9
24 S R Q P T B A E D C Ù Ù Ù Ù Ù Ù Ù Ù Ù Ù Example 1 24.1 Polygons For Free Distribution 84

24.1 Polygons Example 1 - Educational Publications … 24.pdfx x E A D C B 800 1000 x x 160 0 0 140 120 0 2x 3x Exercise 24.1 160 + 95 + 140 + 100 + 100 + o o o o o x = 720o 595 +

  • Upload
    doandat

  • View
    216

  • Download
    0

Embed Size (px)

Citation preview

Page 1: 24.1 Polygons Example 1 - Educational Publications … 24.pdfx x E A D C B 800 1000 x x 160 0 0 140 120 0 2x 3x Exercise 24.1 160 + 95 + 140 + 100 + 100 + o o o o o x = 720o 595 +

24

SR

Q

P

T

B

A E

D

C

Ù Ù Ù Ù Ù

Ù Ù Ù Ù Ù

Example 1

24.1 Polygons

For Free Distribution84

Page 2: 24.1 Polygons Example 1 - Educational Publications … 24.pdfx x E A D C B 800 1000 x x 160 0 0 140 120 0 2x 3x Exercise 24.1 160 + 95 + 140 + 100 + 100 + o o o o o x = 720o 595 +

O

A

E

D

CB

Ù Ù Ù Ù Ù

Ù Ù Ù Ù Ù

The sum of the interior angles of a polygon with n sides is (2n - 4) right angles'

24.2 Sum of the interior angles of a polygon

For Free Distribution85

Page 3: 24.1 Polygons Example 1 - Educational Publications … 24.pdfx x E A D C B 800 1000 x x 160 0 0 140 120 0 2x 3x Exercise 24.1 160 + 95 + 140 + 100 + 100 + o o o o o x = 720o 595 +

A

B C

D

E

Example 1

Example 2

Q

R S

T

UP095 0100

0140

0100

0160

0x

For Free Distribution86

Page 4: 24.1 Polygons Example 1 - Educational Publications … 24.pdfx x E A D C B 800 1000 x x 160 0 0 140 120 0 2x 3x Exercise 24.1 160 + 95 + 140 + 100 + 100 + o o o o o x = 720o 595 +

xx

E

A

D

C

B

080

0100

xx

016

0

0

140

0

120

2x

3x

Exercise 24.1

o o o o o o160 + 95 + 140 + 100 + 100 + x = 720o o 595 + x = 720

o o x = 720 - 595o x = 125

1. Find the sum of the interior angles of the polygons given below(i) Pentagon (ii) Heptagon (iii) Decagon (iv) Dodecagon

2. For a square, find (i) the sum the of interior angles(ii) the value of one interior angle

3. For a regular hexagon, find (i) the sum of the interior angles (ii) the value of one interior angle

o4. In a quadrilateral, two interior angles are equal. The other two angles are 100 and o80 . Find the value of each equal angle.

o 5. Find the number of sides in a regular polygon in which one interior angle is144

6. Find the value of an interior angle of a regular polygon with 15 sides.07. One interior angle of a quadrilateral is 90 . If the other three angles are equal, find

the value of each.

8. Find the value of x

9. Find the magnitude of the remaining angles of the polygon.

10. In a certain polygon, when one vertex is joined to the rest of the vertices, 5 triangles are formed.

(i) What is the sum of interior angles of the polygon?(ii) Find the number of sides in the polygon.

For Free Distribution87

Page 5: 24.1 Polygons Example 1 - Educational Publications … 24.pdfx x E A D C B 800 1000 x x 160 0 0 140 120 0 2x 3x Exercise 24.1 160 + 95 + 140 + 100 + 100 + o o o o o x = 720o 595 +

x x

xx

F

A

E

D

CB

060060

A

G

B

H

CE

D

F

rs

BPA

q

x

yE

Da b

t C

u

24.3 Exterior angles of a polygon

11. When all the vertices of a polygon are joined to a point P inside the polygon, 6 triangles are formed. Find the(i) number of sides it has(ii) The sum of the angles at P (iii) The sum of the angles of 6 triangles(iv) The sum of the interior angles of the polygon.

12. Consider the hexagon ABCDEF given in the figure

(i) Find the magnitude of ABF and AFB(ii) Show thar ABF is an equilateral triangle.(iii) Find the value of x (iv) Show that BCEF is a square

When a side of a polygon is produced, the angle

between the produced part and the adjacent side is

called the exterior angle.

GAB, HBC, ECD, FDA are the exterior angles.

At any vertex of the pentagon ABCDE, the exterior and interior angles lie on the same straight line.

oThus x + y = 180op + q = 180or + s = 180ou + t = 180oa + b = 180

Accordingly, Ù Ù Ù Ù

< <

For Free Distribution88

Page 6: 24.1 Polygons Example 1 - Educational Publications … 24.pdfx x E A D C B 800 1000 x x 160 0 0 140 120 0 2x 3x Exercise 24.1 160 + 95 + 140 + 100 + 100 + o o o o o x = 720o 595 +

Then the sum of the interior angles and the exterior o angles at all 5 vertices = 180 ´ 5o (x + y) + (p + q) + (r + s) + (u + t) + (a + b) = 180 ´ 5 o (x + p + r + u + a ) + (y + q + s + t + b ) = 900

o o (x + p + r + u + a ) + 540 = 900

o o\ (x + p + r + u + a ) = 900 - 540o\ The sum of the exterior angles of the pentagon = 360

o This shows that the sum of the exterior angles of the above polygons is 360

Find the value of x

o o o o x + 100 + 110 + 70 = 360o o x + 280 = 360

o o x = 360 - 280ox = 80

0100

0110

070

x

Check and see whether the sum of the exterior angles of a quadrilateral and a ohexagon are also 360

Complete the table given below. Hence build up a relationship for the sum of the exterior angles of any polygon

Activity 1

Activity 2

0 The sum of exterior angles of any polygon is 360

Example 3

(as the sum of the interioroangles of a pentagon is 540 )

Name of the polygon

num

ber

of

side

snu

mbe

r of

ve

rtic

es

Sum of the exterior and interior angles

No, of triangles formed by a vertex with the other vertices

Sum of the interior angles

Sum of the exterior angles

Triangle

Quadrilateral

Pentagon

Hexagon

Heptagon

Octagon

3

4

5

.....

.....

.....

3

.....

.....

.....

.....

.....

o o180 ´ 3 = 540 1 o o180 ´1 =180 o o o 540 - 180 = 360

For Free Distribution89

Page 7: 24.1 Polygons Example 1 - Educational Publications … 24.pdfx x E A D C B 800 1000 x x 160 0 0 140 120 0 2x 3x Exercise 24.1 160 + 95 + 140 + 100 + 100 + o o o o o x = 720o 595 +

24.4 Magnitude of an exterior angle of a regular polygon

Activity 3

Name of the polygon

No: of sides

No: of exterior angles

Sum of the exterior angles

magnitude of one exterior angle

Equilateral triangle

Square

Regular pentagon

Regular hexagon

Regular heptagon

Regular polygon with n sides

3

4

5

...

...

...

3

...

...

...

...

...

03600360

3

0= 120

Example 5

Find the values of a, x, y, p, in the diagram o ox + 110 = 180

o ox = 180 - 110ox = 70

o o100 + p = 180o o p = 180 - 100

o p = 80

o oy + 80 = 180o oy = 180 - 80o y = 100

o o x + p + 80 + a + a = 360o o o o o o 70 + 80 + 80 + 2a = 360 (as x = 70 and p = 80 )

o o230 + 2a = 360o o2a = 360 - 230o2a = 130

o a = = 65

In a regular polygon, all the interior angles are equal. Therefore all the exterior angles too are equal.

Build up a relationship to find the magnitude of an exterior angle of a regular polygon by completing the table given below.

0130

2

0a

0a

080

y

P0100

0110

x

For Free Distribution90

Page 8: 24.1 Polygons Example 1 - Educational Publications … 24.pdfx x E A D C B 800 1000 x x 160 0 0 140 120 0 2x 3x Exercise 24.1 160 + 95 + 140 + 100 + 100 + o o o o o x = 720o 595 +

The magnitude of an exterior angle of any regular polygon = o360

The number of sides in the polygon

Example 5

Example 6

Exercise 24.2

Find the magnitude of an exterior angle of a regular polygon with 12 sidesoSum of all the exterior angles = 360

Number of sides in the regular polygon = 12

magnitude of one exterior angle =

o= 30

0The exterior angle of a regular polygon is 72 . Find the number of sides in the polygonoSum of all the exterior angles = 360

omagnitude of one exterior angle = 72

\ Number of sides =

= 5

1. Find the(i) magnitude of one exterior angle,(ii) magnitude of one interior angle, of a regular polygon with 6 sides.

2. Find the(i) magnitude of one exterior angle,(ii) magnitude of one interior angle, of a regular heptagon.

3. Find the(i) magnitude of one exterior angle,(ii) magnitude of one interior angle, of a square.

4. Find the(i) magnitude of one exterior angle,(ii) magnitude of one interior angle, of an equilateral triangle.

5. Find the(i) magnitude of one exterior angle,(ii) magnitude of one interior angle, of a regular polygon with 8 sides.

o6. The exterior angle of a regular polygon is 60 . Find(i) the number of sides.

0360

12

0360

72

For Free Distribution91

Page 9: 24.1 Polygons Example 1 - Educational Publications … 24.pdfx x E A D C B 800 1000 x x 160 0 0 140 120 0 2x 3x Exercise 24.1 160 + 95 + 140 + 100 + 100 + o o o o o x = 720o 595 +

Can there be a regular polygon with one exterior angle equal to 064 ? Give reasons for your answer.

b

3x

a7x

c 5x

0120bB C

c

a

D0100

050

efE

A

(ii) the magnitude of one interior angle,

o7. For a regular polygon in which one exterior angle is 36 , find(i) the number of sides.(ii) the magnitude of one interior angle.

o8. For a regular polygon in which one exterior angle is 45 , find(i) the number of sides.(ii) magnitude of one interior angle.

o9. For a regular polygon in which one exterior angle is 20 , find(i) the number of sides.(ii) the magnitude of one interior angle.

o10. For a regular polygon in which one interior angle is140 , find(i) the magnitude of one exterior angle. (ii) the number of sides.

11. In a regular polygon, an interior angle is equal to twice the exterior angle, find(i) the magnitude of an exterior angle(ii) the magnitude of an interior angle (iii) the number of sides(iv) the sum of the interior angles.

12. According to the information given in the diagram, find the magnitude of the angles a, b, c, d, e, f.

13. According to the information given in the diagram,(i) find the magnitude of x(ii) find the magnitude of the angles a, b, c,

o14. The magnitude of one exterior angle of a regular polygon is 40 . Find(i) the number of sides in the polygon(ii) the magnitude of an interior angle(iii) the sum of the interior angles of the polygon.

For Free Distribution92