geometry section 5-2 1112

SECTION 5-2Medians and Altitudes of Triangles

Thursday, March 1, 2012

ESSENTIAL QUESTIONS

How do you identify and use medians in triangles?

How do you identify and use altitudes in triangles?


VOCABULARY1. Median:

2. Centroid:

3. Altitude:

4. Orthocenter:


VOCABULARY1. Median: A segment in a triangle that connects a vertex to

the midpoint of the opposite side

2. Centroid:

3. Altitude:

4. Orthocenter:




2. Centroid: The point of concurrency where the medians intersect in a triangle (always inside)

3. Altitude:

4. Orthocenter:





3. Altitude: A segment in a triangle that connects a vertex to the opposite side so that the side and altitude are perpendicular

4. Orthocenter:





3. Altitude: A segment in a triangle that connects a vertex to the opposite side so that the side and altitude are perpendicular

4. Orthocenter: The point of concurrency where the altitudes of a triangle intersect


5.7 - CENTROID THEOREM

The centroid of a triangle is two-thirds the distance from each vertex to the midpoint of the opposite side


5.7 - CENTROID THEOREM

The centroid of a triangle is two-thirds the distance from each vertex to the midpoint of the opposite side

If G is the centroid of ∆ABC, then

AG =

23

AF , BG =23

BD, and CG =23

CE


Special Segments and Points in TrianglesSpecial Segments and Points in TrianglesSpecial Segments and Points in TrianglesSpecial Segments and Points in TrianglesSpecial Segments and Points in Triangles

Name Example Point of Concurrency

Special Property Example

Perpendicular Bisector

CircumcenterCircumcenter is equidistant from

each vertex

Angle Bisector

IncenterIn center is

equidistant from each side

Median CentroidCentroid is two-

thirds the distance from vertex to

opposite midpoint

Altitude OrthocenterAltitudes are concurrent at orthocenter


EXAMPLE 1

In ∆XYZ, P is the centroid and YV = 12. Find YP and PV.


EXAMPLE 1


YP =

23

YV


EXAMPLE 1


YP =

23

YV

YP =

23

(12)


EXAMPLE 1


YP =

23

YV

YP =

23

(12)

YP = 8


EXAMPLE 1


YP =

23

YV

YP =

23

(12)

YP = 8

PV = YV − YP


EXAMPLE 1


YP =

23

YV

YP =

23

(12)

YP = 8

PV = YV − YP

PV =12 − 8


EXAMPLE 1


YP =

23

YV

YP =

23

(12)

YP = 8

PV = YV − YP

PV =12 − 8

PV = 4


EXAMPLE 2

In ∆ABC, CG = 4. Find GE.


EXAMPLE 2


CG =

23

CE


EXAMPLE 2


CG =

23

CE

4 =

23

CE


EXAMPLE 2


CG =

23

CE

4 =

23

CE

CE = 6


EXAMPLE 2


CG =

23

CE

4 =

23

CE

CE = 6

GE = CE −CG


EXAMPLE 2


CG =

23

CE

4 =

23

CE

CE = 6

GE = CE −CG

GE = 6 − 4


EXAMPLE 2


CG =

23

CE

4 =

23

CE

CE = 6

GE = CE −CG

GE = 6 − 4

GE = 2


EXAMPLE 3

An artist is designing a sculpture that balances a triangle on tip of a pole. In the artistʼs design on the coordinate plane, the vertices are located at (1, 4), (3, 0), and (3, 8). What are the coordinates of the point where the artist should place

the pole under the triangle so that is will balance?


EXAMPLE 3

(1, 4), (3, 0), (3, 8)


EXAMPLE 3

x

y (1, 4), (3, 0), (3, 8)


EXAMPLE 3

x

y

We need to find the centroid, so we start by finding the midpoint

of our vertical side.

(1, 4), (3, 0), (3, 8)


EXAMPLE 3

x

y



M =

3 + 32

,0 + 8

2

⎛⎝⎜

⎞⎠⎟

(1, 4), (3, 0), (3, 8)


EXAMPLE 3

x

y



M =

3 + 32

,0 + 8

2

⎛⎝⎜

⎞⎠⎟ =

62

,82

⎛⎝⎜

⎞⎠⎟

(1, 4), (3, 0), (3, 8)


EXAMPLE 3

x

y



M =

3 + 32

,0 + 8

2

⎛⎝⎜

⎞⎠⎟ =

62

,82

⎛⎝⎜

⎞⎠⎟

= 3,4( )

(1, 4), (3, 0), (3, 8)


EXAMPLE 3

x

y


EXAMPLE 3

x

y

Next, we need the distance from the opposite vertex to this

midpoint.


EXAMPLE 3

x

y


midpoint.

d = (1− 3)2 + (4 − 4)2


EXAMPLE 3

x

y


midpoint.

d = (1− 3)2 + (4 − 4)2

= (−2)2 + 02


EXAMPLE 3

x

y


midpoint.

d = (1− 3)2 + (4 − 4)2

= (−2)2 + 02

= 4


EXAMPLE 3

x

y


midpoint.

d = (1− 3)2 + (4 − 4)2

= (−2)2 + 02

= 4 = 2


EXAMPLE 3

x

y


EXAMPLE 3

x

y The centroid P is 2/3 of this distance from the vertex.


EXAMPLE 3

x


P = 1+

23

(2),4⎛⎝⎜

⎞⎠⎟


EXAMPLE 3

x


P = 1+

23

(2),4⎛⎝⎜

⎞⎠⎟

P = 1+

43

,4⎛⎝⎜

⎞⎠⎟


EXAMPLE 3

x


P = 1+

23

(2),4⎛⎝⎜

⎞⎠⎟

P = 1+

43

,4⎛⎝⎜

⎞⎠⎟

P =

73

,4⎛⎝⎜

⎞⎠⎟


EXAMPLE 4The vertices of ∆HIJ are H(1, 2), I(−3, −3), and J(−5, 1).

Find the coordinates of the orthocenter of ∆HIJ.




x

y




x

y

H

I

J




x

y

To find the orthocenter, find the intersection of two altitudes.

H

I

J




x

y


H

I

J Let’s find the equations for the altitudes coming from I and H.




x

y


H

I


m JI( ) = 1+ 3

−5 + 3




x

y


H

I


m JI( ) = 1+ 3

−5 + 3 =

4−2




x

y


H

I


m JI( ) = 1+ 3

−5 + 3 =

4−2 = −2




x

y


H

I


m JI( ) = 1+ 3

−5 + 3 =

4−2

⊥ m =12 = −2




x

y


H

I


m JI( ) = 1+ 3

−5 + 3 =

4−2

⊥ m =12

m HJ( ) = 2 −1

1+ 5

= −2




x

y


H

I


m JI( ) = 1+ 3

−5 + 3 =

4−2

⊥ m =12

m HJ( ) = 2 −1

1+ 5 =

16

= −2




x

y


H

I


m JI( ) = 1+ 3

−5 + 3 =

4−2

⊥ m =12

m HJ( ) = 2 −1

1+ 5 =

16

⊥ m = −6

= −2




x

y

H

I

J

Altitude through H




x

y

H

I

J ⊥ m =

12

, (1,2)

Altitude through H




x

y

H

I

J ⊥ m =

12

, (1,2)

Altitude through H

y − 2 =

12

(x −1)




x

y

H

I

J ⊥ m =

12

, (1,2)

Altitude through H

y − 2 =

12

(x −1)

y − 2 =

12

x −12




x

y

H

I

J ⊥ m =

12

, (1,2)

Altitude through H

y − 2 =

12

(x −1)

y − 2 =

12

x −12

y =

12

x +32




x

y

H

I

J

Altitude through I




x

y

H

I

J

⊥ m = −6, (−3,−3)

Altitude through I




x

y

H

I

J

⊥ m = −6, (−3,−3)

Altitude through I

y + 3 = −6(x + 3)




x

y

H

I

J

⊥ m = −6, (−3,−3)

Altitude through I

y + 3 = −6(x + 3)

y + 3 = −6x −18




x

y

H

I

J

⊥ m = −6, (−3,−3)

Altitude through I

y + 3 = −6(x + 3)

y + 3 = −6x −18

y = −6x − 21


EXAMPLE 4


EXAMPLE 4

−6x − 21 =

12

x +32


EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x


EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x

−21 =

132

x +32


EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x

−21 =

132

x +32

−

32

−32


EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x

−21 =

132

x +32

−

32

−32

−

452

=132

x


EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x

−21 =

132

x +32

−

32

−32

−

452

=132

x

213

−452

⎛⎝⎜

⎞⎠⎟=

132

x⎛⎝⎜

⎞⎠⎟

213


EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x

−21 =

132

x +32

−

32

−32

−

452

=132

x

213

−452

⎛⎝⎜

⎞⎠⎟=

132

x⎛⎝⎜

⎞⎠⎟

213

x = −

4513


EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x

−21 =

132

x +32

−

32

−32

−

452

=132

x

213

−452

⎛⎝⎜

⎞⎠⎟=

132

x⎛⎝⎜

⎞⎠⎟

213

x = −

4513

−6 −

4513

⎛⎝⎜

⎞⎠⎟− 21 =

12

−4513

⎛⎝⎜

⎞⎠⎟+

32


EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x

−21 =

132

x +32

−

32

−32

−

452

=132

x

213

−452

⎛⎝⎜

⎞⎠⎟=

132

x⎛⎝⎜

⎞⎠⎟

213

x = −

4513

−6 −

4513

⎛⎝⎜

⎞⎠⎟− 21 =

12

−4513

⎛⎝⎜

⎞⎠⎟+

32

−

313

= −3

13


EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x

−21 =

132

x +32

−

32

−32

−

452

=132

x

213

−452

⎛⎝⎜

⎞⎠⎟=

132

x⎛⎝⎜

⎞⎠⎟

213

x = −

4513

−6 −

4513

⎛⎝⎜

⎞⎠⎟− 21 =

12

−4513

⎛⎝⎜

⎞⎠⎟+

32

−

313

= −3

13

−

4513

,−3

13

⎛⎝⎜

⎞⎠⎟


CHECK YOUR UNDERSTANDING

p. 337 #1-4


PROBLEM SET


PROBLEM SET

p. 338 #5-31 odd, 49, 53

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