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help

This topic has 8 expert replies and 4 member replies
sana.noor Legendary Member Default Avatar
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help

Post Fri Aug 23, 2013 11:12 pm
In the figure Shown,two identical squares are inscibed in the rectangle.if the area of the rectangle is 36 Sq. Units,then what is the perimeter of each Square??



i dont have its answer

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Post Fri Jul 07, 2017 3:27 pm
The squares just take up half the rectangle. There are a few ways to see that. For example, if you divide up the picture into a grid of 8 squares, as I did below, you can see that half of each grid zone is taken up by part of a square, and the other half is taken up by non-square.

So the area of the squares is half the total area of the rectangle, so the area of the two squares is 36/2 = 18, and the area of each square is 9. So the squares measure 3 by 3, and the perimeter of each is 12.
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Post Tue Jul 04, 2017 11:57 pm
hotcool030 wrote:
I solved it in a very easy way.
Lets take side of square is x. You can see from figure, two diagonals of squares = length of rectangle.
And one diagonal of square = width of rectangle.
So, as Length x Width = 36,
we can say (2 * root2x)* (root2x) = 36
x = 3
Perimeter of square = 12
Smile
Nice approach! For anyone needing a visual, this is the same approach the great Brent uses above.

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Post Sat Aug 24, 2013 12:21 am
Hi sana.noor,

In the drawing that you provided, you should notice that:

the width of the rectangle = 2(square's diagonal)
the height of the rectangle = 1(square's diagonal)

Using square "rules", the diagonal = x(root2)

So the area of the rectangle = (width)(height) = 2(x(root2))^2 = 36

Let's simplify:
(x(root2))^2 = 18
x^2(2) = 18
x^2 = 9
x = 3

So the perimeter of each square = 3(4) = 12

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Post Sat Aug 24, 2013 5:36 am
sana.noor wrote:


In the figure shown, two identical squares are inscribed in the rectangle. If the area of the rectangle is 36, what is the perimeter of each square?

Here's a slightly different approach:
Since the height of the rectangle and the diagonal of a square are the same length, let's let x = height of rectangle


Since the width of the rectangle is equal to the length of two square diagonals, then the width of the rectangle = 2x.


The area of the rectangle is 36.
So, (base)(height) = 36
(2x)(x) = 36
2x²= 36
x²= 18
x = √18

NOTE: There's no need to simplify √18 at this point (you'll see why shortly)

If the height of the rectangle is √18, then the length of the red line (shown below) must equal √18/(2)


Likewise, the other red line has length √18/(2)


If we let y = the length of the hypotenuse, then the Pythagorean Theorem states that...


Now solve this equation for y.


If y = 3, then the perimeter of one square = (4)(3) = 12

Cheers,
Brent

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happy888 Newbie | Next Rank: 10 Posts Default Avatar
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Post Sun Jan 10, 2016 3:07 pm
Hi Brent,

I didn't understand why 1/4th length of the rectangle is squareroot18/2

Thanks

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Post Sun Jan 10, 2016 8:33 pm
Hi happy888,

The 'key' to this whole question is in realizing that the width and height of the rectangle can be expressed in terms of the DIAGONAL of each of the SQUARES. Are you able to follow the 'set-up' of the calculations involved?

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Post Sun Jan 10, 2016 9:38 pm
happy888 wrote:
Hi Brent,

I didn't understand why 1/4th length of the rectangle is squareroot18/2

Thanks
Once we know that the height of the rectangle is √18, then HALF of that is (√18)/2


I hope that helps.

Cheers,
Brent

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happy888 Newbie | Next Rank: 10 Posts Default Avatar
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Post Wed Jan 13, 2016 4:40 am
Hi,

Got it

Thanks Smile

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hotcool030 Newbie | Next Rank: 10 Posts Default Avatar
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Post Sun Jul 02, 2017 4:34 am
I solved it in a very easy way.
Lets take side of square is x. You can see from figure, two diagonals of squares = length of rectangle.
And one diagonal of square = width of rectangle.
So, as Length x Width = 36,
we can say (2 * root2x)* (root2x) = 36
x = 3
Perimeter of square = 12
Smile

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Post Fri Aug 18, 2017 1:22 pm
Neat follow up question for anyone who wants to delve further into similar ideas:

http://www.beatthegmat.com/triangle-inside-a-rectangle-t16268.html

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hopperyelena Newbie | Next Rank: 10 Posts Default Avatar
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Post Fri Nov 24, 2017 4:52 am
thanks..

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Post Thu Dec 07, 2017 3:57 pm
No prob!

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