chapter4001 and 4002 traingles

Holt McDougal Geometry

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



Warm Up

Classify each angle as acute, obtuse, or right.

3. 4.

5.

6. If the perimeter is 47, find x and the lengths

of the three sides.

rightacute

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



auxiliary line

corollary

interior

exterior

interior angle

exterior angle

remote interior angle

Vocabulary



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.



Acute Triangle

Three acute angles

Triangle Classification By Angle Measures



Equiangular Triangle

Three congruent acute angles




Right Triangle

One right angle




Obtuse Triangle

One obtuse angle




Classify BDC by its angle measures.

Example 1A: Classifying Triangles by Angle Measures

DBC is an obtuse angle.

DBC is an obtuse angle. So BDC is an obtuse triangle.



Classify ABD by its angle measures.

Example 1B: Classifying Triangles by Angle Measures

ABD and CBD form a linear pair, so they are supplementary.

Therefore mABD + mCBD = 180°. By substitution, mABD + 100° = 180°. So mABD = 80°. ABD is an acute triangle by definition.



Classify FHG by its angle measures.

Check It Out! Example 1

EHG is a right angle. Therefore mEHF +mFHG = 90°. By substitution, 30°+ mFHG = 90°. So mFHG = 60°.

FHG is an equiangular triangle by definition.



Equilateral Triangle

Three congruent sides

Triangle Classification By Side Lengths



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.



Classify EHF by its side lengths.

Example 2A: Classifying Triangles by Side Lengths

From the figure, . So HF = 10, and EHF is isosceles.



Classify EHG by its side lengths.

Example 2B: Classifying Triangles by Side Lengths

By the Segment Addition Postulate, EG = EF + FG = 10 + 4 = 14. Since no sides are congruent, EHGis scalene.



Classify ACD by its side lengths.


From the figure, . So AC = 15, and ACD is isosceles.



Find the side lengths of JKL.

Example 3: Using Triangle Classification

Step 1 Find the value of x.

Given.

JK = KL Def. of segs.

4x – 10.7 = 2x + 6.3Substitute (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 = 7Add 4 and subtract 2y from both sides.



Find the side lengths of equilateral FGH.

Check It Out! Example 3 Continued

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



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

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?

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?

Example 4: Application Continued

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



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 mXYZ.

Example 1A: Application

mXYZ + mYZX + mZXY = 180° Sum. Thm

mXYZ + 40 + 62 = 180Substitute 40 for mYZX and

62 for mZXY.

mXYZ + 102 = 180 Simplify.

mXYZ = 78° Subtract 102 from both sides.




enforcement to investigate the collision. Use the diagram

drawn from the information collected to find mYWZ.

Example 1B: Application

mYXZ + mWXY = 180° Lin. Pair Thm. and Add. Post.

62 + mWXY = 180 Substitute 62 for mYXZ.

mWXY = 118° Subtract 62 from both sides.

Step 1 Find mWXY.

118°




enforcement to investigate the collision. Use the diagram drawn from the information collected

to find mYWZ.

Example 1B: Application Continued

Step 2 Find mYWZ.

118°

mYWX + mWXY + mXYW = 180° Sum. Thm

mYWX + 118 + 12 = 180 Substitute 118 for mWXY and 12 for mXYW.

mYWX + 130 = 180 Simplify.

mYWX = 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.



One of the acute angles in a right triangle measures 2x°. What is the measure of the other

acute angle?

Example 2: Finding Angle Measures in Right Triangles

mA + mB = 90°

2x + mB = 90 Substitute 2x for mA.

mB = (90 – 2x)° Subtract 2x from both sides.

Let the acute angles be A and B, with mA = 2x°.

Acute s of rt. are comp.



The measure of one of the acute angles in a right triangle is 63.7°. What is the measure of

the other acute angle?

Check It Out! Example 2a

mA + mB = 90°

63.7 + mB = 90 Substitute 63.7 for mA.

mB = 26.3° Subtract 63.7 from both sides.

Let the acute angles be A and B, with mA = 63.7°.




The measure of one of the acute angles in a right triangle is x°. What is the measure of the

other acute angle?

Check It Out! Example 2b

mA + mB = 90°

x + mB = 90 Substitute x for mA.

mB = (90 – x)° Subtract x from both sides.

Let the acute angles be A and B, with mA = x°.




The measure of one of the acute angles in a right

triangle is 48 . What is the measure of the other

acute angle?

Check It Out! Example 2c

mA + mB = 90° Acute s of rt. are comp.

2°5

Let the acute angles be A and B, with mA = 48 . 2°5

Subtract 48 from both sides.2 5

Substitute 48 for mA.2 548 + mB = 90

2 5

mB = 41 3°5



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

3 is an interior angle.

4 is an exterior angle.

The remote interior angles of 4 are 1

and 2.

Find the missing angle measures.



Find mB.

Example 3: Applying the Exterior Angle Theorem

mA + mB = mBCD Ext. Thm.

15 + 2x + 3 = 5x – 60 Substitute 15 for mA, 2x + 3 for mB, and 5x – 60 for mBCD.

2x + 18 = 5x – 60 Simplify.

78 = 3xSubtract 2x and add 60 to

both sides.

26 = x Divide by 3.

mB = 2x + 3 = 2(26) + 3 = 55°



Find mACD.


mACD = mA + mB Ext. Thm.

6z – 9 = 2z + 1 + 90 Substitute 6z – 9 for mACD, 2z + 1 for mA, and 90 for mB.

6z – 9 = 2z + 91 Simplify.

4z = 100Subtract 2z and add 9 to both

sides.

z = 25 Divide by 4.

mACD = 6z – 9 = 6(25) – 9 = 141°



Find mK and mJ.

Example 4: Applying the Third Angles Theorem

K J

mK = mJ

4y2 = 6y2 – 40

–2y2 = –40

y2 = 20

So mK = 4y2 = 4(20) = 80°.

Since mJ = mK, mJ = 80°.

Third s Thm.

Def. of s.

Substitute 4y2 for mK and 6y2 – 40 for mJ.

Subtract 6y2 from both sides.

Divide both sides by -2.




Find mP and mT.

P T

mP = mT

2x2 = 4x2 – 32

–2x2 = –32

x2 = 16

So mP = 2x2 = 2(16) = 32°.

Since mP = mT, mT = 32°.

Third s Thm.

Def. of s.

Substitute 2x2 for mP and 4x2 – 32 for mT.

Subtract 4x2 from both sides.

Divide both sides by -2.



Lesson Quiz

Classify each triangle by its angles and sides.

1. MNQ

2. NQP

3. MNP

4. Find the side lengths of the triangle.

acute; equilateral

obtuse; scalene

acute; scalene

29; 29; 23

chapter4001 and 4002 traingles

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