Trong tam giác vuông \(IAB,\) ta có:
\(A{B^2} = A{I^2} + I{B^2}\) (định lý Pi-ta-go)
\(\begin{array}{l} \Rightarrow I{B^2} = A{B^2} - A{I^2} = 25 - 9 = 16\\ \Rightarrow IB = 4(cm)\\AC = 2AI = 2.3 = 6(cm)\\BD = 2IB = 2.4 = 8(cm)\\{S_{ABCD}} = \dfrac{1}{2}AC.BD = \dfrac{1}{2}6.8\\ = 24(c{m^2})\end{array}\)
