BTGmoderatorDC wrote:A rectangle has sides x and y and diagonal z. What is the perimeter of the rectangle?
(1) x - y = 7.
(2) z = 13.
Source: Princeton Review
\[? = 2\left( {x + y} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{\,\,? = x + y\,\,}\]
$$\left( 1 \right)\,\,\,x - y = 7\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {8,1} \right)\,\,\,\,\, \Rightarrow \,\,\,? = 9\,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {9,2} \right)\,\,\,\,\, \Rightarrow \,\,\,? = 11\,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,{x^2} + {y^2} = {z^2} = {13^2}\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {5\,\,;\,\,12} \right)\,\,\,\,\, \Rightarrow \,\,\,? = 17\,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {{{13} \over {\sqrt 2 }}\,\,;\,\,{{13} \over {\sqrt 2 }}} \right)\,\,\,\,\, \Rightarrow \,\,\,? = {{2 \cdot 13} \over {\sqrt 2 }} \ne 17\, \hfill \cr} \right.$$
$$\left( {1 + 2} \right)\,\,\,\left\{ \matrix{
\,x - y = 7 \hfill \cr
\,{x^2} + {y^2} = {13^2} \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{x+y\,\, > \,\,0} \,\,\,\,\,x + y = 17\,\,\,\,\,\left[ {\,5,12,13\,\,{\rm{shortcut}}\,} \right]\,\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,$$
$${\rm{POST - MORTEM}}\,\,:\,\,\,\left\{ \matrix{
\,x - y = 7\,\,\,\mathop \Rightarrow \limits^{{\rm{squaring}}} \,\,\,{x^2} + {y^2} - 2xy = {7^2} \hfill \cr
\,{x^2} + {y^2} = {13^2} \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,2xy = {13^2} - {7^2} = \left( {13 + 7} \right)\left( {13 - 7} \right)\,\,\,\,\, \Rightarrow \,\,\,\,2xy = 120\,$$
$${x^2} + {y^2} + \underline {2xy} = {13^2} + \underline {2xy} = {13^2} + 120\,\,\,\,\, \Rightarrow \,\,\,\,\,{\left( {x + y} \right)^2} = 289 = {17^2}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{x + y\,\, > \,\,0} \,\,\,\,\,\,\,x + y = 17\,\,\,\,$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
