Sprog
Areal af en cirkel
Formler og lommeregner til beregning areal af en cirkel for forskellige kildedata. En tabel med formler til areal af en cirkel. Vores lommeregner hjælper dig med at beregne arealet af en cirkel gratis online eller tjekke allerede udførte beregninger.
Tabel med formler for cirkelområdet (i slutningen af siden)
- Beregning (vise) (skjult)
- noter (vise) (skjult)
1
Cirkelområde gennem radius
π ⋅ r 2 = A r e a \displaystyle{ \pi \cdot r^2 = Area} π ⋅ r 2 = A r e a = π ⋅ 0 2 = 0 \displaystyle{ = \pi \cdot 0^2 = 0} = π ⋅ 0 2 = 0
r - radius
2
Cirkelareal gennem diameter
π ⋅ D 2 4 = A r e a \displaystyle{ \frac{ \pi \cdot D^2}{4} = Area } 4 π ⋅ D 2 = A r e a = π ⋅ 0 2 4 = 0 \displaystyle{ = \frac{\pi \cdot 0^2}{4} = 0} = 4 π ⋅ 0 2 = 0
D - diameter
3
Cirkelområde langs omkredsen
l 2 4 ⋅ π = A r e a \displaystyle{ \frac{l^2}{4 \cdot \pi} = Area } 4 ⋅ π l 2 = A r e a = 0 2 4 ⋅ π = 0 \displaystyle{ = \frac{0^2}{4 \cdot \pi} = 0} = 4 ⋅ π 0 2 = 0
4
Areal af en cirkel gennem en firkant, der er indskrevet i en cirkel
π ⋅ a 2 2 = A r e a \displaystyle{ \frac{\pi \cdot a^2}{2} = Area } 2 π ⋅ a 2 = A r e a = π ⋅ 0 2 2 = 0 \displaystyle{ = \frac{\pi \cdot 0^2}{2} =0 } = 2 π ⋅ 0 2 = 0
a - siden
5
Areal af en cirkel, der er indskrevet i en firkant
π ⋅ ( A 2 ) 2 = π ⋅ A 2 4 = A r e a \displaystyle{ \pi \cdot (\frac{A}{2})^2 = \frac{\pi \cdot A^2}{4} = Area } π ⋅ ( 2 A ) 2 = 4 π ⋅ A 2 = A r e a = π ⋅ 0 2 4 = 0 \displaystyle{ = \frac{\pi \cdot 0^2}{4} = 0 } = 4 π ⋅ 0 2 = 0
A - siden
6
Areal af en cirkel beskrevet nær en vilkårlig trekant
Denne formel er kun anvendelig, hvis en cirkel kan beskrives omkring en trekant, det vil sige, at alle tre hjørner af trekanten skal ligge på en cirkellinie. Trekanten i dette tilfælde kan være en hvilken som helst.
For at beregne arealet af en cirkel beregnes foreløbigt trekantens semiperimeter a + b + c 2 = p \displaystyle{\frac{a+b+c}{2} = p} 2 a + b + c = p
π ⋅ ( a ⋅ b ⋅ c 4 ⋅ p ⋅ ( p − a ) ⋅ ( p − b ) ⋅ ( p − c ) ) 2 = A r e a \displaystyle{ \pi \cdot \left( \frac{a \cdot b \cdot c}{4 \cdot \sqrt{p \cdot (p-a) \cdot (p-b) \cdot (p-c)} } \right) ^2 = Area} π ⋅ ( 4 ⋅ √ p ⋅ ( p − a ) ⋅ ( p − b ) ⋅ ( p − c ) a ⋅ b ⋅ c ) 2 = A r e a = π ⋅ ( 0 ⋅ 0 ⋅ 0 4 ⋅ 0 ⋅ ( 0 − 0 ) ⋅ ( 0 − 0 ) ⋅ ( 0 − 0 ) ) 2 = N a N \displaystyle{ =\pi \cdot \left( \frac{0 \cdot 0 \cdot 0}{4 \cdot \sqrt{0 \cdot (0-0) \cdot (0-0) \cdot (0-0)} } \right)^2 = NaN} = π ⋅ ( 4 ⋅ √ 0 ⋅ ( 0 − 0 ) ⋅ ( 0 − 0 ) ⋅ ( 0 − 0 ) 0 ⋅ 0 ⋅ 0 ) 2 = N a N
a - siden
b - siden
c - siden
7
Området med cirklen beskrevet nær en ligesidet trekant
π ⋅ a 2 3 = A r e a \displaystyle{ \pi \cdot \frac{a^2}{3} = Area } π ⋅ 3 a 2 = A r e a = π ⋅ 0 2 3 = 0 \displaystyle{= \pi \cdot \frac{0^2}{3} = 0 } = π ⋅ 3 0 2 = 0
a - siden
8
Areal af en cirkel beskrevet nær en ligesidet trekant, beregnet af trekantens højde
π ⋅ ( 2 ⋅ h 3 ) 2 = A r e a \displaystyle{ \pi \cdot \left( \frac{2 \cdot h}{3} \right) ^2 = Area } π ⋅ ( 3 2 ⋅ h ) 2 = A r e a = π ⋅ ( 2 ⋅ 0 3 ) 2 = 0 \displaystyle{ = \pi \cdot \left( \frac{2 \cdot 0}{3} \right) ^2 = 0} = π ⋅ ( 3 2 ⋅ 0 ) 2 = 0
h - højden
9
Areal af en cirkel beskrevet i nærheden af en ensartet trekant
π ⋅ ( a 4 4 ⋅ a 2 − b 2 ) = A r e a \displaystyle{ \pi \cdot \left( \frac{a^4}{4 \cdot a^2-b^2} \right) = Area } π ⋅ ( 4 ⋅ a 2 − b 2 a 4 ) = A r e a = π ⋅ ( 0 4 4 ⋅ 0 2 − 0 2 ) = N a N \displaystyle{ = \pi \cdot \left( \frac{0^4}{4 \cdot 0^2-0^2} \right) = NaN} = π ⋅ ( 4 ⋅ 0 2 − 0 2 0 4 ) = N a N
a - siden
b - basen
10
Areal af en cirkel beskrevet nær en højre trekant
π 4 ⋅ ( a 2 + b 2 ) = A r e a \displaystyle{ \frac{\pi}{4} \cdot \left( a^2 + b ^2 \right) = Area } 4 π ⋅ ( a 2 + b 2 ) = A r e a = π 4 ⋅ ( 0 2 + 0 2 ) = 0 \displaystyle{ = \frac{\pi}{4} \cdot \left( 0^2 + 0 ^2 \right) = 0} = 4 π ⋅ ( 0 2 + 0 2 ) = 0
a - siden
b - siden
11
Areal af en cirkel, der er indskrevet i en ensartet trekant
π ⋅ b 2 4 ⋅ ( 2 ⋅ a − b 2 ⋅ a + b ) = A r e a \displaystyle{ \pi \cdot \frac{b^2}{4} \cdot \left( \frac{2 \cdot a - b}{2 \cdot a + b} \right) = Area } π ⋅ 4 b 2 ⋅ ( 2 ⋅ a + b 2 ⋅ a − b ) = A r e a = π ⋅ 0 2 4 ⋅ ( 2 ⋅ 0 − 0 2 ⋅ 0 + 0 ) = N a N \displaystyle{ = \pi \cdot \frac{0^2}{4} \cdot \left( \frac{2 \cdot 0 - 0}{2 \cdot 0 + 0} \right) = NaN } = π ⋅ 4 0 2 ⋅ ( 2 ⋅ 0 + 0 2 ⋅ 0 − 0 ) = N a N
a - siden
b - basen
12
Areal af en cirkel, der er indskrevet i en ensartet trekant, beregnet af siderne af trekanten og vinklen mellem dem
π ⋅ b 2 4 ( sin α ∘ 1 + sin α 2 ∘ ) 2 = A r e a \displaystyle{ \pi \cdot \frac{b^2}{4} \left( \frac{\sin \alpha ^\circ}{1+\sin \frac{\alpha}{2}^\circ} \right)^2 = Area } π ⋅ 4 b 2 ( 1 + sin 2 α ∘ sin α ∘ ) 2 = A r e a = π ⋅ 0 2 4 ( sin 0 ∘ 1 + sin 0 2 ∘ ) 2 = 0 \displaystyle{ = \pi \cdot \frac{0^2}{4} \left( \frac{\sin 0 ^\circ}{1+\sin \frac{0}{2}^\circ} \right)^2 = 0} = π ⋅ 4 0 2 ( 1 + sin 2 0 ∘ sin 0 ∘ ) 2 = 0
b - siden
α - vinkel mellem sider
13
Areal af en cirkel indskrevet i en højre trekant
π ⋅ ( a + b − c 2 ) 2 = A r e a \displaystyle{ \pi \cdot \left( \frac{a + b - c}{2} \right)^2 = Area } π ⋅ ( 2 a + b − c ) 2 = A r e a = π ⋅ ( 0 + 0 − 0 2 ) 2 = 0 \displaystyle{ = \pi \cdot \left( \frac{0 + 0 - 0}{2} \right)^2 = 0} = π ⋅ ( 2 0 + 0 − 0 ) 2 = 0
a - siden
b - siden
c - siden
14
Areal af en cirkel, der er indskrevet i en retvinklet trekant, beregnet ved siden og hjørnet
π ⋅ b 2 4 ( sin α ∘ + cos α ∘ − 1 cos α ∘ ) 2 = A r e a \displaystyle{ \pi \cdot \frac{b^2}{4} \left( \frac{\sin \alpha ^\circ + \cos \alpha ^\circ -1}{\cos \alpha ^\circ} \right)^2 = Area } π ⋅ 4 b 2 ( cos α ∘ sin α ∘ + cos α ∘ − 1 ) 2 = A r e a = π ⋅ 0 2 4 ( sin 0 ∘ + cos 0 ∘ − 1 cos 0 ∘ ) 2 = 0 \displaystyle{ = \pi \cdot \frac{0^2}{4} \left( \frac{\sin 0 ^\circ + \cos 0 ^\circ -1}{\cos 0 ^\circ} \right)^2 =0 } = π ⋅ 4 0 2 ( cos 0 ∘ sin 0 ∘ + cos 0 ∘ − 1 ) 2 = 0
b - siden
α - basisvinkel
15
Areal af en cirkel, der er indskrevet i en ligesidet trekant
π ⋅ a 2 1 2 = A r e a \displaystyle{ \pi \cdot \frac{a^2}{12} = Area } π ⋅ 1 2 a 2 = A r e a = π ⋅ 0 2 1 2 = 0 \displaystyle{ = \pi \cdot \frac{0^2}{12} = 0} = π ⋅ 1 2 0 2 = 0
a - siden
16
Areal af en cirkel, der er indskrevet i en isosceles trapezoid, beregnet ud fra basen af trapezoidet og vinklen ved basen
π ⋅ b 2 4 ⋅ ( t g α ∘ 2 ) 2 = A r e a \displaystyle{ \pi \cdot \frac{b^2}{4} \cdot \left( tg\frac{\alpha ^\circ}{2} \right)^2 =Area } π ⋅ 4 b 2 ⋅ ( t g 2 α ∘ ) 2 = A r e a = π ⋅ 0 2 4 ⋅ ( t g 0 ∘ 2 ) 2 = 0 \displaystyle{ = \pi \cdot \frac{0^2}{4} \cdot \left( tg\frac{0 ^\circ}{2} \right)^2 = 0 } = π ⋅ 4 0 2 ⋅ ( t g 2 0 ∘ ) 2 = 0
b - siden
α - basisvinkel
17
Området med cirklen beskrevet i nærheden af den ensartede trapezoid, beregnet på siderne af trapezoidet, dens diagonal og base
For at beregne arealet af en cirkel beregnes foreløbigt trekantens semiperimeter ABC a + d + c 2 = p \displaystyle{ \frac{a+d+c}{2} = p} 2 a + d + c = p
π ⋅ ( a ⋅ d ⋅ c 4 ⋅ p ⋅ ( p − a ) ⋅ ( p − d ) ⋅ ( p − c ) ) 2 = A r e a \displaystyle{ \pi \cdot \left( \frac{a \cdot d \cdot c}{4 \cdot \sqrt{p \cdot (p-a) \cdot (p-d) \cdot (p-c)} } \right) ^2 = Area } π ⋅ ( 4 ⋅ √ p ⋅ ( p − a ) ⋅ ( p − d ) ⋅ ( p − c ) a ⋅ d ⋅ c ) 2 = A r e a = π ⋅ ( 0 ⋅ 0 ⋅ 0 4 ⋅ 0 ⋅ ( 0 − 0 ) ⋅ ( 0 − 0 ) ⋅ ( 0 − 0 ) ) 2 = N a N \displaystyle{ =\pi \cdot \left( \frac{0 \cdot 0 \cdot 0}{4 \cdot \sqrt{0 \cdot (0-0) \cdot (0-0) \cdot (0-0)} } \right)^2 =NaN } = π ⋅ ( 4 ⋅ √ 0 ⋅ ( 0 − 0 ) ⋅ ( 0 − 0 ) ⋅ ( 0 − 0 ) 0 ⋅ 0 ⋅ 0 ) 2 = N a N
a - siden
c - siden
d - diagonal
18
Området med cirklen beskrevet nær rektanglet
π ⋅ ( a 2 + b 2 4 ) = A r e a \displaystyle{ \pi \cdot \left( \frac{a^2 + b^2}{4} \right) = Area } π ⋅ ( 4 a 2 + b 2 ) = A r e a = π ⋅ ( 0 2 + 0 2 4 ) = 0 \displaystyle{ = \pi \cdot \left( \frac{0^2 + 0^2}{4} \right) =0 } = π ⋅ ( 4 0 2 + 0 2 ) = 0
a - siden
b - siden
19
Areal af en cirkel beskrevet nær en almindelig polygon
π ⋅ ( a 2 ⋅ sin 1 8 0 ∘ N ) 2 = A r e a \displaystyle{ \pi \cdot \left( \frac{a}{2 \cdot \sin \frac{180 ^\circ}{N} } \right)^2 = Area } π ⋅ ( 2 ⋅ sin N 1 8 0 ∘ a ) 2 = A r e a = π ⋅ ( 0 2 ⋅ sin 1 8 0 ∘ 0 ) 2 = N a N \displaystyle{ = \pi \cdot \left( \frac{0}{2 \cdot \sin \frac{180 ^\circ}{0} } \right)^2 =NaN } = π ⋅ ( 2 ⋅ sin 0 1 8 0 ∘ 0 ) 2 = N a N
a - siden
N - antallet af sider af polygonen
20
Areal af en cirkel beskrevet nær en almindelig sekskant
π ⋅ a 2 = A r e a \displaystyle{ \pi \cdot a^2 = Area } π ⋅ a 2 = A r e a = π ⋅ 0 2 = 0 \displaystyle{ = \pi \cdot 0^2 = 0 } = π ⋅ 0 2 = 0
a - siden
Bemærk:
Hvis vinklen i kildedataene er angivet i radianer, kan du bruge formlen til konvertering til grader: 1 radian × (180/π)° = 57,296°
Tabel med formler for cirkelområdet