Geometry Ch. 5 Test Review

advertisement
Geometry
Ch. 5 Test Review
5-1 Midsegment
Solve for x
X=10
X=9
X=7
5-2 Perpendicular Bisector / Angle Bisector
Solve for x
X=5
X=5/2
5-3 Incenter
Solve for x.
X= -1
5-3 Graph the points.
Find Circumcenter. (0,0)
Find Orthocenter. (4,-3)
5-3 & 5-4 Point of Concurrency
Name it!
Altitude
Median
Perpendiculuar
Bisector
5-3 Draw an angle bisector!
5-3-5-4 Point of Concurrency
Name it!
Bisectors Form _________________
incenter
 Perpendicular Bisectors Form _____________
circumcenter
centroid
 Medians Form __________________
orthocenter
 Altitudes Form _________________
 Angle
5-3-5-4 Point of Concurrency
Name the line!
5-4 Centroid
5
10
12
36
5-6 List the SIDES in order.
Smallest to largest.
54
5-6 Determine the SHORTEST side?
Not
EG
Look for next small
L
X
M
67
LS
O
S
47
M
DG
5-6 Write sides in order from smallest to
largest
DG, ED, EG/ EG, FG, EF
L
X
M
67
LS
O
S
47
M
5-6 Longest side of triangle?
 In triangle ABC,
m<A =
 Solve
2x + 20, m<B = 4x – 30, m<C = x + 50.
for x. Find LONGEST side of triangle ABC.
2x + 20 + 4x – 30 + x + 50 = 180
7x + 40 = 180
7x = 140
x = 20
Continuation… Longest side..
 In triangle ABC,
m<A =
 Solve
2x + 20, m<B = 4x – 30, m<C = x + 50.
for x. Find LONGEST side of triangle ABC.
C
m<A = 2(20)+20 = 60
70
m<B = 4(20) – 30 = 50 A 60 50 B
m<C = 20 + 50 = 70
AB
5-6 Which lengths could be SIDES of a
triangle?
2.5,
6,
8.5, 5.5
5, 11
5x,
8x, 12x
No, 2.5+5.5>8.5
No, 6+5>11
Yes, 5x+8x>12x
Small side + small side > 3rd side
5-6 Triangle Inequality
Find range of values.
 If lengths of sides of a triangle are 2k+3 and
,
4k-3
then the third side must be greater than ________
8k+3
and less than _________
6k
Small side + small side > 3rd side
2k+3 + 6k > x
2k+3 + x > 6k
OR
-2k -3
-2k -3
8k+3 > x
x > 4k-3
x < 8k+3
Greater than
Less than
5-7 Hinge Theorem
& Converse of Hinge Theorem
Fill in with <, >, or =. By which theorem?
Big
AEB >
BDC
so
Big
By Hinge Theorem
5-7 Hinge Theorem
& Converse of Hinge Theorem
Fill in with <, >, or =. By which theorem?
Big
Big
AB<ED
so
By Converse of
Hinge Theorem
Download