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# How do you find the perimeter and area of a rectangle with width of $2\sqrt 7 - 2\sqrt 5$ and length of $3\sqrt 7 + 3\sqrt 5$?

Last updated date: 19th Sep 2024
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Hint:Start by mentioning all the formulas that are necessary in these types of questions. Then next start by evaluating the perimeter of the rectangle then further evaluate the area of the rectangle.The perimeter and area of the rectangle are given by $2(l + b)$ and $A = l \times b$.

First we will start off by mentioning the formula for the perimeter of the rectangle, which is given by $2(l + b)$ where $l$ is the length of the rectangle and $b$ is the breadth or width of the rectangle. Now, we will substitute the values of the terms in the above mentioned formula,
$P = 2(l + b) \\ \Rightarrow P = 2((3\sqrt 7 + 3\sqrt 5 ) + (2\sqrt 7 - 2\sqrt 5 )) \\ \Rightarrow P = 2(3\sqrt 7 + 2\sqrt 7 + 3\sqrt 5 - 2\sqrt 5 ) \\ \therefore P = 2(5\sqrt 7 + \sqrt 5 )$
Hence, the value of the perimeter of the rectangle will be $2(5\sqrt 7 + \sqrt 5 )$.
$A = l \times b$ where $l$ is the length of the rectangle and $b$ is the breadth or width of the rectangle.Now, we will substitute the values of the terms in the above mentioned formula,
$A = l \times b \\ \Rightarrow A = (3\sqrt 7 + 3\sqrt 5 ) \times (2\sqrt 7 - 2\sqrt 5 ) \\ \Rightarrow A = (3\sqrt 7 \times 2\sqrt 7 ) + (3\sqrt 7 \times - 2\sqrt 5 ) + (3\sqrt 5 \times 2\sqrt 7 ) + (3\sqrt 5 \times - 2\sqrt 5 ) \\ \Rightarrow A = 6 \times 7 - 6\sqrt {35} + 6\sqrt {35} - 6 \times 5 \\ \Rightarrow A = 42 - 30 \\ \therefore A = 12$
Hence, the value of the area of the rectangle will be $12$ sq. units.
Therefore, the area and perimeter of the rectangle will be $12$ sq. units and $2(5\sqrt 7 + \sqrt 5 )$ units.