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We characterize conformally removable sets in the plane with the aid of the recent developments in the theory of metric surfaces. We prove that a compact set in the plane is -removable if and only if there exists a quasiconformal map from the plane onto a metric surface that maps the given set to a set of linear measure zero. The statement fails if we consider maps into the plane rather than metric surfaces. Moreover, we prove that a set is -removable (resp. -removable) if and only if every homeomorphism from the plane onto a metric surface (resp. reciprocal metric surface) that is quasiconformal in the complement of the given set is quasiconformal everywhere.

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