2020-I-18

$\begin{array}{rcll}
\angle UTV & = & \angle WTU & \text{(common $\angle$)} \\
\angle VUT & = & \angle UWT & \text{($\angle$s in alt. segment)} \\
\angle UVT & = & 180^\circ – \angle UTV – \angle VUT & \text{($\angle$ sum of $\Delta$)} \\
& = & 180^\circ – \angle WTU – \angle UWT & \text{(proved)} \\
& = & \angle WUT & \text{($\angle$ sum of $\Delta$)}
\end{array}$

Therefore, $\Delta UTV \sim \Delta WTU\ \text{(A.A.A.)}$.

1. By the result of (a), we have

   $\begin{array}{rcll}
   \dfrac{TU}{TW} & = & \dfrac{TV}{TU} & \text{(corr. sides, $\sim \Delta$s)} \\
   \dfrac{TU}{WV + TV} & = & \dfrac{TV}{TU} \\
   \dfrac{780}{WV + 325} & = & \dfrac{325}{780} \\
   608400 & = & 325 WV + 105625 \\
   325 WV & = & 502775 \\
   WV & = & 1547 \text{ cm}
   \end{array}$

   Therefore, the circumference of $C$

   $\begin{array}{cl}
   = & WV \times \pi \\
   = & 1547 \pi \text{ cm}
   \end{array}$

2. Since $VW$ is a diameter of $C$, $\angle VUW = 90^\circ$ $\text{($\angle$ in semi-circle)}$. Hence, by applying the Pythagoras Theorem to $\Delta UVW$, we have

   $\begin{array}{rcl}
   UV^2 + UW^2 & = & WV^2 \ \ldots \unicode{x2460}
   \end{array}$

   Also by the result of (a), we have

   $\begin{array}{rcll}
   \dfrac{UV}{WU} & = & \dfrac{TV}{TU} & \text{(corr. sides, $\sim \Delta$s)} \\
   \dfrac{UV}{WU} & = & \dfrac{325}{780} \\
   UV & = & \dfrac{5}{12} WU \ \ldots \unicode{x2461}
   \end{array}$

   Sub. $\unicode{x2461}$ into $\unicode{x2460}$, we have

   $\begin{array}{rcl}
   (\dfrac{5}{12}WU)^2 + WU^2 & = & WV^2 \\
   \dfrac{25}{144} WU^2 + WU^2 & = & (1547)^2 \\
   WU & = & 1428 \text{ cm}
   \end{array}$

   Sub. $WU = 1428$ into $\unicode{x2461}$, we have

   $\begin{array}{rcl}
   UV & = & \dfrac{25}{144} \times 1428 \\
   & = & 595\text{ cm}
   \end{array}$

   Hence, the perimeter of $\Delta UVW$

   $\begin{array}{cl}
   = & 1547 + 595 + 1425 \\
   = & 3570 \text{ cm} \\
   = & 35.5\text{ m} \\
   > & 35 \text{ m}
   \end{array}$

   Therefore, I agree with the claim.