hình 9 khó

D

depvazoi

Kẻ $DK \perp AB (K \in AB)$
$=> DK//AC$ và $\Delta AKD$ vuông cân tại K
$=> \dfrac{AK}{AB}=\dfrac{CD}{BC}$
và $\Delta BKD \sim \Delta BAC$
$=> \dfrac{KD}{AC}=\dfrac{BD}{CD}$
$=>\dfrac{AK}{AB}+\dfrac{KD}{AC}=\dfrac{CD}{BC}$$+\dfrac{BD}{CD}=1$
$=> \dfrac{1}{AB}+\dfrac{1}{AC}=\dfrac{1}{DK}=\dfrac{ \sqrt{2}}{AD}$
$=>AD\approx 2,997$
Ta có:
$\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}$
$=>AH\approx 2,927$
Áp dụng định lý Pytago, ta được:
$HD\approx 0,644$
$=>S_{AHD}=\dfrac{AH.HD}{2}=\dfrac{2,927.0,644}{2}=0,942494(đvS)$
 
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