5-6: Inequalities Involving 2 5-6: Inequalities Involving 2 TrianglesTriangles

Expectation:G1.2.2 Construct and justify arguments

and solve problems involving angle measure, side length, perimeter and area of all types of triangles.

In the triangles below, 2 pairs of In the triangles below, 2 pairs of corresponding sides are congruent. corresponding sides are congruent.

Compare the third sides.Compare the third sides.

2.67 2.67

2.172.17

82° 32°

x

y

SAS Inequality (Hinge SAS Inequality (Hinge Theorem) Theorem)

If two sides of one triangle are congruent to 2 sides of a second triangle, but the included angle of the first is greater than the included angle of the second, then the 3rd side of the first is _______________ than the 3rd side of the second.

Two pairs of corresponding sides are Two pairs of corresponding sides are congruent in the triangles below. congruent in the triangles below.

Compare the included angles for the Compare the included angles for the given congruent sides.given congruent sides.

SSS Inequality SSS Inequality

If two sides of one triangle are congruent to 2 sides of a second triangle, but the third side of the first is longer than the third side of the second, then the included angle between the pair of congruent sides of the first is ______________ than the included angle between the pair of congruent sides of the second.

Use the figure to compare the Use the figure to compare the indicated measures. indicated measures.

A

ED

CB 9

106

9

8

10

8

a. m∠ADB and m∠DBC

Use the figure to compare the Use the figure to compare the indicated measures. indicated measures.

A

ED

CB 9

106

9

8

10

8

b. m∠EAD and m∠BCD

In ΔABC and ΔDEF below (not drawn to scale) AB, In ΔABC and ΔDEF below (not drawn to scale) AB, BC, DE and EF are all 12 units long. If m∠A = 30° BC, DE and EF are all 12 units long. If m∠A = 30°

and m∠F = 50°, compare AC and DF.and m∠F = 50°, compare AC and DF.

A. AC = DFB. AC > DFC. AC < DFD. not enough information given to

determine an answer.

A C

B

D F

E

Given : QC bisects PD and m∠PCQ > m∠DCQ

Prove: m∠D > m∠P

PC

D

Q

Assignment:

pages 277,

# 15 - 29 (odds)