A circle is shown. Chords A C and B D intersect at point E. The length of A E is x, the length of E C is x + 12, the length of B
1 answer:
Complete Question:
A circle is shown. Chords A C and B D intersect at point E. The length of A E is x, the length of E C is x + 12, the length of B E is x + 2, and the length of E D is x + 5. BE is 2 units longer than AE, DE is 5 units longer than AE, and CE is 12 units longer than AE. What is BD?
*The circle is attached below in the attachment
Answer:
BD = 11
Step-by-step explanation:
Given:
AE = x
EC = x + 12
BE = x + 2
ED = x + 5
Required:
Length of BD
Solution:
First, create an equation to find the value of (x + 2)(x + 5) = (x)(x + 12) => intersecting chords theorem
x(x + 5) + 2(x + 5) = x(x + 12)
x² + 5x + 2x + 10 = x² + 12x
Add like terms
x² + 7x + 10 = x² + 12x
Subtract x² from each side
x² + 7x + 10 - x² = x² + 12x - x²
7x + 10 = 12x
10 = 12x - 7x
10 = 5x
10/5 = 5x/5
2 = x
x = 2
Next, find BD:
BD = BE + ED
BD = x + 2 + x + 5
Add like terms
BD = 2x + 7
Plug in the value of x
BD = 2(2) + 7
BD = 4 + 7
BD = 11
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