Given CD is an altitude such that AD=BC , AB=3 cm and CD= √2 cm.
Let AD=x, Since given AB=3
AD+DB=3
x+DB = 3
DB = 3-x
Since ΔBCD is rght angle triangle, let's apply Pythagoras theorem
[tex]BC^{2} = DB^{2} +CD^{2}[/tex]
[tex]BC^{2} = (3-x)^{2} +(\sqrt{2} )^{2}[/tex]
[tex]BC^{2} =(3-x)^{2} +2[/tex]
Since given AD=BC,let us plugin BC=x in above step.
[tex]x^{2} =(3-x)^{2} +2[/tex]
[tex]x^{2} =9-6x+x^{2} +2[/tex]
6x=11
x=[tex]\frac{11}{6}[/tex]
Now we know AD=x=[tex]\frac{11}{6}[/tex] and given CD=√2.
Let us apply Pythagoras theorem for ΔACD
[tex]AC^{2} =AD^{2} +DC^{2}[/tex]
[tex]AC^{2} = (\frac{11}{6} )^{2} +(\sqrt{2} )^{2}[/tex]
[tex]AC^{2} =\frac{193}{36}[/tex]
[tex]AC=\sqrt{\frac{193}{36} }[/tex] = 2.315cm
