MathRevolution
[Math Revolution GMAT math practice question]
Attachment:
11.11.png
The above figure shows a sector of a circle. What is the area of the sector?
\(1) x = 120^o\)
\(2) AB=6√3\)
(We assume - as part of the definition of a sector a circle - that the "origin" of the angle x shown in the figure is the center of the circle.)
The variable R will denote the radius of the circle. All angles are measured in degrees.
\(? = \frac{{x\,}}{{360\,}}\left( {\pi {R^{\,2}}} \right)\)

\(x = 120\,\,\,\left( {{\text{both}}\,\,{\text{figures}}} \right)\)

\(AB = 6\sqrt 3 \,\,\,\left( {{\text{both}}\,\,{\text{figures}}} \right)\)
\(x = 90\,\,\,\mathop {\,\,\, \Rightarrow \,\,\,\,}\limits^{L\,,\,L\,,\,L\sqrt 2 } \,\,\,R\sqrt 2 = 6\sqrt 3 \,\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,\frac{{\sqrt 2 }}{2}\,} \,\,\,\,\,R = 3\sqrt 6 \,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,?\,\,\, = \,\,\,\frac{1}{4}\,\left( {\pi \, \cdot 9 \cdot 6} \right)\)
\(x = 120\,\,\,\mathop \Rightarrow \limits_{\left( * \right)}^{30\,,\,60\,,\,90\,\,} \,\,\,\,?\,\,\, = \,\,\,\frac{1}{3}\,\left( {\pi \, \cdot 36} \right)\,\,\, = 12\pi \,\,\, \ne \,\,\,\,\frac{1}{4}\,\left( {\pi \, \cdot 9 \cdot 6} \right)\,\,\,\,\,\,\)
\(\left( * \right)\,\,30\,,\,60\,,\,90\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered}\\
\,L\sqrt 3 \,\,\, = \,\,\frac{{6\sqrt 3 }}{2} \hfill \\\\
\,2L = R \hfill \\ \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,R = 6\)
\(\left( {1 + 2} \right)\,\,\,?\,\, = \,\,12\pi \,\,\,\left( {{\text{shown}}\,\,{\text{above}}} \right)\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.