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