|
Exercise |
Solution |
| 2A1 + A2 = area of one semicircle | (1) | |
| 4A1 + 4A2 = area of square | (2) |
| That is, |
|
2A1 + A2 = |
(3) | |
| 4A1 + 4A2 = 2² = 4 | (4) |
| (2) - 2 × (1): |
|
2A2 = 4 - |
||
|
A2 = 2 - |
(5) |
| According to (3), |
|
2A1 = |
||
|
A1 = ( |