Question

de = 20 in, fh = 8 in, fg = 10 in
df =

pq = 30 ft, rq = 27 ft, rt = 18 ft
st =

uy = 27 yd, wx = 45 yd, vw = 50 yd, uv = 36 yd
vy =, vx =

Explanation:

1. For the first - triangle problem (finding \(DF\)):

• Explanation:
• Step 1: Use similar - triangles property
• Triangles \(DFE\) and \(FGH\) are similar. The ratio of their corresponding sides is equal. We know that \(\frac{DF}{FG}=\frac{DE}{FH}\).
• Step 2: Substitute the given values
• Given \(DE = 20\) in, \(FH = 8\) in, and \(FG = 10\) in. Substituting into the proportion \(\frac{DF}{10}=\frac{20}{8}\).
• Step 3: Solve for \(DF\)
• Cross - multiply: \(8\times DF=20\times10\), so \(8DF = 200\). Then \(DF=\frac{200}{8}=25\) in.

1. For the second - triangle problem (finding \(ST\)):

• Explanation:
• Step 1: Use the basic proportionality theorem (Thales' theorem)
• Since \(ST\parallel PQ\), we have \(\frac{RT}{RQ}=\frac{ST}{PQ}\).
• Step 2: Substitute the given values
• Given \(PQ = 30\) ft, \(RQ = 27\) ft, and \(RT = 18\) ft. Substituting into the proportion \(\frac{18}{27}=\frac{ST}{30}\).
• Step 3: Solve for \(ST\)
• Cross - multiply: \(27\times ST=18\times30\), so \(27ST = 540\). Then \(ST=\frac{540}{27}=20\) ft.

1. For the third - triangle problem (finding \(VY\) and \(VX\)):

• Explanation for finding \(VY\):
• Step 1: Use the similarity of triangles \(UVY\) and \(WVX\)
• The ratio of corresponding sides of similar triangles is equal. Let \(\frac{VY}{VX}=\frac{UV}{VW}=\frac{UY}{WX}\). First, we find the ratio \(\frac{UY}{WX}=\frac{27}{45}=\frac{3}{5}\). Also, \(\frac{UV}{VW}=\frac{36}{50}=\frac{18}{25}\). Since the triangles are similar, we can use \(\frac{VY}{VX}=\frac{UY}{WX}\). Let \(VY = x\) and \(VX = y\). We know that \(\frac{x}{y}=\frac{3}{5}\), or \(y=\frac{5}{3}x\).
• Step 2: Use the property of similar - triangle side - length relationships
• We can also use the fact that if we consider the similarity of the two triangles, we can set up a proportion based on the side - length ratios. Since \(\frac{UY}{WX}=\frac{VY}{VX}\), and we know \(UY = 27\) yd and \(WX = 45\) yd. Let's assume the similarity ratio \(k=\frac{UY}{WX}=\frac{3}{5}\).
• We know that \(\frac{VY}{VW}=\frac{UY}{WX}\). Substituting the values, \(\frac{VY}{50}=\frac{27}{45}\). Cross - multiply: \(45\times VY=27\times50\), so \(VY = 30\) yd.
• Step 3: Find \(VX\)
• Since \(\frac{VY}{VX}=\frac{UY}{WX}\), and \(VY = 30\) yd, \(\frac{30}{VX}=\frac{27}{45}\). Cross - multiply: \(27VX=30\times45\), so \(VX = 50\) yd.

Answer:

DF = 25 in
ST = 20 ft
VY = 30 yd
VX = 50 yd