Originally Posted by

**Soroban** Hello, ur5pointos2slo!

$\displaystyle A$ is the peak of the notch.

$\displaystyle \angle BAC = 90^o$ . (Doesn't look like it in the diagram.)

$\displaystyle O$ is the center of the circle.

Radii $\displaystyle OB = OC = OD = 1$ cm.

$\displaystyle \angle ABO = \angle ACO = 90^o$

$\displaystyle \Delta ABO\text{ and }\Delta ACO$ are isosceles right triangles.

. . Hence: .$\displaystyle AO = \sqrt{2}$

Then: .$\displaystyle AD \:=\:\sqrt{2} + 1$

Therefore, the depth of the notch is: .$\displaystyle (\sqrt{2} + 1) - 1.3 \;=\;\boxed{\sqrt{2}-0.3\text{ cm.}}$