Question

1. an obstruction lies between points a and b as shown in figure 23 - 39. distances ae, ed, be, and ec are measured and marked off as shown. distance cd is measured. determine distance ab.

figure 23 - 39
line segments a d, and b c intersects at the point e, such that a e is 30 meters, and d e is 12 meters; b e is 25 meters and c e is 10 meters. obstruction lies between a and b, shown in dotted lines.

Explanation:

Step1: Prove similarity of triangles

Since \(\angle AEB=\angle DEC\) (vertically - opposite angles) and \(\frac{AE}{ED}=\frac{30}{12}=\frac{5}{2}\), \(\frac{BE}{EC}=\frac{25}{10}=\frac{5}{2}\). By the Side - Angle - Side (SAS) similarity criterion, \(\triangle AEB\sim\triangle DEC\).

Step2: Use the property of similar triangles

For similar triangles \(\triangle AEB\) and \(\triangle DEC\), the ratio of corresponding sides is equal. That is \(\frac{AB}{CD}=\frac{AE}{ED}=\frac{BE}{EC}\). We know \(CD = 15\) m (given in the figure from \(C\) to \(D\)). Let \(AB=x\).
Since \(\frac{AB}{CD}=\frac{5}{2}\), substituting \(CD = 15\) m, we get \(x=\frac{5\times15}{2}\).
\(x=\frac{75}{2}=37.5\) m.

Answer:

\(37.5\) m