Question

the remains of an ancient ball court include a rectangular playing alley with a perimeter of about 48 M. the length of the alley is 5 times the width. find the length and the width of the playing alley​

Answer 1

Answer:

Length is 20 m and width is 4 m.

Step-by-step explanation:

Given:

Perimeter of the rectangular alley, $P=48\textrm{ m}$

Length is 5 times the width.

Let width be $x$.

So, as per question,

Length,$l = 5x$

Now, perimeter of rectangle is given as:

$P=2(l+b)$

Plug in 48 for $P$, $5x$ for $l$ and $x$ for b.

$48=2(5x+x)\\48=2(6x)\\48=12x\\x=\frac{48}{12}=4$

Therefore, width is 4 m.

Length is $5x=5\times 4=20$ m.

Answer 2

The dimensions of the rectangular playing alley with a perimeter of 48 m are:

• Length = 20 m
• Width = 4 m

What is the Perimeter of a Rectangle?

Perimeter of a rectangle = 2(length + width)

Given the following dimensions of the rectangular playing alley:

• Perimeter = 48 m
• Width = x
• Length = 5x

Plug in the values into the perimeter formula of a rectangle and find x:

48 = 2(5x + x)

48 = 2(6x)

48 = 12x

x = 4

Width = x = 4 m

Length = 5x = 5(4) = 20 m

Therefore, the dimensions of the rectangular playing alley with a perimeter of 48 m are:

• Length = 20 m
• Width = 4 m

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