Question

from unit 1, lesson 12
a map of a rectangular park has a length of 4 inches and a width of 6 inches. it uses a scale of 1 inch for every 30 miles.
a. what is the actual area of the park?
show how you know.
b. the map needs to be reproduced at a different scale so that it has an area of 6 square inches and can fit in a brochure. at what scale should the map be reproduced so that it fits on the brochure?
show your reasoning.

Explanation:

Step1: Find the actual length and width

Given the scale \(1\) inch \( = 30\) miles.
The length on the map \(l_m=4\) inches, so the actual length \(l = 4\times30=120\) miles.
The width on the map \(w_m = 6\) inches, so the actual width \(w=6\times30 = 180\) miles.

Step2: Calculate the actual area

The area of a rectangle \(A=l\times w\).
Substitute \(l = 120\) and \(w = 180\) into the formula: \(A=120\times180=21600\) square - miles.

Step3: For part (b)

Let the new length on the map be \(x\) inches and the new width be \(y\) inches. We know that \(xy = 6\). Let the scale be \(1\) inch \(=k\) miles.
The actual length \(L=120\) miles and the actual width \(W = 180\) miles. Also, \(L=xk\) and \(W = yk\). Then \(xyk^{2}=LW\).
Since \(LW=21600\) and \(xy = 6\), we substitute into the equation \(6k^{2}=21600\).
Divide both sides by \(6\): \(k^{2}=\frac{21600}{6}=3600\).
Take the square - root of both sides: \(k = 60\) (we take the positive value since \(k\) represents a scale factor).

Answer:

a. The actual area of the park is \(21600\) square - miles.
b. The scale should be \(1\) inch for every \(60\) miles.