W tym zadaniu wykorzystamy twierdzenia o trójkątach o kątach:
- 45°, 45°, 90°
- 30°, 60°, 90°
W każdym przykładzie najpierw wypiszemy dane boki a następnie porzebne nam boki do obliczenia pola
Wzór na pole trójkąta:
[tex] \frac{a \times h}{2} [/tex]
a)
[tex]2a = 10 \\ a = 5 \\ a \sqrt{3} = 5 \sqrt{3} [/tex]
Potrzebne nam dane:
[tex]a = 5 \\ h = a \sqrt{3} = 5 \sqrt{3} [/tex]
Zatem pole
[tex] \frac{5 \times 5 \sqrt{3} }{2} = \frac{25 \sqrt{3} }{2} = 12.5 \sqrt{3} [/tex]
b)
[tex]a = 5 \\ a \sqrt{2} = 5 \sqrt{2} [/tex]
Potrzebne nam dane:
[tex]a = 5 \\ h = a = 5[/tex]
Zatem pole
[tex] \frac{5 \times 5}{2} = \frac{25}{2} = 12.5[/tex]
c)
[tex]a = 6 \\ 2a = 12 \\ a \sqrt{3} = 6 \sqrt{3} [/tex]
Potrzebne nam dane:
[tex]a = 6 \\ h = a \sqrt{3} = 6 \sqrt{3} [/tex]
Zatem pole
[tex] \frac{6 \times 6 \sqrt{3} }{2} = \frac{36 \sqrt{3} }{2} = 18 \sqrt{3} [/tex]
d)
[tex]a = 1 \\ a \sqrt{2} = \sqrt{2} [/tex]
Potrzebne nam dane:
[tex]a = 1 \\ h = a = 1[/tex]
Zatem pole
[tex] \frac{1 \times 1}{2} = \frac{1}{2} = 0.5[/tex]
e)
[tex]a \sqrt{3} = 9 \\ a = \frac{9}{ \sqrt{3} } = \frac{9 \sqrt{3} }{3} = 3 \sqrt{3} \\ 2a = 3 \sqrt{3} \times 2 = 6 \sqrt{3} [/tex]
Potrzebne nam dane:
[tex]a = 3 \sqrt{3} \\ h = a \sqrt{3} = 3 \sqrt{3} \times \sqrt{3} = 3 \sqrt{9} = 3 \times 3 = 9[/tex]
Zatem pole
[tex] \frac{3 \sqrt{3} \times 9 }{2} = \frac{27 \sqrt{3} }{2} = 13.5 \sqrt{3} [/tex]
f)
[tex]a \sqrt{2 } = 6 \sqrt{2} \\ a = 6[/tex]
Potrzebne nam dane:
[tex]a = 6 \\ h = a = 6[/tex]
Zatem pole
[tex] \frac{6 \times 6}{2 } = \frac{36}{2} = 18[/tex]
g)
[tex]a \sqrt{3} = 8 \\ a = \frac{8}{ \sqrt{3} } = \frac{8 \sqrt{3} }{3} = 2 \frac{2}{3} \sqrt{3} \\ 2a = 2 \frac{2}{3} \sqrt{3} \times 2 = 4 \frac{4}{3} \sqrt{3} = 5 \frac{1}{3} \sqrt{3} [/tex]
Potrzebne nam dane:
[tex]a = 2 \frac{2}{3} \sqrt{3} \\ h = a \sqrt{3} = 8[/tex]
Zatem pole
[tex] \frac{ 2 \frac{2}{3} \sqrt{3} \times 8}{2} = 2 \frac{2}{3} \sqrt{3} \times 4 = 8 \frac{8}{3} \sqrt{3} = 10 \frac{2}{3} \sqrt{3} [/tex]
h)
[tex] a \sqrt{2} = 5 \\ a = \frac{5}{ \sqrt{2} } = \frac{5 \sqrt{2} }{2} = 2.5 \sqrt{2} [/tex]
Potrzebne nam dane:
[tex]a = 2.5 \sqrt{2} \\ h = a = 2.5 \sqrt{2} [/tex]
Zatem pole
[tex] \frac{2.5 \sqrt{2} \times 2.5 \sqrt{2} }{2} = \frac{6.25 \sqrt{4} }{2} = \frac{6.25 \times 2}{2} = 6.25[/tex]
Myślę że pomogłem ;)
