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# Squares I, II and III above are arranged along the x-axis as shown.

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Math Expert
Joined: 02 Sep 2009
Posts: 43380
Squares I, II and III above are arranged along the x-axis as shown. [#permalink]

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15 Dec 2017, 01:11
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Difficulty:

25% (medium)

Question Stats:

68% (00:48) correct 32% (00:54) wrong based on 38 sessions

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Squares I, II and III above are arranged along the x-axis as shown. What is the area of square II?

(A) 9
(B) 16
(C) 25
(D) 49
(E) 64

[Reveal] Spoiler:
Attachment:

2017-12-15_1259_001.png [ 6.12 KiB | Viewed 432 times ]
[Reveal] Spoiler: OA

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Joined: 22 May 2016
Posts: 1261
Squares I, II and III above are arranged along the x-axis as shown. [#permalink]

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15 Dec 2017, 13:29
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KUDOS
Bunuel wrote:

Squares I, II and III above are arranged along the x-axis as shown. What is the area of square II?

(A) 9
(B) 16
(C) 25
(D) 49
(E) 64

[Reveal] Spoiler:
Attachment:
2017-12-15_1259_001.png

1) all sides of a square are equal

2) square II's side length = (x-coordinate of III's lower left corner) - (x-coordinate of I's lower right corner)

2) side of square I = 3
-- points on a vertical line (y-axis here) have the same x-coordinates, but different y-coordinates; for length, use the unequal coordinates
-- Length = absolute value of difference between $$y_2$$ and $$y_1 = (3 - 0) = 3$$

3) Square I's lower right vertex/corner is at (3,0); it must be 3 away from origin on the x-axis

4) Lower left corner of square III is at (7,0)
-- upper and lower left corners have the same x-coordinate
-- upper left corner of III is (7,8)
-- so its lower left corner is at (7,0)

5) Square II, between lower right corner of I and lower left corner of III on the x-axis, has side length (7 - 3) = 4

Area of square II = $$s^2 = 4^2 = 16$$

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Formerly genxer123

Squares I, II and III above are arranged along the x-axis as shown.   [#permalink] 15 Dec 2017, 13:29
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