12. 如图,在$ △ ABC $中,$ AD ⊥ BC $,$ BE ⊥ AC $,$ BC = 12 $,$ AC = 9 $,$ AD = 6 $,求$ BE $的长.

答案
12. $ \because $ 在 $ △ ABC $ 中,$ AD ⊥ BC $,$ BE ⊥ AC $,$ \therefore S_{△ ABC} = \frac{1}{2} BC · AD = \frac{1}{2} AC · BE $. $ \therefore BC · AD = AC · BE $. $ \because BC = 12 $,$ AC = 8 $,$ AD = 6 $,$ \therefore BE = \frac{BC · AD}{AC} = \frac{12 × 6}{8} = 9 $
13. 如图,在$ △ ABC $中,$ AE ⊥ BC $于点$ E $,$ AD $为边$ BC $上的中线,$ DF $为$ △ ABD $的边$ AB $上的中线,$ AB = 5\mathrm{cm} $,$ AC = 3\mathrm{cm} $,$ △ ABC $的面积为$ 12\mathrm{cm}^{2} $.求:
(1)$ △ ABD $与$ △ ACD $的周长的差;
(2)$ △ ABD $和$ △ ADF $的面积.

(1)$ △ ABD $与$ △ ACD $的周长的差;
(2)$ △ ABD $和$ △ ADF $的面积.
答案
13. (1) $ \because AD $ 为边 $ BC $ 上的中线,$ \therefore BD = CD $,$ \therefore △ ABD $ 与 $ △ ACD $ 的周长的差 $ = ( AB + AD + BD ) - ( AC + AD + CD ) = AB - AC = 2 \mathrm{cm} $ (2) $ \because AD $ 为边 $ BC $ 上的中线,$ \therefore S_{△ ABD} = \frac{1}{2} × S_{△ ABC} = 6 \mathrm{cm}^2 $,$ S_{△ ADF} = \frac{1}{2} × △ ABD = 3 \mathrm{cm}^2 $
14. 如图,在$ △ AMH $中,$ AN $、$ ME $、$ MF $分别为$ △ AMH $、$ △ AMN $、$ △ MHE $的中线,且$ △ AMH $的面积为$ 80\mathrm{cm}^{2} $.求:
(1)$ △ AME $与$ △ AHE $的面积的和;
(2)$ △ MEF $的面积.

(1)$ △ AME $与$ △ AHE $的面积的和;
(2)$ △ MEF $的面积.
答案
14. (1) $ \because AN $、$ ME $ 分别为 $ △ AMH $、$ △ AMN $ 的中线,$ △ AMH $ 的面积为 $ 80 \mathrm{cm}^2 $,$ \therefore S_{△ AME} = \frac{1}{2} S_{△ AMN} = \frac{1}{2} × \frac{1}{2} S_{△ AMH} = 20 \mathrm{cm}^2 $,$ S_{△ AHE} = \frac{1}{2} S_{△ AHN} = \frac{1}{2} × \frac{1}{2} S_{△ AMH} = 20 \mathrm{cm}^2 $. $ \therefore S_{△ AHE} + S_{△ AME} = 40 \mathrm{cm}^2 $ (2) $ \because MF $ 为 $ △ MHE $ 的中线,$ S_{△ MHE} = S_{△ AMH} - ( S_{△ AHE} + S_{△ AME} ) = 40 \mathrm{cm}^2 $,$ \therefore S_{△ MEF} = \frac{1}{2} S_{△ MEH} = 20 \mathrm{cm}^2 $
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