Question

in the diagram, dg = 15, gf = 5, eh = 12, and de = 8. to prove that △dfe ~ △gfh by the sss similarity theorem using the information provided in the diagram, it would be enough additional information to know that hf is 2 units and gh is 3 units. hf is 3 units and gh is 2 units. hf is 4 units and gh is 2 units. hf is 3 units and gh is 4 units.

Explanation:

Step1: Calculate the ratio of \(DF\) to \(GF\)

We know that \(DF = DG+GF=15 + 5=20\), and \(GF = 5\). So the ratio \(\frac{DF}{GF}=\frac{20}{5} = 4\)

Step2: Calculate the ratio of \(EF\) to \(HF\)

We know that \(EF=EH + HF\). Let's check each option for the ratio \(\frac{DE}{GH}\) and \(\frac{EF}{HF}\)

• Option 1: If \(HF = 2\) and \(GH=3\), \(EF=12 + 2=14\), \(\frac{DE}{GH}=\frac{8}{3}\), \(\frac{EF}{HF}=\frac{14}{2}=7\), \(\frac{DF}{GF}

eq\frac{DE}{GH}
eq\frac{EF}{HF}\)

• Option 2: If \(HF = 3\) and \(GH = 2\), \(EF=12+3 = 15\), \(\frac{DE}{GH}=\frac{8}{2}=4\), \(\frac{EF}{HF}=\frac{15}{3}=5\), \(\frac{DF}{GF}

eq\frac{DE}{GH}
eq\frac{EF}{HF}\)

• Option 3: If \(HF = 4\) and \(GH=2\), \(EF=12 + 4=16\), \(\frac{DE}{GH}=\frac{8}{2}=4\), \(\frac{EF}{HF}=\frac{16}{4}=4\), \(\frac{DF}{GF}=\frac{20}{5}=4\), \(\frac{DF}{GF}=\frac{DE}{GH}=\frac{EF}{HF}\)
• Option 4: If \(HF = 3\) and \(GH = 4\), \(EF=12+3=15\), \(\frac{DE}{GH}=\frac{8}{4}=2\), \(\frac{EF}{HF}=\frac{15}{3}=5\), \(\frac{DF}{GF}

eq\frac{DE}{GH}
eq\frac{EF}{HF}\)

Answer:

HF is 4 units and GH is 2 units.