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chords ac and bd intersect at e, with bd = 7 units, de = 2 units, and a…
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Question

chords ac and bd intersect at e, with bd = 7 units, de = 2 units, and ae = 4 units. what is the length of segment ce? units

Explanation:

Step1: Recall the Intersecting Chords Theorem

The Intersecting Chords Theorem states that if two chords \( AC \) and \( BD \) intersect at a point \( E \), then \( AE \times CE = DE \times BE \).

Step2: Find the length of \( BE \)

We know that \( BD = 7 \) units and \( DE = 2 \) units. Since \( BD = BE + DE \), we can solve for \( BE \):
\( BE = BD - DE = 7 - 2 = 5 \) units.

Step3: Apply the Intersecting Chords Theorem

We know \( AE = 4 \) units, \( DE = 2 \) units, and \( BE = 5 \) units. Let \( CE = x \). According to the theorem:
\( AE \times CE = DE \times BE \)
Substitute the known values:
\( 4 \times x = 2 \times 5 \)

Step4: Solve for \( x \) (length of \( CE \))

Simplify the right - hand side: \( 2\times5 = 10 \)
So we have the equation \( 4x = 10 \)
Divide both sides by 4: \( x=\frac{10}{4}=\frac{5}{2} = 2.5 \)

Answer:

\( 2.5 \) (or \( \frac{5}{2} \))