Question

if ( jk = 25 ), ( jl = 21 ), ( kl = 14 ), and ( ml = 20 ), calculate ( nl ). image not set to scale.

Explanation:

Step1: Use the SAS (Side - Angle - Side) similarity criterion

Since \(\angle JLK=\angle NLM\) (vertically opposite angles) and \(\angle J=\angle N\) (given), by the SAS similarity criterion, \(\triangle JKL\sim\triangle NML\).

Step2: Set up the proportion

For similar triangles \(\triangle JKL\) and \(\triangle NML\), the ratios of corresponding sides are equal. So, \(\frac{JK}{NL}=\frac{JL}{ML}=\frac{KL}{ML}\). Using \(\frac{JK}{NL}=\frac{JL}{ML}\), we substitute the given values \(JK = 25\), \(JL=21\), and \(ML = 20\).

$$ \frac{25}{NL}=\frac{21}{20} $$

Step3: Solve for \(NL\)

Cross - multiply: \(21\times NL=25\times20\). Then \(NL=\frac{25\times20}{21}=\frac{500}{21}\approx23.81\). But wait, we made a mistake above. The correct proportion from similar triangles \(\triangle JKL\) and \(\triangle NML\) (by ASA, since \(\angle J=\angle N\), \(\angle JLK=\angle NLM\)) is \(\frac{JK}{MN}=\frac{JL}{ML}=\frac{KL}{NL}\). Using \(\frac{KL}{NL}=\frac{JL}{ML}\) (because \(\triangle JKL\sim\triangle NML\) by ASA, \(\angle J=\angle N\), \(\angle JLK = \angle NLM\)).
Substitute \(KL = 14\), \(JL = 21\), \(ML=20\) into \(\frac{KL}{NL}=\frac{JL}{ML}\)

$$ \frac{14}{NL}=\frac{21}{20} $$

Cross - multiply: \(21\times NL=14\times20\)

$$ NL=\frac{14\times20}{21}=\frac{280}{21}=\frac{40}{3}\approx13.33 $$

Answer:

\(NL=\frac{40}{3}\)