已知三高求面積-例20081010-1

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屬性

三邊長 $\left\{ \begin{align} & a=\frac{16\sqrt{15}}{15} \\ & b=\frac{16\sqrt{15}}{10} \\ & c=\frac{32\sqrt{15}}{15} \\ \end{align} \right.$
三角形面積 $\Delta =\frac{16\sqrt{15}}{5}$
三角形外接圓半徑 $R=\frac{64}{15}$
三角形內切圓半徑 $r=\frac{4}{3}$

$h_{a}=6,h_{b}=4,h_{c}=3\,$
$\Delta =\frac{1}{2}ah_{a}=\frac{1}{2}bh_{b}=\frac{1}{2}ch_{c}$
$\Rightarrow \left\{ \begin{align} & a=\frac{2\Delta }{h_{a}}=\frac{2\Delta }{6}=\frac{\Delta }{3} \\ & b=\frac{2\Delta }{h_{b}}=\frac{2\Delta }{4}=\frac{\Delta }{2} \\ & c=\frac{2\Delta }{h_{c}}=\frac{2\Delta }{3} \\ \end{align} \right.$
$\Rightarrow s=\frac{a+b+c}{2}=\frac{\frac{\Delta }{3}+\frac{\Delta }{2}+\frac{2\Delta }{3}}{2}$
$=\frac{\frac{2+3+4}{6}\Delta }{2}=\frac{2+3+4}{12}\Delta =\frac{3}{4}\Delta$
又， $\Delta =\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}=\sqrt{\frac{3}{4}\Delta \frac{5}{12}\Delta \frac{1}{4}\Delta \frac{1}{12}\Delta }=\frac{\sqrt{15}}{48}\Delta ^{2}$
$\Rightarrow \frac{\sqrt{15}}{48}\Delta =1\Rightarrow \Delta =\frac{48}{\sqrt{15}}=\frac{48\sqrt{15}}{15}=\frac{16\sqrt{15}}{5}\#$
$\Rightarrow \left\{ \begin{align} & a=\frac{\Delta }{3}=\frac{16\sqrt{15}}{15}\# \\ & b=\frac{\Delta }{2}=\frac{16\sqrt{15}}{10}\# \\ & c=\frac{2\Delta }{3}=\frac{32\sqrt{15}}{15}\# \\ \end{align} \right.$
$R=\frac{a}{2\sin A}=\frac{\frac{16\sqrt{15}}{15}}{2\times \frac{\sqrt{15}}{8}}=\frac{64}{15}\#$
$rs=\Delta \Rightarrow r=\frac{\Delta }{s}=\frac{\Delta }{\frac{3}{4}\Delta }=\frac{4}{3}\#$

附： $\left\{ \begin{align} & \sin A=\frac{h_{c}}{b}=\frac{3}{\frac{16\sqrt{15}}{10}}=\frac{30}{16\sqrt{15}}=\frac{\sqrt{15}}{8}\# \\ & \sin B=\frac{h_{a}}{c}=\frac{6}{\frac{32\sqrt{15}}{15}}=\frac{6\times 15}{32\sqrt{15}}=\frac{6\sqrt{15}}{32}\# \\ & \sin C=\frac{h_{b}}{a}=\frac{4}{\frac{16\sqrt{15}}{15}}=\frac{4\times 15}{16\sqrt{15}}=\frac{\sqrt{15}}{4}\# \\ \end{align} \right.$