Problem

There is an isosceles trapezoid with a bottom side of $18 \ cm$, lateral sides of $32.8 \ cm$, and a top side of $3.6 \ cm$. Find the diameter of the circle inscribed in the top and lateral sides as shown.

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Solution

If you draw the diagram according to the problem statement, it will actually look like the diagram belowis:

Let the lengths of bottom side $CD$, lateral side $BC \ (=DA)$, and top side $AB$ be $x, y$, and $z$ respectively.

First, it turns out that

$$DE=\frac{x-z}{2}. \qquad [1]$$

By setting $∠ADE=θ$, we find that

$$\cos⁡θ=\frac{DE}{DA}=\frac{(x-z)/2}{y}=\frac{x-z}{2y}. \quad [2]$$

Next, from the figure, the following can be said:

$$BF=\frac{AB}{2}=\frac{z}{2}. \quad [3]$$

And

$$∠OBF=\frac{180-θ}{2}=90°-\frac{θ}{2},$$

$$∴ \ \tan∠OBF=\tan⁡(90°-\frac{θ}{2})=\cot\frac{θ}{2}. \quad [4]$$

From the half-angle formula, we get

$$\tan^2⁡ \frac{θ}{2}=\frac{1-\cos⁡θ}{1+\cos⁡θ}=\frac{1-(x-z)/2y}{1+(x-z)/2y}=\frac{2y-x +z}{2y+x-z},$$

$$∴ \ \tan\frac{θ}{2}=\sqrt{\frac{2y-x+z}{2y+x-z}}. \quad [5]$$

Therefore, from $[4]$ and $[5]$, we know that

$$\tan∠OBF=\frac{1}{\sqrt\frac{2y-x+z}{2y+x-z}}. \quad [6]$$

Also, from the figure and $[3]$, we get the following:

$$\tan∠OBF=\frac{FO}{BF}=\frac{d/2}{z/2}=\frac{d}{z}. \quad [7]$$

From $[6]$ and $[7]$, it turns out that

$$\frac{d}{z}=\frac{1}{\sqrt\frac{2y-x+z}{2y+x-z}},$$

$$∴ \ d=\frac{z}{\sqrt\frac{2y-x+z}{2y+x-z}}. \quad[8]$$

According to the problem statement, $x=18, \ y=32.8,$ and $\ z=3.6$, so by substituting these into $[8]$, we have

$$d=\frac{3.6}{\sqrt\frac{2×32.8-18+3.6}{2×32.8+18-3.6}}=\frac{3.6}{\sqrt{0.64}}=\frac{3.6}{0.8}=4.5.$$

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$$Ans.\quad 4.5 \ cm$$

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Reference

Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), p.51; pp.306-307.