第117页 - 苏科版八年级（初二）数学学习与评价答案（上下册）

B

$解：(1)原式=\frac {\sqrt {2}}{10}；(2)原式=\sqrt {\frac {40}{9}}=\frac {2}{3}\sqrt {10}；$

$(3)原式=\frac {\sqrt {6x}}{4x}；(4)原式=\frac {\sqrt {7}}{77}$

$解：(1)原式=\frac {1}{\sqrt {3}}=\frac {\sqrt {3}}{3}；(2)原式=\frac {1}{2}×\sqrt {\frac {27}{12}}=\frac {1}{2}×\sqrt {\frac {9}{4}}=\frac {3}{4}$

$(3)原式=\frac {a\sqrt {2a}}{2a}=\frac {\sqrt {2a}}{2}；(4)原式=\frac {\sqrt {6ab}}{2b²}$

$解：10\sqrt {50}÷5\sqrt {14}=\frac {10}{7}\sqrt {7}$

$解：(1)4\sqrt {3}×2÷2\sqrt {6}=2\sqrt {2}(\mathrm {cm})$

$(2)∵点E、F、G、H分别是菱形ABCD的四边的中点$

$∴EH=\frac {1}{2}BD，EF=\frac {1}{2}AC，且EH⊥EF$

$∴S _{四边形EFCH} =\frac {1}{2}S_{ 菱形ABCD }=2\sqrt {3}(\mathrm {cm}²)$