Aufgabe 10
Teil a
AB = a = BC²
CA = b
AC² = b²
BD = √(51405² + 522758²) = 522959
Pythagoras: a² + b² = c²
29 + 25 = 54
54 = 54
Seite 178
11 12 D A von oben b (²²) |B| = (4)² + 3² + 2²² = 29
|BC| = 3² + 4² + 0² = 25
|CA| = (-1)² + (-7)² + (-2)² = 54
B A(51212) B (11514) C (41914) (²2)
Ja, die Diagonalen sind gleich lang. Das Dreieck ist rechtwinklig.
|AC²| = (-1)² + 7² + 2² = 54
|BD³²| = (-7)² + 1² + (-2)² = 54
Teil b
5² = a + |AC²| = 1 + 49 = 50
Aufgabe 13
Teil a
A(2/3/17), B(4/15/15), C(6/7/13)
AB = (2/3/17) - (4/15/15) = (-2/12/2)
BC = (6/7/13) - (4/15/15) = (2/8/-2)
CA = (2/3/17) - (6/7/13) = (-4/4/4)
Aufgabe 14
Teil a
A(3/12/14), B(5/7/1), C(11/r/s)
AB² = (5/7/1) - (3/12/14) = (2/-5/-13)
BC² = (11/r/s) - (5/7/1) = (6/(r-7)/(s-1))
CA² = (11/r/s) - (3/12/14) = (8/(r-4)/(s-2))
AB² + BC² = CA²
4 + 36/(r-7)² + 169/(s-1)² = 64/(r-4)² + 196/(s-2)²
Aufgabe 16
Teil a
r(2) + s(4) = 22
r + 4s = 10
2r + 9s = 22
2r + 8s = 20
2r + 9s = 22
S = 2
AB = BC, deswegen kollinear
15 = 17
22 = 2r + 8(2)
22 = 2r + 16
2r = 6
r = 3
s = 1
Teil b
r(-2) + s(2) = 3
3r + 5s = 1
-2r - 4s = 3
6r + 10s = 2
-6r - 12s = 9
2s = -11
s = -5.5
3r - 27.5 = 1
3r = 28.5
r = 9.5
