Exercise 63 Page 581

Let's begin with recalling the Incenter Theorem. The angle bisectors of a triangle intersect at a point called the incenter that is equidistant from the sides of the triangle. Since we are given that S is the incenter of triangle PLJ, SQ and KS are congruent.

As △ JSK is a right triangle, we can use the Pythagorean Theorem to evaluate the length of KS. According to this theorem, the sum of the squared legs of a right triangle is equal to its squared hypotenuse. KS^2+JK^2=JS^2 Let's substitute 8 for JK and 10 for JS, and solve for KS.

KS^2+JK^2=JS^2

SubstituteII

JK= 8, JS= 10

KS^2+ 8^2= 10^2

CalcPow

Calculate power

KS^2+64=100

SubEqn

LHS-64=RHS-64

KS^2=36

▼

sqrt(LHS)=sqrt(RHS)

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(KS^2)=sqrt(36)

SqrtPowToNumber

sqrt(a^2)=a

KS=sqrt(36)

CalcRoot

Calculate root

KS=6

The length of KS is 6. As SQ is congruent to this segment, it will also has a length of 6.