Odpowiedź:
a)
[tex]h = \frac{a \sqrt{3} }{2} = \frac{12 \sqrt{3} }{2} = 6 \sqrt{3 } \\ r = \frac{1}{3} h = \frac{1}{3} \times 6 \sqrt{3} = 2 \sqrt{3} [/tex]
b)
[tex]r = \frac{1}{3} \times 3 = 1 \: cm \\ \frac{a \sqrt{3} }{2} = 3 \\ a = 3 \times \frac{2}{ \sqrt{3} } = \frac{6}{ \sqrt{3} } = \frac{ 6 \sqrt{3} }{3} = 2 \sqrt{3} [/tex]
c)
[tex] \frac{a \sqrt{3} }{6} + 4 = \frac{a \sqrt{3} }{2} \\ \frac{a \sqrt{3} }{2} - \frac{a \sqrt{3} }{6} = 4 \\ \frac{3a \sqrt{3} }{6} - \frac{a \sqrt{3} }{6} = 4 \\ \frac{2a \sqrt{3} }{6} = 4 \\ \frac{a \sqrt{3} }{3} = 4 \\ a = 4 \times \frac{3}{ \sqrt{3} } = \frac{12}{ \sqrt{3} } = \frac{12 \sqrt{3} }{3} = 4 \sqrt{3} \\ \\ h = \frac{4 \sqrt{3} \times \sqrt{3} }{2} = 6 \\ \\ r = \frac{1}{3} \times 6 = 2[/tex]
d)
[tex] \frac{a \sqrt{3} }{2} + 12 = a \\ a - \frac{a \sqrt{3} }{2} = 12 \\ \frac{2a - a \sqrt{3} }{2} = 12 \\ a(2 - \sqrt{3} ) = 6 \\ a = \frac{6}{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } = \frac{12 + 6 \sqrt{3} }{4 - 3} = 12 + 6 \sqrt{3} \\ \\ h = a - 12 = 12 + 6 \sqrt{3} - 12 = 6 \sqrt{3} [/tex]
e)r=a-6
[
[tex] = 6\frac{a \sqrt{3} }{6} + 6 = a \\ a - \frac{a \sqrt{3} }{6} = 6 \\ \frac{6a - a \sqrt{3} }{6} \\ \frac{a(6 - \sqrt{3}) }{6} = 6 \\ a(6 - \sqrt{3} ) = 36 \\ a = \frac{36}{6 - \sqrt{3} } \times \frac{6 + \sqrt{3} }{6 + \sqrt{3} } = \frac{36(6 + \sqrt{3}) }{36 - 3} = \frac{12(6 + \sqrt{3} )}{11} \\ \\ r = a - 6 = \frac{72 + 12\sqrt{3} }{11} - \frac{66}{11} = \frac{6 + 12 \sqrt{3} }{11} [/tex]