Title: Lesson 6-5
Author: Tim Heppner
Last saved by: MHE
Company: MCGRAW-HILL

Geometry	Chapter 7

Lesson 7-5

Example 1  Perimeters of Similar Triangles
If △LMN ~ △QRS, QR = 40, RS = 41, SQ = 9, 	
and LM = 9, find the perimeter of △LMN.
Let x represent the perimeter of △LMN. 
The perimeter of △QRS = 40 + 41 + 9 or 90 units.

	LM/QR	= (perimeter of LMN)/(perimeter of QRS)	Proportional Perimeter Theorem
	9/40	= x/90	Substitution
	810	= 40x	Cross products
	20.25	= x	Divide each side by 40.

The perimeter of △LMN is 20.25 units.

Example 2  Write a Proof
Write a paragraph proof.	
Given:	△ABC ~ △LMN
 and  are angle bisectors.
Prove:	AD/LP = CA/NL

Paragraph Proof:	Because corresponding angles of similar triangles are congruent, ∠C ≅ ∠N and ∠CAB ≅ ∠NLM. Since ∠CAB and ∠NLM are bisected, we know that 1/2m∠CAB = 1/2m∠NLM or m∠CAD = m∠NLP. This makes ∠CAD ≅ ∠NLP and △CAD ~ △NLP by AA Similarity. Thus, AD/LP = CA/NL.

Example 3  Medians of Similar Triangles
In the figure △LMN ~ △QRS.  is a median of	
△LMN, and  is a median of △QRS. Find RC if 
MN = 20, MD = 15 and RS = 15.
Let x represent RC.
	MD/RC	= MN/RS	Write a proportion.
	15/x	= 20/15	MD = 15, RC = x, MN = 20, RS = 15
	20x	= 225	Cross products
	x	= 11.25	Divide each side by 20.

Thus, RC = 11.25

Example 4  Solve Problems with Similar Triangles
PHOTOGRAPHY  The drawing below illustrates the position of the camera and the distance from the lens of the camera to the film. Find the height of the sculpture.

△ABC and △EFC are similar. The distance from the lens to the film in the camera is 
CH = 64 mm.  and  are medians of △ABC and △EFC, respectively. If two triangles are similar, then the measures of the corresponding medians are proportional to the measures of the corresponding sides. This leads to the proportion AB/EF = GC/HC .
	AB/EF	= GC/HC	Write the proportion.
	(x m)/(35 mm)	= (6 m)/(64 mm)	AB = x m, EF = 35 mm, GC = 6 m, HC = 64 mm
	64x	= 210	Cross products
	x	≈ 3.3	Divide each side by 64.

The sculpture is about 3.3 meters tall.