# Question Video: Using Equations to Find the Area of a Rectangle given Its Width and Perimeter as Mixed Numbers Mathematics • 7th Grade

Find the area of the rectangle with a width of 4 7/9 inches and a perimeter of 22 2/3 inches.

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### Video Transcript

Find the area of a rectangle with a width of four and seven-ninths inches and a perimeter of 22 and two-thirds inches.

Our first step is to calculate the length of the rectangle, in this case 𝑥. We are told that the width is four and seven-ninths and the perimeter is 22 and two-thirds. Four and seven-ninths is equal to 43 ninths as an improper or top heavy fraction.

We work out the numerator 43 by multiplying four by nine and adding seven. In a similar way, we can work out that the perimeter is 68 thirds. 22 multiplied by three plus two is equal to 68.

The perimeter of a rectangle can be calculated by adding the two lengths and two widths. In this case, 𝑥 plus 𝑥 plus 43 ninths plus 43 ninths is equal to 68 thirds.

Simplifying the left-hand side by grouping like terms gives us two 𝑥 plus 86 ninths equals 68 thirds. Subtracting 86 ninths from both sides of the equation gives us two 𝑥 is equal to 118 ninths, as 68 thirds minus 86 ninths is equal to 118 ninths.

Dividing both sides of this equation by two gives us a value of 𝑥 of 59 ninths. This means that the length of the rectangle is 59 ninths. The area of a rectangle can be calculated by multiplying the length by the width. In this case, we need to multiply 59 ninths by 43 ninths. This gives us an answer of 2537 over 81 or 2537 divided by 81.

Converting this back into a mixed fraction or mixed number gives us thirty-one and twenty-six eighty-firsts, 31 and 26 over 81. This means that the area of a rectangle with width four and seven-ninth inches and a perimeter of 22 and two-thirds inches is thirty-one and twenty-six eighty-firsts inches squared.

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