Chapter4001 and 4002 traingles

4-1 Classifying Triangles
Hon Geom Drill 12/11/14
• Take out any hw you have and then
complete the drill on your own.
Use your homework to help you
answer the questions. There is a
front and back
Holt McDougal Geometry

Warm Up
Classify each angle as acute, obtuse, or right.
3. 4.
5.
right
6. If the perimeter is 47, find x and the lengths
of the three sides.
acute
x = 5; 8; 16; 23
obtuse

1. Take a look at the three triangles I have provided
and answer the following questions:
What do you notice?
What do you wonder?

2. List any information you already know about triangles, their angles, and
classify triangles

1. Classify triangles by their angle measures
and side lengths.
2. Use triangle classification to find angle
measures and side lengths.
3. Find the measures of interior and
exterior angles of triangles.
4. Apply theorems about the interior
and exterior angles of triangles.
Objectives

acute triangle
equiangular triangle
right triangle
obtuse triangle
equilateral triangle
isosceles triangle
scalene triangle
Vocabulary

Vocabulary
auxiliary line
corollary
interior
exterior
interior angle
exterior angle
remote interior angle

Recall that a triangle ( ) is a polygon
with three sides. Triangles can be
classified in two ways: by their angle
measures or by their side lengths.

B
A
C
AB, BC, and AC are the sides of ABC.
A, B, C are the triangle's vertices.

Triangle Classification By Angle Measures
Acute Triangle
Three acute angles

Equiangular Triangle
Three congruent acute angles

Right Triangle
One right angle

Obtuse Triangle
One obtuse angle

Example 1A: Classifying Triangles by Angle Measures
Classify BDC by its angle measures.
DBC is an obtuse angle.
DBC is an obtuse angle. So BDC is an obtuse
triangle.

Example 1B: Classifying Triangles by Angle Measures
Classify ABD by its angle measures.
ABD and CBD form a linear pair, so they are
supplementary.
Therefore mABD + mCBD = 180°. By substitution,
mABD + 100° = 180°. So mABD = 80°. ABD is an
acute triangle by definition.

Check It Out! Example 1
Classify FHG by its angle measures.
EHG is a right angle. Therefore mEHF +mFHG = 90°.
By substitution, 30°+ mFHG = 90°. So mFHG = 60°.
FHG is an equiangular triangle by definition.

Triangle Classification By Side Lengths
Equilateral Triangle
Three congruent sides

Isosceles Triangle
At least two congruent sides

Scalene Triangle
No congruent sides

Remember!
When you look at a figure, you cannot assume
segments are congruent based on appearance.
They must be marked as congruent.

Example 2A: Classifying Triangles by Side Lengths
Classify EHF by its side lengths.
From the figure, . So HF = 10, and EHF is
isosceles.

Example 2B: Classifying Triangles by Side Lengths
Classify EHG by its side lengths.
By the Segment Addition Postulate, EG = EF + FG =
10 + 4 = 14. Since no sides are congruent, EHG
is scalene.

Classify ACD by its side lengths.
From the figure, . So AC = 15, and ACD is
isosceles.

Example 3: Using Triangle Classification
Find the side lengths of JKL.
Step 1 Find the value of x.
Given.
JK = KL Def. of  segs.
4x – 10.7 = 2x + 6.3
Substitute (4x – 10.7) for
JK and (2x + 6.3) for KL.
2x = 17.0
x = 8.5
Add 10.7 and subtract 2x
from both sides.
Divide both sides by 2.

Find the side lengths of JKL.
Example 3 Continued
Step 2 Substitute 8.5 into
the expressions to find the
side lengths.
JK = 4x – 10.7
= 4(8.5) – 10.7 = 23.3
KL = 2x + 6.3
= 2(8.5) + 6.3 = 23.3
JL = 5x + 2
= 5(8.5) + 2 = 44.5

Find the side lengths of equilateral FGH.
Step 1 Find the value of y.
Given.
FG = GH = FH Def. of  segs.
3y – 4 = 2y + 3
Substitute
(3y – 4) for FG and
(2y + 3) for GH.
y = 7
Add 4 and subtract
2y from both sides.

Check It Out! Example 3 Continued
Find the side lengths of equilateral FGH.
Step 2 Substitute 7 into the expressions to find the
side lengths.
FG = 3y – 4
= 3(7) – 4 = 17
GH = 2y + 3
= 2(7) + 3 = 17
FH = 5y – 18
= 5(7) – 18 = 17

Example 4: Application
A steel mill produces roof supports by welding
pieces of steel beams into equilateral
triangles. Each side of the triangle is 18 feet
long. How many triangles can be formed from
420 feet of steel beam?
The amount of steel needed to make one triangle
is equal to the perimeter P of the equilateral
triangle.
P = 3(18)
P = 54 ft

Example 4: Application Continued
A steel mill produces roof supports by welding
pieces of steel beams into equilateral
triangles. Each side of the triangle is 18 feet
long. How many triangles can be formed from
420 feet of steel beam?
To find the number of triangles that can be made
from 420 feet of steel beam, divide 420 by the
amount of steel needed for one triangle.
420  54 = 7 triangles
7
9
There is not enough steel to complete an eighth
triangle. So the steel mill can make 7 triangles
from a 420 ft. piece of steel beam.

An auxiliary line is a line that is added to a
figure to aid in a proof.
An auxiliary
line used in the
Triangle Sum
Theorem

Example 1A: Application
After an accident, the positions
of cars are measured by law
enforcement to investigate the
collision. Use the diagram
drawn from the information
collected to find mXYZ.
mXYZ + mYZX + mZXY = 180° Sum. Thm
mXYZ + 40 + 62 = 180
Substitute 40 for mYZX and
62 for mZXY.
mXYZ + 102 = 180 Simplify.
mXYZ = 78° Subtract 102 from both sides.

Example 1B: Application
After an accident, the positions
of cars are measured by law
collision. Use the diagram
drawn from the information
collected to find mYWZ.
Step 1 Find mWXY.
mYXZ + mWXY = 180° Lin. Pair Thm. and  Add. Post.
62 + mWXY = 180 Substitute 62 for mYXZ.
mWXY = 118° Subtract 62 from both sides.
118°

Example 1B: Application Continued
After an accident, the positions of
cars are measured by law
collision. Use the diagram drawn
from the information collected
to find mYWZ.
Step 2 Find mYWZ.
118°
mYWX + mWXY + mXYW = 180° Sum. Thm
mYWX + 118 + 12 = 180 Substitute 118 for mWXY and
12 for mXYW.
mYWX + 130 = 180 Simplify.
mYWX = 50° Subtract 130 from both sides.

A corollary is a theorem whose proof follows
directly from another theorem. Here are two
corollaries to the Triangle Sum Theorem.

Example 2: Finding Angle Measures in Right Triangles
One of the acute angles in a right triangle
measures 2x°. What is the measure of the other
Let the acute angles be A and B, with mA = 2x°.
acute angle?
mA + mB = 90°
Acute s of rt. are comp.
2x + mB = 90 Substitute 2x for mA.
mB = (90 – 2x)° Subtract 2x from both sides.

Check It Out! Example 2a
The measure of one of the acute angles in a
right triangle is 63.7°. What is the measure of
the other acute angle?
Let the acute angles be A and B, with mA = 63.7°.
mA + mB = 90°
63.7 + mB = 90 Substitute 63.7 for mA.
mB = 26.3° Subtract 63.7 from both sides.

Check It Out! Example 2b
The measure of one of the acute angles in a
right triangle is x°. What is the measure of the
Let the acute angles be A and B, with mA = x°.
other acute angle?
mA + mB = 90°
x + mB = 90 Substitute x for mA.
mB = (90 – x)° Subtract x from both sides.

Check It Out! Example 2c
The measure of one of the acute angles in a right
triangle is 48 . What is the measure of the other
2°
5
Let the acute angles be A and B, with mA = 48 .
2
5 48 + mB = 90
acute angle?
mA + mB = 90° Acute s of rt. are comp.
2°
5
Substitute 48 for mA.
Subtract 48 2 from both sides.
5
2
5
3°
5
mB = 41

The interior is the set of all points inside the
figure. The exterior is the set of all points
outside the figure.
Interior
Exterior

An interior angle is formed by two sides of a
triangle. An exterior angle is formed by one
side of the triangle and extension of an adjacent
side.
Interior
Exterior
4 is an exterior angle.
3 is an interior angle.

Each exterior angle has two remote interior
angles. A remote interior angle is an interior
angle that is not adjacent to the exterior angle.
Interior
Exterior
4 is an exterior angle.
The remote interior
angles of 4 are 1
3 is an interior angle.
and 2.

Find the missing angle measures.

Example 3: Applying the Exterior Angle Theorem
Find mB.
mA + mB = mBCD Ext.  Thm.
15 + 2x + 3 = 5x – 60 Substitute 15 for mA, 2x + 3 for
mB, and 5x – 60 for mBCD.
2x + 18 = 5x – 60 Simplify.
78 = 3x
Subtract 2x and add 60 to
both sides.
26 = x Divide by 3.
mB = 2x + 3 = 2(26) + 3 = 55°

Find mACD.
mACD = mA + mB Ext.  Thm.
6z – 9 = 2z + 1 + 90 Substitute 6z – 9 for mACD,
2z + 1 for mA, and 90 for mB.
6z – 9 = 2z + 91 Simplify.
4z = 100
Subtract 2z and add 9 to both
sides.
z = 25 Divide by 4.
mACD = 6z – 9 = 6(25) – 9 = 141°

Example 4: Applying the Third Angles Theorem
Find mK and mJ.
K  J
mK = mJ
4y2 = 6y2 – 40
–2y2 = –40
y2 = 20
So mK = 4y2 = 4(20) = 80°.
Since mJ = mK, mJ = 80°.
Third s Thm.
Def. of  s.
Substitute 4y2 for mK and 6y2 – 40 for mJ.
Subtract 6y2 from both sides.
Divide both sides by -2.

Find mP and mT.
P  T
mP = mT
2x2 = 4x2 – 32
–2x2 = –32
x2 = 16
So mP = 2x2 = 2(16) = 32°.
Since mP = mT, mT = 32°.
Third s Thm.
Def. of  s.
Substitute 2x2 for mP and 4x2 – 32 for mT.
Subtract 4x2 from both sides.
Divide both sides by -2.

acute; equilateral
obtuse; scalene
acute; scalene
Lesson Quiz
Classify each triangle by its angles and sides.
1. MNQ
2. NQP
3. MNP
4. Find the side lengths of the triangle.
29; 29; 23

Chapter4001 and 4002 traingles

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