Question

1. solve for the missing side lengths given the following information

oq = 20 qn = 15 pq = 16 and pm = 10

what is the length of om:_________

what is the length of mn:_________

Explanation:

Step1: Identify Similar Triangles

From the diagram, \(PQ \parallel MN\) (implied by the right angle at \(P\) and the triangle structure), so \(\triangle OPQ \sim \triangle OMN\) by the Basic Proportionality Theorem (Thales' theorem).

Step2: Find \(OM\) Length

We know \(OP = OM - PM\). Let \(OM = x\), then \(OP = x - 10\). By similarity, \(\frac{OQ}{ON}=\frac{OP}{OM}\). \(ON = OQ + QN = 20 + 15 = 35\). So \(\frac{20}{35}=\frac{x - 10}{x}\). Cross - multiply: \(20x=35(x - 10)\). Expand: \(20x = 35x-350\). Subtract \(20x\): \(0 = 15x - 350\). Add \(350\): \(15x=350\)? Wait, no, correction: Wait, \(OQ = 20\), \(ON=OQ + QN = 20 + 15 = 35\), \(OP=OM - PM\), let \(OM = y\), then \(OP=y - 10\). \(\frac{OQ}{ON}=\frac{OP}{OM}\) \(\frac{20}{35}=\frac{y - 10}{y}\) \(20y=35(y - 10)\) \(20y=35y - 350\) \(350 = 35y-20y\) \(15y = 350\)? No, that's wrong. Wait, maybe the right angle is at \(P\) and \(Q\) is on \(ON\), \(P\) is on \(OM\). So \(\triangle PQO\) and \(\triangle NMO\) are similar? Wait, \(PQ = 16\), \(PM = 10\), \(OQ = 20\), \(QN = 15\). Let's use the formula for similar triangles: \(\frac{OQ}{ON}=\frac{PQ}{MN}=\frac{OP}{OM}\). \(ON=20 + 15 = 35\), \(OP=OM - 10\). So \(\frac{20}{35}=\frac{OM - 10}{OM}\) \(20OM=35(OM - 10)\) \(20OM=35OM-350\) \(350 = 15OM\) \(OM=\frac{350}{15}=\frac{70}{3}\)? No, that can't be. Wait, maybe I mixed up the sides. Wait, maybe \(PQ\) is parallel to \(MN\), so \(\frac{OQ}{QN}=\frac{OP}{PM}\)? Wait, no, the correct proportion for similar triangles (if \(PQ\parallel MN\)) is \(\frac{OQ}{ON}=\frac{OP}{OM}=\frac{PQ}{MN}\). Wait, \(OQ = 20\), \(ON=OQ + QN = 35\), \(OP=OM - PM\), \(PM = 10\), let \(OM = x\), \(OP=x - 10\). So \(\frac{20}{35}=\frac{x - 10}{x}\) \(20x=35x - 350\) \(350 = 15x\) \(x=\frac{350}{15}=\frac{70}{3}\approx23.33\)? No, that seems odd. Wait, maybe the triangle is such that \(OP\) and \(OQ\) are corresponding sides, and \(OM\) and \(ON\) are corresponding sides. Wait, another approach: Let's assume that \(\triangle OPQ\sim\triangle OMN\), so \(\frac{OP}{OM}=\frac{OQ}{ON}\). \(OP = OM - PM=OM - 10\), \(ON=OQ + QN = 20 + 15 = 35\). So \(\frac{OM - 10}{OM}=\frac{20}{35}\) Cross - multiply: \(35(OM - 10)=20OM\) \(35OM-350 = 20OM\) \(35OM-20OM=350\) \(15OM = 350\) \(OM=\frac{350}{15}=\frac{70}{3}\approx23.33\). Now for \(MN\): \(\frac{PQ}{MN}=\frac{OQ}{ON}\), \(PQ = 16\), so \(\frac{16}{MN}=\frac{20}{35}\) \(20MN=16\times35\) \(MN=\frac{16\times35}{20}=\frac{560}{20}=28\). Wait, but let's check the first part again. If \(OM=\frac{70}{3}\approx23.33\), \(OP=\frac{70}{3}-10=\frac{70 - 30}{3}=\frac{40}{3}\), and \(\frac{OP}{OM}=\frac{\frac{40}{3}}{\frac{70}{3}}=\frac{40}{70}=\frac{4}{7}\), \(\frac{OQ}{ON}=\frac{20}{35}=\frac{4}{7}\), so that's correct.

Wait, but maybe the problem is using the geometric mean or right - triangle altitude? Wait, the right angle at \(P\), so \(PQ\) is perpendicular to \(OM\), and \(MN\) is perpendicular to \(OM\) (since \(PQ\parallel MN\) and \(PQ\perp OM\)). So \(\triangle OPQ\) and \(\triangle OMN\) are right - triangles and similar.

So, length of \(OM\): Let's re - do the proportion. \(\frac{OQ}{ON}=\frac{OP}{OM}\), \(OQ = 20\), \(ON=20 + 15 = 35\), \(OP=OM - 10\). Let \(OM = x\), then \(\frac{20}{35}=\frac{x - 10}{x}\) \(20x=35x - 350\) \(350 = 15x\) \(x=\frac{350}{15}=\frac{70}{3}\approx23.33\). For \(MN\), \(\frac{PQ}{MN}=\frac{OQ}{ON}\), \(PQ = 16\), so \(\frac{16}{MN}=\frac{20}{35}\) \(MN=\frac{16\times35}{20}=28\).

Wait, but maybe I made a mistake in the proportion. Let's think again. If \(PQ\parallel MN\), then \(\frac{OQ}{QN}=\frac{OP}{PM}\). \(OQ = 20\), \(QN = 15\), \(PM = 10\), so \(\frac{20}{15}=\frac{OP}{10}\) \(OP=\frac{20\times10}{15}=\frac{40}{3}\approx13.33\). Then \(OM=OP + PM=\frac{40}{3}+10=\frac{40 + 30}{3}=\frac{70}{3}\approx23.33\), which matches the previous result. And for \(MN\), \(\frac{PQ}{MN}=\frac{OP}{OM}\), \(PQ = 16\), \(OP=\frac{40}{3}\), \(OM=\frac{70}{3}\), so \(\frac{16}{MN}=\frac{\frac{40}{3}}{\frac{70}{3}}=\frac{40}{70}=\frac{4}{7}\), \(MN = 16\times\frac{7}{4}=28\).

Answer:

Length of \(OM\): \(\frac{70}{3}\) (or approximately \(23.33\))
Length of \(MN\): \(28\)