91Ó°ÊÓ

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

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 $∠ACI≈∠FBI$

By using law of sines

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

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

Thus,$∠ACI=∠FBI=67.98°$

$$\begin{gathered} m\stackrel{Áåœ}{CH}=2\mathrm{Ï€°ù}\frac{180°-∠\mathrm{CAI}}{360°} \\ =2\mathrm{Ï€}×6.5\frac{180°-67.98°}{360°} \\ =12.71 \end{gathered}$$

and

$$\begin{gathered} m\stackrel{Áåœ}{FG}=2\mathrm{Ï€°ù}\frac{180°-∠FBI}{360°} \\ =2\mathrm{Ï€}×2.5\frac{180°-67.98°}{360°} \\ =4.89 \end{gathered}$$

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