Tính các độ dài AB, AC biết tam giác ABC vuông tại A có BC=26 cm, AB:AC=5:12

\(a.\)

Qua \(E\) , kẻ \(IK\perp BC\)

Dễ chứng minh : \(BK=AI\) , \(KC=ID\)

Ta có :

\(EC^2=EK^2+KC^2\) \(\left(1\right)\)

\(EK^2=EB^2-BK^2=16-AI^2\) \(\left(2\right)\)

\(KC^2=ID^2=ED^2-EI^2=4-EI^2\) \(\left(3\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right)\)

\(\Rightarrow EC^2=\left(16-AI^2\right)+\left(4-EI^2\right)\)

\(=20-\left(AI^2+EI^2\right)\) \(\left(4\right)\)

Ta lại có : \(AI^2+EI^2=AE^2=9\) \(\left(5\right)\)

Từ \(\left(4\right),\left(5\right)\)

\(\Rightarrow EC^2=20-9=11\)

\(\Rightarrow EC=\sqrt{11}\)

\(\Rightarrow x=\sqrt{11}\)

\(b.\)

\(AC^2-AB^2=HC^2-HB^2=32^2=18^2=1024-324=700\)

\(AC^2+AB^2=BC^2=\left(18+32\right)^2=2500\)

Từ đó : \(AC^2=1600\)

\(\Rightarrow AC=\sqrt{1600}\)

\(\Rightarrow AC=40\)

\(\Rightarrow x=40\)

\(c.\)

Kẻ \(AH\perp BC\)

Ta có : \(HB=3,5\)

\(AH=49-12,25=36,75\)

\(\Rightarrow x^2=36,75+11,5^2=169\)

\(\Rightarrow x=\sqrt{169}\)

\(\Rightarrow x=13\)