Question

Three parallel lines are cut by two transversals as shown in the given figure. If AB=2 cm, BC=4 cm and DE=1.5 cm, then the length of EF is:

• A. 2 cm
• B. 3 cm
• C. 3.5 cm
• D. 4 cm

Answer 1

The Correct Option is B

To solve this problem, we leverage the geometric principle that when three parallel lines are intersected by two transversals, the corresponding segments they form on one transversal are proportional to the segments they form on the other transversal. Given the segments AB = 2 cm, BC = 4 cm on one transversal and DE = 1.5 cm on the other transversal, we want to find the length of segment EF.

According to the principle of proportionality:

\(\frac{AB}{BC} = \frac{DE}{EF}\) 

Substitute the known values:

\(\frac{2}{4} = \frac{1.5}{EF}\)

Simplify the left side:

\(\frac{1}{2} = \frac{1.5}{EF}\)

To find EF, cross-multiply and solve for EF:

\(EF = 1.5 \times 2\)

Therefore, \(EF = 3 \, \text{cm}\).

The length of EF is therefore 3 cm.