Properties of parallelogram applies to rectangles, rhombi and squares. In a parallelogram, Opposite sides of a parallelogram are parallel. A diagonal of a parallelogram divides it into two congruent triangles. Opposite sides of a parallelogram are congruent. Opposite angles of a parallelogram are congruent. If one angle of a parallelogram is right, then all the angles are right. Consecutive angles of a parallelogram are supplementary. Diagonals of a parallelogram bisect each other. https://www.youtube.com/channel/UCOuMfD4sggCh7XeiAHlus6Q

1. Are you ready?
Learning is FUN.

2. APPLICATION of the
PROPERTIES of
PARALLELOGRAMS

3. Things you need to
remember…

4.  A diagonal of a parallelogram divides it
into two congruent triangles.
 Opposite sides of a parallelogram are
congruent.
 Opposite angles of a parallelogram are
congruent.

5.  Consecutive angles of a parallelogram
are supplementary.
 If one angle in a parallelogram is right,
then all angles are right.
 The diagonals of a parallelogram
bisect each other.

6. EXAMPLES
Part 1

7. LADY is a parallelogram.
A
R
D
Y
L
1. If LA = 15, find YD.
2. If m∠LYD = 108, find m∠LAD.
3. If m∠LYA = 73 and m∠AYD = 51, find
m∠YDA.
4. If RA = 7, find YA.
5. If m∠LAY = 65 and m∠ALY = 55, find
m∠DAY.

8. A
R
D
Y
L
1. If 𝐋𝐀 = 𝟏𝟓, find 𝐘𝐃.
Solution:
Opposite sides of a parallelogram
are congruent. Hence,
𝐘𝐃 = 𝐋𝐀 = 15.

9. 2. If m∠LYD = 108, find m∠LAD.
Solution:
Opposite angles of a parallelogram
are congruent. Hence,
m∠LAD = m∠LYD = 108
A
R
D
Y
L

10. 3. If m∠LYA = 73 and m∠AYD = 51, find m∠YDA.
Solution:
Consecutive angles of a parallelogram
are supplementary. We have,
m∠LYD + m∠YDA = 180.
But m∠LYD = m∠LYA + m∠AYD = 73 + 51 = 124.
Therefore, m∠YDA = 180 – m∠LYD
m∠YDA = 180 – 124 = 56
A
R
D
Y
L

11. A
R
D
Y
L
4. If 𝐑𝐀 = 𝟕, find 𝐘𝐀.
Solution:
The diagonals f a parallelogram
bisect each other. Thus, YR ≅ RA or
YR = RA. Therefore,
YA = 2RA
YA = 2(7)
𝐘𝐀 = 𝟏𝟒

12. 5. If m∠LAY = 65 and m∠ALY = 55, find m∠DAY.
A
R
D
Y
L
Solution:
Consecutive angles of a parallelogram
are supplementary. We have,
m∠ALY + m∠LAD = 180.
But m∠LAD = m ∠LAY + m∠DAY = 65 + m∠DAY.
Therefore, m∠DAY = 180 – (m∠ALY + m∠LAY)
m∠DAY = 180 – (55 + 65)
m∠DAY = 180 – 120 = 60

13. EXAMPLES
Part 2

14. LADY is a parallelogram. Find the value of x.
A
R
D
Y
L
1. LY = 5 − 3𝑥 and AD = 2𝑥 + 15
2. m∠YLA = 7x – 32 and m∠ADY = 3x + 40
3. m∠ALY = 4x + 51 and m∠LYD = 5x + 48
4. LR = 2𝑥 + 5 and RD = 𝑥 + 9.

15. 1. 𝐋𝐘 = 𝟓 − 𝟑𝒙 and 𝐀𝐃 = 𝟐𝒙 + 𝟏𝟓
A
R
D
Y
L
Solution:
Opposite sides of a parallelogram are congruent.
Therefore,
LY = AD
5 – 3x = 2x + 15
–3x – 2x = 15 – 5
–5x = 10
x = – 2

16. 2. m∠YLA = 7x – 32 and m∠ADY = 3x + 40
Solution:
Opposite angles of a parallelogram are congruent.
Therefore,
m∠YLA = m∠ADY
7x – 32 = 3x + 40
7x – 3x = 40 + 32
4x = 72
x = 18
A
R
D
Y
L

17. 3. m∠ALY = 4x + 51 and m∠LYD = 5x + 48
Solution:
Consecutive angles of a parallelogram are
supplementary. Therefore,
m∠ALY + m∠LYD = 180
4x + 51 + 5x + 48 = 180
9x + 99 = 180
9x = 180 – 99
9x = 81
A
R
D
Y
L
x = 9

18. 4. 𝐋𝐑 = 𝟐𝒙 + 𝟓 𝐚𝐧𝐝 𝐑𝐃 = 𝒙 + 𝟗
A
R
D
Y
L
Solution:
Diagonals of a parallelogram bisect each other.
Therefore,
LR = RD
2x + 5= x + 9
2x - x = 9 – 5
x = 4

19. EXAMPLES
Part 3

20. Find the value of x and y.
Given:
PE = 2𝑥 EA = 9 − 𝑥
LE = 5𝑦 − 11 YE = 2𝑦 + 1
Solution:
Solve for x.
PE = EA
2x = 9 – x
2x + x = 9
3x = 9
x = 3
L
A
Y
P
E
Solve for y.
LE = YE
5y – 11 = 2y + 1
5y – 2y = 1 + 11
3y = 12
y = 4

21. If 𝐍𝐄 = 𝟏𝟐, 𝐄𝐓 = 𝟏𝟖, 𝐎𝐓 = 𝟑𝒚 − 𝒙, and 𝐍𝐎 = 𝒙 + 𝟐𝒚,
find the values of x and y and the perimeter of
parallelogram NOTE.
O
18 T
E
N x + 2y
12
3y – x

22. Solution:
NO = ET 𝑥 + 2𝑦 = 18
NE = OT 12 = 3𝑦 − 𝑥
We can find the values of x and y using the elimination
method.

23. By elimination method, we eliminate
first the variable x using addition.
𝑥 + 2𝑦 = 18
+ −𝑥 + 3𝑦 = 12
5y = 30
y = 6
Solution:
𝑥 + 2𝑦 = 18 (eq 1)
12 = 3𝑦 − 𝑥 −𝑥 + 3𝑦 = 12 (eq 2)
Use the value of y to find the value of x by
substituting to any of the two equation.
y = 6
𝑥 + 2𝑦 = 18
x + 2(6) = 18
x + 12 = 18
x = 18 – 12
x = 6

24. Solution:
To find the perimeter of parallelogram NOTE, we just substitute the value
of x and y to find the measures of each side of the parallelogram then add all the
measure of each side.
NO = 𝑥 + 2𝑦 = 6 + 2(6) = 6 + 12 = 18 ET = 18
OT = 3𝑦 − 𝑥 = 3 6 − 6 = 18 − 6 = 12 NE = 12
Perimeter = NO + OT + ET + NE
Perimeter = 18 + 12 + 18 + 12
Perimeter = 60
Therefore, the values of x and y
are 6 and 6, and the perimeter
of parallelogram NOTE is 60.

25. THANK
YOU
HAVE A NICE
DAY!
I
MATh