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A rectangle has sides x and y and diagonal z. What is the

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A rectangle has sides x and y and diagonal z. What is the

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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

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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.

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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
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).

(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!

-Jay
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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.
We are given that a rectangle has sides x and y and diagonal z, thus:

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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