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?

Thursday, March 1, 2012

VOCABULARY1. Median:

2. Centroid:

3. Altitude:

4. Orthocenter:

Thursday, March 1, 2012

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

the midpoint of the opposite side

2. Centroid:

3. Altitude:

4. Orthocenter:

Thursday, March 1, 2012

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

the midpoint of the opposite side

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

3. Altitude:

4. Orthocenter:

Thursday, March 1, 2012

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

the midpoint of the opposite side

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

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:

Thursday, March 1, 2012

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

the midpoint of the opposite side

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

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

Thursday, March 1, 2012

5.7 - CENTROID THEOREM

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

Thursday, March 1, 2012

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

Thursday, March 1, 2012

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

Thursday, March 1, 2012

EXAMPLE 1

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

Thursday, March 1, 2012

EXAMPLE 1

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

YP =

23

YV

Thursday, March 1, 2012

EXAMPLE 1

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

YP =

23

YV

YP =

23

(12)

Thursday, March 1, 2012

EXAMPLE 1

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

YP =

23

YV

YP =

23

(12)

YP = 8

Thursday, March 1, 2012

EXAMPLE 1

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

YP =

23

YV

YP =

23

(12)

YP = 8

PV = YV − YP

Thursday, March 1, 2012

EXAMPLE 1

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

YP =

23

YV

YP =

23

(12)

YP = 8

PV = YV − YP

PV =12 − 8

Thursday, March 1, 2012

EXAMPLE 1

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

YP =

23

YV

YP =

23

(12)

YP = 8

PV = YV − YP

PV =12 − 8

PV = 4

Thursday, March 1, 2012

EXAMPLE 2

In ∆ABC, CG = 4. Find GE.

Thursday, March 1, 2012

EXAMPLE 2

In ∆ABC, CG = 4. Find GE.

CG =

23

CE

Thursday, March 1, 2012

EXAMPLE 2

In ∆ABC, CG = 4. Find GE.

CG =

23

CE

4 =

23

CE

Thursday, March 1, 2012

EXAMPLE 2

In ∆ABC, CG = 4. Find GE.

CG =

23

CE

4 =

23

CE

CE = 6

Thursday, March 1, 2012

EXAMPLE 2

In ∆ABC, CG = 4. Find GE.

CG =

23

CE

4 =

23

CE

CE = 6

GE = CE −CG

Thursday, March 1, 2012

EXAMPLE 2

In ∆ABC, CG = 4. Find GE.

CG =

23

CE

4 =

23

CE

CE = 6

GE = CE −CG

GE = 6 − 4

Thursday, March 1, 2012

EXAMPLE 2

In ∆ABC, CG = 4. Find GE.

CG =

23

CE

4 =

23

CE

CE = 6

GE = CE −CG

GE = 6 − 4

GE = 2

Thursday, March 1, 2012

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?

Thursday, March 1, 2012

EXAMPLE 3

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

Thursday, March 1, 2012

EXAMPLE 3

x

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

Thursday, March 1, 2012

EXAMPLE 3

x

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

Thursday, March 1, 2012

EXAMPLE 3

x

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

Thursday, March 1, 2012

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)

Thursday, March 1, 2012

EXAMPLE 3

x

y

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

of our vertical side.

M =

3 + 32

,0 + 8

2

⎛⎝⎜

⎞⎠⎟

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

Thursday, March 1, 2012

EXAMPLE 3

x

y

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

of our vertical side.

M =

3 + 32

,0 + 8

2

⎛⎝⎜

⎞⎠⎟ =

62

,82

⎛⎝⎜

⎞⎠⎟

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

Thursday, March 1, 2012

EXAMPLE 3

x

y

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

of our vertical side.

M =

3 + 32

,0 + 8

2

⎛⎝⎜

⎞⎠⎟ =

62

,82

⎛⎝⎜

⎞⎠⎟

= 3,4( )

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

Thursday, March 1, 2012

EXAMPLE 3

x

y

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

of our vertical side.

M =

3 + 32

,0 + 8

2

⎛⎝⎜

⎞⎠⎟ =

62

,82

⎛⎝⎜

⎞⎠⎟

= 3,4( )

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

Thursday, March 1, 2012

EXAMPLE 3

x

y

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

of our vertical side.

M =

3 + 32

,0 + 8

2

⎛⎝⎜

⎞⎠⎟ =

62

,82

⎛⎝⎜

⎞⎠⎟

= 3,4( )

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

Thursday, March 1, 2012

EXAMPLE 3

x

y

Thursday, March 1, 2012

EXAMPLE 3

x

y

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

midpoint.

Thursday, March 1, 2012

EXAMPLE 3

x

y

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

midpoint.

d = (1− 3)2 + (4 − 4)2

Thursday, March 1, 2012

EXAMPLE 3

x

y

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

midpoint.

d = (1− 3)2 + (4 − 4)2

= (−2)2 + 02

Thursday, March 1, 2012

EXAMPLE 3

x

y

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

midpoint.

d = (1− 3)2 + (4 − 4)2

= (−2)2 + 02

= 4

Thursday, March 1, 2012

EXAMPLE 3

x

y

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

midpoint.

d = (1− 3)2 + (4 − 4)2

= (−2)2 + 02

= 4 = 2

Thursday, March 1, 2012

EXAMPLE 3

x

y

Thursday, March 1, 2012

EXAMPLE 3

x

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

Thursday, March 1, 2012

EXAMPLE 3

x

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

P = 1+

23

(2),4⎛⎝⎜

⎞⎠⎟

Thursday, March 1, 2012

EXAMPLE 3

x

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

P = 1+

23

(2),4⎛⎝⎜

⎞⎠⎟

P = 1+

43

,4⎛⎝⎜

⎞⎠⎟

Thursday, March 1, 2012

EXAMPLE 3

x

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

P = 1+

23

(2),4⎛⎝⎜

⎞⎠⎟

P = 1+

43

,4⎛⎝⎜

⎞⎠⎟

P =

73

,4⎛⎝⎜

⎞⎠⎟

Thursday, March 1, 2012

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

Find the coordinates of the orthocenter of ∆HIJ.

Thursday, March 1, 2012

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

Thursday, March 1, 2012

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

H

I

J

Thursday, March 1, 2012

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

H

I

J

Thursday, March 1, 2012

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

H

I

J

Thursday, March 1, 2012

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

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

H

I

J

Thursday, March 1, 2012

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

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

H

I

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

Thursday, March 1, 2012

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

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

H

I

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

m JI( ) = 1+ 3

−5 + 3

Thursday, March 1, 2012

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

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

H

I

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

m JI( ) = 1+ 3

−5 + 3 =

4−2

Thursday, March 1, 2012

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

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

H

I

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

m JI( ) = 1+ 3

−5 + 3 =

4−2 = −2

Thursday, March 1, 2012

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

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

H

I

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

m JI( ) = 1+ 3

−5 + 3 =

4−2

⊥ m =12 = −2

Thursday, March 1, 2012

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

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

H

I

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

m JI( ) = 1+ 3

−5 + 3 =

4−2

⊥ m =12

m HJ( ) = 2 −1

1+ 5

= −2

Thursday, March 1, 2012

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

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

H

I

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

m JI( ) = 1+ 3

−5 + 3 =

4−2

⊥ m =12

m HJ( ) = 2 −1

1+ 5 =

16

= −2

Thursday, March 1, 2012

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

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

H

I

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

m JI( ) = 1+ 3

−5 + 3 =

4−2

⊥ m =12

m HJ( ) = 2 −1

1+ 5 =

16

⊥ m = −6

= −2

Thursday, March 1, 2012

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

H

I

J

Altitude through H

Thursday, March 1, 2012

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

H

I

J ⊥ m =

12

, (1,2)

Altitude through H

Thursday, March 1, 2012

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

H

I

J ⊥ m =

12

, (1,2)

Altitude through H

y − 2 =

12

(x −1)

Thursday, March 1, 2012

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

H

I

J ⊥ m =

12

, (1,2)

Altitude through H

y − 2 =

12

(x −1)

y − 2 =

12

x −12

Thursday, March 1, 2012

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

H

I

J ⊥ m =

12

, (1,2)

Altitude through H

y − 2 =

12

(x −1)

y − 2 =

12

x −12

y =

12

x +32

Thursday, March 1, 2012

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

H

I

J

Altitude through I

Thursday, March 1, 2012

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

H

I

J

⊥ m = −6, (−3,−3)

Altitude through I

Thursday, March 1, 2012

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

H

I

J

⊥ m = −6, (−3,−3)

Altitude through I

y + 3 = −6(x + 3)

Thursday, March 1, 2012

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

H

I

J

⊥ m = −6, (−3,−3)

Altitude through I

y + 3 = −6(x + 3)

y + 3 = −6x −18

Thursday, March 1, 2012

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

H

I

J

⊥ m = −6, (−3,−3)

Altitude through I

y + 3 = −6(x + 3)

y + 3 = −6x −18

y = −6x − 21

Thursday, March 1, 2012

EXAMPLE 4

Thursday, March 1, 2012

EXAMPLE 4

−6x − 21 =

12

x +32

Thursday, March 1, 2012

EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x

Thursday, March 1, 2012

EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x

−21 =

132

x +32

Thursday, March 1, 2012

EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x

−21 =

132

x +32

−

32

−32

Thursday, March 1, 2012

EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x

−21 =

132

x +32

−

32

−32

−

452

=132

x

Thursday, March 1, 2012

EXAMPLE 4

−6x − 21 =

12

x +32

+6x +6x

−21 =

132

x +32

−

32

−32

−

452

=132

x

213

−452

⎛⎝⎜

⎞⎠⎟=

132

x⎛⎝⎜

⎞⎠⎟

213

Thursday, March 1, 2012

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

Thursday, March 1, 2012

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

Thursday, March 1, 2012

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

Thursday, March 1, 2012

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

⎛⎝⎜

⎞⎠⎟

Thursday, March 1, 2012

CHECK YOUR UNDERSTANDING

p. 337 #1-4

Thursday, March 1, 2012

PROBLEM SET

Thursday, March 1, 2012

PROBLEM SET

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

"Education's purpose is to replace an empty mind with an open one." – Malcolm Forbes

Thursday, March 1, 2012