Cho tam giác $ABC$  cân tại $A.$ Kẻ $AH$  vuông góc với $BC$  tại $H.$  Cho $BH = 2cm,AB = 4cm.$

Tính $AH.$

\(10\,\,\left( {cm} \right)\)

\(12\,\,\left( {cm} \right)\)     

\(11\,\,\left( {cm} \right)\)     

\(13\,\left( {cm} \right)\)

\(\sqrt {10} \,\,\left( {cm} \right)\)

\(\sqrt {13} \,\,\left( {cm} \right)\)

\(\sqrt {12} \,\,\left( {cm} \right)\)

\(12\,\left( {cm} \right)\)

\(\sqrt {10} \,\,\left( {cm} \right)\)

\(\sqrt {13} \,\,\left( {cm} \right)\)

\(\sqrt {12} \,\,\left( {cm} \right)\)

\(12\,\left( {cm} \right)\)

\(M{N^2} = M{P^2} - N{P^2}\)

\(M{P^2} = M{N^2} + N{P^2}\)

\(N{P^2} = M{N^2} + M{P^2}\)      

\(M{N^2} = M{P^2} + N{P^2}\)

\(BC = 6\,dm\)    

\(BC = \sqrt {23} \,dm\)

\(BC = 4\,dm\)

\(BC = \sqrt {32} \,dm\)

\(9\,cm;\,12\,cm\)    

\(10\,cm;\,16\,cm.\)

\(12\,cm;\,16\,cm\)

\(12\,cm;\,14\,cm\)

\(AH = 4,8\,cm;\,AB = 6\,cm\)

\(AH = 8,4\,cm;\,AB = 6\,cm\)

\(AH = 4\,cm;\,AB = 6\,cm\)

\(AH = 5\,cm;\,AB = 6\,cm\)