If two chords AB and CD of a circle intersect at point E, and AE = 3 cm, EB = 4

₹0.0

Question: If two chords AB and CD of a circle intersect at point E, and AE = 3 cm, EB = 4 cm, what is the length of CE if ED = 6 cm?

Options:

Correct Answer: 2 cm

Solution:

Using the intersecting chords theorem, AE * EB = CE * ED. Thus, 3 * 4 = CE * 6, which gives CE = 2 cm.

If two chords AB and CD of a circle intersect at point E, and AE = 3 cm, EB = 4 cm, what is the length of CE if ED = 6 cm?

If two chords AB and CD of a circle intersect at point E, and AE = 3 cm, EB = 4 cm, what is the length of CE if ED = 6 cm?

Step 1: Identify the lengths given in the problem. We have AE = 3 cm, EB = 4 cm, and ED = 6 cm.
Step 2: Use the intersecting chords theorem, which states that the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord.
Step 3: Write the equation based on the theorem: AE * EB = CE * ED.
Step 4: Substitute the known values into the equation: 3 * 4 = CE * 6.
Step 5: Calculate the left side of the equation: 3 * 4 = 12.
Step 6: Now the equation looks like this: 12 = CE * 6.
Step 7: To find CE, divide both sides of the equation by 6: CE = 12 / 6.
Step 8: Calculate CE: CE = 2 cm.

Intersecting Chords Theorem – The theorem states that if two chords intersect inside a circle, the products of the lengths of the segments of each chord are equal.