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phatthemkem

1) ta có
2S$ABC$=$AH.BC$=$BK.AC$
\Rightarrow $\frac{BC}{AC}$=$\frac{BK}{AH}$=$\frac{12}{10}$= $\frac{6}{5}$ (1)
Đặt $BH=x$, ta có: $BC=2x, AC=\sqrt{x^2+100}$
\Rightarrow $(1)$\Leftrightarrow $\frac{2x}{\sqrt{x^2+100}}$=$\frac{6}{5}$
\Leftrightarrow $10x$= $6\sqrt{x^2+100}$
\Leftrightarrow $100^2$= $36x^2+3600$
\Leftrightarrow $64x^2$= $3600$
\Leftrightarrow $x=7,5$
\Rightarrow $BC=15 cm, AB=AC=12,5 cm$
 
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