Benchmarks

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总结帝笔记—初三寒假班第三讲

54

7. 如图, 梯形 $ABCD$ 中, $AB\; \backslash parallel\; CD$ . 若 $AB=1$ , $CD=2$ , $AC=6$ , $S\_\{\backslash triangle\; AOB\}=1$ . 则 $AO=$ __________
$S\_\{\backslash text\{梯形\}ABCD\}=$ __________

解: $\backslash because\; AB\; \backslash parallel\; CD$
$\backslash therefore\; \backslash triangle\; AOB\; \backslash sim\; \backslash triangle\; COD$
即 $\backslash frac\{AO\}\{CO\}\; =\; \backslash frac\{AB\}\{CD\}$ . $\backslash because\; AB=1$ , $CD=2$ , $AC=6$
$\backslash therefore\; \backslash frac\{AO\}\{CO\}\; =\; \backslash frac\{1\}\{2\}\; \backslash Rightarrow\; CO\; =\; 2AO$

又 $\backslash because\; AC\; =\; AO\; +\; CO\; =\; 3AO\; \backslash Rightarrow\; 3AO\; =\; 6$ . $AO\; =\; 2$

又 $\backslash because\; \backslash frac\{S\_\{\backslash triangle\; AOB\}\}\{S\_\{\backslash triangle\; COD\}\}\; =\; \backslash left(\backslash frac\{AB\}\{CD\}\backslash right)^2\; =\; \backslash left(\backslash frac\{1\}\{2\}\backslash right)^2\; =\; \backslash frac\{1\}\{4\}\; =\; \backslash frac\{1\}\{S\_\{\backslash triangle\; COD\}\}$

$\backslash therefore\; S\_\{\backslash triangle\; COD\}\; =\; 4$

$\backslash because\; S\_\{\backslash triangle\; AOB\}$ 与 $S\_\{\backslash triangle\; BOC\}$ (不同底, 同高)
且底之比为 $OA:OC\; =\; 1:2$

$\backslash therefore\; S\_\{\backslash triangle\; AOB\}\; :\; S\_\{\backslash triangle\; BOC\}\; =\; 1:2$ . $\backslash because\; S\_\{\backslash triangle\; AOB\}=1$

$\backslash therefore\; S\_\{\backslash triangle\; BOC\}\; =\; 2$ . 同理: $S\_\{\backslash triangle\; AOD\}$ 与 $S\_\{\backslash triangle\; COD\}$ (同底、同高)
且底之比为 $OA:OC\; =\; 1:2$ . $\backslash therefore\; S\_\{\backslash triangle\; AOD\}\; :\; S\_\{\backslash triangle\; COD\}\; =\; 1:2$

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