# Thread: Word problem

1. ## Word problem

C is a point on line segment AB such that AB*BC=AC^2. If AB is 6cm, find the length of BC to the nearest millimeter.

I substituted 6 for AB and re-arranged the equation but it didn't do much.

2. Originally Posted by sinjid9
C is a point on line segment AB such that AB*BC=AC^2. If AB is 6cm, find the length of BC to the nearest millimeter.

I substituted 6 for AB and re-arranged the equation but it didn't do much.

The equation you've got has two variables in it. You need another equation to solve for both variables. Don't forget to draw a picture. What realtionship do AC and CB have to AB?

3. Hello sinjid9!

Did you make a sketch?

$C$ is a point on line segment $AB$ such that: . $AB\cdot BC \:=\:AC^2$ .[1]

If $AB$ is 6 cm, find the length of $BC$ to the nearest millimeter.
Code:
      A           C       B
*-----------*-------*
:    6-x    :   x   :

Let $x \:=\: BC$ . . (Note that: . $0 \leq x \leq 6$)

Then: $6-x \:=\:AC$

Substitute into [1]: . $6x \:=\:(6-x)^2 \quad\Rightarrow\quad x^2 - 18x + 36 \:=\:0$

Quadratic Formula: . $x \;=\; \frac{18 \pm\sqrt{18^2-4\cdot36}}{2(1)} \;=\;\frac{18 \pm6\sqrt{5}}{2} \;=\;9\pm3\sqrt{5}$

Therefore: . $x \;=\; 3(3-\sqrt{5} \;=\;2.291796068\text{ cm} \;\approx\;\boxed{2.3\text{ mm}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

A geometric interpretation . . .

Given a line segment $AB$, find an interior point $C$

Code:
      A           C       B
*-----------*-------*
:     x     :   y   :

so that these two areas are equal.

Code:
      *-------------------*
|                   |
y |                   | y
|                   |
*-----------*-------*
|     x     |   y
|           |
x |           |
|           |
|           |
*-----------*
x