Задание 16, стр 26

Question

Найдите периметры фигур.

Gimli · Accepted Answer

\begin{enumerate}[label=(alph*)]
\item $P = 4\sqrt{6} + 3\sqrt{6}+ 5\sqrt{6} = 12\sqrt{6}.$
\item $$P = 3\sqrt{12}+6\sqrt{3}+3\sqrt{27} = 3\sqrt{4}\cdot \sqrt{3}+6\sqrt{3}+3\sqrt{9}\sqrt{3} = 3\cdot 2\sqrt{3} + 6\sqrt{3} + 3\cdot 3\sqrt{3} = 6\sqrt{3} + 6\sqrt{3} + 9\sqrt{3} = $$
$$=21\sqrt{3}.$$
\item $$P = 5\sqrt{8} + 9\sqrt{8}+15\sqrt{2}+17\sqrt{2} = 14\sqrt{8}+32\sqrt{2} = 14\sqrt{4}\cdot \sqrt{2} + 32\sqrt{2} = 14\cdot 2\sqrt{2}+32\sqrt{2} = 28\sqrt{2}+$$ $$+32\sqrt{2} =  60\sqrt{2}.$$
\item $$P = 2\sqrt{18}+6\sqrt{8}+9\sqrt{2}+15\sqrt{2} = 2\sqrt{9}\cdot \sqrt{2} + 6\sqrt{4}\cdot \sqrt{2} + 9\sqrt{2} + 15\sqrt{2} = 2\cdot 3\sqrt{2}+6\cdot 2\sqrt{2}+9\sqrt{2}+$$
$$+15\sqrt{2} = 6\sqrt{2}+12\sqrt{2}+9\sqrt{2} + 15\sqrt{2} = 42\sqrt{2}.$$
\end{enumerate}

Задание 16, стр 26

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Задание 16, стр 26

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