topper97 wrote:
The radius of a circle is r yards. Is the area of the circle at least r square yards? (1 yard = 3 feet)
(1) The diameter of the circle is more than 2 feet.
(2) If the radius of the same circle is f feet, the area of the circle is more than 2f square feet.
This is MUCH harder than I expected. Excellent question (kudos)!
\(\pi {r^2}\,\,\mathop \ge \limits^? \,\,\,r\,\,\,\,\left[ {{\rm{yard}}{{\rm{s}}^{\rm{2}}}} \right]\,\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{r\,\, > \,\,0} \,\,\,\,\,\,\,\pi r\,\,\,\mathop \ge \limits^? \,\,\,1\,\,\,\,\,\left[ {{\rm{yards}}} \right]\)
\(\left( 1 \right)\,\,\,2r\,\,{\rm{yards}}\,\,\, > \,\,\,2\,\,{\rm{ft}}\,\,\left( {{{\,1\,\,{\rm{yard}}\,} \over {3\,\,{\rm{ft}}}}} \right)\,\,\,\,\,\,\left[ {{\rm{yards}}} \right]\,\,\,\,\,\, \Leftrightarrow \,\,\,\,r > \,\,{1 \over 3}\,\,\,\,\left[ {{\rm{yards}}} \right]\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{ \cdot \,\,\pi } \,\,\,\,\,\,\pi r > 1\,\,\,\,\,\left[ {{\rm{yards}}} \right]\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)
\(\left( 2 \right)\,\,\,\pi {f^2}\,\, > \,\,2f\,\,\,\,\left[ {{\rm{fee}}{{\rm{t}}^{\rm{2}}}} \right]\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{f\,\, > \,\,0} \,\,\,\,\,\pi f > 2\,\,\,\,\,\,\left[ {{\rm{feet}}} \right]\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\pi f\,\,{\rm{ft}}\,\,\,\left( {{{\,1\,\,{\rm{yard}}\,} \over {3\,\,{\rm{ft}}}}} \right) > \,\,\,2\,\,\,\left( {{{\,1\,\,{\rm{yard}}\,} \over {3\,\,{\rm{ft}}}}} \right)\,\,\,\,\,\,\left[ {{\rm{yard}}} \right]\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,{{\pi f} \over 3} > {2 \over 3}\,\,\,\,\left[ {{\rm{yard}}} \right]\)
\({{\pi f} \over 3} > {2 \over 3}\,\,\,\,\left[ {{\rm{yard}}} \right]\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{r\,\, = \,\,{f \over 3}\,\,!!} \,\,\,\,\,\pi r > {2 \over 3}\,\,\,\,\left[ {{\rm{yard}}} \right]\,\,\,\,\,\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\pi r = {3 \over 4}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,\pi r = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)