Réponse:
[tex]d \: apres \: le \: theoreme \: de \: pythagore \: dans \: le \: triagl \: rectangle \: eir \: on \: a \: \\ \: \\ {er}^{2} = {ri}^{2} + {ei}^{2} \\ er = \sqrt{ {10.5}^{2} + {6}^{2} } \\ er = 12.09cm \\ \\ d \: apres \: pythagore \: dans \: le \: triangle \: rectangle \: ern \: on \: a \: \\ {nr}^{2} = {ne}^{2} - {er}^{2} \\ nr = \sqrt{ {13.5}^{2} - {12.09}^{2} } \\ nr = 6cm \: [/tex]
