McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
4. Parallel Lines and Proportional Parts
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Exercise 8 Page 577

x=5 and y=8

Practice makes perfect

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

We can see that two segments on the left are congruent, so 3y = 12y+20. Let's solve this equation to find y.
3y=1/2y+20
â–Ľ
Solve for y
6y=y+40
5y=40
y=8
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. 3y/12y+20 = 20-3x/2x-5 Since we know that two segments on the left side are congruent, we know that two segments on the right side are also congruent. Let's solve the equation 20-3x = 2x-5 to find x.
20-3x = 2x-5
â–Ľ
Solve for x
25-3x=2x
25=5x
5=x
x=5
We found that x=5 and y=8.