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What is the area of a rectangle whose length is twice its wi [#permalink]
Expert's post 00:00

Question Stats: 71% (01:56) correct 28% (01:27) wrong based on 14 sessions
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) $$\frac{2}{3}$$

D) $$\frac{4}{3}$$

E) $$\frac{8}{9}$$
[Reveal] Spoiler: OA

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Founder  Joined: 18 Apr 2015
Posts: 8485
Followers: 166

Kudos [?]: 1843 , given: 7861

Re: What is the area of a rectangle whose length is twice its wi [#permalink]
Expert's post
The key information here from which is good to start is that

whose perimeter is equal to that of a square whose area is 1

PR (perimeter rectangle) = PS (perimeter square)

The area of the square is 1. Now to have a square area = 1, one side of the square must be 1 as well. Area square is $$1^2=1$$

So its perimeter is $$1 \times 4 = 4$$; $$PS = 4$$

The area of a rectangle is $$PR = \ell \times w$$

In our case, the length is twice its width; $$PR = \ell + 2w + 2we + \ell = 2\ell + 4w$$

\ell = w

we do have that PR = 6w

$$PR=PS >>>>>>> 4 = 6w >>> w=\frac{4}{6} = \frac{2}{3}$$

$$2w = \frac{4}{3}$$

Area rectangle $$\ell \times w = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}$$
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Re: What is the area of a rectangle whose length is twice its wi [#permalink]
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KUDOS
Carcass wrote:
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) $$\frac{2}{3}$$

D) $$\frac{4}{3}$$

E) $$\frac{8}{9}$$

Let length be l and breadth be b
Given that L=2B
For Square; side =1
so, its perimeter will be 4
Thus, 2(L+B)=4; which simplifies to B=2/3 (After plugging in L=2B)
Which implies; => L=4/3
Thus Area of rectangle=LxB => 8/9 Re: What is the area of a rectangle whose length is twice its wi   [#permalink] 05 Sep 2019, 08:38
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