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Wednesday, May 3, 2023
AIRIL AHMAD
Wednesday, May 3, 2023
AIRIL AHMAD
POLYGONBy
Airil Bin Ahmad
Wednesday, May 3, 2023
AIRIL AHMAD
Polygon is a plane figure bounded by straight lines
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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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AIRIL AHMAD
Wednesday, May 3, 2023
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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
4. Find the value of x:
1390
x0
1130
1310
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5. Find the value of x:
1020
1250
x0650
6. Find the value of x:
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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AIRIL AHMAD
Wednesday, May 3, 2023
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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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AIRIL AHMAD
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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AIRIL AHMAD
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
Airil Bin Ahmad
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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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AIRIL AHMAD
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
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