Question

cho \(\Delta\) ABC \(\perp\) A đường cao AH chia cạnh huyền 2 loại BH = 9, CH = 16. tính AB, AC, BC

mn giuwsp mik vs ạ

Answer 1

Ta có \(\widehat{A^1}+\widehat{A^2}\) = 90⁰

\(\widehat{B}+\widehat{C}=90^0\); \(\widehat{A^1}+\widehat{C}=90^0\)

=> \(\widehat{A^1}=\widehat{B}\) ; \(\widehat{A^2}=\widehat{C}\)

Xét Δ ABH và ΔCAH có:  \(\widehat{A^1}=\widehat{B}\) ; \(\widehat{A^2}=\widehat{C}\)

=> Δ ABH ∼ ΔCAH (g.g)

=> \(\dfrac{BH}{AH}=\dfrac{AH}{CH}\) => AH² = BH.CH = 9 . 16 = 144

=> AH = 12

=> AB = \(\sqrt{BH^2+AH^2}\) = 15

=> AC = \(\sqrt{CH^2+AH^2}\) = 20

=> BC = 9 + 16 = 25

=> AB = 15;AC = 20;BC = 25