Download Section 4.5 - Isosceles and Equilateral Triangles

K
A
Warm Up
J
1. Write a congruence statement
for the triangles.
L
B
C
ABCLKJ
2. Which Congruence Theorem supports your
conclusion? SSA. There are 2 congruent sides and
1 angle. The angle is not between the sides.
3. Which of the following statements is true by
CPCTC?
C  K
𝐽𝐿 ≅ 𝐴𝐶
4.5 Isosceles and Equilateral
Triangles
Target: SWBAT use and apply
properties of isosceles and equilateral
triangles.
Vocabulary
Isosceles triangle
legs of an isosceles triangle
vertex angle
base
base angles
Equilateral and Equiangular triangles
Recall that an isosceles triangle has at least two
congruent sides. The congruent sides are called the
legs. The vertex angle is the angle formed by the
legs. The side opposite the vertex angle is called the
base, and the base angles are the two angles that
are connected by the base.
A
3 is the vertex angle.
1 and 2 are the base angles.
Sides 𝐴𝐶𝑎𝑛𝑑 𝐵𝐶 are the legs
Side 𝐴𝐵 is the base
B
C
4-3
Pg 250
4-4
Pg 251
- Page 252
A corollary is a theorem that can be proved easily
using another theorem.
Page 252
Corollary to
Theorem 4-3
Page 252
Corollary to
Theorem 4-4
Assignment #38: Pages 254-256
Foundation:
10-13
Core:
14, 16-19, 22, 28, 30-31
Challenge:
32
Example 2A: Finding the Measure of an Angle
Find mF.
mF = mD = x°
Isosc. ∆ Thm.
mF + mD + mA = 180 ∆ Sum Thm.
Substitute the
x + x + 22 = 180 given values.
Simplify and subtract
2x = 158 22 from both sides.
x = 79 Divide both
sides by 2.
Thus mF = 79°
Example 2B: Finding the Measure of an Angle
Find mG.
mJ = mG Isosc. ∆ Thm.
(x + 44) = 3x
44 = 2x
Substitute the
given values.
Simplify x from
both sides.
Divide both
sides by 2.
Thus mG = 22° + 44° = 66°.
x = 22
Check It Out! Example 2A
Find mH.
mH = mG = x°
Isosc. ∆ Thm.
mH + mG + mF = 180 ∆ Sum Thm.
Substitute the
x + x + 48 = 180 given values.
Simplify and subtract
2x = 132 48 from both sides.
x = 66 Divide both
sides by 2.
Thus mH = 66°
Check It Out! Example 2B
Find mN.
mP = mN Isosc. ∆ Thm.
(8y – 16) = 6y
2y = 16
y = 8
Substitute the
given values.
Subtract 6y and
add 16 to both
sides.
Divide both
sides by 2.
Thus mN = 6(8) = 48°.
Example 3A: Using Properties of Equilateral
Triangles
Find the value of x.
∆LKM is equilateral.
Equilateral ∆  equiangular ∆
(2x + 32) = 60
2x = 28
x = 14
The measure of each  of an
equiangular ∆ is 60°.
Subtract 32 both sides.
Divide both sides by 2.
Example 3B: Using Properties of Equilateral
Triangles
Find the value of y.
∆NPO is equiangular.
Equiangular ∆  equilateral ∆
5y – 6 = 4y + 12
y = 18
Definition of
equilateral ∆.
Subtract 4y and add 6 to
both sides.
Check It Out! Example 3
Find the value of JL.
∆JKL is equiangular.
Equiangular ∆  equilateral ∆
4t – 8 = 2t + 1
2t = 9
t = 4.5
Definition of
equilateral ∆.
Subtract 4y and add 6 to
both sides.
Divide both sides by 2.
Thus JL = 2(4.5) + 1 = 10.
Lesson Quiz: Part I
Find each angle measure.
1. mR
28°
2. mP
124°
Find each value.
3. x
5. x
20
4. y
26°
6