Answer
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Hint: We will use the area of the rectangle to solve this question. The area of the rectangle is given by A = lb, where l is the length of the rectangle and b is the breadth of the rectangle.
Complete step-by-step answer:
Given that each side of a rectangle is increased by 50% and we have to determine the increased area in percentage.
To do this we will assume variables of length and breadth of the given rectangle.
Let the length of the rectangle is x and let the breadth of the rectangle is y.
We know that,
The area of rectangle A= $l \times b$.
Substituting the value of length and breadth as variable x and y respectively.
So, area of rectangle A = $x \times y$.
According to the given conditions in the question, the length and breadth are increased by 50%.
Let the new length be l’.
New length of rectangle l’ will be,
\[\begin{align}
& l'=\dfrac{50x}{100}+x \\
& \Rightarrow l'=\dfrac{3x}{2} \\
\end{align}\]
Let the new breadth be b’.
New breadth of rectangle b’ will be,
\[\begin{align}
& b'=\dfrac{50y}{100}+y \\
& \Rightarrow b'=\dfrac{3y}{2} \\
\end{align}\]
Then we calculate the new Area of the rectangle. Let it be A’.
New area of the rectangle using length l’ and breadth b’ we have,
\[\begin{align}
& A'=\left( \dfrac{3x}{2} \right)\left( \dfrac{3y}{2} \right) \\
& \Rightarrow A'=\dfrac{9xy}{4} \\
\end{align}\]
Now we calculate the percentage increase in area would be,
% Increase in area \[=\left( \dfrac{A'-A}{A} \right)100\] .
\[\Rightarrow \] % Increase in area \[=\left( \dfrac{\dfrac{9xy}{4}-xy}{xy} \right)100\]
\[\Rightarrow \] % Increase in area \[=\left( \dfrac{5xy}{4xy} \right)100\]
\[\Rightarrow \] % Increase in area \[=\left( \dfrac{5}{4} \right)100\]
\[\Rightarrow \] % Increase in area \[=125%\]
Therefore, we got the percentage increase in the area given by 125%.
Hence, we obtain the answer as 125% which is option (C).
Note: The possibility of error in this question is using old length and breadth given by x and y to determine the new area and the new percentage change in area. Always go for using new length l’ and new breadth b’ to determine the new area A’.
Complete step-by-step answer:
Given that each side of a rectangle is increased by 50% and we have to determine the increased area in percentage.
To do this we will assume variables of length and breadth of the given rectangle.
Let the length of the rectangle is x and let the breadth of the rectangle is y.
We know that,
The area of rectangle A= $l \times b$.
Substituting the value of length and breadth as variable x and y respectively.
So, area of rectangle A = $x \times y$.
According to the given conditions in the question, the length and breadth are increased by 50%.
Let the new length be l’.
New length of rectangle l’ will be,
\[\begin{align}
& l'=\dfrac{50x}{100}+x \\
& \Rightarrow l'=\dfrac{3x}{2} \\
\end{align}\]
Let the new breadth be b’.
New breadth of rectangle b’ will be,
\[\begin{align}
& b'=\dfrac{50y}{100}+y \\
& \Rightarrow b'=\dfrac{3y}{2} \\
\end{align}\]
Then we calculate the new Area of the rectangle. Let it be A’.
New area of the rectangle using length l’ and breadth b’ we have,
\[\begin{align}
& A'=\left( \dfrac{3x}{2} \right)\left( \dfrac{3y}{2} \right) \\
& \Rightarrow A'=\dfrac{9xy}{4} \\
\end{align}\]
Now we calculate the percentage increase in area would be,
% Increase in area \[=\left( \dfrac{A'-A}{A} \right)100\] .
\[\Rightarrow \] % Increase in area \[=\left( \dfrac{\dfrac{9xy}{4}-xy}{xy} \right)100\]
\[\Rightarrow \] % Increase in area \[=\left( \dfrac{5xy}{4xy} \right)100\]
\[\Rightarrow \] % Increase in area \[=\left( \dfrac{5}{4} \right)100\]
\[\Rightarrow \] % Increase in area \[=125%\]
Therefore, we got the percentage increase in the area given by 125%.
Hence, we obtain the answer as 125% which is option (C).
Note: The possibility of error in this question is using old length and breadth given by x and y to determine the new area and the new percentage change in area. Always go for using new length l’ and new breadth b’ to determine the new area A’.
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