Dany jest kwadrat o boku 1 oraz odcinek a (0≤a≤1) tak jak na rysunku. Oblicz pole zielonego
trójkąta.
| 1 | ||
tgα = | ||
| a |
| h | ||
tgα = | ||
| x |
| h | ||
tg(45o) = 1 = | −−−> h = a−x | |
| a−x |
| a−x | 1 | a2 | ||||
tgα = | = | −−> x = a(a−x) −−−> x + ax = a2 −−−> x = | ||||
| x | a | a+1 |
| a2 | a2 + a − a2 | a | ||||
h = a − x = a − | = | = | ||||
| a+1 | a+1 | a+1 |
| a2 | ||
Pmałego Δ = | ||
| 2(a+1) |
| a | ||
PΔ prostokątnego = | ||
| 2 |
| a2+a − a2 | a | |||
różnica = | = | |||
| 2(a+1) | 2(a+1) |
| 1 | 1 | a | 1−a | |||||
Pszukane = | − 2*różnica = | − | = | |||||
| 2 | 2 | a+1 | 2(a+1) |
Kolega nie obrazi się, że napiszę drugie rozw. bez trygonometrii?
| 1 | ||
1) Pz= | −1*x | |
| 2 |
| a | ||
ΔAKS∼ΔDCS w skali k= | ⇔ | |
| 1 |
| AE | a | ||
= | |||
| DE | 1 |
| x | a | ||
=a⇔x= | |||
| 1−x | a+1 |
| 1 | a | |||
Pz= | −1* | |||
| 2 | a+1 |
| 1−a | ||
Pz= | ||
| 2(a+1) |
| 1 | ||
S= | −2P1 | |
| 2 |
| a√2 | ||
Odcinek e dwusiecznej ma długość e= | ||
| a+1 |
| a | ||
2P1= e*sin45o = | ||
| a+1 |
| a−1 | ||
S= | ||
| 2(a+1) |
| 1−a | ||
S= | ||
| 2(a+1) |
