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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: 50730
Squares I, II and III above are arranged along the x-axis as shown.  [#permalink]

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15 Dec 2017, 01:11
00:00

Difficulty:

15% (low)

Question Stats:

82% (01:14) correct 18% (01:36) wrong based on 77 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

Attachment:

2017-12-15_1259_001.png [ 6.12 KiB | Viewed 793 times ]

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

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15 Dec 2017, 13:29
1
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

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 = $$s^2 = 4^2 = 16$$

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