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Bunuel

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


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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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
Bunuel

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


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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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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