Rework Problem 37 if the belt is crossed, as shown in the figure.

Here, we can form two right triangles namely $∆ACI$and $∆BFI$.

To solve the length of the belt required to join the two, we only solve the measure of the lines and arcs in red as shown below, and multiply resulting value by 2.

Hence, $Belt=2\left(CI+FI+m\stackrel{⏜}{CH}+m\stackrel{⏜}{FG}\right)$

Let $BI=xandAI=24-x$

since the two angles of right triangles are congruent, then they are similar. Therefore,

$$\begin{gathered} \frac{x}{24-x}=\frac{2.5}{6.5} \\ 6.5x=60-2.5x \\ 9x=60 \\ x=\frac{20}{3} \end{gathered}$$

then$AI=24-\frac{20}{3}=\frac{52}{3}$

Using Pythagoras Theorem,

$$\begin{gathered} C{I}^2=A{I}^2-A{C}^2 \\ CI=\sqrt{{\left(\frac{52}{3}\right)}^2-{\left(6.5\right)}^2} \end{gathered}$$

$CI=\sqrt{\frac{9295}{36}}=16.07$

and

$$\begin{gathered} F{I}^2=B{I}^2-B{F}^2 \\ FI=\sqrt{{\left(\frac{20}{3}\right)}^2-{\left(2.5\right)}^2} \\ FI=\sqrt{\frac{1375}{36}}=6.18 \end{gathered}$$

Since $∆ACI$and $∆FBI$are similar. So $\angle ACI\approx \angle FBI$

By using law of sines

$\frac{\sin A}{a}=\frac{\sin B}{b}$

$$\begin{gathered} \frac{\sin \angle FBI}{FI}=\frac{\sin 90°}{BI} \\ \sin \angle FBI=FI\times \frac{\sin 90°}{BI} \\ =\frac{6.18\times \sin 90°}{{\displaystyle \raisebox{1ex}{$20$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=0.927 \\ \angle FBI={\sin }^{-1}\left(0.927\right)=67.98° \end{gathered}$$

Thus,$\angle ACI=\angle FBI=67.98°$

$$\begin{gathered} m\stackrel{⏜}{CH}=2\mathrm{\pi r}\frac{180°-\angle \mathrm{CAI}}{360°} \\ =2\mathrm{\pi }\times 6.5\frac{180°-67.98°}{360°} \\ =12.71 \end{gathered}$$

and

$$\begin{gathered} m\stackrel{⏜}{FG}=2\mathrm{\pi r}\frac{180°-\angle FBI}{360°} \\ =2\mathrm{\pi }\times 2.5\frac{180°-67.98°}{360°} \\ =4.89 \end{gathered}$$

$$\begin{gathered} Belt=2\left(CI+FI+m\stackrel{⏜}{CH}+m\stackrel{⏜}{FG}\right) \\ =2\left(16.07+6.18+12.71+4.89\right)=79.69 \end{gathered}$$