Rastis wrote:If p is the perimeter of rectangle Q, what is the value of p?

1) Each diagonal of rectangle Q has a length of 10

2) The area of rectangle Q is 48.

Let L = length, W = width, D = diagonal.
L² + W² = D².

Statement 1: Each diagonal of rectangle Q has a length of 10
Thus, L² + W² = 10² = 100.

Case 1: L=8 and W=6, with the result that L² + W² = 8²+ 6² = 100.
In this case, p = 8+8+6+6 = 28.

Case 2: L=1 and W=√99, with the result that L² + W² = 1²+ √99² = 100.
In this case, p = 1+1+√99+√99 = 2 + 2√99.

Since p can be different values, INSUFFICIENT.

Statement 2: The area of rectangle Q is 48
Case 1 also satisfies statement 2.

Case 1: L=8 and W=6, with result that A = 8*6 = 48.
In this case, p=28.

Case 3: L=1 and W=48, with the result that A = 1*48 = 48.
In this case, p = 1+1+48+48 = 98.

Since p can be different values, INSUFFICIENT.

Statements combined:
Only Case 1 satisfies both statements.
Thus, p=28.
SUFFICIENT.

The correct answer is C.

What do we need to know to work out P? Lets call the length of the rectangle L and the width W. To work out the perimeter we need to know L and W as P = 2.L + 2.W.

Statement 1: diagonal (lets call this D) = 10. Now we know that L^2 + W^2 = 10^2 = 100. The problem though is that we cannot use this alone to get L and W. INSUFFICIENT.

Statement 2: area of the rectangle is 48. L.W = 48. This doesn't help us either as we cannot determine L and W from just this equation.

Statements 1 & 2 together:

We know that:
(1) L^2 + W^2 = 100
(2) L.W = 48 --> we can rearrange this so that L = 48/W

Using (2), (1) can become: (48/W)^2 + W^2 = 100, which is solvable for W. One trick here is not to go ahead and waste time solving this. It is not necessary. We know we can get W and hence we know we can get L from (2). Thus combined the statements are sufficient. DONT WASTE TIME IN THE EXAM SOLVING EQUATIONS YOU DONT NEED TO SOLVE --> the question asks if it is sufficient not what the actual numbers are!