# Thread: geometry extra credit help

1. ## geometry extra credit help

help?

2. no congruent triangles ... you have some similar triangles, though.

remember that the sides of similar triangles are proportional.

3. so theyre similar and right (they're both perpendicular to the ground)
but where do i go from there?

4. Hello, natbat77!

On a level plot of ground there stand two vertical posts, one 6 ft tall, the other 10 ft tall.
From the top of each pole a rope is stretched to the foot of the other pole.
How far above ground is the point where the ropes cross?
Code:
* C
* |
*   |
*     |
*       |
*         |
A *           *           | 10
|   *   P *             |
|       *               |
6 |     * :   *           |
|   *   :h      *       |
| *     :           *   |
B * - - - * - - - - - - - * D
a   Q       b

The poles are: . $AB = 6,\;\;CD = 10$
The ropes $AD$ and $BC$ cross at $P$.
$PQ = h$ is perpendicular to $BD.$
Let $a = BQ,\;\;b = QD$

In similar right triangles $PQD\text{ and }ABD\!:\;\;\frac{h}{b} \:=\: \frac{6}{a+b} \quad\Rightarrow\quad b \:=\:\frac{ah}{6-h}$ .[1]

In similar right triangles $PQB \text{ and }CDB\!:\;\;\frac{h}{a} \:=\:\frac{10}{a+b} \quad\Rightarrow\quad b \:=\:\frac{10a - ah}{h}$ .[2]

Equate [1] and [2]: . $\frac{ah}{6-h} \:=\:\frac{10a-ah}{a} \quad\Rightarrow\quad ah^2 \:=\:60a - 6ah - 10ah + ah^2$

. . $16ah \:=\:60a \quad\Rightarrow\quad h \:=\:\frac{60a}{16a} \quad\Rightarrow\quad\boxed{ h \:=\:\frac{15}{4}\text{ ft}}$

5. thank you so much!