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The combined area of the two black squares is equal to 1000 [#permalink]

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07 Feb 2012, 17:34

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The combined area of the two black squares is equal to 1000 square units. A side of the larger black square is 8 units longer than a side of the smaller black square. What is the combined area of the two white rectangles in square units?

Side of smaller square be x Side of larger square will be x+8

Area of smaller square \(x^2\) Area of larger square \((x+8)^2\)

So the sum of the areas = 1000 which brings me to the equation \(x^2\)+8x-468 . I don't think we an simplify this equation any further, well I can't and therefore need help.

The combined area of the two black squares is equal to 1000 square units. A side of the larger black square is 8 units longer than a side of the smaller black square. What is the combined area of the two white rectangles in square units?

(A) 928 (B) 936 (C) 948 (D) 968 (E) 972

I guess the black squares are the red squares on the diagram.

The length of a smaller square - x; The length of a larger square - x+8;

The area of entire square - x^2+(x+8)^2=1000 --> 2x^2+16x=936; The combined area of the two white rectangles - x(x+8)+x(x+8)=2x^2+16x --> look up: 2x^2+16x=936.

Re: The combined area of the two black squares is equal to 1000 [#permalink]

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27 May 2013, 23:21

enigma123 wrote:

Attachment:

Square.PNG

The combined area of the two black squares is equal to 1000 square units. A side of the larger black square is 8 units longer than a side of the smaller black square. What is the combined area of the two white rectangles in square units?

(A) 928 (B) 936 (C) 948 (D) 968 (E) 972

Let

Side of smaller square be x Side of larger square will be x+8

Area of smaller square \(x^2\) Area of larger square \((x+8)^2\)

So the sum of the areas = 1000 which brings me to the equation \(x^2\)+8x-468 . I don't think we an simplify this equation any further, well I can't and therefore need help.

X^2 +26x- 18x-468=0 ie X=18 and X= -26.

Discard negative value thus X=18.

Hope it helps...
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Do not forget to hit the Kudos button on your left if you find my post helpful.

Re: The combined area of the two black squares is equal to 1000 [#permalink]

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29 May 2013, 09:25

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enigma123 wrote:

Attachment:

Square.PNG

The combined area of the two black squares is equal to 1000 square units. A side of the larger black square is 8 units longer than a side of the smaller black square. What is the combined area of the two white rectangles in square units?

Side of smaller square be x Side of larger square will be x+8

Area of smaller square \(x^2\) Area of larger square \((x+8)^2\)

So the sum of the areas = 1000 which brings me to the equation \(x^2\)+8x-468 . I don't think we an simplify this equation any further, well I can't and therefore need help.

Let smaller square be of side x then bigger square has a side of x+8 (x)^2 +(x+8)^2=1000 2x^2+8x-468=1000 (x+26)(x-18)=0 x=18 Now,the entire figure is a square of side 18+26=44 Total are=44^2=1936 White are=total area-coloured area=1936-1000=936

Re: The combined area of the two black squares is equal to 1000 [#permalink]

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29 May 2013, 21:32

sum of area of black squares(1000) + sum of area of white rectangles(X) = area of the larger square (Y)

From the options, X is an integer, hence Y (= X + 1000) is an integer too.

As Y is a square number, From the properties of square numbers, Y can only end with 0,1,4,6,9 or 25.

From the given options, only B (ending 6) is a possible solution, as +1000 doesn't change unit digit value. Number ending 2 or 8 + 1000 (as in all other options) can never be a square number.

Besides, all of the above ways are equally correct.
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Re: The combined area of the two black squares is equal to 1000 [#permalink]

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28 Oct 2014, 09:04

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Re: The combined area of the two black squares is equal to 1000 [#permalink]

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04 Dec 2015, 21:30

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Re: The combined area of the two black squares is equal to 1000 [#permalink]

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30 Dec 2016, 05:17

Assume the larger red square has a side of length x + 4 units and the smaller red square has a side of length x - 4 units. This satisfies the condition that the side length of the larger square is 8 more than that of the smaller square.

Therefore, the area of the larger square is (x + 4)2 or x^2 + 8x + 16. Likewise, the area of the smaller square is (x - 4)^2 or x^2 - 8x + 16. Set up the following equation to represent the combined area:

It is possible, but not necessary, to solve for the variable x here.

The two white rectangles, which are congruent to each other, are each x + 4 units long and x - 4 units high. Therefore, the area of either rectangle is (x + 4)(x - 4), or x^2 - 16. Their combined area is 2(x^2 - 16), or 2x^2 - 32.

Since we know that 2x^2 = 968, the combined area of the two white rectangles is 968 - 32, or 936 square units. The correct answer is B.
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