Question

Find the sides of the square whose area is \[1\,\,{{\left( are \right)}^{2}}\]. What will be the area if (i) the sides of squares are doubled (ii) sides of the square are halved?

Answer 1

Hint: Consider a variable (say x cm) for the length of the side of the square. Area of Square is \[{{x}^{2}}\,cm\] and equate with a given area of the square then solve for x. In the first part of the question, we have to find the area when sides are doubled that means \[2x\] and in the second part sides of the squares are halved that means \[\dfrac{x}{2}\]. After substituting this value on Formula, we get the required answer for the first and second part of this problem.

Complete step-by-step solution:
According to the question, the area of the square is \[1\,\,{{\left( are \right)}^{2}}\]. We have to find the sides of square

According to the given figure, consider the ABCD is a square whose sides are considered to be x are
That means \[side=x\,are\]
We know that area of square \[={{\left( side \right)}^{2}}\]
So, the area of the square length of whose sides are ‘x’ are will be \[{{x}^{2}}\,\,{{\left( are \right)}^{2}}\]
But according to the question, the area of the square will be \[=1\,\,{{\left( are \right)}^{2}}\]
So, \[{{x}^{2}}=1\,\,{{\left( are \right)}^{2}}\]
Taking positive square root of both sides of the equation, we will get
\[\Rightarrow x=1\,\,are\]
Hence the required side of the given square is 1 are
Now, we have to solve the first part, that is, we have to find the area in which the sides of the square are doubled.
So, new side length of the square is 2x that means
\[\Rightarrow side=2x\,\,are---(1)\]
Here, as we know that value of x is 1 are that is \[x=1\,\,are\] substitute the value on above equation (1) we get:
\[\Rightarrow side=\left( 2\times 1 \right)\,\,are\]
By simplifying this we get the value of side length of square
\[\Rightarrow side=2\,\,are\]
Now, by applying the formula, area of square \[={{\left( side \right)}^{2}}\]
By substituting the value of new side length of square on above formula we get:
\[\Rightarrow \text{Area of square}={{\left( 2\,\,are \right)}^{2}}\]
By simplifying this we get the value of area that is
\[\Rightarrow \text{Area of square}=4\,\,{{\left( are \right)}^{2}}\]
Now, we have to solve the second part, that is, we have to find the area in which the sides of the square are halved.
So, new side length of the square is \[\dfrac{x}{2}\]that means
\[\Rightarrow side=\dfrac{x}{2}\,\,are---(2)\]
Here, as we know that value of x is 1 are that is \[x=1\,\,are\] substitute the value on above equation (2) we get:
\[\Rightarrow side=\left( \dfrac{1}{2} \right)\,\,are\]
By simplifying this we get the value of side length of square
\[\Rightarrow side=\dfrac{1}{2\,}\,are\]
Now, by applying the formula, area of square \[={{\left( side \right)}^{2}}\]
By substituting the value of new side length of square on above formula we get:
\[\Rightarrow \text{Area of square}={{\left( \dfrac{1}{2}\,\,are \right)}^{2}}\]
By simplifying this we get the value of area that is
\[\Rightarrow \text{Area of square}=\dfrac{1}{4}\,\,{{\left( are \right)}^{2}}\]
Hence, required area is \[4\,\,{{\left( are \right)}^{2}}\] when sides are doubled and area is \[\dfrac{1}{4}\,\,{{\left( are \right)}^{2}}\] when sides are halved.

Note: While solving the equation \[{{x}^{2}}=1\,\,{{\left( are \right)}^{2}}\] in the solution, here Most of the students might make mistake while considering the negative value of side length of square that is wrong because side length cannot be negative. We have taken only the positive square root because ‘x’ is the length of a side and the length of the side of a square cannot be negative.