Transcript Jeopardy - Chapter_5_Review

Vocabulary
Truths About
Triangles
Midsegments
Inequalities
Relationships
In Triangles
100
100
100
100
100
200
200
200
200
200
300
300
300
300
300
400
400
400
400
400
500
500
500
500
500
Vocabulary 100
A segment whose
endpoints are at the
vertex of a triangle and
the midpoint of the side
opposite is a…
Vocabulary 100
Median
Vocabulary 200
A perpendicular segment
from a vertex to the line
containing the side
opposite the vertex is
called a(n)…
Vocabulary 200
Altitude
Vocabulary 300
A point where three lines
intersects is called a(n)…
Vocabulary 300
Point of
Concurrence
Vocabulary 400
The point of concurrency
of the angle bisectors of a
triangle is called the…
Vocabulary 400
Incenter
Vocabulary 500
The point of concurrency
of the altitudes of a
triangle is called the…
Vocabulary 500
Orthocenter
Truths About Triangles 100
The largest angle of a
triangle is across from
the _________ side.
Truths About Triangles 100
Longest
Truths About Triangles 200
Given points A 1,3
B 5,1 and C 4,4  does
point C lie on the
perpendicular bisector of
segment AB?
Truths About Triangles 200
AC  1 4   3  4 
2
AC   3   1
2
2
2
BC   5  4   1 4 
2
BC  1   3
2
AC  9 1
BC  1 9
AC  10
BC  10
2
2
Since AC = BC, point C is on the perpendicular bisector
because of the perpendicular bisector theorem – point C is
equidistant from the endpoints of the segment AB.
Truths About Triangles 300
The vertices of a triangle lie
at  0, 4  ,  0,0  and  4,0 
Find the center of a circle
that would be circumscribed
about this triangle.
Truths About Triangles 300
midpoint  0, 4  and  4, 0 
x1  x 2 y1  y 2 

,
midpoint  

2 
 2
0  4 4  0 

,


2 
 2
4 4 

 , 
 2 2
  2, 2 
Truths About Triangles 400
Given A  0,6 B  0,0  and
C5, 0 find the coordinates
of the endpoints of the
midsegment that is
parallel to BC.
Truths About Triangles 400
midpoint A  0, 6  and B  0, 0 
midpoint A  0, 6  and C  5, 0 
x1  x 2 y1  y 2 

midpoint  
,

2 
 2
00 60
 
,

2
2


0 6

 , 
2 2
  0, 3
x1  x 2 y1  y 2 

midpoint  
,

2 
 2
05 60
 
,

2
2


5 6

 , 
2 2
5 

  , 3
2 
Truths About Triangles 500
P  2,1 , Q  1,5 and R  2,  2  are vertices
of triangle PQR.
 What are the coordinates of T if QT is
a median of the triangle?
 What is the slope of PA if PA is the
altitude from P?
 Tell why or why not PA is a
perpendicular bisector.
Truths About Triangles 500
 T is the midpoint of PR
x1  x 2 y1  y 2   2  2 1 2   0 1  1

midpoint T  
,
,

   ,    0, 
2   2
2  2 2   2 
 2
 PA  QR , so find the slope of QR , and
take the opposite reciprocal:
slopeQR
y 2  y1
2  5
7



7
x 2  x1 2   1 1
slopePA
1

7
x1  x 2 y1  y 2   1  2  5   2    3

,
,
 ,
 Midpoint of QR  


2  
2
2   2
 2
3
1
PA is the
1
2
slope

3 3 
perpendicular
3
P  2,1 to  , 
 2 2
2 7
bisector
2
3

2
Midsegments 100
Find the
value of x.
Midsegments 100
1
x  18
2
x 9
2x  18
x 9
Midsegments 200
Find the
value of x.
Midsegments 200
Equilateral
Triangle
x  55
x  10
5
60°
5
Midsegments 300
Find the lengths
of AC,CB, and AB.
Midsegments 300
AC  6
CB  5
AB  7
7
6
5
Midsegments 400
Find the
values of
x and y.
3x - 6
y
x
2x + 1
Drawing not to scale.
Midsegments 400
2  x   3x  6
2  y   2x  1
2x  3x  6
2y  2  6   1
 x  6
2y  12  1
2y  13
x6
13
y
2
Midsegments 500
Marita is designing a kite. The kites
diagonals are to measure 64 cm and
90 cm. She will use ribbon to connect
the midpoints of its sides that form a
pretty rectangle inside the kite. How
much ribbon will Marita need to make
the rectangle connecting the
midpoints?
Midsegments 500
The red segments are
midsegments of the diagonal
that measures 64 cm, so they
measure 32 cm. The green
segments are midsegments of
the diagonal that measure 90
cm, so they measure 45 cm. So
the perimeter is
P  2  32   2  45
P  64  90
P  154
Inequalities 100
If a = b + c and c > 0,
then a > b is which
property of inequality?
Inequalities 100
Comparison Property
of Inequality
Inequalities 200
Two sides of a triangle
have measure of 12 meters
and 22 meters what are the
possible measures of the
rd
3 side?
Inequalities 200
22  12  34
22 12  10
10  x  34
Inequalities 300
Can a triangle have lengths of
2 yards, 9 yards, and 15 yards?
Inequalities 300
2  9 15
2 15  9
15  9  2
No!
Inequalities 400
If KL = x – 4 , LM = x + 4 and
KM = 2x – 1, and the
perimeter of the triangle is 27,
find the order of the angles
from smallest to largest.
Inequalities 400
 x  4    x  4    2 x 1  27
4 x 1  27
4 x  28
x 7
M, K, L
KL  x  4  7  4  3
LM  x  4  7  4  11
KM  2x 1  2  7  1  14 1  13
Inequalities 500
State the Exterior Angle
Inequality Theorem.
Inequalities 500
The measure of an exterior angle
of a triangle is greater than each
of its remote interior angles.
Relationships in Triangles 100
If a point lies on the
perpendicular bisector of a
segment, what holds true
about its distance from the
endpoints of the segment?
Relationships in Triangles 100
The point is equidistant
from the endpoints of the
segment.
Relationships in Triangles 200
Solve for x.
Relationships in Triangles 200
x  5  2x  7
5  x 7
12  x
Relationships in Triangles 300
Point C is the centroid of
triangle DEF. If GF, G being
the midpoint of segment DE,
is 9 meters long, what is the
length of CF?
Relationships in Triangles 300
2
CF   GF 
3
2
CF   9 
3
CF  6
Relationships in Triangles 400
Find the slope of the altitude
drawn from vertex A.
Relationships in Triangles 400
Find the slope of BC. The slope of the altitude drawn
from vertex A will have a slope that is the opposite
reciprocal of the slope of BC.
mBC
y 2  y1

x 2  x1
80

4  20
8

16
1

2
So the slope of the
altitude drawn from
vertex A is 2.
Relationships in Triangles 500
Find the equation of the line that is the
perpendicular bisector of segment CA.
Relationships in Triangles 500
Step 1: Find the midpoint of CA.
x  x 2 y1  y 2   3  0 16  25   3 41
midpoint T   1
,
,

 , 
2   2
2  2 2 
 2
Step 2: Find the slope of CA.
slope CA 
y 2  y1 16  25 9


 3
x 2  x1
30
3
Step 3: The slope of the perpendicular bisector of CA
is the opposite reciprocal of the slope of CA. So the
slope of the perpendicular bisector equals 1 .
3
Step 4: Write the equation using point-slope
form y  y1  m x  x1  . Therefore the answer is:
41 1
3
y   x  
2
3
2