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Question

Answers

$

{\text{A}}{\text{. 23 }}{{\text{m}}^2} \\

{\text{B}}{\text{. 11 }}{{\text{m}}^2} \\

{\text{C}}{\text{. 54 }}{{\text{m}}^2} \\

{\text{D}}{\text{. 10 }}{{\text{m}}^2} \\

$

Answer
Verified

Hint: Here, we will proceed by obtaining one equation in one variable according to the problem statement in order to find the value of that variable. Here, we will also use the formula i.e., Area of the rectangle = (Length)(Breadth).

Complete step-by-step answer:

Given, Ratio of the original length to the original breadth of the rectangle = 3 : 2

i.e., $\dfrac{{{\text{Original Length}}}}{{{\text{Original Breadth}}}} = \dfrac{3}{2} = \dfrac{{3x}}{{2x}}$ (say)

$ \Rightarrow $Original Length of the rectangle = 3x m

$ \Rightarrow $Original Breadth of the rectangle = 2x m

Now, when all the sides of the rectangle are extended by 1 m, we have

Final Length of the rectangle = (3x+1) m

Final Breadth of the rectangle = (2x+1) m

So, $\dfrac{{{\text{Final Length}}}}{{{\text{Final Breadth}}}} = \dfrac{{3x + 1}}{{2x + 1}}{\text{ }} \to {\text{(1)}}$

It is given that the ratio of the final length to the final breadth of the rectangle after extending each side by 1 m is 10 : 7

i.e., $\dfrac{{{\text{Final Length}}}}{{{\text{Final Breadth}}}} = \dfrac{{10}}{7}$

By substituting equation (1) in the above equation, we get

$ \Rightarrow \dfrac{{3x + 1}}{{2x + 1}} = \dfrac{{10}}{7}$

By cross multiplying the above equation, we get

$

\Rightarrow 7\left( {3x + 1} \right) = 10\left( {2x + 1} \right) \\

\Rightarrow 21x + 7 = 20x + 10 \\

\Rightarrow 21x - 20x = 10 - 7 \\

\Rightarrow x = 3 \\

$

$ \Rightarrow $Original Length of the rectangle = 3x = 3(3) = 9 m

$ \Rightarrow $Original Breadth of the rectangle = 2x = 2(3) = 6 m

As we know that the area of any rectangle is given by

Area of the rectangle = (Length)(Breadth)

$ \Rightarrow $Area of the original rectangle = (Original Length of the rectangle)(Original Breadth of the rectangle)

$ \Rightarrow $Area of the original rectangle = (9)(6) = 54 ${{\text{m}}^2}$.

Hence, option C is correct.

Note: In this particular problem, the ratio given are in the most simplified form so in order to obtain the values of the numerator and denominator of that ratio we will multiply both the numerator and denominator with same number (in this case it is x) which is the same ratio as the given ratio because this number can be cancelled.

Complete step-by-step answer:

Given, Ratio of the original length to the original breadth of the rectangle = 3 : 2

i.e., $\dfrac{{{\text{Original Length}}}}{{{\text{Original Breadth}}}} = \dfrac{3}{2} = \dfrac{{3x}}{{2x}}$ (say)

$ \Rightarrow $Original Length of the rectangle = 3x m

$ \Rightarrow $Original Breadth of the rectangle = 2x m

Now, when all the sides of the rectangle are extended by 1 m, we have

Final Length of the rectangle = (3x+1) m

Final Breadth of the rectangle = (2x+1) m

So, $\dfrac{{{\text{Final Length}}}}{{{\text{Final Breadth}}}} = \dfrac{{3x + 1}}{{2x + 1}}{\text{ }} \to {\text{(1)}}$

It is given that the ratio of the final length to the final breadth of the rectangle after extending each side by 1 m is 10 : 7

i.e., $\dfrac{{{\text{Final Length}}}}{{{\text{Final Breadth}}}} = \dfrac{{10}}{7}$

By substituting equation (1) in the above equation, we get

$ \Rightarrow \dfrac{{3x + 1}}{{2x + 1}} = \dfrac{{10}}{7}$

By cross multiplying the above equation, we get

$

\Rightarrow 7\left( {3x + 1} \right) = 10\left( {2x + 1} \right) \\

\Rightarrow 21x + 7 = 20x + 10 \\

\Rightarrow 21x - 20x = 10 - 7 \\

\Rightarrow x = 3 \\

$

$ \Rightarrow $Original Length of the rectangle = 3x = 3(3) = 9 m

$ \Rightarrow $Original Breadth of the rectangle = 2x = 2(3) = 6 m

As we know that the area of any rectangle is given by

Area of the rectangle = (Length)(Breadth)

$ \Rightarrow $Area of the original rectangle = (Original Length of the rectangle)(Original Breadth of the rectangle)

$ \Rightarrow $Area of the original rectangle = (9)(6) = 54 ${{\text{m}}^2}$.

Hence, option C is correct.

Note: In this particular problem, the ratio given are in the most simplified form so in order to obtain the values of the numerator and denominator of that ratio we will multiply both the numerator and denominator with same number (in this case it is x) which is the same ratio as the given ratio because this number can be cancelled.

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