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Pb = 2Pp
Pp = a²√3/4
Pb = 3aH
3aH = a²√3/2
H = a√3/6
z tw. Pitagorasa:
D² = H² + a² => D = a√39/6
d² = (a/2)² + H² = a√12/6
h = a√3/2
z tw. cosinusów:
h² = D² + d² - 2Ddcosα
3a²/4 = 39a²/36 + 12a²/36 - 2a²√468/36 *cosα |*(36/a²)
27 = 51 - 2√468 *cosα
cosα = 24/2√468 = 12/√468 = 1/√13 = √13/13
zadanie 303
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z tw. Pitagorasa:
d = a√2
D = c + e
D² = a² + d² => D = a√3
a² = h² + c²
d² = h² + e²
a² - d² = c² - e²
a² - 2a² = (c - e)(c + e) = (c - e)D
- a² = ca√3 - ea√3
a/√3 = e - c = e + c - 2c = D - 2c
2c = D - a/√3 = a√3 - a/√3 = a3√3/3 - a√3/3 = a2√3/3
c = a√3/3
e = 2a√3/3
zadanie 305
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najkrótsza przekątna podstawy to podwojona wysokość trójkąta równobocznego o boku a:
2h = a√3
z tw. Pitagorasa:
D² = (2a)² + H²
(4 * 2h)² = (2a)² + H²
48a² = 4a² + H²
H = 2a√11
V = Pp * H
Pp = 6 a²√3/4
V = 3a²√3/2 * 2a√11 = a³*6√33
jak masz pytania to pisz na pw