Post on 29-Dec-2015
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4.1 Classifying Triangles
Triangle – A figure formed when three noncollinear points are connected by segments.
E
DF
Angle
SideVertex
The sides are DE, EF, and DF.The vertices are D, E, and F.The angles are D, E, F.
Triangles Classified by AnglesAcute Obtuse Right
60º
50º
70º
All acute anglesOne obtuse angle
One right angle
120º
43º
17º
30°
60º
Triangles Classified by Sides
Scalene Isosceles Equilateral
no sidescongruent at least two
sides congruent
all sidescongruent
Classify each triangle by its angles and by its sides.
60°
60° 60°A B
C
45°
45°
E
F G
EFG is a right
isosceles triangle.
ABC is an acute
equilateral triangle
Try These:1. ABC has angles that
measure 110, 50, and 20. Classify the triangle by its angles.
2. RST has sides that measure 3 feet, 4 feet, and 5 feet. Classify the triangle by its sides.
Adjacent Sides- share a vertex ex. The sides DE & EF are adjacent to E.
E
D F
Opposite Side- opposite the vertex ex. DF is opposite E.
Parts of Isosceles Triangles The angle formed by the congruent sides is called the vertex angle.
leg leg The congruent sides are called legs.
The side opposite the vertex is the base.
base anglebase angle
The two angles formed by the base and one of the congruent sides are called base angles.
Base Angles Theorem
If two sides of a triangle are congruent, then the angles opposite them are congruent.
If , thenACAB CB
Converse of Base Angles Theorem
If two angles of a triangle are congruent, then the sides opposite them are congruent.
If , thenCB ACAB
EXAMPLE 1 Apply the Base Angles Theorem
P
R
Q
(30)°
Find the measures of the angles.SOLUTION
Since a triangle has 180°, 180 – 30 = 150° for the other two angles.
Since the opposite sides are congruent, angles Q and P must be congruent.
150/2 = 75° each.
EXAMPLE 4 Apply the Base Angles Theorem
Find the value of x. Then find the measure of each angle.
P
RQ(20x-4)°
(12x+20)° SOLUTION
Since there are two congruent sides, the angles opposite them must be congruent also. Therefore, 12x + 20 = 20x – 4
20 = 8x – 4
24 = 8x
3 = x
Plugging back in,
And since there must be 180 degrees in the triangle,
564)3(20
5620)3(12
Rm
Pm
685656180Qm
EXAMPLE 5 Apply the Base Angles Theorem
Find the value of x. Then find the measure of each angle.
P
R
Q(11x+8)° (5x+50)°
EXAMPLE 6 Apply the Base Angles Theorem
Find the value of x. Then find the length of the labeled sides.
P
R
Q(80)° (80)°
SOLUTION
Since there are two congruent sides, the angles opposite them must be congruent also. Therefore, 7x = 3x + 40
4x = 40
x = 10
7x 3x+40
Plugging back in,
QR = 7(10)= 70PR = 3(10) + 40 = 70
EXAMPLE 7 Apply the Base Angles Theorem
Find the value of x. Then find the length of the labeled sides.
P
RQ
(50)°
(50)°
10x – 2
5x+3
Triangle Sum TheoremTriangle Sum TheoremThe measures of the three interior angles
in a triangle add up to be 180º.
x°
y° z°
x + y + z = 180°
54°
67°
R
S T
m R + m S + m T = 180º 54º + 67º + m T = 180º
121º + m T = 180º
m T = 59º
Find in RST.m T
85° x°55°
y°
A
B
C
D
E m D + m DCE + m E = 180º55º + 85º + y = 180º
140º + y = 180º
y = 40º
Find the value of each variable in DCE
The measure of the exterior angle is equal to the sum of two nonadjacent interior angles
1
2 3
m1+m2 =m3
Exterior Angle TheoremExterior Angle Theorem