body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

4. Parallel Lines and Proportional Parts

Continue to next subchapter

Exercise 26 Page 579

Write a proportion using the Triangle Proportionality Theorem.

x=10 and y=3

Practice makes perfect

Let's analyze the given figure. We will start by finding y.

We can see that two segments on the right are congruent, so 2y+1 = 5y-8. Let's solve this equation to find y.

2y+1 = 5y-8

▼

Solve for y

LHS+8=RHS+8

2y+9 = 5y

LHS-2y=RHS-2y

9 = 3y

.LHS /3.=.RHS /3.

3=y

Rearrange equation

y=3

Since we are given a triangle with a line that is parallel to one of its sides, we can use the Triangle Proportionality Theorem.

If a segment parallel to one of the sides of a triangle is drawn between the other sides, the segment divides the other two sides proportionally.

Let's write a proportion using the expressions for the lengths of the segments. 15x+3/4x-35 = 2y+1/5y-8 Since we know that two segments on the right side are congruent, we know that two segments on the left side are also congruent. Let's solve the equation 15x+3 = 4x-35 to find x.

1/5x+3 = 4x-35

▼

Solve for x

LHS+35=RHS+35

1/5x+38 = 4x

LHS * 5=RHS* 5

x+190 = 20x

LHS-x=RHS-x

190=19x

.LHS /19.=.RHS /19.

10=x

Rearrange equation

x=10

We found that x=10 and y=3.

Subchapter links

Exercises p.577-581