Length of Wire and Area of Squares: Solving for the Length of Each Piece

[mathdad]

A piece of wire is cut into two pieces.
Each piece is then bent to form a square. If the sum of the areas of the two squares is 5L^2/128, how long is each piece of the wire?

My Reasoning:

I see the words square and sum of the squares.

A = s^2, where s is side of the square.

Sum of the areas of the squares is given to be 5L^2/128.

Area of first square + area of square 2 = 5L^2/128.

Do I divide 5L^2/128 by 2 to find the length of each piece of wire?

 

[Greg]

Does the problem give a relationship between the two pieces of wire?

 

[I like Serena]

A piece of wire is cut into two pieces.
Each piece is then bent to form a square. If the sum of the areas of the two squares is 5L^2/128, how long is each piece of the wire?

My Reasoning:

I see the words square and sum of the squares.

A = s^2, where s is side of the square.

Sum of the squares is given to be 5L^2/128.

Area of first square + area of square 2 = 5L^2/128.

Do I divide 5L^2/128 by 2 to find the length of each piece of wire?

Hey RTCNTC,

Apparently it is not given how the piece of wire is cut into pieces.
And I assume the total length of the wire is $L$?

So let's assume we cut off a piece of (as yet unknown) length $x$.
Then we're left with a piece of length $x$ and a piece of length $L-x$.
If we bend both of those pieces into squares, what would their areas be?

 

[mathdad]

Does the problem give a relationship between the two pieces of wire?

If it does, I can't see the connection.

 

[mathdad]

Hey RTCNTC,

Apparently it is not given how the piece of wire is cut into pieces.
And I assume the total length of the wire is $L$?

So let's assume we cut off a piece of (as yet unknown) length $x$.
Then we're left with a piece of length $x$ and a piece of length $L-x$.
If we bend both of those pieces into squares, what would their areas be?

If we bend x and L - x into squares, their areas would be as follows:

For x: x^2

For L - x: (L - x)^2

The sum of the areas is the following:

x^2 + (L - x)^2

Do I equate x^2 + (L - x)^2 = 5L^2/128?

Do I then solve for x or L?

 

[MarkFL]

If you have a piece of wire whose length is $a$, and we bend it into a square, then each side of the square will be:

\(\displaystyle \frac{a}{4}\)

And so what will the area of the resulting square be?

Now use this formula for area in terms of the length of wire for your two segments, one of length $x$, and the other of length $L-x$, and then find their sum.

Equate this sum to the given area, and solve for $x$ (in terms of $L$) to determine the length of both pieces of wire. You will find two possibilities for $x$, but you should be able to show that when you choose one root of the resulting quadratic equation as the value of $x$, then $L-x$ will be equal to the other root.

 

[mathdad]

If you have a piece of wire whose length is $a$, and we bend it into a square, then each side of the square will be:

\(\displaystyle \frac{a}{4}\)

And so what will the area of the resulting square be?

Now use this formula for area in terms of the length of wire for your two segments, one of length $x$, and the other of length $L-x$, and then find their sum.

Equate this sum to the given area, and solve for $x$ (in terms of $L$) to determine the length of both pieces of wire. You will find two possibilities for $x$, but you should be able to show that when you choose one root of the resulting quadratic equation as the value of $x$, then $L-x$ will be equal to the other root.

Each side of the square is a/4.

The area is (side)^2.

A = (a/4)^2

A = a^2/16

First Area:

(x/2)^2 or x^2/4.

Second Area:

(L - x)^2 or L^2 -2Lx + x^2.

Adding the two area:

[(x^2/4)] + [x^2 - 2Lx + x^2] = given area.

I am confused because there are two variables on the left side when adding the two areas. We have L and x on the left side. Which variable do I solve for?

 

[MarkFL]

Each side of the square is a/4.

The area is (side)^2.

A = (a/4)^2

A = a^2/16

First Area:

(x/2)^2 or x^2/4.

Second Area:

(L - x)^2 or L^2 -2Lx + x^2.

Adding the two area:

[(x^2/4)] + [x^2 - 2Lx + x^2] = given area.

I am confused because there are two variables on the left side when adding the two areas. We have L and x on the left side. Which variable do I solve for?

I solved for $x$. :D

But check your work...the areas aren't correct yet. ;)

 

[I like Serena]

Each side of the square is a/4.

The area is (side)^2.

A = (a/4)^2

A = a^2/16

First Area:

(x/2)^2 or x^2/4.

Shouldn't that be (x/4)^2 or x^2/16?

Second Area:

(L - x)^2 or L^2 -2Lx + x^2.

Shouldn't that be ((L-x)/4)^2?

 

[mathdad]

Shouldn't that be (x/4)^2 or x^2/16?
Shouldn't that be ((L-x)/4)^2?

A square has 4 sides. Let one side be x.

One side ÷ 2 = x/2. Its area is represented by (x/2)^2 or x^2/4.

This is my reasoning.

Same thing for a square whose side is (L - x).

Divide (L - x)/2 and then square that side.

[(L - x)^2/2]^2 = ?

 

[I like Serena]

A square has 4 sides. Let one side be x.

One side ÷ 2 = x/2. Its area is represented by (x/2)^2 or x^2/4.

This is my reasoning.

Didn't we pick x to be the length of the piece we cut off from the string with length L?
If so, then one side has length x/4, since indeed, a square has 4 sides.

 

[mathdad]

Didn't we pick x to be the length of the piece we cut off from the string with length L?
If so, then one side has length x/4, since indeed, a square has 4 sides.

Can you set up the sum of the areas to equal the given area? I can take it from there.

 

[I like Serena]

Can you set up the sum of the areas to equal the given area? I can take it from there.

It's:
$$x^2/16 + (L-x)^2/16 = 5L^2/128$$
Solve for x.

 

[mathdad]

I got it. Thank you everyone.

 

[mathdad]

It's:
$$x^2/16 + (L-x)^2/16 = 5L^2/128$$
Solve for x.

When I solve for x, I get x in terms of L but not a specific integer answer. I need to find how long each piece of the wire actually is not an expression or function in terms of L.

 

[MarkFL]

When I solve for x, I get x in terms of L but not a specific integer answer. I need to find how long each piece of the wire actually is not an expression or function in terms of L.

The best you can do is get $x$ in terms of $L$. :D

 

[mathdad]

The best you can do is get $x$ in terms of $L$. :D

The answer is x in terms of L? Ok.

 

[MarkFL]

The answer is x in terms of L? Ok.

No, you will find that $x$ is in terms of $L$ when you solve the resulting quadratic. :D

 

[mathdad]

No, you will find that $x$ is in terms of $L$ when you solve the resulting quadratic. :D

Ok. I will work on this and post my reply.