Question

What happens to the area of the rectangle when length is halved and breadth is doubled?

Answer 1

Hint: To solve the question we need to have the knowledge of Mensuration more precisely know the area of the rectangle. The formula for the area of a rectangle is the product of length and breadth of the rectangle. $A=l\times b$. We will be putting the condition on the formula and checking the dependencies of the values on the area of the rectangle.

Complete step by step answer:
The question asks us to find the change in the area of the rectangle, if the condition for the length and the breadth changes as length is halved and breadth is doubled. We know that the area of the rectangle is:
$\text{Area = length }\!\!\times\!\!\text{ breadth}$
Consider the original length and the breadth if the rectangle be $l$ and $b$ respectively. Soo the area thus become as:
$\Rightarrow A=l\times b$
Now, applying the condition given the new length and the breadth of the rectangle becomes $\dfrac{l}{2}$ and $2b$ respectively. The area we get on putting the values in the formula of the area of the rectangle we get:
$\Rightarrow A'=\dfrac{l}{2}\times 2b$
On calculating the above expression we get:
$\Rightarrow A'=\dfrac{2lb}{2}$
Since the number $2$ is common in both the numerator and the denominator so the number gets cancelled. So on changing the fraction in the lowest term we get:
$\Rightarrow A'=lb$
Since $A=A'$ , so the area of the rectangle remains the same on changing the dimensions in the given order.
$\therefore $ The area of the rectangle remains the same when length is halved and breadth is doubled.

Note: The above solution will be for the case of the area of the rectangle when dimensions are changed. The question may ask us to state the change in the perimeter of the rectangle. So for that case we will take the formula for the perimeter which is $2\left( l+b \right)$. On checking the change in the perimeter by changing in the dimensions of the rectangle in the same way as given in the question then we see that new perimeter is
 $\Rightarrow 2\left( \dfrac{l}{2}+2b \right)$
$\Rightarrow 2\left( \dfrac{l+4b}{2} \right)$
$\Rightarrow l+4b$
So the new perimeter changes to $l+4b$.