Problem

As shown in the figure, two regular pentagons, one large and one small, are placed on a line, adjacent to each other.
When you know the area of ​​an isosceles triangle in which base is one side of the larger regular pentagon and the angle of its vertex, find the lengths of the sides of the larger and smaller regular pentagons.

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Solution

Let the area of ​​the isosceles triangle be $S$, and $∠A=θ$.

Let the length of the side of the larger regular pentagon be $a$. Then,

$$BC×AD=a×AD=2S,$$

$$∴ \ AD=\frac{2S}{a}.$$

Since $∠BAD=θ/2$,

$$tan⁡\frac{θ}{2}=\frac{BD}{AD}=\frac{a/2}{2S/a}=\frac{a^2}{4S},$$

$$a^2=4S×tan\frac{θ}{2},$$

$$∴ \ a=2\sqrt{S×tan\frac{θ}{2}}.$$

Let the length of the side of the smaller regular pentagon be $b$.

Since $∠FCE=∠FEC=∠FGH=∠FHG=72°$, $△FCE$ and $△FGH$ are isosceles triangles.

Since $EF=GF$,

$$∆FCE≡∆FGH.$$

Therefore,

$$FC=FH.$$

Since $FC=FH=EF=GF$,

$$b=\frac{a}{2}=\sqrt{S×tan\frac{θ}{2}}.$$

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Reference

Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), pp.44-45; pp.327-328.