note math pmr polygon

Wednesday, May 3, 2023

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POLYGONBy

Airil Bin Ahmad


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Polygon is a plane figure bounded by straight lines

D:\Airil\MathsT3\2 Polygons\1 polygons.gsp


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1200

300300

2. There are 2 types of polygon: a. Irregular Polygon A polygon in which not all the sides are of equal length or not all the interior angles are of equal sides.


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b. Regular Polygon A polygon in which all the sides are of equal length and all the interior angles are of equal size.


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ba

cd Angles a, b, c and d are interior angles.

Interior Angle


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3.The name of polygon :

No.of side (n)

Name of figure No.of triangle Sum of interior angles

345678910

TriangleQuadilateralPentagonHexagonHeptagonOctagonNonagonDecagon

1

23

3600

1800

45678

7200900010800

12600

14400

5400

n -2 (n -2) x 180 3 – 2 =

4 – 2 =5 – 2 =

(3 -2) x 180 =(4 -2) x 180 = (5 -2) x 180 =


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a0

c0

b0

a0 + b0 + c0 = 1800


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a0 c0

b0

a0 + b0 + c0 = 1800

x0

z0

y0

x0 + y0 + z0 = 1800 0


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a0

c0

b0

a0 + b0 + c0 = 1800

q0 s0

r0

q0 + r0 + s0 = 1800

x0

z0

y0

x0 + y0 + z0 = 1800

0


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MATHEMATICS GAMES


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Exercise:

1. Find the sum of the interior angle of each of the

following regular polygons:

a. 12-sides b. 26-sides

c. 22-sides d. 36-sides


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Exercise: 2. Find the value of x:

1000

1200

x0

1250

1100

x0 +

x + 455

x = 5400- 455

1200

1000

11001250

+ + + = 540

= 540

x = 850


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3. Find the value of x:

1020

1250

x01290

1180


1390

x0

1130

1310


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1020

1250

x0650


1090

x0

1130

1010

1020

2390


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7. Find the value of x: 8. Find the value of x:

2x0

2x0

1150 1030

1030

2390

10201250

x0950

x0


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SUM OF EXTERIOR ANGLES

By

Airil Bin Ahmad


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f

e

g

hAngles e, f, g and h are exterior angles.


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ae

d

c

b

Exterior Angle


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ae

d

c

ba+b+c+d+e=3600


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SUM OF EXTERIOR ANGLES = 3600


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Solving Problem Example 1: Find the value of x:

+ 360

x + 2850 = 3600 800

600

x0

750

700

600

800

700

750

x0

+ + + =

x = 3600

- 2850

x = 750

x0

600

800

700

750


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Solving Problem Example 2: Find the value of x:

+ 360

x + 2710 = 3600 800

650

x0

700

560

650

800

560

700

x0

+ + + =

x = 3600

- 2710

x = 890


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Example 3: Find the value of x:

x +

= 3600800

x0x0

650

800

560

700

1150

1240

1100

650

+ + + + = 3600

2710

x = 3600

- 2710

X = 890


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EXERCISE PG 28 EX 2.2C NO 1, 2 AND 3


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NO. OF SIDESBy

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No.of sides(n)=

360 ÷ exterior angle


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Example 1: Find the no. of sides if :

1 a. Exterior angle = 600

n = 3600 ÷ 600

= 6 = Hexagon


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Example 2: Find the no. of sides if :

1b. Interior angle = 1400

n = 3600 ÷ 400

= 9 = Nonagon

Exterior angle = 1800 – 140=40


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Exercise: Find the no. of sides in each of the regular polygons.

1.

450

2.

360

3.

400

4.

200


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Exercise: Find the no. of sides in each of the regular polygons.

5.

1600

6.

1440

7.

1700

8.

1080


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SOLVING PROBLEM OF POLYGONS

By

Airil Bin Ahmad


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EXERCISE:

1.

X0

Y0

2. X0

Y0


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EXERCISE:3.

X0

Y0

4.

X0

Y0


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5. In the regular polygon, find x0 and y0

x0

y0

6. In the regular polygon, find x0 and y0

x0

y0

note math pmr polygon

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