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A piece of wire 24 cm long has the shape of a rectangle. Given that the width is w cm,...

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sam-129 | Student, Undergraduate | (Level 2) eNoter

Posted September 20, 2010 at 10:55 PM via web

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A piece of wire 24 cm long has the shape of a rectangle. Given that the width is w cm, show that the area, A cm^2.....

, of the rectangle is given by the function A = 36 - (6-w)^2.

Please could you explain how or write in steps? Thanks in advance.

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william1941 | College Teacher | (Level 3) Valedictorian

Posted September 20, 2010 at 11:18 PM (Answer #1)

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The shape of the wire is a rectangle. Let the length of the rectangle be L. The width is given as w.

As the rectangle is made with a wire that is 24 centimetres long, 24 = 2*L + 2*w

=> 2*L= 24 - 2*w

=> L = 12 -w

The area of the rectangle is length*width

= (12- w)*w

Now A= 36 - ( 6-w)^2

Subtracting  (12- w)*w

=> 36 - 36 +w^2 -12w - 12w - w^2

=> 0

Therefore A = 36 - (6 -w)^2

 

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krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted September 20, 2010 at 11:14 PM (Answer #2)

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The perimeter of the rectangle = length of wire = 24 cm (Given)

Width = w (given)

Then length of the rectangle = (Perimeter - 2*width)/2

= (24 - 2w)/2 = 12 - w

Then area of the rectangle is given by:

Area = A = Width*Length

= w*(12 - w)

= 12 w - w^2

= 36 - 36 + 12w - w^2

= 36 - (36 - 12w + w^2)

= 36 - (6 - w)^2

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neela | High School Teacher | (Level 3) Valedictorian

Posted September 20, 2010 at 11:18 PM (Answer #3)

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Since the shape of the wire of 24 cm  is bent into a rectangle, the perimeter p of the wire is 24cm.

So  P = 2l(l+w), where l is length and w is width of the rectangle.

Therefore  P=2l+2w. O

l = (p-2w)/2 = (24-2w)/2 = 12-w.

Now Area A = l*w. Or

A =  l*w

A = (12-w)w

A = 12w -w^2

A = 36 - 36+12w-w^2. 36 is added and subtracted.

A = 36 - (6^2-2*6w+w^2)

A = 36 - (6-w)^2.

So we proved what is given

 

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted September 20, 2010 at 11:43 PM (Answer #4)

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We are given that the wire length = 24

Then the wire lengt is the perimeter of the rectangle;

==> P = 2*L + 2*w = 24

==> L = (24-2w)/2

==> L = 12-w ........(1)

We know that the area of the rectangle is:

 A = L*w

Substitute with L = 12-w

==> A = (12-w)*w

           = 12w-w^2

Now add and subtract 36:

==> A = 36 + 12w-w^2 - 36

           = 36 - (w^2 -12w +36)

           = 36 - (w-6)^2

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted September 20, 2010 at 11:25 PM (Answer #5)

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If the length of the rectangle is 24 and the width is w, then the width of the rectangle could be found from the formua of perimeter of the rectangle.

The perimeter of a rectangle is the sum between twice length and twice width.

P = 2*24 + 2*w

We'll factorize 2 and we'll get:

P = 2(24+w)

The area of the rectangle is:

A = 36 - (6-w)^2

We'll expand the square and we'll get:

36 - 36 + 12w - w^2

We'll eliminate like terms and we'll get:

A = 12w - w^2

We also know that the area of the rectangle is the product of the length and the width.

24w = 12w - w^2

We'll move all terms to one side:

w^2 - 12w + 24w = 0

w^2 + 12w = 0

We'll factorize by w:

w(w+12)=0

Since the width cannot be 0 or negative, the relation for area doesn't hold.

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