Area of ΔABC = \(rac{1}{2}\) x 3 x 4 cm2 = 6 cm2. Also, AC = \(\sqrt{3^2+4^2}\) = 5 cm.∴ Area of ΔABC = \(rac{1}{2}\) x BD x AC ⇒ 6 = \(rac{1}{2}\) BD x 5 ⇒ BD = \(rac{12}{5}\) cm.Now in ΔABD, AD = \(\sqrt{AB^2+BD^2}\) = \(\sqrt{3^2-\big(rac{12}{5}\big)^2}\) = \(\sqrt{9-rac{144}{25}}\)= \(\sqrt{rac{225-144}{25}}\) = \(\sqrt{rac{81}{25}}\) = \(rac{9}{5}\) cm∴ Area of ΔABD = \(rac{1}{2}\)x AD x BD = \(rac{1}{2}\) x \(rac{9}{5}\) x \(rac{12}{5}\) = \(rac{54}{25}\) cm2.