$ABEF$ is a square, and its area is $4\times$ area$(BDA) = 4\times\frac{1}{2}(BD\times DA) = 2BD^2 = 24$cm$^2$, so $BD=\sqrt{12}=2\sqrt{3}$cm.

The angle $ABD$ is $45^{\circ}$ so the angle $CBD$ is $45^{\circ}-15^{\circ}=30^{\circ}$.

Therefore $\tan{30^{\circ}}=\frac{1}{\sqrt{3}}=\frac{CD}{BD}=\frac{CD}{2\sqrt{3}}$ so $CD=2$cm.

*This problem is taken from the UKMT Mathematical Challenges.*

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