Width - Shaded Portion of Square / Partial Solution

[Guest]

Another variation of the problem.....
This is the same diagram as before with the cente square unshaded and the 4 corner squares unshaded.
Leaving 4 rectangles or squares shaded. And this fits the equations in your solution better.

This time Let the sides of the unshaded corners = X/2
Then the sides of the centre square is (6 - X/2 - X/2) equals (6 - X)
And the shaded area is still 4/9 of 36 = 16 sq inches.

<x/2><----(6-x)----><x/2>

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This means each shaded rectangle is (X/2) by (6 - X) ....and there are 4 of them..... So the area equation now is .......

4*(X/2)*(6 - X) = 16

(2X)*(6 - X) = 16

12X - 2X^2 = 16

6X -X^2 = 8

-X^2 + 6X - 8 = 0

OR multiply across by -1 gives.....

X^2 -6X + 8 = 0

(x - 4) (X - 2) = 0

gives .... X = 4 OR X = 2

So the size of the unshaded corners is a square of 2 inches OR a square of 1 inch.

And the shaded rectangles are 4 by 1 OR 2 by 2 inches....same as we got before in my first solution of this diagram.

When X = 4 the corners are 2 by 2 squares, and the shaded areas are 2 by 2 squares, and the centre square is 2 by 2 square.

When X = 2 the corners are 1 by 1 squares, and the shaded areas are 4 by 1 squares, and the centre square is 4 by 4 square.


It seems I had to derive the question from the solution, rather than the other way around.
In your solution you found a value for X , but did not know what to do with it because you did not understand the question what it was asking. Therefore you did not get the answer.

In all of my 3 solution I provided a diagram to illustrate the question I was solving. Then when I let X equal an unknown at least I knew what part of the square I was talking about. And when I got a value for X I could then refer to the diagram and find what part it referred to.

[Guest]

Another variation of the problem.....
This is the same diagram as before with the cente square unshaded and the 4 corner squares unshaded.
Leaving 4 rectangles or squares shaded. And this fits the equations in your solution better.

This time Let the sides of the unshaded corners = X/2
Then the sides of the centre square is (6 - X/2 - X/2) equals (6 - X)
And the shaded area is still 4/9 of 36 = 16 sq inches.

<x/2><----(6-x)----><x/2>

....xxxxxxxxxxxxxxxx....
....xxxxxxxxxxxxxxxx....
....xxxxxxxxxxxxxxxx....
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
....xxxxxxxxxxxxxxxx....
....xxxxxxxxxxxxxxxx....
....xxxxxxxxxxxxxxxx....

This means each shaded rectangle is (X/2) by (6 - X) ....and there are 4 of them..... So the area equation now is .......

4*(X/2)*(6 - X) = 16

(2X)*(6 - X) = 16

12X - 2X^2 = 16

6X -X^2 = 8

-X^2 + 6X - 8 = 0

OR multiply across by -1 gives.....

X^2 -6X + 8 = 0

(x - 4) (X - 2) = 0

gives .... X = 4 OR X = 2

So the size of the unshaded corners is a square of 2 inches OR a square of 1 inch.

And the shaded rectangles are 4 by 1 OR 2 by 2 inches....same as we got before in my first solution of this diagram.

When X = 4 the corners are 2 by 2 squares, and the shaded areas are 2 by 2 squares, and the centre square is 2 by 2 square.

When X = 2 the corners are 1 by 1 squares, and the shaded areas are 4 by 1 squares, and the centre square is 4 by 4 square.


It seems I had to derive the question from the solution, rather than the other way around.
In your solution you found a value for X , but did not know what to do with it because you did not understand the question what it was asking. Therefore you did not get the answer.

In all of my 3 solution I provided a diagram to illustrate the question I was solving. Then when I let X equal an unknown at least I knew what part of the square I was talking about. And when I got a value for X I could then refer to the diagram and find what part it referred to.

[Guest]

Another variation of the problem.....
This is the same diagram as before with the cente square unshaded and the 4 corner squares unshaded.
Leaving 4 rectangles or squares shaded. And this fits the equations in your solution better.

This time Let the sides of the unshaded corners = X/2
Then the sides of the centre square is (6 - X/2 - X/2) equals (6 - X)
And the shaded area is still 4/9 of 36 = 16 sq inches.

<x/2><----(6-x)----><x/2>

....xxxxxxxxxxxxxxxx....
....xxxxxxxxxxxxxxxx....
....xxxxxxxxxxxxxxxx....
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
....xxxxxxxxxxxxxxxx....
....xxxxxxxxxxxxxxxx....
....xxxxxxxxxxxxxxxx....

This means each shaded rectangle is (X/2) by (6 - X) ....and there are 4 of them..... So the area equation now is .......

4*(X/2)*(6 - X) = 16

(2X)*(6 - X) = 16

12X - 2X^2 = 16

6X -X^2 = 8

-X^2 + 6X - 8 = 0

OR multiply across by -1 gives.....

X^2 -6X + 8 = 0

(x - 4) (X - 2) = 0

gives .... X = 4 OR X = 2

So the size of the unshaded corners is a square of 2 inches OR a square of 1 inch.

And the shaded rectangles are 4 by 1 OR 2 by 2 inches....same as we got before in my first solution of this diagram.

When X = 4 the corners are 2 by 2 squares, and the shaded areas are 2 by 2 squares, and the centre square is 2 by 2 square.

When X = 2 the corners are 1 by 1 squares, and the shaded areas are 4 by 1 squares, and the centre square is 4 by 4 square.


It seems I had to derive the question from the solution, rather than the other way around.
In your solution you found a value for X , but did not know what to do with it because you did not understand the question what it was asking. Therefore you did not get the answer.

In all of my 3 solution I provided a diagram to illustrate the question I was solving. Then when I let X equal an unknown at least I knew what part of the square I was talking about. And when I got a value for X I could then refer to the diagram and find what part it referred to.

[Guest]

Another variation of the problem.....
This is the same diagram as before with the cente square unshaded and the 4 corner squares unshaded.
Leaving 4 rectangles or squares shaded. And this fits the equations in your solution better.

This time Let the sides of the unshaded corners = X/2
Then the sides of the centre square is (6 - X/2 - X/2) equals (6 - X)
And the shaded area is still 4/9 of 36 = 16 sq inches.

<x/2><----(6-x)----><x/2>

....xxxxxxxxxxxxxxxx....
....xxxxxxxxxxxxxxxx....
....xxxxxxxxxxxxxxxx....
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
xxxx................xxxx
....xxxxxxxxxxxxxxxx....
....xxxxxxxxxxxxxxxx....
....xxxxxxxxxxxxxxxx....

This means each shaded rectangle is (X/2) by (6 - X) ....and there are 4 of them..... So the area equation now is .......

4*(X/2)*(6 - X) = 16

(2X)*(6 - X) = 16

12X - 2X^2 = 16

6X -X^2 = 8

-X^2 + 6X - 8 = 0

OR multiply across by -1 gives.....

X^2 -6X + 8 = 0

(x - 4) (X - 2) = 0

gives .... X = 4 OR X = 2

So the size of the unshaded corners is a square of 2 inches OR a square of 1 inch.

And the shaded rectangles are 4 by 1 OR 2 by 2 inches....same as we got before in my first solution of this diagram.

When X = 4 the corners are 2 by 2 squares, and the shaded areas are 2 by 2 squares, and the centre square is 2 by 2 square.

When X = 2 the corners are 1 by 1 squares, and the shaded areas are 4 by 1 squares, and the centre square is 4 by 4 square.


It seems I had to derive the question from the solution, rather than the other way around.
In your solution you found a value for X , but did not know what to do with it because you did not understand the question what it was asking. Therefore you did not get the answer.

In all of my 3 solution I provided a diagram to illustrate the question I was solving. Then when I let X equal an unknown at least I knew what part of the square I was talking about. And when I got a value for X I could then refer to the diagram and find what part it referred to.

[Guest]

I seem to have a problem posting

[Guest]

problem posting

[Guest]

Sorry. I hadn't noticed a second page was created, I kept posting the same several times thinking it was not posted.

And I looked at your link....Good maths book.....and it shows that you typed in the problem wrongly in the first place.

As I said before you did not make the proper use of brackets to set out your work when typing text on a single line.

But at least the book uses the same methods and got the same answers as my 3 example solutions

[Guest]

And Now the textbook version......which is what you originally asked about.

The textbook diagrams are same as my first diagram posted.....

My first example.... Let the corner square equal X.

The textbook Lets the sides of the centre square equal X

<(6-X)/2><----(X)----><(6-X)/2>

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Then the corner squares are (6 -X)/2

The equations for the textbook should be typed as.......

Total shaded area is .... 4 * X * (6 - X) / 2

Hence the equation is .... 4 * X * (6 - X) / 2 = 4/9*36 = 16

Read as 4 times X times (6 - X) in brackets divided by 2 equals 4/9 of 36 equals 16

and written all together as 4X(6 - X)/2 = 16

This gives 12X - 2X^2 = 16

OR -2X^2 + 12X = 16

OR -2X^2 + 12X - 16 = 0

OR -X^2 + 6X - 8 = 0

OR X^2 - 6X + 8 = 0

Factorise to solve......

(X - 2)(X - 4) = 0

gives .... X = 2 OR X = 4

This means the centre square is 4 inches OR 2 inches.....both work as before

This means the corner squares are (6 - 2)/2 = 4/2 = 2 inches OR (6 - 4) = 2/2 = 1 inch as before
Both roots of the quadratic are positive and valid for the question problem.
When the corner square is 1 inch the centre square is 4 inches so the recrangle shaded is 4 inches by 1 inch
When the corner square is 2 inches the centre square is 2 inches and the shaded rectangles is actually square 2 inches by 2 inches

[Guest]

I understand it now. Thanks for your detailed and thorough explanation, patience and understanding.