Consider the figure.
Find AB if BC = 3, BD = 5, and AD = 5.
AB= ?

[tex]AB=3\sqrt{10}[/tex]
In right angled triangle ΔBCD we have:
Using the Pythagorean Theorem we have:
[tex]BD^2=BC^2+CD^2[/tex]
[tex]5^2=3^2+CD^2\\\\25=9+CD^2\\\\CD^2=25-9\\\\CD^2=16\\\\CD=\pm 4[/tex]
But as a length of a side can't be negative.
Hence, we have:
[tex]CD=4[/tex]
Also, In right angled triangle ΔBCA using the Pythagorean Theorem we have:
[tex]AB^2=AC^2+BC^2\\\\AB^2=(AD+CD)^2+BC^2\\\\AB^2=(5+4)^2+3^2\\\\AB^2=9^2+3^2\\\\AB^2=81+9\\\\AB^2=90\\\\AB=\pm 3\sqrt{10}[/tex]
( Since, on taking square root on both the side we have negative or positive term but the length of a side can't be negative.
Hence, we took answer as a positive value)
Hence, we have:
[tex]AB=3\sqrt{10}[/tex]