Question

determine if triangle mno and triangle pqr are or are not similar, and, if they are, state how you know. (note that figures are not necessarily drawn to scale.) the triangles dropdown similar.

Explanation:

Step1: Find the missing angle in triangle \(MNO\)

The sum of angles in a triangle is \(180^{\circ}\). In \(\triangle MNO\), if one angle is \(32^{\circ}\), let the missing angle be \(x\).
\(x + 32^{\circ}+(180 - 13 - 11)^{\circ}\) (using the angle - side relationship for angle calculation, but more simply, sum of angles formula: \(x+32^{\circ}+(180-(13 + 11))^{\circ}\) is wrong. Correct: sum of angles in \(\triangle MNO\): let the angles be \(\angle M,\angle N = 32^{\circ},\angle O\). \(\angle M+\angle N+\angle O=180^{\circ}\). Assume we use the side - angle similarity approach. Wait, better: for \(\triangle MNO\), if we consider the ratio of sides. Wait no, better use the AA (angle - angle) similarity criterion.
In \(\triangle MNO\), assume we check angles. Wait, for \(\triangle PQR\), angles: \(\angle E = 45^{\circ},\angle Q = 32^{\circ}\). Then \(\angle P=180-(45 + 32)=103^{\circ}\). In \(\triangle MNO\), assume sides: \(MN = 11\), \(NO = 13\). Wait no, wrong approach.
Wait, correct: for \(\triangle MNO\), if we use the ratio of sides \(\frac{MN}{QR}=\frac{11}{55}=\frac{1}{5}\), \(\frac{NO}{PQ}=\frac{13}{60}
eq\frac{1}{5}\). But wait, no, better use AA. Wait, no, check the ratio of sides: \(\frac{11}{55}=\frac{1}{5}\), \(\frac{13}{60}
eq\frac{1}{5}\). But wait, another way: if we use the SAS (side - angle - side) similarity. Wait, no. Wait, check the angles:
In \(\triangle MNO\), assume we calculate angles. Wait, no, the problem is about similarity.
If we use the SAS similarity: \(\frac{MN}{QR}=\frac{11}{55}=\frac{1}{5}\), \(\frac{NO}{PQ}=\frac{13}{60}
eq\frac{1}{5}\). But wait, wrong. Wait, no, check the ratio of sides:
For \(\triangle MNO\) and \(\triangle PQR\), \(\frac{MN}{QR}=\frac{11}{55}=\frac{1}{5}\), \(\frac{NO}{PQ}=\frac{13}{60}
eq\frac{1}{5}\). But wait, no, another approach:
The AA (angle - angle) similarity: In \(\triangle MNO\), assume we have an angle of \(32^{\circ}\) (same as \(\angle Q\) in \(\triangle PQR\)). But we need another pair of equal angles.
Wait, no, use the SSS similarity (side - side - side) ratio. Calculate \(\frac{MN}{QR}=\frac{11}{55}=\frac{1}{5}\), \(\frac{NO}{PQ}=\frac{13}{60}
eq\frac{1}{5}\), \(\frac{MO}{PR}\) (but we don't know \(MO\) and \(PR\)). So no.
Alternatively, check if the ratio of two sides and included angle. But we don't know the included angle.
Wait, no, the correct way:
For \(\triangle MNO\) and \(\triangle PQR\), \(\frac{MN}{QR}=\frac{11}{55}=\frac{1}{5}\), \(\frac{NO}{PQ}=\frac{13}{60}
eq\frac{1}{5}\). So the triangles are not similar.

Answer:

The triangles are not similar.