Chapter 7.4 Similarity in Right Triangles
In this study guide you will learn about the similarity in right triangles, the geometric
mean, and corollaries of Theorem 7-3
**THEOREM 7-3**
The altitude to the hypotenuse of a right triangle divides the triangles into two
triangles that are similar to the original triangle and to each other
To Prove Theorem 7-3
This is ∆ABC with line CD as the altitude to the hypotenuse. The altitude creates 3
similar triangles, ∆ABC ~ ∆ADC ~ ∆BDC
Need Proof?
Look, both triangles ∆ADC and ∆BDC share an angle with ∆ABC and both triangles and
∆ABC have 90˚ angles. So, by the AA~ postulate, ∆ADC is similar to ∆ABC. Now,
because triangles ∆ADC and ∆BDC are both similar to ∆ABC, they are also similar to
each other.
Now...
The geometric mean is x in
*ALL NUMBERS ARE POSITIVE*
To find x, you must take the square root of the product of a and b
So when simplified by using cross multiplication x=√ab
Here’s an example:
To find the geometric mean of 4 and 18,
√√√√√√√√√√
Have some practice
Now there are two corollaries of Theorem 7-3 that have to do with the geometric mean
Ready?
**Corollary 1 to Theorem 7-3**
The length of the altitude to the hypotenuse of a right triangle is the geometric
mean of lengths of the segments of the hypotenuse
So in ∆ABC, the Altitude to hypotenuse AB is line CD which is the geometric mean of
lines AD and BD
Need Proof?
∆ACD ~ ∆CBD-----------------------Theorem 7-3
AD CD
Similar corresponding
CD
BD
sides of similar triangles
are proportional
Have some practice:
Look to the triangle above for reference. The two numbers given are the values of lines
AD and DB, find CD
**Corollary 2 to Theorem 7-3**
The altitude to the hypotenuse of a right triangle separates the hypotenuse so
that the length of each leg of the triangle is the geometric mean of the length of
the adjacent hypotenuse segment and the length of the hypotenuse
So in ∆ABC
AND
Need Proof?
∆ACD ~ ∆ABC-----------------------Theorem 7-3
Similar triangles have corresponding
sides that are proportional
∆CBD ~ ∆ABC-----------------------Theorem 7-3
Similar triangles have corresponding
sides that are proportional
Just to get these corollaries implanted in your brain, do these problems
Now you should have these mastered corollaries, so now you have to apply them. YAY
Now try this problem
Need Help?
Here are some more simple problems for applying the corollaries
And here are some more complicated problems
Here are some more complicated problems for applying the corollaries
More practice/review can be found in the textbook (pg.391-396)
ANSWERS ANSWERS ANSWERS
Answers to geometric mean problems
1. 4 and 9
2. 4 and 10
4*9=36
4*10=40
√36= 6
√40= 2√10
3. 4 and 12
4*12=48
√48= 4√3
4. 3 and 48
3*48=144
√144= 12
Answers to Corollary 1 problems
5. 7 and 56
6. 5 and 125
7*56=392
5*125=625
√392= 14√2
√625= 5√5
7. 9 and 24
9*24=216
√216= 6√6
8. 7 and 9
7*9=63
√63= 3√7
Answers to proportion problems
9. r/h=S/h
10. c/a=a/R
11. C/b=b/s
12. r/A=A/c
13. r/h=H/s
14. s/b=B/c
Answer to the simple problems
15. x/6 = 4/6
16. 10/x = x/40
x=4
x=20
19. 16/x=x/9
x=12
20. 144/x=x/25
x=60
Answer to the harder problems
34. y=√9*7 x/9=9+7/x
z/7=7+9/z
y=3√7
x=12
z=4√7
35. y=√6*24
y=12
36. 6=√x*9
36=x*9
x=4
x/24=30/x
x=12√5
y/4=4+9/y
y=2√13
z/6=30/z
z=6√5
z/9=9+4/z
z=3√13
17. 4/x=x/4+21
x=10
18. 9/x= 9=x/9+3
x=6√3