Question

A piece of wire $8m$ in length is cut into two pieces, and each piece is bent into a square. Where the cut in the wire should be made if the sum of the areas of these squares is to be $2{m^2}$ ? How many meters from the left end?

Answer 1

Hint:
Analyse the question carefully and notice that the length of the wire initially will be the same as the length of two pieces that are used to make squares. Assume some variable $'f'$ for the side of one square. Use the relation to find sides of both the squares in terms of $'f'$. Now add the areas and equate it to $2{m^2}$. Solve this equation for the only unknown $'f'$.

Complete step by step solution:
Let’s first try to analyse the situation given. Here in this problem, a $8m$ long piece of wire is cut into two pieces by a single cut. Then these new pieces are bent to form two squares with a total sum of areas of these two as $2{m^2}$ . With that given information we need to calculate the length of cut made on the $8m$ long wire from the left end.
There’s one thing that you should observe that during this whole operation of cutting and bending, there is one thing that remained constant. The total length of wire before cutting/bending does not change after the operation completes. So, we can say that the length of wire initially will be equal to the length of the closed boundary of the two new squares, i.e. perimeter of the two squares.
$ \Rightarrow $ The total length of wire $ = $ Sum of perimeters of both the squares
Let us assume that the length of side one of the squares is $'f'$ meters.
As we know that the perimeter of a square is four times its length of the side. Therefore, we can write the perimeter of the first square as $ = 4 \times f = 4f{\text{ m}}$
So, the perimeter of second of square $ = $ Total length of wire $ - $ Perimeter of the first square
$ \Rightarrow $ The perimeter of the second square $ = \left( {8 - 4f} \right){\text{ m}}$
Therefore, the length of the side of the second square $ = \dfrac{{{\text{Perimeter}}}}{4} = \dfrac{{\left( {8 - 4f} \right)}}{4} = \left( {2 - f} \right){\text{ m}}$
Hence, we got the lengths of sides of the two squares as $f{\text{ and }}\left( {2 - f} \right)$ meters.
According to the question, the sum of areas of these two squares is $2{\text{ }}{{\text{m}}^2}$
But we know that the area of a square is given by squaring its length of the side. Therefore, the above statement gives us a relation:
$ \Rightarrow {f^2} + {\left( {2 - f} \right)^2} = 2$
From this relation, we can easily find the value of the unknown $'f'$ . Let’s expand the square using the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
$ \Rightarrow {f^2} + {\left( {2 - f} \right)^2} = 2 \Rightarrow {f^2} + {2^2} - 4f + {f^2} = 2 \Rightarrow 2{f^2} - 4f + 2 = 0$
Now after taking $2$ common from the left-side, we can again use the identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ . This will give us:
$ \Rightarrow 2\left( {{f^2} - 2f + 1} \right) = 0 \Rightarrow {\left( {f - 1} \right)^2} = 0$
Therefore, we get the two equal roots of this quadratic equation, i.e. $f = 1$
So, the length of sides of the first square will be $f = 1{\text{ m}}$ and for the second $\left( {2 - f} \right) = 2 - 1 = 1{\text{ m}}$
And the length of cut made in the total length of the wire will be equal to the perimeter of the first square, i.e. $4 \times f = 4 \times 1 = 4{\text{ m}}$.

Hence, the cut made on the total length of $8{\text{ m}}$ wire divides the wire into two equal pieces and these two equal pieces are used to make two squares of equal dimensions.

Note:
In questions like this, always try to find the constraints or limitations in the whole operation. This will help you establish a relationship between the unknowns like the length of wire in this case. An alternate approach to the same problem can be taken by assuming the perimeter of the first square as some variable. Then find the perimeter of other squares and areas in terms of that.