\(ABCD\) is a square with a side \(y,\) and \(JKLM\) is a side \(x.\) If Rectangle \(S\) (not shown) with length

Vincen wrote: ↑

Sun May 03, 2020 12:02 pm

square 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

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.

Say the width of Rectangle S = w; thus, its area = (x + y)*w

=> (x + y)*w = y^2 – x^2

w = y – x.

The correct answer: D

Hope this helps!

-Jay
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Vincen wrote: ↑

Sun May 03, 2020 12:02 pm

square 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.

Let’s let w = the width of Rectangle S. Then the area of rectangle S is A = w * (x + y).

We have two expressions for the area of Rectangle S, so we equate them:

y^2 - x^2 = w * (x + y)

(y - x)(y + x) = w * (x + y)

y - x = w

Answer: D