Hello, algebra2!
Code:
A 2 H
*---*
2/ | \
B *---*P * G
2| | |
| | |
C *---*Q * F
2\ | /
*---*
D 2 E
We have a regular octagon with side 2.
Draw $\displaystyle BP \perp AD \text{ and }CQ \perp AD$
Draw $\displaystyle DH.$
We have isosceles right triangle $\displaystyle APB$ with hypotenuse 2.
. . Hence: .$\displaystyle AP = \sqrt{2}$
Similarly: .$\displaystyle QD = \sqrt{2}$
Hence: .$\displaystyle AD \;=\;AP + PQ + QD \;=\;\sqrt{2} + 2 + \sqrt{2}$
. . Therefore: .$\displaystyle \boxed{AD \;=\;2 + 2\sqrt{2}}$
In right triangle $\displaystyle H\!AD\!:\;\;DH^2 \:=\:AH^2 + AD^2 \;=\;2^2 + (2+2\sqrt{2})^2 \;=\;16 + 8\sqrt{2}$
. . Therefore: .$\displaystyle \boxed{DH \:=\:2\sqrt{4 + 2\sqrt{2}}}$