Question:
Two vertical poles \(AB=15\) m and \(CD=10\) m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m) above the line AC is :
Two vertical poles \(AB=15\) m and \(CD=10\) m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P (in m) above the line AC is :
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Use similar triangles (or coordinates): the height of the intersection of the two diagonals is \(\frac{h_1h_2}{h_1+h_2}\).
Updated On: Nov 5, 2026
- 6
- 10/3
- 20/3
- 5
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The Correct Option is A
Solution and Explanation
Let the foot of the perpendicular from P to AC be Q, with \(PQ=h\), \(AQ=x\), \(QC=y\).
\(\triangle AQP\sim\triangle ACD\): \(\frac{h}{10}=\frac{x}{x+y}\). \(\triangle CQP\sim\triangle CAB\): \(\frac{h}{15}=\frac{y}{x+y}\).
Adding: \(\frac{h}{10}+\frac{h}{15}=1\Rightarrow h\cdot\frac{5}{30}=1\Rightarrow h=6\).
Answer: (1) 6
\(\triangle AQP\sim\triangle ACD\): \(\frac{h}{10}=\frac{x}{x+y}\). \(\triangle CQP\sim\triangle CAB\): \(\frac{h}{15}=\frac{y}{x+y}\).
Adding: \(\frac{h}{10}+\frac{h}{15}=1\Rightarrow h\cdot\frac{5}{30}=1\Rightarrow h=6\).
Answer: (1) 6
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