How is this not an A dim chord?
Diminished chords are built with a root, minor 3rd and a diminished 5th. Since there's no B#, do I play a C instead of a Cb?
- Steve4PhysicsLv 74 months ago
The root form of Adim is A, C, D# and F# (you just move up a minor-third to get to the next note).
There are 4 notes because it also includes the double-diminished (double-flattened) 7th of the scale.
(Note, the chords Adim, Cdim, D#dim and F#dim, for example, all have the same notes, so are really the same chord.)
In your diagram, you have only 3 notes. There are *many* chords which contain these 3 notes - as shown on the right of the diagram.
The chord you show (A, D#, B) is not Adim, but if you raise the B to a C, you get A dim.
Hope that helps. (By the way, it’s not a Physics question!)