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The area of a rectangle and the square of its perimeter are in the

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Joined: 01 Nov 2017
Posts: 67
Location: India
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The area of a rectangle and the square of its perimeter are in the  [#permalink]

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New post 26 Feb 2019, 23:40
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Question Stats:

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The area of a rectangle and the square of its perimeter are in the ratio 1 : 25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio

(A) 1:4
(B) 2:9
(C) 1:3
(D) 3:8
(E) 4:9
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V
Joined: 02 Aug 2009
Posts: 8023
The area of a rectangle and the square of its perimeter are in the  [#permalink]

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New post 27 Feb 2019, 05:32
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raghavrf wrote:
The area of a rectangle and the square of its perimeter are in the ratio 1 : 25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio

(A) 1:4
(B) 2:9
(C) 1:3
(D) 3:8
(E) 4:9


PROPER METHOD
Let the length and breadth be x and y..
so Area = xy and perimeter = 2(x+y)..
Area: square of perimeter = 1:25 => \(xy:(2(x+y))^2=1:25..........4(x+y)^2=25xy.......\frac{x^2+y^2+2xy}{xy}=\frac{25}{4}\)..
\(\frac{x^2}{xy}+\frac{y^2}{xy}+\frac{2xy}{xy}=\frac{25}{4}\)..
\(\frac{x}{y}+\frac{y}{x}+2=\frac{25}{4}.............\)..
Let the ratio we are looking for x/y be a
\(a+\frac{1}{a}=\frac{25}{4}-\frac{17}{4}..........\frac{a^2+1}{a}=\frac{17}{4}..........4a^2+4=17a..........4a^2-17a+4=0\)..
\(4a^2-16a-a+4=0...........4a(a-4)-1(a-4)=0......(4a-1)(a-4)=0\)..
so either a=4 or a=\(\frac{1}{4}\).....

thus A

SHORTCUT
Or you can just SUBSTITUTE the choices to check the answer..
A 1:4, so length =1 and breadth = 4..
Area = 1*4=4 Perimeter = 2(1+4)=10..
Thus \(\frac{A}{P}=\frac{4}{10^2}=\frac{1}{25}\)... TRUE

A
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The area of a rectangle and the square of its perimeter are in the   [#permalink] 27 Feb 2019, 05:32
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