a) \(∆ABC\) có \(MN // BC\) (gt)
\( \Rightarrow \dfrac{MN}{CB} = \dfrac{AK}{AH}\) (kết quả bài tập 10) (định lý TaLet)
Mà \(AK = KI = IH\).
Nên \(\dfrac{AK}{AH} = \dfrac{1}{3}\)
\( \Rightarrow \dfrac{MN}{CB} = \dfrac{1}{3}\)
\( \Rightarrow MN = \dfrac{1}{3}BC = \dfrac{1}{3}.15 = 5\, cm\).
\(∆ABC\) có \(EF // BC\) (gt)
\( \Rightarrow \dfrac{EF}{BC} = \dfrac{AI}{AH} = \dfrac{2}{3}\) (định lý TaLet)
\(\Rightarrow EF = \dfrac{2}{3}.15 =10 \,cm\).
b) Áp dụng kết quả ở câu b của bài 10 ta có:
\(\eqalign{
& {S_{AMN}} = {1 \over 2}.AK.MN \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over 2}.{1 \over 3}AH.{1 \over 3}BC \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over 9}.\left( {{1 \over 2}AH.BC} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over 9}.{S_{ABC}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over 9}.270 = 30\,c{m^2} \cr} \)
\(\eqalign{
& {S_{AEF}} = {1 \over 2}.AI.EF \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over 2}.{2 \over 3}AH.{2 \over 3}BC \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {4 \over 9}.\left( {{1 \over 2}AH.BC} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {4 \over 9}.{S_{ABC}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {4 \over 9}.270 = 120\,c{m^2} \cr} \)
Do đó \({S_{MNEF}} = {S_{AEF}} - {S_{AMN}} = 120 - 30 \)\(\,= 90c{m^2}\)
