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19 Jun 2020, 04:26
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
80% (02:28) correct
20%
(02:06)
wrong
based on 51
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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
Are You Up For the Challenge: 700 Level Questions
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yashikaaggarwal
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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
Posted from my mobile device
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
Posted from my mobile device
19 Jun 2020, 07:28
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Bunuel
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
