- By
**sen** - Follow User

- 115 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' 4-1' - sen

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Warm Up

Classify each angle as acute, obtuse, or right.

1.2.

3.

4. If the perimeter is 47, find x and the lengths of the three sides.

right

acute

obtuse

x = 5; 8; 16; 23

Objectives

Classify triangles by their angle measures and side lengths.

Use triangle classification to find angle measures and side lengths.Find the measures of interior and exterior angles of triangles.

Apply theorems about the interior and exterior angles of triangles.

Vocabulary

acute triangle Corollary

equiangular triangle Legs

right triangle Adjacent

obtuse triangle Exterior

equilateral triangle Interior

isosceles triangle Hypotenuse

scalene triangle base

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.

C

A

AB, BC, and AC are the sides of ABC.

B

A, B, C are the triangle\'s vertices.

In a right triangle, the two sides making the right angle are called Legs.

The side opposite of the right angle is called the Hypothenuse.

In an isosceles triangle, the two sides that are congruent are called the Legs.

The third side is the called the base.

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

Example 1A: Classifying Triangles by Angle Measures

Classify BDC by its angle measures.

B is an obtuse angle.

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

Example 1B: Classifying Triangles by Angle Measures

Classify ABD by its angle measures.

ABD andCBD form a linear pair, so they are supplementary.

FHG is an equiangular triangle by definition.

Check It Out! Example 1

Classify FHG by its angle measures.

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

Remember!

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

When the sides of the triangle are extended, the original angles are the interior angles.

The angles that are adjacent (next to) the interior angles are the exterior angles.

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

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

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

Sum. Thm

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 mXYZ.

mXYZ + mYZX + mZXY = 180°

Substitute 40 for mYZX and 62 for mZXY.

mXYZ + 40+ 62= 180

mXYZ + 102= 180

Simplify.

mXYZ = 78°

Subtract 102 from both sides.

118°

Example 1B: 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 mYWZ.

Step 1 Find mWXY.

mYXZ + mWXY = 180°

Lin. Pair Thm. and Add. Post.

62 + mWXY = 180

Substitute 62 for mYXZ.

mWXY = 118°

Subtract 62 from both sides.

118°

Sum. Thm

Example 1B: Application Continued

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 mYWZ.

Step 2 Find mYWZ.

mYWX + mWXY + mXYW = 180°

Substitute 118 for mWXY and 12 for mXYW.

mYWX + 118+ 12= 180

mYWX + 130= 180

Simplify.

Subtract 130 from both sides.

mYWX = 50°

Sum. Thm

Check It Out! Example 1

Use the diagram to find mMJK.

mMJK + mJKM + mKMJ = 180°

Substitute 104 for mJKM and 44 for mKMJ.

mMJK + 104+ 44= 180

mMJK + 148= 180

Simplify.

Subtract 148 from both sides.

mMJK = 32°

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

Acute s of rt. are comp.

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 acute angle?

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

mA + mB = 90°

2x+ mB = 90

Substitute 2x for mA.

mB = (90 – 2x)°

Subtract 2x from both sides.

Acute s of rt. are comp.

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 mA = 63.7°.

mA + mB = 90°

63.7 + mB = 90

Substitute 63.7 for mA.

mB = 26.3°

Subtract 63.7 from both sides.

Acute s of rt. are comp.

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 other acute angle?

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

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 = 48 .

Acute s of rt. are comp.

48 + mB = 90

Substitute 48 for mA.

mB = 41

Subtract 48 from both sides.

3° 5

2° 5

2° 5

2 5

2 5

2 5

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 acute angle?

mA + mB = 90°

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

Exterior

Interior

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.

4 is an exterior angle.

Exterior

Interior

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.

4 is an exterior angle.

The remote interior angles of 4 are 1 and 2.

Exterior

Interior

3 is an interior angle.

Example 3: Applying the Exterior Angle Theorem

Find mB.

mA + mB = mBCD

Ext. Thm.

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

15 + 2x + 3= 5x – 60

2x + 18= 5x – 60

Simplify.

Subtract 2x and add 60 to both sides.

78 = 3x

26 = x

Divide by 3.

mB = 2x + 3 = 2(26) + 3 = 55°

Check It Out! Example 3

Find mACD.

mACD = mA + mB

Ext. Thm.

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

6z – 9 = 2z + 1+ 90

6z – 9= 2z + 91

Simplify.

Subtract 2z and add 9 to both sides.

4z = 100

z = 25

Divide by 4.

mACD = 6z – 9 = 6(25) – 9 = 141°

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

Example 2A: Classifying Triangles by Side Lengths

Classify EHF by its side lengths.

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

Example 2B: Classifying Triangles by Side Lengths

Classify EHGby its side lengths.

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

Check It Out! Example 2

Classify ACD by its side lengths.

Example 3: Using Triangle Classification

Find the side lengths of JKL.

Step 1 Find the value of x.

Given.

JK = KL

Def. of segs.

Substitute (4x – 10.7) for JK and (2x + 6.3) for KL.

4x – 10.7 = 2x + 6.3

Add 10.7 and subtract 2x from both sides.

2x = 17.0

x = 8.5

Divide both sides by 2.

Example 3 Continued

Find the side lengths of JKL.

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

Check It Out! Example 3

Find the side lengths of equilateral FGH.

Step 1 Find the value of y.

Given.

FG = GH = FH

Def. of segs.

Substitute

(3y – 4) for FG and (2y + 3) for GH.

3y – 4 = 2y + 3

Add 4 and subtract 2y from both sides.

y = 7

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

420 54 = 7 triangles

7 9

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.

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.

Check It Out! Example 4a

Each measure is the side length of an equilateral triangle. Determine how many 7 in. triangles can be formed from a 100 in. piece of steel.

The amount of steel needed to make one triangle is equal to the perimeter P of the equilateral triangle.

P = 3(7)

P = 21 in.

100 7 = 14 triangles

2 7

Check It Out! Example 4a Continued

Each measure is the side length of an equilateral triangle. Determine how many 7 in. triangles can be formed from a 100 in. piece of steel.

To find the number of triangles that can be made from 100 inches of steel, divide 100 by the amount of steel needed for one triangle.

There is not enough steel to complete a fifteenth triangle. So the manufacturer can make 14 triangles from a 100 in. piece of steel.

Check It Out! Example 4b

Each measure is the side length of an equilateral triangle. Determine how many 10 in. triangles can be formed from a 100 in. piece of steel.

The amount of steel needed to make one triangle is equal to the perimeter P of the equilateral triangle.

P = 3(10)

P = 30 in.

Check It Out! Example 4b Continued

Each measure is the side length of an equilateral triangle. Determine how many 10 in. triangles can be formed from a 100 in. piece of steel.

To find the number of triangles that can be made from 100 inches of steel, divide 100 by the amount of steel needed for one triangle.

100 10 = 10 triangles

The manufacturer can make 10 triangles from a 100 in. piece of steel.

Lesson Quiz I

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

33 °

4.1 Triangles and Angles

2 3

Lesson Quiz: Part II

5. The measure of one of the acute angles in a right triangle is 56 °. What is the measure of the other acute angle?

6. Find mABD.

124°

Lesson Quiz: Part III

7. The diagram is a map showing John\'s house, Kay\'s house, and the grocery store. What is the angle the two houses make with the store?

30°

Download Presentation

Connecting to Server..