$BF_3$ has a planar triangular structure which arises from the $sp^2$ hybrid orbitals. (ground sytate) B 2s :

↿⇂

2p :

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(excited state) B 2s :

↿

2p :

↿

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These three $sp^2$ hybrid orbitals are directed towards the corners of triangle and $BF_3$ has a trigonal structure. $BH_4^{–}$ may be assumed to be made of a central B atom, 3H atoms and onehydride ion $H^{–}$. In order to accommodate the 3H atoms and one $H^{–}$ ion, B undergoes $sp^3$ hybridisation yielding four orbitals, 3 of which contain one $e^{–}$ each and one is empty. The fourth, empty orbital accomodates the $H^{–}$ ion. Thus, the structure of $BH_4^{–}$ is tetrahedral.