2021 AMC 10B Problem 20

Let $F$ be the unlabeled point joined to $A,B,$ and $C.$ Because all the drawn segments have length $2,$ triangles $ABF$ and $CBF$ are equilateral and lie on opposite sides of $BF.$ Hence $\angle ABC=120^\circ,$ so $$[ABC]=\frac12\cdot2\cdot2\sin120^\circ=\sqrt3.$$ The same reasoning on the other side gives $[ADE]=\sqrt3.$ Also, the Law of Cosines in $\triangle ABC$ gives $AC^2=12,$ and similarly $AD^2=12.$ Thus the altitude of isosceles triangle $ACD$ to its base $CD=2$ is $$\sqrt{(\sqrt{12})^2-1^2}=\sqrt{11}.$$ Therefore $[ACD]=\frac12\cdot2\cdot\sqrt{11}=\sqrt{11}.$ The pentagon’s total area is $$\sqrt{12}+\sqrt{11},$$ so $m+n=12+11=23.$ Thus, the answer is D .