Answer
AB = $\sqrt 41 \approx 6.40$
Work Step by Step
In given figure 23, we can see two right triangles BCD and BCA, both right angled at C.
Given-
BC = 4, BD = 5 , AD = 2
In right triangle BCD, applying Pythagoras theorem-
$ BD^{2} $ = $ BC^{2} $ + $ DC^{2} $
Therefore- $ DC^{2} $ = $ BD^{2} $- $ BC^{2} $
$ DC^{2} $ = $ 5^{2} $- $ 4^{2} $ = 25-16=9
$ DC = \sqrt 9$ = 3
Now we can calculate AC as-
AC = AD + DC = 2 + 3 = 5
Now applying Pythagoras theorem in right triangle BCD-
$ AB^{2} $ = $ BC^{2} $ + $ AC^{2} $
$ AB^{2} $ = $ 4^{2} $ + $ 5^{2} $
$ AB^{2} $ = 16 + 25 = 41
Therefore AB = $\sqrt 41 \approx 6.40$
