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# In the figure above, each square is contained within the area of the l

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In the figure above, each square is contained within the area of the l [#permalink]

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15 Jul 2015, 03:25
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In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices?

A. 20
B. 30
C. 40
D. 50
E. 60

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:

squares.gif [ 2.08 KiB | Viewed 3381 times ]
[Reveal] Spoiler: OA

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Re: In the figure above, each square is contained within the area of the l [#permalink]

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15 Jul 2015, 03:47
1
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Let Area of larger square be x

Therefore side of smaller square is x/_/2

Given:

(X^2 / 2)* (1/4) = 5/2

X^2 = 20

Which is area of larger Square

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Re: In the figure above, each square is contained within the area of the l [#permalink]

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15 Jul 2015, 04:59
1
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Bunuel wrote:

In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices?

A. 20
B. 30
C. 40
D. 50
E. 60

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
squares.gif

Solution -

Lets say, area of the shaded square A.
A/4 = 5/2
A = 4 × (5/2)
A = 10

So, the area of the shaded square is equal to 10, and each side is equal to √10.

Diagonal of shaded square is side of a largest square = √(side^2+side^2) = √((√10)^2+(√10)^2) = √20

Area of largest square = √20 * √20 = 20. ANS A
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In the figure above, each square is contained within the area of the l [#permalink]

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15 Jul 2015, 05:51
Expert's post
1
This post was
BOOKMARKED
Bunuel wrote:

In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices?

A. 20
B. 30
C. 40
D. 50
E. 60

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
squares.gif

Area of Smallest Square = 2^2 = 4

area of the shaded square / the area of the smallest square = 5/2
i.e. area of the shaded square = (5/2)*4 = 10

Since the Shaded Square is bisecting the sides of Largest square therefore the Area of Largest Square will be twice the area of Shaded Square
Area of Shaded Square = (1/2) Area of Biggest Square

i.e. Area of Biggest Square = 2 * (10) = 20
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Re: In the figure above, each square is contained within the area of the l [#permalink]

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15 Jul 2015, 07:22
3
KUDOS
let side length of larger square be x
so, side length of shaded square is (x/2)*√2 or x/√2

now, (x/√2)^2 : 2^2 = 5 : 2
or, x^2 = 4*5 = 20

area of larger square is 20
ans: A
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Re: In the figure above, each square is contained within the area of the l [#permalink]

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15 Jul 2015, 22:44
1
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Area of shaded region = 5/2 *4 = 10
So side =√10

Therefore diagonal of shaded square =√20 which is the side of larger square so the area of larger square = (√20)^2 = 20

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Re: In the figure above, each square is contained within the area of the l [#permalink]

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17 Jul 2015, 16:58
Bunuel wrote:

In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices?

A. 20
B. 30
C. 40
D. 50
E. 60

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
squares.gif

Area of the Smallest Sq = 4

Ratio of Shaded to Smallest sq = 5/2 = x/4 =>Area of shaded sq. x = 10.
If Area = 10, Side =$$\sqrt{10}$$
Draw a diagonal to the Shaded square, x.

x^2 = 10 + 10 = 20 => Area of the big square (Side ^ 2)

or in other way,
Since the shaded sq. bisects the vertices of big sq., the area of the big square will be twice the shaded sq. = 2 * 10 = 20.

Option A

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Re: In the figure above, each square is contained within the area of the l [#permalink]

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19 Jul 2015, 13:09
Expert's post
1
This post was
BOOKMARKED
Bunuel wrote:

In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices?

A. 20
B. 30
C. 40
D. 50
E. 60

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
squares.gif

800score Official Solution:

The area of the smallest square must be equal to 2^2 = 4. Since the ratio of the area of the shaded square (it consists of the shaded region and the smallest square) to the area of the smallest square is 5/2, we can solve for the area of the shaded square by solving for A in the following equation:
A/4 = 5/2
A = 4 × (5/2)
A = 10

So, the area of the shaded square is equal to 10, and each side is equal to √10. Now, all we have to do is recognize that the triangles made at each corner of the large square where, two sides are equal to half the side length and the hypotenuse is equal to the edge length of the shaded square, are right triangles and to use the Pythagorean Theorem to solve for half the edge length. Let each side of the triangle be equal to x, then 2x^2 = 10, so x^2 = 5, and x = √5. This means that the entire side length of the larger square is twice the length of a side of that triangle, or 2√5. Finally, the area of a square is equal to a side squared, so the answer must be (2√5 )^2 = 20, or answer choice (A).
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In the figure above, each square is contained within the area of the l [#permalink]

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22 Jul 2015, 11:26
Bunuel wrote:
Bunuel wrote:

In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices?

A. 20
B. 30
C. 40
D. 50
E. 60

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
squares.gif

800score Official Solution:

The area of the smallest square must be equal to 2^2 = 4. Since the ratio of the area of the shaded square (it consists of the shaded region and the smallest square) to the area of the smallest square is 5/2, we can solve for the area of the shaded square by solving for A in the following equation:
A/4 = 5/2
A = 4 × (5/2)
A = 10

So, the area of the shaded square is equal to 10, and each side is equal to √10. Now, all we have to do is recognize that the triangles made at each corner of the large square where, two sides are equal to half the side length and the hypotenuse is equal to the edge length of the shaded square, are right triangles and to use the Pythagorean Theorem to solve for half the edge length. Let each side of the triangle be equal to x, then 2x^2 = 10, so x^2 = 5, and x = √5. This means that the entire side length of the larger square is twice the length of a side of that triangle, or 2√5. Finally, the area of a square is equal to a side squared, so the answer must be (2√5 )^2 = 20, or answer choice (A).

Hello Bunuel ,
Can i not take as area of shaded region + area of small square = 10+4 = 14
so area of square (shaded region + small square)=14
so its side is root(14)
since this particular square bisects the larger square
So area of larger square =2(area of this square)= 2*14=28
Is this not correct ?
problem is i could not understand how area of shaded region can be interpreted to get the side

Thanks

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Re: In the figure above, each square is contained within the area of the l [#permalink]

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30 Oct 2015, 07:50
Hi
I have the Same Problem with this one. Since the area of These shaded Square is 10 you have to add the small Square to get the whole area. Thats 14 then and one side is the suare Root of 14.
Why is that wrong?

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Re: In the figure above, each square is contained within the area of the l [#permalink]

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30 Oct 2015, 08:09
Bunuel wrote:

In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices?

A. 20
B. 30
C. 40
D. 50
E. 60

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
squares.gif

The area of the small square is 4. If the area of the shaded square has a a ratio of 5/2 that of the small square, then its area is 10.
We can therefore determine each side of the shaded square has a length of \sqrt{10}
If each side of the shaded squre bisects the side of the large square, the triangle created by side of the shaded square and 1/2 each side of the large square is a 45:45:90 suqare with the shaded square's side as the hypotenuse. The proportion of the sides of a 45:45:90 triangle are 1:1:\sqrt{2} and, in this case the hypotenuse is \sqrt{10}, so the sides of the large square are each 2\sqrt{5}. The area, therefore, is 2\sqrt{5} squared, or 20.

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Re: In the figure above, each square is contained within the area of the l [#permalink]

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Re: In the figure above, each square is contained within the area of the l [#permalink]

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15 Sep 2017, 08:58
Bunuel wrote:

In the figure above, each square is contained within the area of the larger square. The smallest square has a side length of 2 and the ratio of the area of the shaded square to the area of the smallest square is 5/2, what is the area of the largest square if the shaded square bisects each side of the largest square at its vertices?

A. 20
B. 30
C. 40
D. 50
E. 60

Kudos for a correct solution.

[Reveal] Spoiler:
Attachment:
squares.gif

this is gonna be tested in gmat?

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Re: In the figure above, each square is contained within the area of the l   [#permalink] 15 Sep 2017, 08:58
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