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Perbandingan Ukuran –ukuran
Sebuah Lingkaran
The Ratios of Measurements of a
Circle.
Consider the following figure.
α = area of sector 1=length of arc 1
β area of sector 2 length of arc 21 α
2 P
β
Example.
The point P is the centre of the circle . If area of sector
PKAL = 20 cm2 then, L
a. find area of sector PQBR A •
b. find area of the circle. K
• R R
Q B
Answer
LPQBR = 75o ↔LPQBR=LPKAL x 75 =20cm2.3=30cm2
LPKAL 50o 50 2
L =360o↔L=LPKAL x 360= 20cm2x 36 =144cm2
LPKAL 50o 50 5
500 P
75o
B
A
β
D
∠AOB=α, ∠ADB=β, αnβ are directed to arc ACB
The point O is centter of the circle and r is length of
radius of the circle.
In ∆ AOD, ∠DAO = ∠ADO, since OA = OD = r
In ∆ BOD, ∠OBD = ∠OBD, since OB = OD = r
α
O
β
So, ∠AOD = 180o - ∠DAO - ∠ADO
= 180o - ∠ADO - ∠ADO
= 180o – 2 ∠ADO
∠BOD = 180o - ∠OBD - ∠ODB
= 180o - ∠ODB -∠ODB
= 180o -2 ∠ODB
∠AOD + ∠BOD + α = 360o
↔ (180o - 2∠ADO) + (180o - 2∠ODB) + α =360o
↔ (180o + 180o) – 2(∠ADO + ∠ODB ) + α =360o
↔ α = 2(∠ADO+∠ODB) = 2β ↔ α=2β ↔ β=1 α
2
↔ α = 2β
Lingkaran ii
Lingkaran ii

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Lingkaran ii

  • 1. Perbandingan Ukuran –ukuran Sebuah Lingkaran The Ratios of Measurements of a Circle. Consider the following figure. α = area of sector 1=length of arc 1 β area of sector 2 length of arc 21 α 2 P β
  • 2. Example. The point P is the centre of the circle . If area of sector PKAL = 20 cm2 then, L a. find area of sector PQBR A • b. find area of the circle. K • R R Q B Answer LPQBR = 75o ↔LPQBR=LPKAL x 75 =20cm2.3=30cm2 LPKAL 50o 50 2 L =360o↔L=LPKAL x 360= 20cm2x 36 =144cm2 LPKAL 50o 50 5 500 P 75o
  • 3. B A β D ∠AOB=α, ∠ADB=β, αnβ are directed to arc ACB The point O is centter of the circle and r is length of radius of the circle. In ∆ AOD, ∠DAO = ∠ADO, since OA = OD = r In ∆ BOD, ∠OBD = ∠OBD, since OB = OD = r α O β
  • 4. So, ∠AOD = 180o - ∠DAO - ∠ADO = 180o - ∠ADO - ∠ADO = 180o – 2 ∠ADO ∠BOD = 180o - ∠OBD - ∠ODB = 180o - ∠ODB -∠ODB = 180o -2 ∠ODB ∠AOD + ∠BOD + α = 360o ↔ (180o - 2∠ADO) + (180o - 2∠ODB) + α =360o ↔ (180o + 180o) – 2(∠ADO + ∠ODB ) + α =360o ↔ α = 2(∠ADO+∠ODB) = 2β ↔ α=2β ↔ β=1 α 2 ↔ α = 2β