Question

intro to triangle proportionality (tues 1/27)
on the map below, $1^{\text{st}}$ st. and $2^{\text{nd}}$ st. are parallel with distances as marked.
what is the distance from city hall to $2^{\text{nd}}$ st. along cedar rd.?
distance from city hall to 2nd street = \boxed{} mi \\ \\ \\ \\ round to the nearest tenth of a mile if necessary.

Explanation:

Step1: Apply Triangle Proportionality Theorem

Let \( x \) be the distance from City Hall to \( 2^{\text{nd}} \) St. along Cedar Rd. By the Triangle Proportionality Theorem (also known as the Basic Proportionality Theorem or Thales' theorem), since \( 1^{\text{st}} \) St. and \( 2^{\text{nd}} \) St. are parallel, we have the proportion:
\[ \frac{2.1}{2.8}=\frac{x - 2.4}{2.4} \]
Wait, no, actually, the correct proportion is based on the segments created by the transversals. Let's re - establish the proportion. The segments on Aspen Rd. are \( 2.1\) mi and \( 2.8 - 2.1=0.7\) mi? No, better to see: the two parallel lines (\(1^{\text{st}}\) St. and \(2^{\text{nd}}\) St.) are cut by two transversals: Cedar Rd. and Aspen Rd. So, the ratio of the segments on one transversal should equal the ratio of the segments on the other transversal.

Let the distance from City Hall to \(1^{\text{st}}\) St. along Cedar Rd. be \(2.4\) mi? Wait, no, looking at the diagram: along Aspen Rd., the distance from City Hall to the intersection with \(1^{\text{st}}\) St. is \(2.1\) mi, and from that intersection to the intersection with \(2^{\text{nd}}\) St. is \(2.8 - 2.1 = 0.7\) mi? No, the total length of Aspen Rd. between the two vertical streets ( \(1^{\text{st}}\) and \(2^{\text{nd}}\) St.) is \(2.8\) mi, and the part from City Hall to \(1^{\text{st}}\) St. along Aspen Rd. is \(2.1\) mi. Along Cedar Rd., the distance from City Hall to \(1^{\text{st}}\) St. is, let's say, \(y\) and from \(1^{\text{st}}\) St. to \(2^{\text{nd}}\) St. is \(2.4\) mi. Wait, actually, the correct proportion is:

Since \(1^{\text{st}}\parallel2^{\text{nd}}\), by the Triangle Proportionality Theorem, \(\frac{\text{Distance from City Hall to }1^{\text{st}}\text{ St. along Aspen Rd.}}{\text{Distance from }1^{\text{st}}\text{ St. to }2^{\text{nd}}\text{ St. along Aspen Rd.}}=\frac{\text{Distance from City Hall to }1^{\text{st}}\text{ St. along Cedar Rd.}}{\text{Distance from }1^{\text{st}}\text{ St. to }2^{\text{nd}}\text{ St. along Cedar Rd.}}\)

The distance from City Hall to \(1^{\text{st}}\) St. along Aspen Rd. is \(2.1\) mi, from \(1^{\text{st}}\) St. to \(2^{\text{nd}}\) St. along Aspen Rd. is \(2.8 - 2.1=0.7\) mi. The distance from \(1^{\text{st}}\) St. to \(2^{\text{nd}}\) St. along Cedar Rd. is \(2.4\) mi. Let \(x\) be the distance from City Hall to \(2^{\text{nd}}\) St. along Cedar Rd., so the distance from City Hall to \(1^{\text{st}}\) St. along Cedar Rd. is \(x - 2.4\) mi.

So the proportion is:
\[ \frac{2.1}{0.7}=\frac{x - 2.4}{2.4} \]

Step2: Solve the proportion

First, simplify \(\frac{2.1}{0.7}\). \(2.1\div0.7 = 3\). So our equation becomes:
\[ 3=\frac{x - 2.4}{2.4} \]
Multiply both sides by \(2.4\):
\[ x-2.4=3\times2.4 \]
\[ x - 2.4 = 7.2 \]
Add \(2.4\) to both sides:
\[ x=7.2 + 2.4 \]
\[ x = 9.6 \]
Wait, that can't be right. Maybe I set up the proportion wrong. Let's try another way. The two parallel lines are \(1^{\text{st}}\) St. and \(2^{\text{nd}}\) St. The transversals are Cedar Rd. and Aspen Rd. So the ratio of the segments on Cedar Rd. should equal the ratio of the segments on Aspen Rd.

Let the distance from City Hall to \(2^{\text{nd}}\) St. along Cedar Rd. be \(x\). The distance from City Hall to \(1^{\text{st}}\) St. along Cedar Rd. is \(x - 2.4\) (if \(2.4\) is the distance from \(1^{\text{st}}\) to \(2^{\text{nd}}\) along Cedar Rd.). The distance from City Hall to \(1^{\text{st}}\) St. along Aspen Rd. is \(2.1\), and from \(1^{\text{st}}\) to \(2^{\text{nd}}\) along Aspen Rd. is \(2.8-2.1 = 0.7\). So:

\(\frac{\text{City Hall to }1^{\text{st}}\text{ (Aspen)}}{\text{1st to 2nd (Aspen)}}=\frac{\text{City Hall to }1^{\text{st}}\text{ (Cedar)}}{\text{1st to 2nd (Cedar)}}\)

\(\frac{2.1}{0.7}=\frac{x - 2.4}{2.4}\)

\(3=\frac{x - 2.4}{2.4}\)

\(x-2.4 = 7.2\)

\(x=9.6\). But that seems large. Wait, maybe the \(2.4\) is the distance from City Hall to \(1^{\text{st}}\) St. along Cedar Rd., and \(x\) is from City Hall to \(2^{\text{nd}}\) St. Then the proportion would be \(\frac{2.1}{2.8}=\frac{2.4}{x}\)

Ah! That's probably the mistake. Let's re - define:

If we consider the two triangles formed, with the vertex at City Hall, and the two bases along \(1^{\text{st}}\) St. and \(2^{\text{nd}}\) St. (which are parallel). Then the ratio of the corresponding sides should be equal. So, the side along Aspen Rd. from City Hall to \(1^{\text{st}}\) St. is \(2.1\) mi, from City Hall to \(2^{\text{nd}}\) St. is \(2.8\) mi. The side along Cedar Rd. from City Hall to \(1^{\text{st}}\) St. is \(2.4\) mi, and from City Hall to \(2^{\text{nd}}\) St. is \(x\) mi.

Then by the Triangle Proportionality Theorem (since \(1^{\text{st}}\parallel2^{\text{nd}}\)), \(\frac{2.1}{2.8}=\frac{2.4}{x}\)

Step3: Solve the new proportion

Cross - multiply:
\[ 2.1x=2.8\times2.4 \]
Calculate \(2.8\times2.4\): \(2.8\times2.4 = 6.72\)

Then,
\[ x=\frac{6.72}{2.1} \]
\(6.72\div2.1 = 3.2\)

Answer:

\(3.2\)