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The lengths of the sides of rectangles ABCD and ABEF, shown above, ar

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Bunuel
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The lengths of the sides of rectangles ABCD and ABEF, shown above, ar [#permalink] New post   19 Jun 2020, 04:26
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The lengths of the sides of rectangles ABCD and ABEF, shown above, are integers (in cm). The area of ABCD is 20 cm^2 and the area of ABEF is 10 cm^2. What is the minimum possible perimeter of the rectangle CDFE in cm?

A. 12
B. 14
C. 20
D. 22
E. 26

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Re: The lengths of the sides of rectangles ABCD and ABEF, shown above, ar [#permalink] New post   19 Jun 2020, 06:11
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Area of ABCD = 20 = (x*y)
Area of ABEF = 10 = (x*z)
Where x is the side both rectangles share.
Therefore,
20/y = 10/z
Z/y = 1/2
Z:Y = 1:2
The value of Z and Y can only be factor of 10 & 20 that is 2,4,5,1,10&20 I.e (1,2),(2,4),(5,10),(10,20)


The minimum perimeter is when the sides of an quadrilateral are equal. And since its a rectangle, the sides has to be almost equal.

=> If (Y,Z) = (2,1) , X = 10 (Not possible)
=> If (Y,Z) = (4,2) , X = 5 (Possible)
=> If (Y,Z) = (10,5) , X = 2 (Not possible)
=> If (Y,Z) = (20,10) , X = 1 (Not possible)

Therefore the value of X,Y,Z = 5,4,&2
The perimeter= 5*2+4*2+2*2
=> 10+8+4 = 22

IMO D

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Re: The lengths of the sides of rectangles ABCD and ABEF, shown above, ar [#permalink] New post   19 Jun 2020, 07:28
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Bunuel wrote:

The lengths of the sides of rectangles ABCD and ABEF, shown above, are integers (in cm). The area of ABCD is 20 cm^2 and the area of ABEF is 10 cm^2. What is the minimum possible perimeter of the rectangle CDFE in cm?

A. 12
B. 14
C. 20
D. 22
E. 26

Are You Up For the Challenge: 700 Level Questions


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Attachment:
2020-06-19_1524.png


CONCEPT: Perimeter of a Quadrilateral is minimum when the quadrilateral approaches a Square shape i.e when adjacent sides are as close in value as possible


10 = {1x10} or {2 x 5}
20 = {1x20} or {2 x 10} or {4 x 5}

OBSERVATION: AB is Common Side of both the rectangles



TRICK: Maximize length of AB so that AF+AD can be minimized to bring its value close to AB



i.e. AB = 5, AF=2 and AD = 4

i.e. EF = AB = 5, and FD = 2+4 = 6

Perimeter = 2*(EF+FD) = 2*(5+6) = 22

Answer: Option D

i.e. If AB = 5
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The lengths of the sides of rectangles ABCD and ABEF, shown above, ar [#permalink] New post   19 Jun 2020, 10:28
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Bunuel wrote:

The lengths of the sides of rectangles ABCD and ABEF, shown above, are integers (in cm). The area of ABCD is 20 cm^2 and the area of ABEF is 10 cm^2. What is the minimum possible perimeter of the rectangle CDFE in cm?

A. 12
B. 14
C. 20
D. 22
E. 26

Are You Up For the Challenge: 700 Level Questions


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Attachment:
2020-06-19_1524.png

For a given area, Square has a minimum perimeter
Here Total area=30
side= \(\sqrt{30}\)
=5.5
Perimeter= 5.5*4
=22
D:)
nick1816, Bunuel please correct me if you find any anomaly here
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nick1816
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Re: The lengths of the sides of rectangles ABCD and ABEF, shown above, ar [#permalink] New post   19 Jun 2020, 11:26
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Bhai your approach is perfect. You just missed the word 'integers'(highlighted below) in the question. Length of the sides of rectangles ABCD and ABEF are integers, so, length of sides of rectangle CDFE must be integer.

Area of rectangle CDFE is 30. Since we have to minimize its Perimeter, choose the length and breadth as close as possible .

30=6*5

minm possible perimeter = 2(6+5) = 22.



satya2029 wrote:
Bunuel wrote:

The lengths of the sides of rectangles ABCD and ABEF, shown above, are integers (in cm). The area of ABCD is 20 cm^2 and the area of ABEF is 10 cm^2. What is the minimum possible perimeter of the rectangle CDFE in cm?

A. 12
B. 14
C. 20
D. 22
E. 26

Are You Up For the Challenge: 700 Level Questions


Show Spoiler
Attachment:
2020-06-19_1524.png

For a given area, Square has a minimum perimeter
Here Total area=30
side= \(\sqrt{30}\)
=5.5
Perimeter= 5.5*4
=22
D:)
nick1816, Bunuel please correct me if you find any anomaly here
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agarwalankur
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Re: The lengths of the sides of rectangles ABCD and ABEF, shown above, ar [#permalink] New post   24 Jun 2020, 10:57
Total area = 20+10= 30 cm^2
Length*Breadth of the whole rectangle=EF*EC = 30 cm^2

we are asked to find least value of of 2(EF+EC)

We know, Arithmetic Mean>= Geometric Mean

=> (EF+EC)/2 >= Square root of EF*EC

=> 2(EF+EC) >= 4*square root of 30.

Perimeter will be least when it is equal to 4*square root of 30.

Square root of 30 is greater than 5 and less than 6. Therefore answer will be between 20 and 24. We have one option. Mark D
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Re: The lengths of the sides of rectangles ABCD and ABEF, shown above, ar [#permalink]
24 Jun 2020, 10:57
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