計算三
(1)
\(\begin{align}
& \Delta APE=\frac{2}{5}b,\Delta CPE=\frac{3}{5}b,\Delta APF=\frac{4}{7}c,\Delta BPF=\frac{3}{7}c \\
& \frac{\frac{2}{5}b+c}{\frac{3}{5}b+a}=\frac{2}{3} \\
& \frac{\frac{4}{7}c+b}{\frac{3}{7}c+a}=\frac{4}{3} \\
& a:b:c=3:4:2 \\
& a=6,b=8,c=4 \\
\end{align}\)
(2)
\(\begin{align}
& \frac{\Delta BDP}{\Delta CDP}=\frac{\overline{BD}}{\overline{CD}}=\frac{\Delta ABP}{\Delta ACP}=\frac{1}{2} \\
& \Delta BDP=2,\Delta CDP=4 \\
& \frac{\overline{AP}}{\overline{PD}}\times \frac{\overline{BP}}{\overline{PE}}\times \frac{\overline{CP}}{\overline{PF}}=\frac{\Delta ABP}{\Delta BDP}\times \frac{\Delta BCP}{\Delta CEP}\times \frac{\Delta CAP}{\Delta AFP}=\frac{4}{2}\times \frac{6}{\frac{24}{5}}\times \frac{8}{\frac{16}{7}}=\frac{35}{4} \\
\end{align}\)
計算四
(1)畫成展開圖就知道怎麼做了
(2)長方體的對角線長
(3)國中生的方法，有請高手
小弟提供一個高中生的方法，有用到餘弦定理
定坐標
A(x，y，z)、E(0，0，0)、F(6，0，0)
EH=5，∠FEH=β
易知H(3，4，0)、G(9，4，0)
\(\begin{align}
& \left\{ \begin{align}
& {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{\overline{AE}}^{2}}=16 \\
& {{\left( x-6 \right)}^{2}}+{{y}^{2}}+{{z}^{2}}={{\overline{AF}}^{2}}=28 \\
& {{\left( x-3 \right)}^{2}}+{{\left( y-4 \right)}^{2}}+{{z}^{2}}={{\overline{AH}}^{2}}=9 \\
\end{align} \right. \\
& x=2,{{\left( y-4 \right)}^{2}}+{{z}^{2}}=8 \\
& \overline{AG}=\sqrt{{{\left( x-9 \right)}^{2}}+{{\left( y-4 \right)}^{2}}+{{z}^{2}}}=\sqrt{57} \\
\end{align}\)
