Question

using the secant and segments theorem
what is the value of x?
○ $x=2$
○ $x=3$
○ $x=4$
○ $x=6$
in the circle: segment ed = 14, segment ab = 21, segment bc = $x$, segment dc = $x+1$

Explanation:

Step1: Apply Secant Segments Theorem

For point \(C\) outside the circle, the theorem states \(CA \times CB = CE \times CD\).
First, define the segments:
\(CA = CB + BA = x + 21\), \(CB = x\), \(CE = CD + DE = (x+1) + 14 = x + 15\), \(CD = x+1\)
Substitute into the theorem:
$$(x + 21)x = (x + 15)(x + 1)$$

Step2: Expand both sides

Expand left and right hand expressions:
Left: \(x^2 + 21x\)
Right: \(x^2 + x + 15x + 15 = x^2 + 16x + 15\)
Equation becomes:
$$x^2 + 21x = x^2 + 16x + 15$$

Step3: Simplify and solve for x

Subtract \(x^2 + 16x\) from both sides:
$$21x - 16x = 15$$
$$5x = 15$$
$$x = \frac{15}{5} = 3$$

Answer:

x=3