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.
- Name two similar triangles.[2]
- If the trapezium area is \(144\text{ cm}^2\), find \(FX\).[2]
Solution:
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Answer:(a) \(\triangle AXB\sim\triangle CXD\) (b) \(4.5\) cm.
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