1. 7.2 Properties of Parallelograms
Objectives:
Use some properties of parallelograms.
Use properties of parallelograms in real-life situations. •

2. Concept: Exterior Angle-Sum Thm.
The sum of the measures of the exterior angles of a polygon, one at each vertex, is 360. For the pentagon

3. Concept: Exterior Angle-Sum Thm.
What is the measure of ONE exterior angle of a regular nonagon?

4. Concept: Parallelogram
A parallelogram is a quadrilateral with both pairs of opposite sides parallel.
Ex:

5. Concept: Opposite Sides Property
The opposite sides of a parallelogram are congruent. PQ≅RS and SP≅QR Q R P S

6. Concept: Opposite Angles Property
The opposite angles of a parallelogram are congruent. Q R
LP ≅ LR and LQ ≅ LS P S

7. Concept: Concsecutive Angles Property
The consecutive angles of a parallelogram are supplementary,
(add up to 180º) Q R
mLP +mLQ = 180°, mLQ +mLR = 180°, mLR + mLS = 180°, mLS + mLP = 180° P S

8. Concept: Using the Properties
Problem #1
PQRS is parallelogram. Find the measures of the given angles:
mLR
mLQ
70º
110º

9. Concept: Diagonals Property
If a quadrilateral is a parallelogram, then its diagonals bisect each other. QM ≅ SM and PM ≅ RM Q R P S

10. Concept: Using the Properties
SOLUTION: JH = FG Opposite sides of a are ≅. JH = 5 Since FG = JH then JH = 5
Find the JH and JK

11. Concept: Using the Properties cont…
SOLUTION: JK = GK Diagonals of a bisect each other GK = 3 Since JK = GK then JK = 3
Find the JH and JK

12. Concept: Using the Properties cont…
Solve for x and y
2y + 3 x
HK = y + 5, KF = x, KG = x + 2, KE = 2y + 3
y + 5
x + 2
1. Label the diagram
y + 5 = x
2y + 3 = x + 2
2. Set up equations

13. Concept: Using the Properties cont…
Solve for x and y
2y + 3 x
HK = y + 5, KF = x, KG = x + 2, KE = 2y + 3
y + 5
x + 2
3. Since there are 2 different variables in each equation, we will need to substitute. (Replace x.)
y + 5 = x
2y + 3 = x + 2
2y + 3 = y

14. Concept: Using the Properties cont…
Solve for x and y
HK = y + 5, KF = x, KG = x + 2, KE = 2y + 3
2y + 3 x
2y + 3 = y
y + 5
x + 2
2y + 3 = y + 7
–y –y
y + 3 = 7
4. Solve for y
–3 = –3
y = 4
Substitute the value you
found for y so you can
find x.
y + 5 = x
4 + 5 = x
9 = x