\(\widehat{A'AH}=60^0\Rightarrow A'H=AH.tan60^0=\frac{a\sqrt{2}}{2}\sqrt{3}=\frac{a\sqrt{6}}{2}\)
Do \(CB=2HB\Rightarrow d\left(C;\left(ABB'A'\right)\right)=2d\left(H;\left(ABB'A'\right)\right)\)
Từ H kẻ \(HM\perp AB\Rightarrow HM=\frac{1}{2}AC=\frac{a}{2}\)
Từ H kẻ \(HN\perp A'M\Rightarrow HN\perp\left(ABB'A'\right)\Rightarrow HN=d\left(H;\left(ABB'A'\right)\right)\)
\(\frac{1}{HN^2}=\frac{1}{A'H^2}+\frac{1}{HM^2}\Rightarrow HN=\frac{A'H.HM}{\sqrt{A'H^2+HM^2}}=\frac{a\sqrt{42}}{14}\)
\(\Rightarrow d\left(C;\left(ABB'A'\right)\right)=\frac{a\sqrt{42}}{7}\)