Exercise 5 Page 736

To find the length of PQ, we can start by adding some of the other missing lengths to the given diagram. When examining the diagram, we notice that the chord and the diameter are perpendicular. Therefore, we know that the diameter bisects the chord. Chord length/Bisected → 10/2= 5 This tells us that the length of both QJ and QK is 5.

Now, let's draw a radius from P to K so that we will have a right triangle. Keep in mind that since the radius is constant, its length is always 6.

We want to find the length of leg PQ within △PQK. To find the value of PQ, we will substitute these values into the Pythagorean Theorem.

a^2+b^2=c^2

SubstituteValues

Substitute values

PQ^2+ 5^2=6^2

▼

Solve for x

CalcPow

Calculate power

PQ^2+25=36

SubEqn

LHS-25=RHS-25

PQ^2=11

SqrtEqn

sqrt(LHS)=sqrt(RHS)

PQ=sqrt(11)

CalcRoot

Calculate root

PQ=3.3166...

RoundDec

Round to 2 decimal place(s)

PQ≈ 3.32