|BD| = a√2
|BE| = |DE| = h1 = wysokość trójkąta równoramiennego BCS, obliczymy ją w następujący sposób:
| a2 | ||
4a2 = | + h2 | |
| 4 |
| a2 | ||
h2 = 4a2 − | ||
| 4 |
| 16a2 − a2 | ||
h2 = | ||
| 4 |
| 15a2 | ||
h2 = | , h∊R+ | |
| 4 |
| √15a | ||
h = | ||
| 2 |
| 1 | ||
P = | a * h | |
| 2 |
| 1 | ||
P = | * 2a * h1 | |
| 2 |
| 1 | 1 | ||
* a * h = | * 2a * h1 / * 2 | ||
| 2 | 2 |
| √15a | ||
a * | = 2a * h1 | |
| 2 |
| a√15 | ||
h1 = | ||
| 4 |
| 15a2 | 15a2 | 15a2 | ||||
2a2 = | + | − 2 * | * cosα | |||
| 16 | 16 | 16 |
| 30a2 | 15a2 | |||
2a2 = | − | cosα / * 16 | ||
| 16 | 8 |
| 1 | ||
cosα = − | ||
| 15 |