Height division in a trapezium

Height division in a trapezium

Secondary 2
4 marks
TGM Original Questions

Core stretch

In trapezium \(ABCD\), \(AB\parallel CD\), \(AB=15\) cm, \(CD=9\) cm, and \(EF\) is its perpendicular height. Diagonals meet at \(X\).

Triangles \(AXB\) and \(CXD\) are similar, and their perpendicular heights are in the same ratio as their corresponding bases.

  1. Name two similar triangles.[2]
  2. If the trapezium area is \(144\text{ cm}^2\), find \(FX\).[2]

Solution:

Solution locked

Sign in to view the step-by-step solution

Please join our E Math programme to search for similar questions for more practice.

E Math Tuition
Answer:(a) \(\triangle AXB\sim\triangle CXD\) (b) \(4.5\) cm.

Need help? Join our JC Math tuition classes.

Learn more