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# If the length of a certain rectangle is 2 greater than the width of

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Math Expert
Joined: 02 Sep 2009
Posts: 42649

Kudos [?]: 135955 [0], given: 12717

If the length of a certain rectangle is 2 greater than the width of [#permalink]

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19 Nov 2017, 08:20
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Difficulty:

25% (medium)

Question Stats:

86% (01:00) correct 14% (00:53) wrong based on 22 sessions

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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.
[Reveal] Spoiler: OA

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Kudos [?]: 135955 [0], given: 12717

PS Forum Moderator
Joined: 25 Feb 2013
Posts: 633

Kudos [?]: 317 [0], given: 39

Location: India
GPA: 3.82
If the length of a certain rectangle is 2 greater than the width of [#permalink]

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19 Nov 2017, 08:33
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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 $$x=6$$. Sufficient

Statement 2: $$x(x+2)=48 => x^2+2x-48=0$$

Same as statement 1 above. we get $$x=6$$. Sufficient

Option D

Kudos [?]: 317 [0], given: 39

If the length of a certain rectangle is 2 greater than the width of   [#permalink] 19 Nov 2017, 08:33
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