McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
4. Parallel Lines and Proportional Parts
Continue to next subchapter

Exercise 23 Page 578

Start with evaluating the greater distance between the second and the third dancer.

1.2 in.

Practice makes perfect

Let's make a simplified diagram that describes the given situation.

We are also given that the length of ID is 1 34 inches. Using this information, we can find the length of HG since the sum of the lengths of IH, HG and GD is equal to the length of ID by the Segment Addition Postulate.
IH+HG+GD=ID
1+HG+ 1/3=1 34
Simplify
HG+1/3=3/4
HG=3/4-1/3
HG=9/12-1/3
HG=9/12-4/12
HG=5/12

Now, let's add this information to our diagram. In this exercise we are asked to find the lower distance between the first two dancers, and this distance corresponds to the length of AB. We can call this length x.

Let's recall the Proportional Parts of Parallel Lines Theorem.

Proportional Parts of Parallel Lines Theorem

If three or more parallel lines intersect two transversals, then they cut off the transversals proportionally.

Since we are given that each of the dancers is parallel, we can write a proportion using the above theorem. AB/BC=IH/HG Let's substitute the lengths of the segments into the above proportion and solve for x. To do this, we will use the Cross Products Property.
AB/BC=IH/HG
x/0.5=1/512
x* 5/12=0.5*1
Solve for x
x* 5/12=0.5
5x/12=0.5
5x=6
x=6/5
x=1.2
The distance between the first two dancers is 1.2 inches.