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# The combined area of the two black squares is equal to 1000 square uni

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

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07 Feb 2012, 16:34
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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?

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

[Reveal] Spoiler:
Attachment:

Square.PNG [ 1.52 KiB | Viewed 6579 times ]

Guys this is how I am trying to solve this:

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.
[Reveal] Spoiler: OA

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

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07 Feb 2012, 16:45
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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?

(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.

[Reveal] Spoiler:
Attachment:

Square.PNG [ 1.52 KiB | Viewed 6534 times ]

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

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27 May 2013, 22:21
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.

Hope it helps...
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Re: The combined area of the two black squares is equal to 1000 square uni [#permalink]

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29 May 2013, 08:25
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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?

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

[Reveal] Spoiler:
Guys this is how I am trying to solve this:

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.

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

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29 May 2013, 18:24
Let a be side of big square
b be side of small square
given a*a +b*b =1000
and a=8+b

so (8+b)^2+b^2=1000
on solving b = 18 and a=26

need to find total area of two rectangles given
we get 44^2-1000 = 936

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

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29 May 2013, 20:32
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 square uni [#permalink]

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07 Dec 2015, 04:33
hi Any quick way to solve and get factor for x2+8x-468
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Re: The combined area of the two black squares is equal to 1000 square uni [#permalink]

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30 Dec 2016, 04:17
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:

(x^2 + 8x +16) + (x^2 - 8x +16) = 1000
2x^2 + 32 = 1000
2x^2 = 968

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

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17 Feb 2017, 07:25
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I solved it in under 2 mins using the below approach:
Let X and x be the sides of big and small squares respectively.

X-x=8
X^2 +x^2= 1000
or (X-x)^2 +2Xx=1000
or 64 +2Xx= 1000
or 2Xx = 936

Now. area of the two white squares are Xx +Xx = 2Xx = 936
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Re: The combined area of the two black squares is equal to 1000 square uni [#permalink]

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22 Jan 2018, 15:46
Hi All,

If you don't see the 'elegant' approach to this question, then you can still solve it with some 'brute force' math (and a bunch of multiplication) - it's not 'pretty', but if you're comfortable doing math by hand then you can still get to the correct answer in 2 minutes:

We're told that the sum of two squares is 1000 and that the two numbers (before they are squared) differ by 8. Since the answer choices to the question are integers, it's likely that the two side lengths of the two squares are also integers. Let's list out some perfect squares and get a sense of how big the numbers can get...

20^2 = 400
30^2 = 900

Since the larger square is clearly bigger, it's area will be more than half of the 1000, so it's side length will fall into the range of 20-30. The side length of the smaller triangle will be 8 LESS... Now let's try to narrow things down a bit...

15^2 = 225
25^2 = 625
17^2 = does NOT end in a 5, so there's no way the sum would be 1000, but this example probably isn't too far away from the correct answer...

26^2 = 676
18^2 = 324

This IS the sum that we're looking for, so these MUST be the dimensions of the two squares. By extension, the dimensions of the two white rectangles are 26x18 and the area of each is 468. Since there are two of those rectangles, their total area = 468(2) = 936.

[Reveal] Spoiler:
B

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Re: The combined area of the two black squares is equal to 1000 square uni   [#permalink] 22 Jan 2018, 15:46
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