Question

In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is

• A. 2.5
• B. 3.5
• C. 0.5
• D. 1.5

Answer 1

The Correct Option is C

To solve the problem, we apply the Power of a Point theorem. This theorem states that if two chords intersect at a point inside a circle, then:

\(AE \times EB = CE \times ED\) 

We are given:

• Radius of the circle = 11 cm
• Therefore, diameter CD = \(2 \times 11 = 22 \text{ cm}\)
• CE = 7 cm, so \(ED = 22 - 7 = 15 \text{ cm}\)
• AB = 20.5 cm

Let \(AE = x\). Then \(EB = 20.5 - x\).

Apply the Power of a Point theorem:

\(x(20.5 - x) = 7 \times 15 = 105\)

Expanding:

\(20.5x - x^2 = 105\) 
Rearranging: \(x^2 - 20.5x + 105 = 0\)

We solve this quadratic using the quadratic formula:

\(x = \frac{-(-20.5) \pm \sqrt{(-20.5)^2 - 4 \cdot 1 \cdot 105}}{2 \cdot 1} = \frac{20.5 \pm \sqrt{420.25 - 420}}{2} = \frac{20.5 \pm \sqrt{0.25}}{2} = \frac{20.5 \pm 0.5}{2}\)

So the two roots are:

• \(x = \frac{21}{2} = 10.5 \text{ cm}\)
• \(x = \frac{20}{2} = 10 \text{ cm}\)

Thus, AE = 10 cm and EB = 10.5 cm.

Difference: \(|EB - AE| = 0.5 \text{ cm}\)

Final Answer: \(\boxed{0.5\ \text{cm}}\)