We know that JK=LM=24, NP= 3x, and NQ= 7x-12. Since NJ is a segment whose endpoints are the center and a point on the circle, it is a radius of ⊙ N.
Now we will determine the length of segment NJ and the radius of ⊙ N. Note that △ JNP is a right triangle, because JK⊥NP. Therefore, by the Pythagorean Theorem we have the following.
NP^2+PJ^2= NJ^2
Let's determine the values of NP and PJ using the properties of chords. Then we will use the above equation to find the radius NJ.
In the same circle or in congruent circles, two chords are congruent if and only if they are equidistant from the center.
Recall that JK=LM=24, so JK and LM are congruent chords. Therefore, these two chords are equidistant from the center N. The distance of JK from N is NP, and the distance of LM from N is NQ.
NP= NQ
Let's substitute 3x for NP and 7x-12 for NQ, and solve the equation for x.
If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc.
Now let's take a look at the diagram.
We can see that RS is the diameter of ⊙ N. Additionally, RS is perpendicular to JK. Therefore, RS bisects JK, and PJ=PK= 12 JK. We know that JK= 24, so let's find PJ.
Let's recall the equation that we have written at the start using the Pythagorean Theorem.
NP^2+PJ^2=NJ^2
We already know that NP= 9 and PJ= 12. We will substitute these values into the above equation and simplify.
