Two vertical poles AB = 15m and CD = 10m are standing apart on a horizontal ground with points A and

Q: Wait, why does the distance between the poles not matter?

That is the beauty of this problem! When you add the two similarity ratios, the ground segments $AE$ and $EC$ combine to form the total distance $AC$. Because $AC$ appears in both the numerator and denominator, it cancels out entirely. The height $h$ is purely a function of the pole heights.

Q: Can I use coordinate geometry instead?

Absolutely! You could place point $A$ at $(0,0)$ and $C$ at $(d,0)$. Then $B$ is $(0,15)$ and $D$ is $(d,10)$. Find the equations of lines $AD$ and $BC$, then solve for their intersection. It works, but similar triangles are much faster and less prone to calculation errors.

Q: Is this formula $\frac{1}{h} = \frac{1}{h_1} + \frac{1}{h_2}$ always true?

Yes, for any two vertical poles of heights $h_1$ and $h_2$ on a horizontal ground, the height of the intersection of the cross-lines is always given by the harmonic mean relation $\frac{1}{h} = \frac{1}{h_1} + \frac{1}{h_2}$. It is a classic result in geometry.

Question

Two vertical poles $AB = 15	ext{m}$ and $CD = 10	ext{m}$ 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 :

Options:

(A) 6

(B) 20/3

(C) 10/3

(D) 5

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a perfectly flat, horizontal stretch of ground. Two vertical poles rise from this ground, standing tall and proud. One pole, let's call it $AB$, reaches a height of $15 	ext{m}$.

A distance away, another pole, $CD$, stands at $10 	ext{m}$. We connect the top of each pole to the base of the other—a line from $A$ to $D$ and another from $C$ to $B$. These lines intersect at a point $P$.

Our mission is to find the height $h$ of this point $P$ above the ground. This is not just a math problem; it is a lesson in the hidden symmetries of our world.

### The Similarity Insight

To solve this, we must look past the complexity and see the triangles. Let $E$ be the point on the ground directly below $P$.

Now, look at $	riangle AEP$ and $	riangle ACD$. Because both $PE$ and $CD$ are vertical, they are parallel. This makes $	riangle AEP$ similar to $	riangle ACD$ by the Angle-Angle (AA) similarity criterion.

From this similarity, we can write the ratio of their corresponding sides:

$$\frac{h}{10} = \frac{AE}{AC}$$

But we are not done. Look at the other side. Consider $	riangle CEP$ and $	riangle CAB$.

Again, because $PE$ is parallel to $AB$, $	riangle CEP$ is similar to $	riangle CAB$. This gives us a second ratio:

$$\frac{h}{15} = \frac{EC}{AC}$$

### The Algebraic Symphony

Now, we have two elegant equations. Let's bring them together. If we add them, we get:

$$\frac{h}{10} + \frac{h}{15} = \frac{AE}{AC} + \frac{EC}{AC}$$

Look closely at the right side. Since $AE$ and $EC$ are segments that make up the entire ground distance $AC$, their sum $AE + EC$ is simply $AC$.

Therefore, the right side simplifies to $\frac{AC}{AC}$, which is exactly $1$. The unknown distance between the poles has vanished, leaving us with a beautiful, simple equation:

$$\frac{h}{10} + \frac{h}{15} = 1$$

### The Final Revelation

We can now solve for $h$ with ease. Factoring out $h$, we get:

$$h \left( \frac{1}{10} + \frac{1}{15} ight) = 1$$

Finding a common denominator of $30$, we have:

$$h \left( \frac{3+2}{30} ight) = 1 \Rightarrow h \left( \frac{5}{30} ight) = 1 \Rightarrow h \left( \frac{1}{6} ight) = 1$$

Thus, the final height is **$h = 6 	ext{m}$**.

This result is profound. The height of the intersection point is independent of the distance between the poles. Whether they are $10$ meters apart or $100$ meters apart, the intersection point $P$ will always be at a height of **$6 	ext{m}$**.

Animated Solution for Mathematics - Trigonometry: Two vertical poles $A B = 15 m$ and $C D = 10 m$ are standing apart on a horizontal ground with points $A$ and $C$ on the ground. If $P$ is the point of intersection of $B C$ and $A D$ , then the height of $P$ (in $m$ ) above the line $A C$ is :

Select Answer:

Visualized Solution

Visualizing the Setup

Defining Point $P$ and Height $h$

Identifying Similar Triangles (Part 1)

Setting up the First Ratio

Identifying Similar Triangles (Part 2)

Setting up the Second Ratio

Combining the Equations

The Geometric Simplification

Solving for $h$ (Factoring)

Solving for $h$ (Fractions)

Final Answer and Conclusion

The Sigma Insight: Heights and Distances

Analyzing the Setup

The Similarity Insight

The Algebraic Symphony

The Final Revelation

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