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?

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

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

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

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

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

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

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

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A geometric interpretation . . .

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

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

so that these two areas are equal.

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