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# The figure above is formed by two overlapping squares, each having

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Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132588 [0], given: 12326

The figure above is formed by two overlapping squares, each having [#permalink]

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27 Sep 2017, 04:47
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Difficulty:

35% (medium)

Question Stats:

97% (00:46) correct 3% (01:34) wrong based on 30 sessions

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The figure above is formed by two overlapping squares, each having sides of 6 centimeters in length. If P and Q are the midpoints of the intersecting sides, what is the area, in square centimeters, of the enclosed region?

(A) 72
(B) 63
(C) 60
(D) 54
(E) 45

[Reveal] Spoiler:
Attachment:

2017-09-27_1107.png [ 2.17 KiB | Viewed 444 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 132588 [0], given: 12326

Director
Joined: 22 May 2016
Posts: 977

Kudos [?]: 336 [0], given: 591

Re: The figure above is formed by two overlapping squares, each having [#permalink]

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27 Sep 2017, 07:17
Bunuel wrote:

The figure above is formed by two overlapping squares, each having sides of 6 centimeters in length. If P and Q are the midpoints of the intersecting sides, what is the area, in square centimeters, of the enclosed region?

(A) 72
(B) 63
(C) 60
(D) 54
(E) 45

[Reveal] Spoiler:
Attachment:
The attachment 2017-09-27_1107.png is no longer available

Attachment:

squaresoverlap.png [ 13.32 KiB | Viewed 324 times ]

Method I - Add square areas and subtract overlap - Figure I

Each square has a side of length 6 cm.

P and Q are midpoints.
Thus the sides of the squares are bisected.
In Figure 1, short sides of overlap = $$\frac{6}{2}$$= 3$$cm$$

The area of each square is (6 * 6) = 36 $$cm^2$$
The area of the overlap is (3 * 3) = 9 $$cm^2$$

Area of enclosed region in square centimeters?

Square + Square - Overlap
36 + 36 - 9 = 63

Method II - Divide region into areas. Find and add the areas of each region

See Figure 2: outline different interior areas of the enclosed figure

P and Q , being midpoints, bisect each side of 6 cm into 3 cm each
The short sides of the interior areas are 3 cm. The long sides are 6 cm

Upper left corner, square's area: (6 * 6) = 36 $$cm^2$$

Area of small square: (3*3) = 9 $$cm^2$$

Bottom rectangle's area: (3*6) = 18 $$cm^2$$

Area of enclosed region in square centimeters?

36 + 9 + 18 = 63

Kudos [?]: 336 [0], given: 591

Senior Manager
Joined: 02 Jul 2017
Posts: 269

Kudos [?]: 79 [0], given: 65

Location: India
Re: The figure above is formed by two overlapping squares, each having [#permalink]

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27 Sep 2017, 07:26
The figure above is formed by two overlapping squares, each having sides of 6 centimeters in length. If P and Q are the midpoints of the intersecting sides, what is the area, in square centimeters, of the enclosed region?

P and Q are midpoints.
So here 2 squares overlap at the 1/4 part of each square.
So Total Area = Area of each square - Common area = 2* Area of square - 1/4 part of square
= 2*(6*6) - 1/4(6*6) = 72-9 = 63

Or we can say Area of overlapped part = square of side 3 => Area = 3*3 =9
So area of given figuer = Area of each square - Common area = 2*(6*6) -9 = 63

Attachments

2 sqaures.docx [13.93 KiB]

Kudos [?]: 79 [0], given: 65

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1684

Kudos [?]: 901 [0], given: 5

Re: The figure above is formed by two overlapping squares, each having [#permalink]

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01 Oct 2017, 12:13
Bunuel wrote:

The figure above is formed by two overlapping squares, each having sides of 6 centimeters in length. If P and Q are the midpoints of the intersecting sides, what is the area, in square centimeters, of the enclosed region?

(A) 72
(B) 63
(C) 60
(D) 54
(E) 45

[Reveal] Spoiler:
Attachment:
2017-09-27_1107.png

The total area of the two squares is 36 + 36 = 72 (if they aren’t overlapping).

The area of the overlap is 3 x 3 = 9.

Thus, the area of the entire figure is 72 - 9 = 63.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 901 [0], given: 5

Re: The figure above is formed by two overlapping squares, each having   [#permalink] 01 Oct 2017, 12:13
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