Question

1. solve each proportion. a) \\(\frac{6}{-32} = \frac{x}{16}\\) b) \\(\frac{9}{x - 7} = \frac{6}{x}\\) each pair of triangles are similar. write and solve a proportion to find the value of \\(x\\). 7. \\(\triangle vut \sim \triangle klm\\) 8. \\(\triangle uvw \sim \triangle ust\\)

Explanation:

Part A: Solve \(\boldsymbol{\frac{6}{-32}=\frac{x}{16}}\)

Step 1: Cross - Multiply

Cross - multiply the proportion \(\frac{6}{-32}=\frac{x}{16}\). According to the property of proportions \(\frac{a}{b}=\frac{c}{d}\) implies \(a\times d = b\times c\). So we have \(6\times16=-32\times x\).
\(96=-32x\)

Step 2: Solve for \(x\)

Divide both sides of the equation \(96 = - 32x\) by \(-32\) to isolate \(x\).
\(x=\frac{96}{-32}=-3\)

Part B: Solve \(\boldsymbol{\frac{9}{x - 7}=\frac{6}{x}}\)

Step 1: Cross - Multiply

Using the cross - multiplication property for the proportion \(\frac{9}{x - 7}=\frac{6}{x}\), we get \(9\times x=(x - 7)\times6\).
\(9x = 6x-42\)

Step 2: Subtract \(6x\) from both sides

Subtract \(6x\) from both sides of the equation \(9x=6x - 42\) to get \(9x-6x=6x - 42-6x\).
\(3x=-42\)

Step 3: Solve for \(x\)

Divide both sides of the equation \(3x=-42\) by \(3\).
\(x=\frac{-42}{3}=-14\)

Problem 7: \(\boldsymbol{\triangle VUT\sim\triangle KLM}\)

Step 1: Set up the proportion

Since \(\triangle VUT\sim\triangle KLM\), the corresponding sides are proportional. So \(\frac{VU}{KL}=\frac{UT}{LM}\). We know that \(VU = 60\), \(KL = 130\), \(UT=x\) and \(LM = 117\). So the proportion is \(\frac{60}{130}=\frac{x}{117}\).

Step 2: Cross - Multiply

Cross - multiply the proportion \(\frac{60}{130}=\frac{x}{117}\), we get \(60\times117 = 130\times x\).
\(7020 = 130x\)

Step 3: Solve for \(x\)

Divide both sides of the equation \(7020 = 130x\) by \(130\).
\(x=\frac{7020}{130}=54\)

Problem 8: \(\boldsymbol{\triangle UVW\sim\triangle UST}\)

Answer:

s:

• Part A: \(x=-3\)
• Part B: \(x = - 14\)
• Problem 7: \(x = 54\)
• Problem 8: \(x=11\)