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daser333 [38]
3 years ago
9

What is the approximate perimeter of the square with the area of 112cm^2.Step by step answers please?

Mathematics
1 answer:
Alex17521 [72]3 years ago
3 0
First find the length of the side by taking the square root of 112.

There are 4 sides so multiply your answer by 4 to get the perimeter.

The answer is 42.332 which rounds to 42.
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The length of a rectangle is 7 more than the width. The area is 744 sqaure yards, find the length and width of the rectangle​
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Answer:

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Given:

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Solution:

Let's assume Width of rectangle be x and Length of rectangle be x + 7 respectively.

Using formula

\\  \:  \:  \:  \:  \pink{ \dashrightarrow \:  \:  \:  \:  \sf { \underbrace{Area_{(Rectangle)} =  Length × Width  }}} \\  \\

On Substituting the required values, we get;

\\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf (x)(x + 7) = 744 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf  {x}^{2}  + 7x = 744 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf  {x}^{2}  + 7x - 744 = 0 \\ \\ \\  \: \: \: \:  \dashrightarrow \: \: \: \: \sf  {x}^{2}  + 31x - 24x - 744 = 0 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf x(x + 31) - 24 (x + 31) = 0 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf (x + 31)(x - 24) = 0 \\ \\ \\ \: \: \: \:  \dashrightarrow \: \: \: \: \sf x = 24 \:  or \:   - 31 \\ \\

As we know that width of the rectangle can't be negative. So x = 24

<u>Hence</u>,

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  • Length of the rectangle = x + 7 = 31 yards

\thereforeLength of rectangle is 31 yards<u> </u>and Width is 24 yards.

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