A rectangle has sides x and y and diagonal z. What is the perimeter of the rectangle?
(1) x - y = 7.
(2) z = 13.
OA C
Source: Princeton Review
A rectangle has sides x and y and diagonal z. What is the
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\[? = 2\left( {x + y} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\boxed{\,\,? = x + y\,\,}\]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
$$\left( 1 \right)\,\,\,x - y = 7\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {8,1} \right)\,\,\,\,\, \Rightarrow \,\,\,? = 9\,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {9,2} \right)\,\,\,\,\, \Rightarrow \,\,\,? = 11\,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,{x^2} + {y^2} = {z^2} = {13^2}\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {5\,\,;\,\,12} \right)\,\,\,\,\, \Rightarrow \,\,\,? = 17\,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x;y} \right) = \left( {{{13} \over {\sqrt 2 }}\,\,;\,\,{{13} \over {\sqrt 2 }}} \right)\,\,\,\,\, \Rightarrow \,\,\,? = {{2 \cdot 13} \over {\sqrt 2 }} \ne 17\, \hfill \cr} \right.$$
$$\left( {1 + 2} \right)\,\,\,\left\{ \matrix{
\,x - y = 7 \hfill \cr
\,{x^2} + {y^2} = {13^2} \hfill \cr} \right.\,\,\,\,\,\,\mathop \Rightarrow \limits^{x+y\,\, > \,\,0} \,\,\,\,\,x + y = 17\,\,\,\,\,\left[ {\,5,12,13\,\,{\rm{shortcut}}\,} \right]\,\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,$$
$${\rm{POST - MORTEM}}\,\,:\,\,\,\left\{ \matrix{
\,x - y = 7\,\,\,\mathop \Rightarrow \limits^{{\rm{squaring}}} \,\,\,{x^2} + {y^2} - 2xy = {7^2} \hfill \cr
\,{x^2} + {y^2} = {13^2} \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,2xy = {13^2} - {7^2} = \left( {13 + 7} \right)\left( {13 - 7} \right)\,\,\,\,\, \Rightarrow \,\,\,\,2xy = 120\,$$
$${x^2} + {y^2} + \underline {2xy} = {13^2} + \underline {2xy} = {13^2} + 120\,\,\,\,\, \Rightarrow \,\,\,\,\,{\left( {x + y} \right)^2} = 289 = {17^2}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{x + y\,\, > \,\,0} \,\,\,\,\,\,\,x + y = 17\,\,\,\,$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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Needless to state that none of the statements alone can work; there are many possibilities when we consider that x, y, and z can be real numbers (not necessarily integers).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.
OA C
Source: Princeton Review
(1) and (2) together
You must remember a few Pythagorean triplets: {3, 4, 5}; {5, 12, 13}; {7, 24, 25}.
We see that in the Pythagorean triplet {5, 12, 13}, the diagonal is 13, which is also the value of diagonal in the given problem. The given values of the other two sides are 5 and 12; we see that 12 and 5 differ by 7 (= x - y). Thus, x = 12, and y = 5. Thus, perimeter = 2(x + y) = 2(12 + 5) = 34. Sufficient.
The correct answer: C
Alternatively, you can apply a traditional approach.
Since z is diagonal, we have
z^2 = x^2 + y^2 ---(1)
From x - y = 7, we have x = y + 7
Thus, 13^2 = (y + 7)^2 + y^2
Upon solving, we get y = 12 and x = 5. Sufficient.
Hope this helps!
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We are given that a rectangle has sides x and y and diagonal z, thus: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.
x^2 + y^2 = z^2
We need to determine the perimeter of the rectangle, which is 2x + 2y.
Statement One Alone:
x - y = 7
Since x - y = 7, (x - y)^2 = 7^2 or x^2 + y^2 - 2xy = 49. However, this does not allow us to determine a unique value for x or y. Statement one alone is not sufficient.
Statement Two Alone:
z = 13
Knowing the value of z does not allow us to determine a unique value for x or y. Statement two alone is not sufficient.
Statements One and Two Together:
From both statements, we have x^2 + y^2 - 2xy = 49 and z = 13. From the stem analysis, we have x^2 + y^2 = z^2. So x^2 + y^2 = 13^2 or x^2 + y^2 = 169. Substitute 169 for x^2 + y^2 in x^2 + y^2 - 2xy = 49, we have:
169 - 2xy = 49
120 = 2xy
60 = xy
y = 60/x
Substitute this in x - y = 7, and we have:
x - 60/x = 7
Multiplying the equation by x, we have:
x^2 - 60 = 7x
x^2 - 7x - 60 = 0
(x - 12)(x + 5) = 0
x = 12 or x = -5
Since x can't be negative, x = 12, and, hence, y = 60/x = 5. So the perimeter of the rectangle is 2(12) + 2(5) = 34.
Answer: C
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