Since QA is a segment whose endpoints are the center and a point on a circle, it is a radius of ⊙ Q.
Now we will determine the length of the segment QA and the radius of ⊙ Q. Let's mark point E on the diagram.
We can see that △ AQE is a right triangle, because AD⊥QE. Therefore, by the Pythagorean Theorem we have the following.
QE^2+AE^2= QA^2
We already know that QE=5. Let's determine the value of AE using the properties of chords. Then, we will use the above equation to find the radius QA.
Finding AE
Note that the length of AE is half the length of AD. Therefore, we will start by determining AD. To find its value we will use the Equidistant Chords Theorem.
Equidistant Chords Theorem
In the same circle or in congruent circles, two chords are congruent if and only if they are equidistant from the center.
From the diagram we know that AD and BC are equidistant from the center Q, so AD and BC are congruent chords. Therefore, their lengths are equal.
AD= BC
Let's substitute 4x+4 for AD and 6x-6 for BC, and solve the equation for x.
Let's recall the equation that we wrote at the start using the Pythagorean Theorem.
QE^2+AE^2=QA^2
We already know that QE= 5 and AE= 12. We will substitute these values into the above equation and simplify.
