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Intern Joined: 08 Jun 2018
Posts: 7
Followers: 0

Kudos [?]: 3 , given: 2

The figure above shows a rectangle inscribed within a square [#permalink] 00:00

Question Stats: 100% (02:06) correct 0% (00:00) wrong based on 2 sessions
Hi, I have uploaded the question as an attachment. I'd really appreciate if someone could tell me how to solve the question

Thanks

Attachment: square.jpg [ 7.64 KiB | Viewed 572 times ]

The figure above shows a rectangle inscribed within a square. Now many times greater is the perimeter of the square than the perimeter of the Inscribed rectangle?

A. $$\sqrt{2}$$

B. $$\frac{2 + \sqrt{2}}{2}$$

C. $$2$$

D. $$2 \sqrt{2}$$

E. it cannot be determined from the information given.

Last edited by Carcass on 10 Jun 2018, 11:31, edited 1 time in total.
Formatted GRE Forum Moderator
Joined: 29 May 2018
Posts: 125
Followers: 0

Kudos [?]: 85  , given: 4

Re: The figure above shows a rectangle inscribed within a square [#permalink]
3
KUDOS
Kahani98 wrote:
Hi, I have uploaded the question as an attachment. I'd really appreciate if someone could tell me how to solve the question

Thanks

Attachment:
The attachment square.jpg is no longer available

The figure above shows a rectangle inscribed within a square. Now many times greater is the perimeter of the square than the perimeter of the Inscribed rectangle?

A. $$\sqrt{2}$$

B. $$\frac{2 + \sqrt{2}}{2}$$

C. $$2$$

D. $$2 \sqrt{2}$$

E. it cannot be determined from the information given.

Refer diagram.

We know that perimeter of a square is 4a and perimeter of a rectangle is 2(l+b).

Now let side of a square a = x+y.

Let c and d be sides of the rectangle.

To get value of c , we have a right triangle value of c = $$\sqrt{2}$$x.

Similarly value of d will be = $$\sqrt{2}$$y.

Now total perimeter of a rectangle will be 2( $$\sqrt{2}$$ x + $$\sqrt{2}$$ y ) .
=> 2($$\sqrt{2}$$(x+y)...

Now perimeter of square is 4(x+y).

Now question asks perimeter of sqaure what times greater than perimeter of rectangle.

4(x+y) = a times * 2($$\sqrt{2}$$)(x+y).

Cancel (x+y) on both sides.

4 = a times * 2($$\sqrt{2}$$) is possible when a is ($$\sqrt{2}$$)

So answer is A.

Hope it clears.
Attachments Geo1.PNG [ 23.18 KiB | Viewed 555 times ] Re: The figure above shows a rectangle inscribed within a square   [#permalink] 11 Jun 2018, 03:33
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