koło
Beata: W ćwiartkę koła o promieniu 2r wpisano kwadrat. Oblicz stosunek pola kwadratu do pola wycinka
koła.
3 gru 17:20
Basia:
x
2+x
2=a
2
2x
2=a
2
a = x
√2
x+a=2r
a+a
√2 = 2r
√2
a(1+
√2)=2
√2r
| | 4*2*r2 | | 8r2 | | 8r2 | |
a2 = |
| = |
| = |
| = |
| | (1+√2)2 | | 1+2√2+2 | | 3+2√2 | |
| 8(3−2√2)r2 | |
| = 8(3−2√2)r2 |
| 9−4*2 | |
P
kwadratu = a
2
P
wycinka =
14*P
koła =
14*π(2r)
2 =
14*π*4r
2 = πr
2
| | a2 | | 8(3−2√2)r2 | | 8(3−2√2) | |
S = |
| = |
| = |
| |
| | πr2 | | πr2 | | π | |
3 gru 17:48
Beata: Dziękuje bardzo
3 gru 17:51