AH=2√5

Lời giải:

Kẻ AH⊥BC, BK⊥CD, đường chéo AC⊥AD

Đặt AH=AB=x⇒AH=x

ΔAHD=ΔBKC (c.h - g.n)
$$\begin{gathered} \Rightarrow \mathrm{D} \mathrm{H} = \mathrm{C} \mathrm{K} = \frac{10 - \mathrm{x}}{2} \\ \mathrm{Vậy} \mathrm{H} \mathrm{C} = \mathrm{H} \mathrm{K} + \mathrm{C} \mathrm{K} = \mathrm{x} + \frac{10 - \mathrm{x}}{2}= \frac{\mathrm{x} + 10}{2} \end{gathered}$$
Áp dụng hệ thức lượng trong ΔADC⊥A
$$\begin{gathered} \mathrm{Có} {\mathrm{AH}}^{2 }= \mathrm{D} \mathrm{H} . \mathrm{H} \mathrm{C} \Rightarrow {\mathrm{x}}^2 =\frac{10-\mathrm{x}}{2}.\frac{\mathrm{x}+10}{2} \\ \Rightarrow 4 {\mathrm{x}}^2 = 100 - {\mathrm{x}}^2 \\ \Rightarrow 5 {\mathrm{x}}^2 = 100 \\ \Rightarrow \mathrm{x} = 2 \sqrt{5} \\ \mathrm{Vậy} \mathrm{A} \mathrm{H} = 2 \sqrt{ 5} \end{gathered}$$