1.
150/360=0,5/ 2 PI r
30 r= 36*0,5
30 r= 18 /:30
r= 3/5
2. Pole wycinka = (x/2) * r², gdzie x - miara łukowa kąta środkowego, r - promieńW twoim zadaniu P=(1/9)*π = π/9 oraz α=135⁰x=(α*π)/180⁰=(135⁰π)/180⁰=(45π)/60=(9π)/12=(3π)/4Zatem podstawiając do wzoru mamy(π/9)=[(3π)/4]/2 * r²(π/9)=(3π)/8 * r² - (3π)/8 * r²= - π/9 /:(-1)(3π)/8 * r² = π/9 /:π(3/8)* r²= (1/9) /:(3/8)r²=(1/9):(3/8)r²=(1/9)*(8/3)r²=(8/27)r=√(8/27)r=[2√2]/[3√3]r=[2√6]/9r=(2/9)*√6
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1.
150/360=0,5/ 2 PI r
30 r= 36*0,5
30 r= 18 /:30
r= 3/5
2. Pole wycinka = (x/2) * r², gdzie x - miara łukowa kąta środkowego, r - promień
W twoim zadaniu P=(1/9)*π = π/9 oraz α=135⁰
x=(α*π)/180⁰=(135⁰π)/180⁰=(45π)/60=(9π)/12=(3π)/4
Zatem podstawiając do wzoru mamy
(π/9)=[(3π)/4]/2 * r²
(π/9)=(3π)/8 * r²
- (3π)/8 * r²= - π/9 /:(-1)
(3π)/8 * r² = π/9 /:π
(3/8)* r²= (1/9) /:(3/8)
r²=(1/9):(3/8)
r²=(1/9)*(8/3)
r²=(8/27)
r=√(8/27)
r=[2√2]/[3√3]
r=[2√6]/9
r=(2/9)*√6