(b) 2.5 cm2.
In ΔABC,
cos A = \(\frac{b^2+c^2-a^2}{2bc}\)
= \(\frac{AC^2+AB^2-BC^2}{2\times{AC}\times{AB}}\)
= \(\frac{4^2+2^2-3^2}{2\times4\times2}\)
= \(\frac{16+4-9}{16}=\frac{11}{16}\)
In ΔBAD,
cos A = \(\frac{AD^2+AB^2-BD^2}{2\times{AD}\times{AB}}\)
⇒ \(\frac{11}{16}\) = \(\frac{4+4-BD^2}{2\times2\times2}\) = \(\frac{8-BD^2}{8}\)
⇒ 11 = 16 – 2BD2 ⇒ 2BD2 = 5 ⇒ BD2 = 2.5
∴ Area of square on BD = (BD)2 = 2.5 cm2