Notice that the secant segment is divided into two congruent segments. Look for a theorem involving a tangent and a secant to a circle.
a=bsqrt(2). See solution.
Practice makes perfect
We are given the figure below. Let's mark the intersection points.
Our mission is to find a. To do so we use the steps below.
Step 1
We notice that RA and RC are tangent and secant to ⊙ M, respectively. They also intersect each other outside the circle at point R.
Step 2
We apply Theorem 10.17, which states that the square of the measure of the tangent is equal to the product of the measures of the secant and its external secant segment.
RA^2 = RB* RC
Step 3
By the Segment Addition Postulate, we write RC as RB+BC.
RA^2 = RB(RB+BC)
Now, we substitute the corresponding measures into the equation above.
a^2 = b(b+c)
Step 4
From the diagram, we notice that RB≅BC, which implies that b=c. Let's substitute this into the latter equation.
a^2 = b(b+b)
⇒
a^2 = 2b^2
Step 5
Finally, by taking the square root on both sides of the equation we will get the value of a.
sqrt(a^2) = sqrt(2b^2) ⇒
a = bsqrt(2)
