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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

4. Parallel Lines and Proportional Parts

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Exercise 41 Page 579

By the Converse Triangle Proportionality Theorem, if a line intersects two sides of a triangle and separates the sides into proportional corresponding segments, then the line is parallel to the third side of the triangle.

x=6

Practice makes perfect

We are given △ADF and we want to determine the value of x so that BC is parallel to DF.

By the Converse Triangle Proportionality Theorem, if a line intersects two sides of a triangle and separates the sides into proportional corresponding segments, then the line is parallel to the third side of the triangle. Let's write a proportion using the expressions for the lengths of the segments. AB/BD = AC/CF ⇕ 12/3x-2=15/3x+2 Let's solve this equation for x.

12/3x-2=15/3x+2

▼

Solve for x

LHS * (3x-2)=RHS* (3x-2)

12=15/3x+2(3x-2)

LHS * (3x+2)=RHS* (3x+2)

12(3x+2)=15(3x-2)

Distribute 12

36x+24=15(3x-2)

Distribute 15

36x+24=45x-30

LHS+30=RHS+30

36x+54=45x

LHS-36x=RHS-36x

54=9x

.LHS /9.=.RHS /9.

6=x

Rearrange equation

x=6

We found that if x=6, ABBD is equal to ACCF, and by this BC is parallel to DF.

Subchapter links

Exercises p.577-581