Question

in the diagram below, \\( \angle omn \cong \angle opq \\). solve for \\( x \\). round your answer to the nearest tenth if necessary.

Explanation:

Step1: Use the basic proportionality theorem (Thales' theorem)

Since \(\angle OMN\cong\angle OPQ\), by the AA (angle - angle) similarity criterion, \(\triangle OPQ\sim\triangle OMN\). Then, the ratios of corresponding sides are equal. That is \(\frac{OP}{OM}=\frac{OQ}{ON}\). Let \(OM = 29.5 + x\), \(OQ=14.7\), \(ON = 14.7 + 12.3=27\). Substituting into the proportion \(\frac{29.5}{29.5 + x}=\frac{14.7}{27}\).

Step2: Cross - multiply

Cross - multiplying gives \(14.7\times(29.5 + x)=29.5\times27\). Expand the left - hand side: \(14.7\times29.5+14.7x=29.5\times27\). Calculate \(14.7\times29.5 = 433.65\) and \(29.5\times27 = 796.5\). So, \(433.65+14.7x=796.5\).

Step3: Solve for \(x\)

Subtract \(433.65\) from both sides: \(14.7x=796.5 - 433.65\). Then \(14.7x = 362.85\). Divide both sides by \(14.7\): \(x=\frac{362.85}{14.7}=24.7\).

Answer:

\(x = 24.7\)