- ΔAHB; ˆH=90^o
AB^2=BH^2+HA^2 ( ĐL pitago )
⇒AH^2=AB^2−BH^2=30^2−BH^2=900−BH^2 (1)
- ΔAHC;H^=90o ; AH⊥BC
AH^2=BH.HC( hệ thức liên quan tới đường cao )
⇔AH^2=32.BH (2)
- Từ (1) và (2) => 900−BH^2=32.BH
⇔BH^2+32.BH−900=0
⇔(BH^2+50.BH)−(18.BH+900)=0
⇔BH.(BH+50)−18(BH+50)=0
⇔(BH+50)(BH−18)=0
⇔[BH+50=0BH−18=0
⇔[BH=−50(loai)BH=18(nhan)
BC = BH + HC = 18 + 32 = 50 cm
- ΔABC ; A^=90o
BC^2=AB^2+AC^2 ( ĐL pitago )
⇒AC^2=BC^2−AB^2=50^2−32^2=1600
⇒AC=√1600=40(cm)
Ta có: \(AB^2=BH\cdot BC\)
\(\Leftrightarrow BH\cdot\left(BH+32\right)=900\)
\(\Leftrightarrow BH^2+32HB-900=0\)
\(\Leftrightarrow BH^2+50HB-18HB-900=0\)
\(\Leftrightarrow BH=18cm\)
\(\Leftrightarrow BC=50cm\)
\(\Leftrightarrow AC=40cm\)
