Benchmarks

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总结帝笔记—初三寒假班第三讲

52

延伸拓展 求 $S\_\{\backslash triangle\; MAD\}\; =\; ?$

$$\backslash therefore\; \backslash frac\{S\_\{\backslash triangle\; MAD\}\}\{S\_\{\backslash triangle\; MBC\}\}\; =\; (\backslash text\{相似比\})^2\; =\; \backslash left(\backslash frac\{4\}\{5\}\backslash right)^2\; =\; \backslash frac\{16\}\{25\}$$

$$\backslash text\{即\; \}\; \backslash frac\{S\_\{\backslash triangle\; MAD\}\}\{S\_\{\backslash triangle\; MAD\}\; +\; S\_\{\backslash text\{四边形\}ABCD\}\}\; =\; \backslash frac\{16\}\{25\}\; \backslash quad\; (\backslash triangle)$$

But. $S\_\{\backslash text\{四边形\}ABCD\}\; =\; ?$

有的同学 可能会说 四边形ABCD为梯形

$$=\; \backslash frac\{1\}\{2\}(AD+BC)\; \backslash times\; \backslash text\{高\}\; =\; \backslash frac\{1\}\{2\}\; \backslash times\; (8+10)\; \backslash times\; \backslash text\{高\}$$

$$=\; 9\; \backslash cdot\; \backslash text\{高\}\; \backslash quad\; \backslash textcircled\{1\}$$

但是. 高是多少呢? 这就是问题所在!

看我的! 如图: 作 $DQ\; \backslash parallel\; AB$ . 交 $BC$ 于 $Q$ .

$\backslash therefore\; AD\; \backslash parallel\; BQ$ . $AB\; \backslash parallel\; DQ$ . $\backslash therefore$ 四边形 $BQDA$ 为平行四边形.

$$\backslash therefore\; BQ\; =\; AD\; =\; 8\; \backslash quad\; \backslash therefore\; CQ\; =\; 2.\; \backslash quad\; DQ\; =\; BA\; =\; 4.$$

在 $\backslash triangle\; DQC$ 中 $\backslash left\backslash \{\; \backslash begin\{array\}\{l\}\; QD\; =\; 4\; \backslash \backslash \; QC\; =\; 2\; \backslash \backslash \; CD\; =\; 5\; \backslash end\{array\}\; \backslash right.$ 那面积呢?

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