Chords AB and CD of a circle Intersect at E. If AE = 9 cm, BE = 12 cm and CE = 3DE, then the length of DE (in cm) is

Options

A. `9/4`
B. `4`
C. 6
D. 7

Answer

To solve the problem, we will use the property of intersecting chords in a circle. The property states that if two chords intersect inside a circle, then the products of the lengths of the segments of each chord are equal.

Given:
- AE = 9 cm
- BE = 12 cm
- CE = 3DE
- Let DE = x cm

### Step-by-step Solution:

1. **Identify the segments of the chords:**
- For chord AB, we have:
- AE = 9 cm
- BE = 12 cm
- For chord CD, we have:
- CE = 3DE = 3x cm
- DE = x cm

2. **Apply the intersecting chords theorem:**
According to the theorem, we have:
\[ AE \times BE = CE \times DE \]
Substituting the known values:
\[ 9 \times 12 = (3x) \times x \]

3. **Calculate the left side:**
\[ 9 \times 12 = 108 \]

4. **Set up the equation:**
\[ 108 = 3x^2 \]

5. **Solve for x:**
- Divide both sides by 3:
\[ 36 = x^2 \]
- Take the square root of both sides:
\[ x = 6 \]

6. **Conclusion:**
The length of DE is:
\[ DE = x = 6 \text{ cm} \]

### Final Answer:
The length of DE is 6 cm.

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