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.

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