Given: ΔABC, CM ⊥ AB AC = 10, CM = 4 AM:BM = 2:5 Find: AB, CB

Answer:  [tex]AB = 7\sqrt{21}\\CB= \sqrt{541}[/tex] Step-by-step explanation:  Given:  Δ ABC, CM⊥ AB     [tex]AC = 10, CM = 4 AM:BM = 2:5[/tex]Now, consider  [tex]AM:BM = 2:5 = 2x:5x[/tex]Let In ΔCMA  H=- 10 , P= 4 , B= 2xBy, Pythagoras theorem,  [tex]H^2=P^2+B^2[/tex]putting values we get, [tex]10^2=4^2+(2x)^2[/tex]⇒ [tex]100=16+4x^2[/tex]⇒[tex]x^2= 21[/tex]⇒[tex]x= \sqrt{21}[/tex]which gives us [tex]AM = 2x= 2\sqrt{21}[/tex] and  [tex]MB = 5x= 5\sqrt{21}[/tex]⇒[tex]AB= 2\sqrt{21}+5\sqrt{21}[/tex]⇒[tex]AB= 7\sqrt{21}[/tex]Now, Let In ΔCMB  H=- ? , P= 4 , B= 5√21By, Pythagoras theorem,  [tex]H^2=P^2+B^2[/tex]putting values we get, [tex]H^2=4^2+(5\sqrt{21})^2[/tex]⇒ [tex]H^2=16+525[/tex]⇒[tex]H^2=541 [/tex]⇒[tex]H= \sqrt{541}[/tex]⇒ [tex]CB= \sqrt{541}[/tex]Therefore,  [tex]AB = 7\sqrt{21}\\CB= \sqrt{541}[/tex]