geometry extra credit help

natbat77 · #1 $Old$ 11-16-2008, 07:42 PM

help?

skeeter · #2 $Old$ 11-16-2008, 07:54 PM

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

remember that the sides of similar triangles are proportional.

natbat77 · #3 $Old$ 11-16-2008, 07:55 PM

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

Soroban · #4 $Old$ 11-16-2008, 09:34 PM

Hello, natbat77!

Quote:

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

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

natbat77 · #5 $Old$ 11-16-2008, 10:10 PM

thank you so much!