x = 360^{0} / 18

We get,

x = 20^{0}

Now, all the angles are,

4x = 4 Ã— 20^{0}

= 80^{0}

5x = 5 Ã— 20^{0}

= 100^{0}

3x = 3 Ã— 20^{0}

= 60^{0}

6x = 6 Ã— 20^{0}

= 120^{0}

Therefore, the angles of the quadrilateral are 80^{0}, 100^{0}, 60^{0} and 120^{0}

**8. If one angle of a pentagon is 120 ^{0} and each of the remaining four angles is x^{0}, find the magnitude of x.**

**Solution:**

Given

One angle of a pentagon = 120^{0}

Number of sides of pentagon n = 5

Let us consider all other equal angle of pentagon be x

The sum of interior angle of polygon is (2n â€“ 4) Ã— 90^{0}

The sum of the interior angle of pentagon can be calculated as,

(2n â€“ 4) Ã— 90^{0} = (2 Ã— 5 â€“ 4) Ã— 90^{0}

= 6 Ã— 90^{0 }

We get,

= 540^{0}

Therefore, the sum of interior angles of pentagon is 540^{0}

Now,

x + x + x + x + 120^{0} = 540^{0}

4x + 120^{0} = 540^{0}

4x = 540^{0} – 120^{0}

4x = 420^{0}

x = 420^{0} / 4

We get,

x = 105^{0}

Hence, the value of x = 105^{0}

**9. The angles of a pentagon are in the ratio 5: 4: 5: 7: 6; find each angle of the pentagon.**

**Solution:**

Let us consider all the angle of pentagon as 5x, 4x, 5x, 7x and 6x

The sum of the interior angle of polygon is (2n â€“ 4) Ã— 90^{0}

The sum of the interior angle of pentagon can be calculated as,

(2n â€“ 4) Ã— 90^{0} = (2 Ã— 5 â€“ 4) Ã— 90^{0}

= 6 Ã— 90^{0}

We get,

= 540^{0}

The sum of interior angles of pentagon = 540^{0}

Hence,

5x + 4x + 5x + 7x + 6x = 540^{0}

27x = 540^{0}

x = 540^{0} / 27

We get,

x = 20^{0}

Thus, each angle,

5x = 5 Ã— 20^{0}

= 100^{0}

4x = 4 Ã— 20^{0}

= 80^{0}

5x = 5 Ã— 20^{0}

= 100^{0}

7x = 7 Ã— 20^{0}

= 140^{0}

6x = 6 Ã— 20^{0}

= 120^{0}

Therefore, all the angles of a pentagon are 100^{0}, 80^{0}, 100^{0}, 140^{0} and 120^{0}

**10. Two angles of a hexagon are 90 ^{0} and 110^{0}. If the remaining four angles arc equal, find each equal angle.**

**Solution:**

Let us consider all the angle of hexagon as x

Number of sides in hexagon n = 6

The sum of interior angle of polygon is (2n â€“ 4) Ã— 90^{0}

The sum of interior angle of hexagon can be calculated as,

(2n â€“ 4) Ã— 90^{0} = (2 Ã— 6 â€“ 4) Ã— 90^{0}

= (12 â€“ 4) Ã— 90^{0}

= 8 Ã— 90^{0}

We get,

= 720^{0}

The sum of interior angles of pentagon is 720^{0}

Hence,

90^{0} + 110^{0} + x + x + x + x = 720^{0}

200^{0} + 4x = 720^{0}

4x = 720^{0} – 200^{0}

4x = 520^{0}

We get,

x = 130^{0}

Hence, the measure of each equal angle = 130^{0}

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