Ans: (a) Yes (b) $56$
-
$\begin{array}{cl}
& \angle QOP \\
= & 213^\circ – 123^\circ \\
= & 90^\circ
\end{array}$Therefore, $\Delta OPQ$ is a right-angled triangle.
- By applying the Pythagoras Theorem to $\Delta OPQ$, we have
$\begin{array}{rcl}
PQ^2 & = & OP^2 + OQ^2 \\
25^2 & = & k^2 +24^2 \\
k & = & \sqrt{625 – 576} \\
& = & 7
\end{array}$Therefore, the perimeter of $\Delta OPQ$
$\begin{array}{cl}
= & 7 + 24 + 25 \\
= & 56
\end{array}$
