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Section 2 - Proportions and Similar Figures

Page history last edited by PBworks 16 years, 2 months ago

 

Chapter 4

Section 2: Proportions and Similar Figures

 

 

Similar figures are figures that are the same shape and have the same measure of angles.

 

    If you have two simlar figures, one is triangle A and the other triangle B, and you have all of the measurments of the sides, except one side of triangle B.......how do you find the measurement of that side?? Here's how.

 

 

 

  1. You find the corresponding side of the missing side of triangle B.
  2. Using a proportion, you take the measurment that you know and put it on the bottom, with a variable on top standing for the missing side of triangle B.
  3. you now have one proportion and you need another one that is equal to it in someway.
  4. The other proportion would come from the triangles as well
  5. You find two sides that correspond with each other and put that into a proportion equaling the other one
Here's what this will look like....
X/6 = 9/3
Here's an example:
You have two triangles. one ABC and the other STU. ABC is similar to STU, but larger.
Here are the measurements:
Triangle ABC measurements
Side AB=18
Side BC=6
                                                                                                                Side AC=X <----- *this tells that you dont know the measurement of this side*
Triangle STU measurements
Side ST=6
Side TU=2
Side US=3
 
 

You now have the measurements. To find X, you use proportions. To get these proportions, you use corresponding sides.    
*AB is corrosponding to ST and BC is to TU and so on and so forth.*
So your proportions should be set up to look like this:          
18/6 = X/3
(18 is on top of 6 and X is on top of 3)
  ***As you learned in the previous section, you will cross multiply forming an equation that looks like this:
  6X=54
....then you just solve for X and you get the missing side to triangle ABC.
**Doing this with triangles is not the only situation where you can use this. You can also use it with map scales.**
Example:   You have a map scale that reads   1 in.= 5 miles
You want to find how many miles it is from your house to your friend's house. on the map, the distance is 7 inches.
How would you use a proportion to solve this?
Here's how:
1in./5mi. = 7in/X mi.
**X = miles to friend's**
put it into an equation: 
1X=35
...solve for X...
X=35 miles.
**you can also use this process in problems dealing with heights and shadows.**
Exapmle: there is a boy that is 3ft tall. his shadow is 5ft. The tree next to him has a shadow of 8ft.
How tall is the tree?
3/5 = X/8
5X=24
X= 4.8
so the tree is 4.8 ft. tall

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