A
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E
Area of the square ABCD = y^2 and the area of the square JLKM = x^2; thus, the area of the shaded region = y^2 – x^2.Vincen wrote: ↑Sun May 03, 2020 12:02 pmsquare in a square.png
\(ABCD\) is a square with a side \(y,\) and \(JKLM\) is a side \(x.\) If Rectangle \(S\) (not shown) with length \(x + y\) has the same area as the shaded region above, what is the width of Rectangle \(S?\)
(A) \(x\)
(B) \(y\)
(C) \(y + x\)
(D) \(y - x\)
(E) \(y^2−x^2\)
[spoiler]OA=D[/spoiler]
Source: Magoosh
The area of ABCD is y^2, and the area of JKLM is x^2. The difference of their areas is y^2 - x^2, and this difference is equal to the area of Rectangle S.Vincen wrote: ↑Sun May 03, 2020 12:02 pmsquare in a square.png
\(ABCD\) is a square with a side \(y,\) and \(JKLM\) is a side \(x.\) If Rectangle \(S\) (not shown) with length \(x + y\) has the same area as the shaded region above, what is the width of Rectangle \(S?\)
(A) \(x\)
(B) \(y\)
(C) \(y + x\)
(D) \(y - x\)
(E) \(y^2−x^2\)
[spoiler]OA=D[/spoiler]
Source: Magoosh
