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In the square above, if the ten nonoverlapping regions have the areas

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In the square above, if the ten nonoverlapping regions have the areas [#permalink]

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In the square above, if the ten nonoverlapping regions have the areas as shown, which of the following is equal to y?

(A) x + t
(B) s + t
(C) a + s + t
(D) b + s + t
(E) a + b + s

[Reveal] Spoiler:
Attachment:
2017-09-26_1116.png
2017-09-26_1116.png [ 6.87 KiB | Viewed 413 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 132588 [1], given: 12326

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Re: In the square above, if the ten nonoverlapping regions have the areas [#permalink]

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New post 26 Sep 2017, 01:58
Bunuel wrote:
Image
In the square above, if the ten nonoverlapping regions have the areas as shown, which of the following is equal to y?

(A) x + t
(B) s + t
(C) a + s + t
(D) b + s + t
(E) a + b + s

[Reveal] Spoiler:
Attachment:
2017-09-26_1116.png


In the given figure, the square with area x^2 has length and width x each. So, for the rectangle with area xy on the left, the width is x and area is xy so length is y. Similarly, for the rectangle on the top with area xy, the length is x and area is xy so width is y. Therefore, the length/width of the square = x + y.
Area of the square = \((x+y)^2 = x^2 + 2xy + y^2\)
From the given figure, the area of the square is equal to the sum of the areas of the ten regions inside it.
So, \(x^2 + 2xy + y^2 = x^2 + 2xy + a^2 + 2ab + s^2 + st + st + t^2\)
On simplifying it, \(y^2 = a^2 + 2ab + (s+t)^2\)
\(y^2 = (a+b)^2 - b^2 + (s+t)^2\)--------- equation 1


Similar to how the outside square's length was found out, we can see in the inside figure has length of side a+b.
Its area = \((a+b)^2\)= sum of the areas of the constituting regions inside
So, \((a+b)^2 = a^2 + 2ab + s^2 + 2st + t^2\)
On simplifying it, \(b^2 = (s+t)^2\)----------- equation 2
Using this in equation 1,
\(y^2 = (a+b)^2\)
y = a+b (since they are non-negative)
Since b = s+t from equation 2 (since they are non-negative), y = a+b = a + s + t

Therefore, the answer is C.

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Re: In the square above, if the ten nonoverlapping regions have the areas [#permalink]

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New post 26 Sep 2017, 03:24
Attachment:
2017-09-26_1116.png
2017-09-26_1116.png [ 8.33 KiB | Viewed 291 times ]


From Square 1, we know that side = x
From Rectangle 2, we know length = y as breadth is the same as the side of the square(x)
From Square 3, we know that side = a
From Rectangle 4, we know length = b as breadth is the same as the side of the square(a)
From Square 5, we know that side = s
From Rectangle 2, we know length = t as breadth is the same as the side of the square(s)

Taking Rectangle 2, Square 3, and Rectangle 4,
we know that the length of Rectangle 2 is equal to sum of side of Square 3 and length of Rectangle 4
y = a + b -> (1)

Similarly taking Rectangle 4, Square 5, and Rectangle 6,
we know that the length of Rectangle 4 is equal to sum of side of Square 5 and length of Rectangle 6
b = s + t -> (2)

Substituting (2) in equation (1), we get the value of y = a + s + t(Option C)
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Kudos [?]: 670 [0], given: 17

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Kudos [?]: 336 [0], given: 591

In the square above, if the ten nonoverlapping regions have the areas [#permalink]

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New post 26 Sep 2017, 16:02
Bunuel wrote:
Image
In the square above, if the ten nonoverlapping regions have the areas as shown, which of the following is equal to y?

(A) x + t
(B) s + t
(C) a + s + t
(D) b + s + t
(E) a + b + s

[Reveal] Spoiler:
Attachment:
The attachment 2017-09-26_1116.png is no longer available

Attachment:
2017-09-26_1116ed.png
2017-09-26_1116ed.png [ 30.96 KiB | Viewed 238 times ]


First of all, I wanna know how people do that much algebra in under a minute. (That time is an average).

I used no algebra equations.

I just sketched the figure quickly and wrote in the names of the sides from their areas.

I started with lower right hand corner \(t^2\), and worked northwest (up and left).

I used the same method and direction for the rectangle sides.

I used capital letters on the diagram that are identical to lower-case counterparts.

FIRST: all the squares, e.g. \(t^2\) = (T * T)

So T and T, then S and S, then A and A, then X and X

SECOND: the rectangle sides, starting again from lower right hand corner

If the area is (AB), and one side is A, the other is B

T and T again, then B and B, then Y and Y

When finished, I took a look at Y and the answer choices, and saw which letters fit.

Y equals? A + S + T

Answer C

Kudos [?]: 336 [0], given: 591

In the square above, if the ten nonoverlapping regions have the areas   [#permalink] 26 Sep 2017, 16:02
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In the square above, if the ten nonoverlapping regions have the areas

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