Two chords AB and CD of a circle intersect at A, AB is the diameter and CD is `bot ` to AB .AE = 2 cm AB = 10 cm,Find value of ED.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Circle and Chords
We have a circle with diameter AB and another chord CD that is perpendicular to AB at point E. We know:
- AE = 2 cm
- AB = 10 cm

### Step 2: Determine the Radius
Since AB is the diameter of the circle:
- The radius (r) = AB / 2 = 10 cm / 2 = 5 cm.

### Step 3: Find the Length of OE
The distance from the center O of the circle to point E can be calculated as follows:
- Since AE = 2 cm, then the distance from O to E (OE) is the radius minus AE:
- OE = OA - AE = 5 cm - 2 cm = 3 cm.

### Step 4: Use the Right Triangle OED
In triangle OED, we have:
- OE = 3 cm (the distance from the center to point E)
- OD = radius = 5 cm (the distance from the center to point D)

### Step 5: Apply the Pythagorean Theorem
We can use the Pythagorean theorem to find ED:
- In triangle OED, we have:
\[ OD^2 = OE^2 + ED^2 \]
Substituting the known values:
\[ 5^2 = 3^2 + ED^2 \]
\[ 25 = 9 + ED^2 \]
\[ ED^2 = 25 - 9 = 16 \]

### Step 6: Solve for ED
Taking the square root of both sides gives:
\[ ED = \sqrt{16} = 4 \text{ cm} \]

### Final Answer
The value of ED is 4 cm.

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