Question

Two poles of height 2 meters and 3 meters are 5 meters apart. The height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is:

• A. 1.2 meters
• B. 1.0 meters
• C. 5.0 meters
• D. 3.0 meters
• E. None of the above

Hint:

When two lines from tops of poles intersect, use similar triangles or the intercept theorem. It often reduces to a simple harmonic-like relation.

Answer 1

The Correct Option is A

Step 1: Understand the geometry.
Let poles be AB and CD.
AB = 2 m, CD = 3 m.
Base BC = 5 m.
Lines AD and CB intersect at E.
We need to find AE (height of intersection from ground).
Step 2: Use intercept theorem (similar triangles).
By property of intersecting diagonals in trapezium-like figures:
Height at intersection = $\dfrac{h_1 \times d_2 + h_2 \times d_1}{d_1 + d_2}$
where $h_1, h_2$ = heights of poles, and $d_1, d_2$ are base partitions.
Alternatively, solve using similarity:
$\dfrac{x}{2} = \dfrac{5-d}{5}$ and $\dfrac{x}{3} = \dfrac{d}{5}$.
Adding gives $x\left(\dfrac{1}{2} + \dfrac{1}{3}\right) = 1$.
So, $x \cdot \dfrac{5}{6} = 1$.
Hence $x = \dfrac{6}{5} = 1.2$.
Step 3: Conclude.
Thus, the intersection height is 1.2 meters. \[ \boxed{1.2 \text{meters}} \]