Bunuel wrote:

If the length of a certain rectangle is 2 greater than the width of the rectangle, what is the perimeter of the rectangle?

(1) The length of each diagonal of the rectangle is 10.

(2) The area of the rectangular region is 48.

let the width be \(x\), so length will be \(x+2\)

hence perimeter \(= 2[x+x+2] = 2(2x+2)\). we need the value of \(x\) to determine the perimeter

Statement 1: implies \(\sqrt{length^2+width^2}=diagonal\)

\(\sqrt{x^2+(x+2)^2}=10\)

or \(x^2+x^2+4x+4=100 => x^2+2x-48=0\)

so \((x-6)(x+8)=0\), hence \(x=6\) or \(-8\)

As width cannot be negative so \(x=6\).

SufficientStatement 2: \(x(x+2)=48 => x^2+2x-48=0\)

Same as statement 1 above. we get \(x=6\).

SufficientOption

D