Question about impedance matching transformers

[Ben321]

Using a circuit simulator program, I was experimenting with different turns ratios of a transformer (it uses an ideal transformer model, including having no resistance for the windings of the transformer), and noticed something interesting when using the transformer for impedance matching. In this experiment I was also just treating the source and load impedance as resistance (not taking into account reactive impedance that would exist in real life). It seems that when matching a source with a resistance that is 10 times the resistance of the load, a 3:1 turns ratio is what works best. That completely threw out what I had assumed would be the case. I had assumed that a 10:1 turns ratio would be needed to match a source to a load when the source had 10 times the resistance of the load.
Why is it that the best turns ratio in this case is 3:1?

[KrudyZ]

Because you are transforming voltages / currents by that ratio, so the perfect ratio is sqrt(10) to get the best power match.
If you had 1 A going through a 10 Ohm load you would have 10 V across for 10 W of power.
For the same power in a 1 Ohm load you need 3.16 V yielding 3.16 A...

[Bud]

Because impedance ratio is proportional to N^2, where N is winding turns ratio.

[Ben321]

Because you are transforming voltages / currents by that ratio, so the perfect ratio is sqrt(10) to get the best power match.
If you had 1 A going through a 10 Ohm load you would have 10 V across for 10 W of power.
For the same power in a 1 Ohm load you need 3.16 V yielding 3.16 A...

That makes sense.