$H.1$
Áp dụng định lý Pytago
$BC^2=AB^2+AC^2$
$⇒BC=\sqrt{AB^2+AC^2}=\sqrt{4^2+6^2}=2\sqrt{13}$
Áp dụng hệ thức lượng trong tam giác vuông
$AB^2=HB.BC⇒HB=\dfrac{AB^2}{BC}=\dfrac{4^2}{2\sqrt{13}}=\dfrac{8\sqrt{13}}{13}=x$
$HC=\dfrac{AC^2}{BC}=\dfrac{6^2}{2\sqrt{13}}=\dfrac{18\sqrt{13}}{13}=y$
$H.2$
$HC=\dfrac{AC^2}{BC}=\dfrac{12^2}{16}=9$
$HB=BC-HC=16-9=7$
$H.3$
$BC=HB+HC=4+9=13$
$AB^2=HB.BC=4.13=52$
$⇒AB=2\sqrt{13}=x$
$AC=\sqrt{HC.BC}=\sqrt{9.13}=3\sqrt{13}=y$
$H.4$
$BC=HB+HC=3+7=10$
$AH^2=HB.HC=3.7=21$
$⇒AH=\sqrt{21}=x$
$AC=\sqrt{HC.BC}=\sqrt{7.21}=7\sqrt{3}=y$
$H.5$
$\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}$
$⇒AH=\sqrt{\dfrac{AB^2.AC^2}{AB^2+AC^2}}=\sqrt{\dfrac{13^2.17^2}{13^2+17^2}}=10,3=x$
$AH.BC=AB.AC⇒BC=\dfrac{AB.AC}{AH}=\dfrac{13.17}{10,3}=21,5=y$
$H.6$
$AH^2=HB.HC⇒HC=\dfrac{AH^2}{HB}=\dfrac{5^2}{4}=6,25=x$
$AC=\sqrt{AH^2+HC^2}=\sqrt{5^2+6,25^2}=\dfrac{5\sqrt{41}}{4}=y$
