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# What is the area of a rectangle whose length is twice its width and wh

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Joined: 02 Sep 2009
Posts: 51035
What is the area of a rectangle whose length is twice its width and wh  [#permalink]

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19 Jul 2018, 23:03
00:00

Difficulty:

35% (medium)

Question Stats:

63% (01:10) correct 38% (01:40) wrong based on 24 sessions

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What is the area of a rectangle whose length is twice its width and whose perimeter is equal to that of a square whose area is 1?

A. 1
B. 6
C. 2/3
D. 4/3
E. 8/9

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Re: What is the area of a rectangle whose length is twice its width and wh  [#permalink]

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19 Jul 2018, 23:16
1
Let l be the length of the rectangle and w be the width of the rectangle

Give l = 2w

Perimeter of this rectangle = 2w + w + 2w + w = 6w

Perimeter is equal to the perimeter of a square whose area is 1

side of a square whose area is 1 = $$\sqrt{1}$$ = 1

perimeter of a square whose side is 1 = $$4 * 1$$ = 4

=> Perimeter of this rectangle = 6w = 4

=> w = $$\frac{2}{3}$$

=> l = $$2 * \frac{2}{3} = \frac{4}{3}$$

Area of the rectangle = l * w = $$\frac{4}{3} * \frac{2}{3} = \frac{8}{9}$$

Hence option E
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Re: What is the area of a rectangle whose length is twice its width and wh  [#permalink]

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20 Jul 2018, 00:11
+1 for E.

Let,
Length = L, Width = W, Area = A, Perimeter = P
L = 2W
A = L*W = 2W*W = 2W^2

P = 2L + 2W
P = 6W

Perimeter of of a square whose area is 1
= S^2 = 1 (S= Side of a square)
= S = 1
Then, P = 4*S = 4

Now,
4 = 6W
W = 4/6
A = 2*W^2 = 2*(4/6)^2
A = 2*(16/36)
A = 8/9

Hence, E.
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Re: What is the area of a rectangle whose length is twice its width and wh &nbs [#permalink] 20 Jul 2018, 00:11
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