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The perimeter of a school volleyball court is 177 ft. and the length is twice the width. What are the dimensions of the volleyball court?

Answer
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Hint: In this problem we need to find the dimensions of the volleyball court. That is, we have to find length and width. For this, we will use the formula of perimeter of rectangle because a volleyball court is shaped like a rectangle. Also we will use the given information that the length is twice the width.

Complete step-by-step answer:
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Let us assume that l is the length (in ft.) and w is the width (in ft.) of a school volleyball court. A volleyball court is shaped like a rectangle. So, we will use the formula of perimeter of the rectangle. We know that the perimeter P of a rectangle is given by P=2(l+w) where l is the length and w is the width of a rectangle. So, in this problem we can say that the perimeter of a school volleyball court is P=2(l+w)(1) .
In the problem it is given that the perimeter of a school volleyball court is 177 (in ft.). So, from (1) we can write 177=2(l+w)(2) .
Also given that the length l is twice the width w . So, we can write l=2w . Now we are going to substitute l=2w in equation (2) , we get
 177=2(2w+w)177=2(3w)177=6w(3)
We can see that the equation (3) is linear equation in one variable. Let us solve this equation and find w . For this, we will divide by 6 on both sides of equation (3) . So, we get
 1776=6w6w=29.5
Now we substitute width w=29.5 ft. in l=2w to find the length l . Therefore, we get l=2(29.5)=59 ft.
Therefore, the length of a school volleyball court is 59 ft. and the width of a school volleyball court is 29.5 ft.

Note: The sum of length of all sides of a rectangle is called the perimeter of a rectangle. In a rectangle, there are two parallel (opposite) sides with equal length. Also there are four right angles. The area of a rectangle is given by A=l×w where l is the length and w is the width of a rectangle.