9.
a = 5 cm
b = c = 6 cm
h = ?
Wysokość opuszczona z wierzchołka dzieli podstawę na dwie równe części (trójkąt równoramienny), więc z twierdzenia Pitagorasa liczę wysokość h:
6² + (5/2)² = h²
36 + 2,5² = h²
36 + 6,25 = h²
h² = 42,25
h = √42,25
h = 6,5 cm
10.
[tex]a = a\sqrt{2}\\oraz\\d = 10\\\\a\sqrt{2} = 10 \ \ /:\sqrt{2}\\\\a = \frac{10}{\sqrt{2}} \cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \ cm\\\\Obw = 4a = 4\cdot5\sqrt{2} = 20\sqrt{2} \ cm[/tex]
11.
[tex]\frac{|AE|}{|AB|} = \frac{|EC|}{|CD|}\\\\Oznaczmy \ |AE| = x\\\\\frac{x}{6} = \frac{x+4}{8}\\\\8x = 6(x+4)\\\\8x = 6x + 24\\\\2x = 24 \ \ /:2\\\\x = 12[/tex]
Odp. Długość odcinka |AE| = 12
12.
3 cm i 5 cm - przyprostokątne
c = ?
3² + 5² = c²
9 + 25 = c²
c² = 34
c = √34 cm
13.
[tex]a = 1\\b = 7\\\\\\\\a^{2}+b^{2} = c^{2}\\\\1^{2}+7^{2} = c^{2}\\\\1+49 = c^{2}\\\\c^{2} = 50\\\\c = \sqrt{50} = \sqrt{25\cdot2}\\\\c = 5\sqrt{2}[/tex]
[tex]sin\alpha = \frac{a}{c} = \frac{1}{5\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{10}\\\\cos\alpfa = \frac{b}{c} = \frac{7}{5\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{7\sqrt{2}}{10}\\\\tg\alpha = \frac{a}{b} = \frac{1}{7}\\\\ctg\alpha = \frac{b}{a} = \frac{7}{1} = 7[/tex]
