The length of a line segment AB is 2 cm. It is divided into two parts at a point C such that AC 2= AB × CB. What is the length of CB?
3 – √5 cm
The question asks us to find the length of a segment CB, where a point C divides a line segment AB of a known length. We are given the total length of the line segment AB and a specific relationship between the lengths of the segments AC and CB, involving the total length AB.
Key Information Provided:
AC2 = AB × CB.We need to determine the length of CB based on this information.
Let's denote the length of the line segment CB as \(x\) cm.
Since C is a point on the line segment AB, the length of AB is the sum of the lengths of AC and CB:
AB = AC + CB
We know AB = 2 cm and we've defined CB = \(x\) cm. So, we can find the length of AC:
\(2 = AC + x\)
\(AC = 2 - x\) cm.
Now, let's use the given condition: \(AC^2 = AB \times CB\).
Substitute the expressions for AC, AB, and CB into this equation:
\((2 - x)^2 = 2 \times x\)
We need to solve the equation \((2 - x)^2 = 2x\) for \(x\). Let's expand and rearrange the equation to form a standard quadratic equation (\(ax^2 + bx + c = 0\)).
\((2 - x)^2 = 2^2 - 2(2)(x) + x^2 = 4 - 4x + x^2\)
\(4 - 4x + x^2 = 2x\)
\(x^2 - 4x - 2x + 4 = 0\)
\(x^2 - 6x + 4 = 0\)
\(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(4)}}{2(1)}\)
\(x = \frac{6 \pm \sqrt{36 - 16}}{2}\)
\(x = \frac{6 \pm \sqrt{20}}{2}\)
\(x = \frac{6 \pm 2\sqrt{5}}{2}\)
\(x = \frac{2(3 \pm \sqrt{5})}{2}\)
\(x = 3 \pm \sqrt{5}\)
This gives us two possible values for the length of CB (\(x\)):
We need to determine which of the two possible values for \(x\) is the correct length for the segment CB.
Remember that CB (\(x\)) is a part of the line segment AB, which has a total length of 2 cm. Therefore, the length of CB must be less than 2 cm.
Let's evaluate the two possible solutions:
We should also check if the corresponding length of AC is positive.
If \(x = 3 - \sqrt{5}\), then \(AC = 2 - x = 2 - (3 - \sqrt{5}) = 2 - 3 + \sqrt{5} = \sqrt{5} - 1\). Since \(\sqrt{5} \approx 2.236\), \(AC \approx 2.236 - 1 = 1.236\) cm, which is positive.
Thus, the only valid length for CB is \(3 - \sqrt{5}\) cm.
Let's verify the condition \(AC^2 = AB \times CB\) using the valid lengths:
Calculate \(AC^2\):
\(AC^2 = (\sqrt{5} - 1)^2 = (\sqrt{5})^2 - 2(\sqrt{5})(1) + 1^2 = 5 - 2\sqrt{5} + 1 = 6 - 2\sqrt{5}\)
Calculate \(AB \times CB\):
\(AB \times CB = 2 \times (3 - \sqrt{5}) = 6 - 2\sqrt{5}\)
Since \(AC^2 = 6 - 2\sqrt{5}\) and \(AB \times CB = 6 - 2\sqrt{5}\), the condition is satisfied.
Therefore, the length of CB is \(3 - \sqrt{5}\) cm.

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