Question

A town is designing a rectangular park that will be 600 feet wide by 1000 feet long. On a scale drawing, the dimensions of the park are 12 inches wide by 20 inches long. A rectangular area of the park for swing sets is 0. 5 inch wide by 2 inches long on the scale drawing. What is the actual length of the swing set area of the park? Enter your answer in the box. The actual length of the swing set area of the park is feet.

Answer 1

Scale drawing takes scaled versions of original lengths. The actual length of the swing set area of the park is 25 ft long and 100 ft wide

How are scale drawings formed?

For a particular scale drawing, it is already specified that all the measurements' some constant scaled version will be taken. For example, let the scale be K feet to s inches.

Then it means

$\rm 1\: ft : \dfrac{s}{k}\: in.$

All feet measurements will then be multiplied by s/k to get the drawing's corresponding lengths.

For the given case, it is given that

Real dimension of a rectangular park : 600 ft by 1000 ft

In drawing, it was of 12 inches by 20 inches.

Let the scale that the drawing is using be k feet to s inches, then

$\rm 1\: ft : \dfrac{s}{k}\: in.$

Since it is given that 600 ft : 12 inches and 1000 ft : 20 inches, thus, we get:

$\rm 1\: ft : \dfrac{s}{k}\: in.\\\\600 \: ft = \dfrac{600\times s}{k} \: in. = 12 \: in.\\\\\dfrac{600 \times s}{k} = 12\\\\\dfrac{s}{k} = \dfrac{12}{600} = 0.02$

Thus, every feet is reduced 0.02 times in drawing.

Now,

it is given that in drawing, the rectangular area for swing sets is 0.5 inch to 2 inch.

Let its real length was L and real width was W. Then we have:
$\rm L \times 0.02 = 0.5 \\\\L = \dfrac{0.5}{0.02} = 25 \:\\\\W \times 0.02 = 2 \\W = \dfrac{2}{0.02} = 100$

Since these are only magnitudes, and unit was converted from feet to inches, so we have L = 25 feet and W = 100 feet

Thus,

The actual length of the swing set area of the park is 25 ft long and 100 ft wide

Learn more about scaling here:
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