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# In the figure above, the large square is divided into two smaller squa

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Math Expert
Joined: 02 Sep 2009
Posts: 46217
In the figure above, the large square is divided into two smaller squa [#permalink]

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28 Sep 2017, 23:47
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Question Stats:

94% (00:48) correct 6% (00:23) wrong based on 31 sessions

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In the figure above, the large square is divided into two smaller squares and two shaded rectangles. If the perimeters of the two smaller squares are 8 and 20, what is the sum of the perimeters of the two shaded rectangles?

(A) 14
(B) 18
(C) 20
(D) 24
(E) 28

Attachment:

2017-09-29_1042.png [ 1003 Bytes | Viewed 894 times ]

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In the figure above, the large square is divided into two smaller squa [#permalink]

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29 Sep 2017, 15:45
Bunuel wrote:

In the figure above, the large square is divided into two smaller squares and two shaded rectangles. If the perimeters of the two smaller squares are 8 and 20, what is the sum of the perimeters of the two shaded rectangles?

(A) 14
(B) 18
(C) 20
(D) 24
(E) 28

Attachment:
2017-09-29_1042.png

I sketched and wrote the values on the figure.

Smallest square with side s:
4s = perimeter
4s = 8
s = 2 = width of both rectangles

Middle-sized square:
4s = 20
s = 5 = length of both rectangles

Each rectangle's perimeter is
2 + 2 + 5 + 5 = 14

Sum of both perimeters is
14 + 14 = 28

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Re: In the figure above, the large square is divided into two smaller squa [#permalink]

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30 Sep 2017, 00:48
Bunuel wrote:

In the figure above, the large square is divided into two smaller squares and two shaded rectangles. If the perimeters of the two smaller squares are 8 and 20, what is the sum of the perimeters of the two shaded rectangles?

(A) 14
(B) 18
(C) 20
(D) 24
(E) 28

Attachment:
2017-09-29_1042.png

The smallest square has perimeter = 8 = 4*side = 4a
i.e. side of the smallest square, a = $$2$$

The Second smallest square has area = 20 = 4*side = 4b
i.e. side of the second smallest square, b = $$5$$

Perimeter of one shaded rectangle = 2(a+b)
Perimeter of both shaded rectangles = 2*2(a+b) = 4(a+b) = $$4(2+5)$$ = 28

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Re: In the figure above, the large square is divided into two smaller squa [#permalink]

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30 Sep 2017, 03:43

Since the squares have a perimeter of 8 and 20,
the sides of the squares are $$(\frac{8}{4})2$$ and $$(\frac{20}{4})5$$.
The rectangles share their lengths and breadth with these 2 squares.

Therefore, the 2 rectangles will have length and breadth 2 and 5 respectively,
and the perimeter of one such rectangle is 2(length + breadth) = 2(2+5) = 14

Hence, the perimeter of the two shaded rectangles is 2*14 = 28(Option E)
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Re: In the figure above, the large square is divided into two smaller squa   [#permalink] 30 Sep 2017, 03:43
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