CHAPTER 5SECTION 4

Inequalities for Sides and Angles of a triangle

WARM-UPDetermine whether triangle STU is congruent to triangle VUT, using the given information. Justify. 1) <S is congruent to <V  2) SU is congruent to VT  3) <STU and <VUT are right angles  4) Given: <STU and <UVT are right angles; SU is congruent to VT. Prove: <S is congruent to <V.

ST

U V

WARM-UPDetermine whether triangle STU is congruent to triangle VUT, using the given information. Justify. 1) <S is congruent to <VYes by LA  2) SU is congruent to VTYes by HL  3) <STU and <VUT are right anglesNo 

ST

U V

WARM-UP4) Given: <STU and <UVT are right angles; SU is congruent to VT. Prove: <S is congruent to <V.

ST

U V

Statements Reasons

<STU and <UVT are right angles; SU is congruent to VT.

Triangle STU and triangle VUT are right triangles

Given

Definition of a right triangle

Congruence of segments is reflexive

TU is congruent to UT

Triangle STU is congruent to triangle VUT

<S is congruent to <V

HL

CPCTC

VOCABULARYTheorem 5-9- If one side of a triangle is longer that another side, then the angle opposite the longer side has a greater measure than the angle opposite the shorter side. Theorem 5-10- If one angle of a triangle has a greater measure than another angle, then the side opposite the greater angle is longer than the side opposite the lesser angle.

Theorem 5-11- The perpendicular segment from a point to a line is the shortest segment from the point to the line.  Corollary 5-1- The perpendicular segment from a point to a plane is the shortest segment from the point to the plane. 

Example 1) Given: m<A is greater than m<DProve: BD is greater than AB. C

B

Statements Reasons

m<A + m<E + m<EBA = 180, m<C + m<D + m<CBD = 180

m<EBA = m<EBD

m<A + m<E = 180 - m<EBA,m<C + m<D = 180 - m<CBD

m<A is greater than m<D 

Angle Sum Theorem

Substitution Property of Equality

Subtraction Property of Equality

m<A + m<D = m<C + m<D

Given

Vertical angles are congruent

BD is greater than ABIf an angle of a triangle is

greater than another, the side opposite the greater angle is longer that the side opposite

the lesser angle.

A

E

D

m<C is greater than m<E  Subtraction Property of Equality

Example 2: Triangle JKL with vertices J(-4, 2), K(4, 3), and L(1, -3). List the angles in order from least measure to greatest measure.

Find the length of each side.The distance formula is d=√((x2 – x1)2 + (y2 – y1)2)

JKd=√((x2 – x1)2 + (y2 – y1)2)d=√((-4 – 4)2 + (2 – 3)2)d=√((-8)2 + (-1)2)d=√(64 + 1)d=√(65)KLd=√((x2 – x1)2 + (y2 – y1)2)d=√((4 – 1)2 + (3 – -3)2)d=√((4 – 1)2 + (3 + 3)2)d=√((3)2 + (6)2)d=√(9 + 36)d=√(45)

JLd=√((x2 – x1)2 + (y2 – y1)2)d=√((-4 – 1)2 + (2 – -3)2)d=√((-1 – 1)2 + (3 + 3)2)d=√((-5)2 + (6)2)d=√(25 + 36)d=√(61)

The shortest side is KL, so the smallest angle is <J. The next shortest side is JL, so the next smallest angle is <K. The greatest side is JK, so the greatest angle is <L.<J, <K, <L

Example 3: Triangle JKL with vertices J(-2, 4), K(-5, -8), and L(6, 10). List the angles in order from least measure to greatest measure.

Find the length of each side.The distance formula is d=√((x2 – x1)2 + (y2 – y1)2)

JKd=√((x2 – x1)2 + (y2 – y1)2)d=√((-2 – -5)2 + (4 – -8)2)d=√((-2 + 5)2 + (4 + 8)2)d=√((-3)2 + (12)2)d=√(9 + 144)d=√(153)

KLd=√((x2 – x1)2 + (y2 – y1)2)d=√((-5 – 6)2 + (-8 – 10)2)d=√((-11)2 + (-18)2)d=√(121 + 324)d=√(445)

JLd=√((x2 – x1)2 + (y2 – y1)2)d=√((-2 – 6)2 + (4 – 10)2)d=√((-8)2 + (-6)2)d=√(64 + 36)d=√(100)d = 10

The shortest side is JL, so the smallest angle is <K. The next shortest side is JK, so the next smallest angle is <L. The greatest side is KL, so the greatest angle is <J.<K, <J, <L

Example 4) Refer to the figure.

A) Which is greater, m<CBD or m<CDB?<CDB because it is across from 16 which is greater than 15.  B) Is m<ADB greater that m<DBA?Yes because it is across from a side that is longer.   C)Which is greater, m<CDA or m<CBA?<CDA because it is across from 10 + 16 = 26 and <CBA is across from 8 + 15 = 23.

BA

D C

10

8

15

1612

Example 5) Refer to the figure.

A) Which side of triangle RTU is the longest?First find all the angles in the triangle.

180 = 110 + m<RUT70 = m<RUT

180 = 30 + 70 + m<TRU180 = 100 + m<TRU80 = m<TRU

The greatest angle is 80 degrees so the longest side is TU.  

UR

T

S

30

70

80 110

B) Name the side of triangle UST that is the longest.Since there can be only one obtuse angle in a triangle, the longest side is across from the obtuse angle.TS is the longest side.

Example 6) Find the value of x and list the sides of Triangle ABC in order from shortest to longest if <A= 3x +20, <B = 2x + 37, and <C = 4x + 15

The angles of a triangle add up to 180.180 = m<A + m<B + m<C180 = 3x + 20 + 2x + 37+ 4x + 15180 = 9x + 72108 = 9x12 = x

Plug 12 in for x in each equation.

m<A = 3x + 20m<A = 3(12) + 20m<A = 36 + 20m<A = 56

BC, AC, AB

m<B = 2x + 37m<B = 2(12) + 37m<B = 24 + 37m<B = 61

m<C = 4x + 15m<C = 4(12) + 15m<C = 48 + 15m<C = 63

Example 7) Find the value of x and list the sides of Triangle ABC in order from shortest to longest if <A= 9x +29, <B = 93 – 5x, and <C = 10x + 2

The angles of a triangle add up to 180.180 = m<A + m<B + m<C180 = 9x + 29 + 93 – 5x + 10x + 2180 = 14x + 12456 = 14x4 = x

Plug 4 in for x in each equation.

m<A = 9x + 29m<A = 9(4) + 29m<A = 36 + 29m<A = 65

AB, BC, AC

m<B = 93 – 5xm<B = 93 – 5(4)m<B = 93 - 20m<B = 73

m<C = 10x + 2m<C = 10(4) + 2m<C = 40 + 2m<C = 42