jjeeyu
jjeeyu
10-03-2019
Mathematics
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Help to solve this question, thanks
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Kamikazeminem
Kamikazeminem
10-03-2019
[tex](a) \: CF \parallel DE \Rightarrow \widehat{ABC} =\widehat{ACB}=\hat{D}=\hat{E}=\widehat{BFC} \\ \Rightarrow \bigtriangleup ABC \cong \bigtriangleup BDE\cong \bigtriangleup BCF\\ (b) \: \Rightarrow \frac{BC}{DE} = \frac{AB}{BD} \Leftrightarrow BC = \frac{AB \times DE}{BD} = \frac{12 \times 14}{21} = 8 \\ \Rightarrow CD = BD - BC = 21 - 8 = 13 \\ (c) \: \bigtriangleup ABC \cong \bigtriangleup BCF\Rightarrow \frac{BC}{AB} = \frac{CF}{BC} \: and \: BC =BF = 8\\ \Leftrightarrow CF = \frac{ {BC}^{2} }{AB} = \frac{ {8}^{2} }{12} = \frac{16}{3} \\ A_{\bigtriangleup BCF} = \sqrt{p(p -BC)(p - BF)(p -CF)} = \frac{1280 \sqrt{3} }{9} \\ A_{\bigtriangleup BDE} = \sqrt{p(p -BD)(p - DE)(p -BE)} = 980 \sqrt{3} \\ \frac{A_{\bigtriangleup BCF}}{A_{CDEF} } = \frac{A_{\bigtriangleup BCF}}{A_{\bigtriangleup BDE} - A_{\bigtriangleup BCF}} = \frac{ \frac{1280 \sqrt{3} }{9} }{980 \sqrt{3} - \frac{1280 \sqrt{3} }{9} } = \frac{64}{377} [/tex]
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