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Square PQRS is inscribed in the square ABCD whose perimeter is four. What is the area of the shaded region:

A: \(\frac{1}{12}\)

B: \(\sqrt{2}/8\)

C: \(\frac{1}{16}\)

D: \(\frac{1}{8}\)

E: \(\sqrt{2}/16\)

There are various ways of approaching it.

Method 1: Area of ABCD is 1. Area of PQRS = (1/2)*(diagonal1)*(diagonal2) Both diagonals of PQRS are 1 each since they are the same lengths as sides of ABCD Area of PQRS = (1/2)*1*1 = 1/2

Area of leftover region = 1 - 1/2 = 1/2 The leftover region after you cut out PQRS is split into 8 equal areas and the red region is one of those 8. Hence area of red region is (1/8)*(1/2) = 1/16

Method 2:

The perimeter of ABCD is 4 so each side is 1. So each half side is 1/2. In triangle APQ, AP and AQ are 1/2 each so \(PQ = 1/\sqrt{2}\) (using Pythegorean theorem) So half of PQ is \(1/2\sqrt{2}\)

Area of red triangle = \((1/2)*(1/2\sqrt{2})*(1/2\sqrt{2}) = 1/16\)
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Re: Square PQRS is inscribed in the square ABCD whose [#permalink]

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30 Jul 2014, 17:53

himanshujovi wrote:

how do we determine that P,Q,R and S are mid-points of the sides of ABCD ?

Refer diagram below:

Draw a square Join both diagonals of the square to get the midpoint (CG) From that midpoint, you can draw a circle with diameter same as that of the side of the square The circle will get inscribed in the square, the point of contact of that would be the midpoints of the side of the square

how do we determine that P,Q,R and S are mid-points of the sides of ABCD ?

A square is a symmetrical figure. All its sides and all angles are equal. When you inscribe another symmetrical figure in it, the final figure will also be symmetrical! Is there any reason why P should be closer to A than to D? Since all points are equivalent, P will be in the middle of A and D. You can use the same argument for all points.
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Re: Square PQRS is inscribed in the square ABCD whose [#permalink]

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30 Jul 2014, 22:32

VeritasPrepKarishma wrote:

himanshujovi wrote:

how do we determine that P,Q,R and S are mid-points of the sides of ABCD ?

A square is a symmetrical figure. All its sides and all angles are equal. When you inscribe another symmetrical figure in it, the final figure will also be symmetrical! Is there any reason why P should be closer to A than to D? Since all points are equivalent, P will be in the middle of A and D. You can use the same argument for all points.

Ok.. At the back of my mind ,I could figure out that the inherent symmetry of the figures in question would have play over here but was not able to recollect any rule or axiom from my study of geometry

Re: Square PQRS is inscribed in the square ABCD whose [#permalink]

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30 Jul 2014, 22:34

PareshGmat wrote:

himanshujovi wrote:

how do we determine that P,Q,R and S are mid-points of the sides of ABCD ?

Refer diagram below:

Draw a square Join both diagonals of the square to get the midpoint (CG) From that midpoint, you can draw a circle with diameter same as that of the side of the square The circle will get inscribed in the square, the point of contact of that would be the midpoints of the side of the square

Hope this clarifies

Sorry but not able to get this logic. karishma's comment is much more intuitive and I could feel it but could not take it as a necessary pre-requisite in this question

Re: Square PQRS is inscribed in the square ABCD whose [#permalink]

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30 Jul 2014, 23:18

himanshujovi wrote:

PareshGmat wrote:

himanshujovi wrote:

how do we determine that P,Q,R and S are mid-points of the sides of ABCD ?

Refer diagram below:

Draw a square Join both diagonals of the square to get the midpoint (CG) From that midpoint, you can draw a circle with diameter same as that of the side of the square The circle will get inscribed in the square, the point of contact of that would be the midpoints of the side of the square

Hope this clarifies

Sorry but not able to get this logic. karishma's comment is much more intuitive and I could feel it but could not take it as a necessary pre-requisite in this question

Never mind... What I meant to say is if you rotate the inscribed square, it will follow the circle & the circle will TOUCH the outer square only on its midpoints

In this way PQRS are the respective midpoints
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Re: Square PQRS is inscribed in the square ABCD whose [#permalink]

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07 Jan 2016, 09:43

VeritasPrepKarishma wrote:

himanshujovi wrote:

how do we determine that P,Q,R and S are mid-points of the sides of ABCD ?

A square is a symmetrical figure. All its sides and all angles are equal. When you inscribe another symmetrical figure in it, the final figure will also be symmetrical! Is there any reason why P should be closer to A than to D? Since all points are equivalent, P will be in the middle of A and D. You can use the same argument for all points.

Unless stated in the question (or maybe I'm misinterpreting the meaning of 'inscribed'), there is actually no reason that the inscribed square should touch on the midpoints of the outer square. Consider the figure below:

The inner square only touches the outer square at 4 points, but they are clearly not the midpoints of the sides of the outer square. In this case, we would not be able to figure out the area of the shaded region without additional information (like the ratio of AP to PD). Likewise, for the problem in question, unless we are given more information to know that the inscribed square touches the outer square at its midpoints, then we cannot solve the problem without making assumptions, something we should rarely ever do on the GMAT.

What is the source of the question? It's certainly not an official GMAT problem...
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Re: Square PQRS is inscribed in the square ABCD whose [#permalink]

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