pole trapezu
hubi: Przekątne trapezu mają długość 15 i 20 a wysokość 12. Oblicz pole tego trapezu
29 maj 22:03
Blee:
| | (a+b+b+c)*12 | |
Ptrapezu = |
| |
| | 2 | |
natomiast:
(a+b)
2 + 12
2 = 20
2
(b+c)
2 + 12
2 = 15
2
29 maj 22:16
Eta:
| | 12 | | 3 | | 4 | |
sinα= |
| = |
| to cosα= |
| |
| | 20 | | 5 | | 5 | |
| | 12 | | 4 | | 3 | |
sinβ= |
| = |
| to cosβ= |
| |
| | 15 | | 5 | | 5 | |
| | 3 | | 3 | | 4 | | 4 | |
sinγ= sin(α+β)= |
| * |
| + |
| * |
| = 1 ⇒ γ=90o |
| | 5 | | 5 | | 5 | | 5 | |
=================
29 maj 22:36
hubi: dziękuję!
30 maj 12:06