Answer:
Step-by-step explanation:
It is given that in △ABC, ∠ABC=90°, BH is an altitude and AB=9 and AC=12, thus using the Pythagoras theorem in △ABC, we get
[tex](AC)^{2}=(AB)^{2}+(BC)^{2}[/tex]
[tex](12)^{2}=(9)^2+(BC)^2[/tex]
[tex]144=81+(BC)^2[/tex]
[tex](BC)^2=63[/tex]
[tex]BC=\sqrt{63}[/tex]
Now, From ΔABC and ΔHBC, we have
∠ABC=∠BHC(each90)
∠ACB=∠HCB (Common)
By AA similarity, ΔABC is similar to ΔHBC.
Thus, using the similarity condition, we get
[tex]\frac{HC}{BC}=\frac{BC}{AC}[/tex]
[tex]\frac{HC}{\sqrt{63}}=\frac{\sqrt{63}}{12}[/tex]
[tex]HC=\frac{63}{12}=\frac{21}{4}[/tex]
