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

Always look for special triangles such as 3-4-5 and 5-12-13.

Statement 2:
Case 1:

Case 2:

Since each case will yield a different perimeter, INSUFFICIENT.

Statement 1:
Case 1:

Case 3:

Since each case will yield a different perimeter, INSUFFICIENT.

Statements combined:
Since only a 5-12-13 triangle has both legs with a difference of 7 and a hypotenuse of 13, only Case 1 is viable:

Thus, the perimeter of the rectangle can be determined.
SUFFICIENT.

The correct answer is C.

For a rectangle, 2 sides are equal, and with diagonal z, we will b having a right angled triangle.
By Pythagoras,
$$z^2=x^2+y^2$$
We are looking for the perimeter of the rectangle,
Let the perimeter = P
$$P=2\left(x+y\right)\ find\ P$$

Statement 1
$$x-y=7$$
This means that x and y can be 1 and 6, 2 and 5, 3 and 4, 0 and 7 respectively and alternatively since there is no definite conclusion, statement 1 is INSUFFICIENT.

Statement 2
z = 13 ; with pythagoras theorem
$$z^2=x^2+y^2$$
$$13^2=x^2+y^2$$
$$169=x^2+y^2$$
Lots of value can also satisfy the equation, hence statement 2 is INSUFFICIENT.

Combining statement 1 and 2 together
Statement 1 : x+y =7
y = 7 - x and z = 13
From pythagoras
$$z^2=x^2+y^2$$
where y = 7- x and z = 13
$$13^2=x^2+\left(7-x\right)^2$$
$$13^2=x^2+\left(7-x\right)\left(7-x\right)$$
$$13^2=x^2+49-7x-7x+x^2$$
$$169=x^2+49-14x+x^2$$
$$169=2x^2-14x+49$$
$$2x^2-14x+49-169=0$$
$$2x^2-14x-120=0\left(Quadratic\ equation\ \right)$$
$$2x^2-24x+10x-120=0$$
$$\left(2x^2-24x\right)+\left(10x-120\right)=0$$
$$2x\left(x-12\right)+10\left(x-12\right)=0$$
$$2x+10=0\ or\ x-12=0$$
$$x=-5\ OR\ x=12$$
(x cannot be a negative integer)
from x + y = 7 where x = 12
12 + y =7
y = 5, the P = 2 (x+y) = 2 (17) = 34
Both statements together are SUFFICIENT.

$$Answer\ is\ Option\ C$$