Đáp án:
a) AB = AM+BM=10 cm
Ta có MN//BC nên theo Talet ta có:
$\begin{array}{l}
\frac{{AM}}{{AB}} = \frac{{MN}}{{BC}} = \frac{{AN}}{{AC}}\\
\Rightarrow \frac{6}{{10}} = \frac{{MN}}{{BC}} = \frac{8}{{AC}}\\
\Rightarrow AC = \frac{{8.10}}{6} = \frac{{40}}{3}\left( {cm} \right)\\
\Rightarrow NC = AC - AN = \frac{{40}}{3} - 8 = \frac{{16}}{3}\left( {cm} \right)\\
Theo\,pytago:MN = \sqrt {A{M^2} + A{N^2}} = \sqrt {{6^2} + {8^2}} = 10\left( {cm} \right)\\
\Rightarrow \frac{6}{{10}} = \frac{{10}}{{BC}}\\
\Rightarrow BC = \frac{{50}}{3}\left( {cm} \right)
\end{array}$
b) AMDN là hình chữ nhật nên AM = DN=6cm
TA có:
$\begin{array}{l}
{S_{BMND}} = {S_{ABC}} - {S_{AMN}} - {S_{DNC}}\\
= \frac{1}{2}.AC.AB - \frac{1}{2}.AM.AN - \frac{1}{2}.DN.NC\\
= \frac{1}{2}.\frac{{40}}{3}.10 - \frac{1}{2}.6.8 - \frac{1}{2}.6.\frac{{16}}{3}\\
= \frac{{80}}{3}\left( {c{m^2}} \right)
\end{array}$
