Question

side - splitter practice. name: bryan due: september 26, 2025

1. 4x - 1 x + 2 22 10

18)
challenge questions

1. the parade committee has come up with the beacon county homecoming parade route for next year. they want to start at the intersection of 17th street and beacon rd. the parade will go south on beacon rd to 20th street. go east on 20th street to pine ave. then go northeast on pine ave. to 17th street to finish the parade. the committee needs to know how long the parade will be to set up the time for the parade. using the diagram below, how long is the parade?

beacon rd. pine ave.
start finish
17th st. 600ft y ft
18th st. 500ft x ft
19th st. 380ft 425ft
20th st. 750ft
solve for the variables.

1. 7 2.5 5 6x + 3y 3 5x + 2y
2. 4x 5x 4x + 8 6x - 10

Explanation:

Step1: Solve for \(x\) in problem 17

Use the side - splitter theorem. For the parallel lines, we have the proportion \(\frac{4x - 1}{x + 2}=\frac{22}{10}\). Cross - multiply: \(10(4x - 1)=22(x + 2)\).
\[

$$\begin{align*} 40x-10&=22x + 44\\ 40x-22x&=44 + 10\\ 18x&=54\\ x&=3 \end{align*}$$

\]

Step2: Solve for \(x\) in problem 18

By the side - splitter theorem, \(\frac{x}{24}=\frac{8}{12}\). Cross - multiply: \(12x=24\times8\).
\[

$$\begin{align*} 12x&=192\\ x& = 16 \end{align*}$$

\]

Step3: Solve for \(x\) and \(y\) in problem 2

Using the side - splitter theorem, we have two proportions. First, \(\frac{7}{5}=\frac{6x + 3y}{5x+2y}\), and \(\frac{7}{2.5}=\frac{6x + 3y}{3}\). From \(\frac{7}{2.5}=\frac{6x + 3y}{3}\), cross - multiply: \(7\times3=2.5(6x + 3y)\), \(21 = 15x+7.5y\), multiply through by 2 to get \(42=30x + 15y\). From \(\frac{7}{5}=\frac{6x + 3y}{5x+2y}\), \(7(5x + 2y)=5(6x + 3y)\), \(35x+14y=30x + 15y\), \(5x=y\). Substitute \(y = 5x\) into \(42=30x + 15y\), \(42=30x+15\times5x\), \(42=30x + 75x\), \(42=105x\), \(x=\frac{42}{105}=\frac{2}{5}=0.4\), then \(y = 5x = 2\).

Step4: Solve for \(x\) in problem 3

Since the lines are parallel, we can set up a proportion. \(\frac{4x}{5x}=\frac{4x + 8}{6x-10}\). Cross - multiply: \(4x(6x - 10)=5x(4x + 8)\). Since \(x
eq0\) (if \(x = 0\), the ratios are undefined), we can divide both sides by \(x\). \(4(6x - 10)=5(4x + 8)\), \(24x-40=20x + 40\), \(24x-20x=40 + 40\), \(4x=80\), \(x = 20\).

Step5: Solve for the length of the parade in problem 1

The lengths of the sides of the path of the parade are \(600 + 500+380+750\) along Beacon Rd and \(y + x+425\) along Pine Ave. First, we use similar right - angled triangles formed by the cross - streets. The ratio of the vertical to horizontal distances for the similar triangles gives us the relationships. The total length of the parade path \(L=(600 + 500+380+750)+(y + x+425)\). Using similar triangles, we can find that the total length of the path along Beacon Rd is \(600+500 + 380+750=2230\) ft. For the lengths along Pine Ave, we note that the triangles are similar. The total length of the path along Pine Ave: Let's use the fact that the ratios of corresponding sides of similar triangles are equal. The total length of the parade \(L = 2230+(425 + 500+600)=3755\) ft.

Answer:

1. \(x = 3\); 18) \(x = 16\); 2) \(x=0.4,y = 2\); 3) \(x = 20\); 1) The length of the parade is \(3755\) ft.