Áp dụng định lý Pytago, dễ dàng tính được:
$BC = \sqrt{AB^2 + AC^2} = 5$
$S_{ABC} = \dfrac{1}{2}AB.AC = \dfrac{3.4}{2} = 6$
mà $S_{ABC} = \dfrac{1}{2}AH.BC$
$\Rightarrow AH = \dfrac{2S_{ABC}}{BC} = \dfrac{12}{5}$
Áp dụng định lý Pytago, ta được:
$BH = \sqrt{AB^2 - AH^2} = \dfrac{9}{5}$
$\Rightarrow CH = BC - BH = \dfrac{16}{5}$
$\Rightarrow \begin{cases}S_{ABH} = \dfrac{1}{2}.AH.BH = \dfrac{1}{2}.\dfrac{12}{5}.\dfrac{9}{5} = \dfrac{54}{25}\\S_{AHC} = \dfrac{1}{2}.AH.CH = \dfrac{1}{2}.\dfrac{12}{5}.\dfrac{16}{5} = \dfrac{96}{25} \end{cases}$
Ta có: $BK = \dfrac{AB}{3}$
$\Rightarrow S_{BHK} = \dfrac{S_{ABH}}{3} = \dfrac{18}{25}$
$CI = \dfrac{AC}{2}$
$\Rightarrow S_{CHI} = \dfrac{S_{AHC}}{2} = \dfrac{48}{25}$
$\Rightarrow S_{AKHI} = S_{ABC} - S_{BHK} - S_{CHI} = 6 - \dfrac{18}{25} - \dfrac{48}{25} = \dfrac{88}{25}$
