Question

consider the two triangles. to prove that \\( \triangle lmn \sim \triangle xyz \\) by the sss similarity theorem using the information provided in the diagram, it would be enough additional information to know that
\\( \bigcirc \\) lm is 3 units and xz is 5 units.
\\( \bigcirc \\) lm is 4 units and xz is 6 units.
\\( \bigcirc \\) lm is 5 units and xz is 3 units.
\\( \bigcirc \\) lm is 6 units and xz is 4 units.

Explanation:

Step1: Recall SSS similarity theorem

For two triangles \(\triangle LMN\) and \(\triangle XYZ\) to be similar by SSS (Side - Side - Side) similarity theorem, the ratios of their corresponding sides must be equal. That is \(\frac{LM}{XY}=\frac{MN}{YZ}=\frac{LN}{XZ}\). We know \(MN = 3\), \(YZ=9\), so \(\frac{MN}{YZ}=\frac{3}{9}=\frac{1}{3}\).

Step2: Check each option

• Option 1: If \(LM = 3\) and \(XZ = 5\), \(XY = 12\). Then \(\frac{LM}{XY}=\frac{3}{12}=\frac{1}{4}

eq\frac{1}{3}\) (from \(MN/YZ\)) and \(\frac{LN}{XZ}\) (since \(LN = 2\)) \(\frac{2}{5}
eq\frac{1}{3}\).

• Option 2: If \(LM = 4\) and \(XZ = 6\), \(XY = 12\). Then \(\frac{LM}{XY}=\frac{4}{12}=\frac{1}{3}\), \(\frac{MN}{YZ}=\frac{3}{9}=\frac{1}{3}\), \(\frac{LN}{XZ}=\frac{2}{6}=\frac{1}{3}\).
• Option 3: If \(LM = 5\) and \(XZ = 3\), \(XY = 12\). Then \(\frac{LM}{XY}=\frac{5}{12}

eq\frac{1}{3}\) (from \(MN/YZ\)) and \(\frac{LN}{XZ}=\frac{2}{3}
eq\frac{1}{3}\).

• Option 4: If \(LM = 6\) and \(XZ = 4\), \(XY = 12\). Then \(\frac{LM}{XY}=\frac{6}{12}=\frac{1}{2}

eq\frac{1}{3}\) (from \(MN/YZ\)) and \(\frac{LN}{XZ}=\frac{2}{4}=\frac{1}{2}
eq\frac{1}{3}\).

Answer:

LM is 4 units and XZ is 6 units.