Exercise 19 Page 333

We are given the lengths of GH and CB. Let's analyze the given triangle.

Since CB connects two midpoints, C and B, it is a midsegment of △ GHJ. According to the Triangle Midsegment Theorem, the segment connecting the midpoints of the two sides of a triangle is parallel to the third side and is half as long. This allows us to write an equation relating CB and GH. CB=1/2 GH ⇒ 4z-3=1/2( 7z-1) Let's solve the equation for z.

4z-3=1/2(7z-1)

▼

Solve for z

MultEqn

LHS * 2=RHS* 2

2(4z-3)=7z-1

Distr

Distribute 2

8z-6=7z-1

SubEqn

LHS-7z=RHS-7z

z-6=-1

AddEqn

LHS+6=RHS+6

z=5

Now, to find the length GA, note that A is the midpoint of GH. This means that the length of GA must be half of that length. Knowing that z= 5, we can substitute this value into the expression 7z-1 to find GH first. 7z-1 ⇒ 7( 5)-1=34 Finally, since GA is half as long, we should divide 34 by 2 to find its length. GA=34/2=17