1.
GRADE 9 MATHEMATICS
QUARTER 3
Module 5
May 2014
2.
FIGURE, PICTURE,
QUOTATION βn ONE
Instruction:
Arrange the jigsaw puzzle to form a picture/figure and
paste it in the manila paper. Shout your yell after you
made this activity.
Present the puzzle, describe the figure and explain
briefly the Mathematics quotation found therein.
3.
Competencies
1. Identifies quadrilaterals that are parallelograms.
2. Determines the conditions that guarantee a
quadrilateral a parallelogram.
3. Uses properties to find measures of angles, sides and
other quantities involving parallelograms.
4. Proves theorems on the different kinds of
parallelograms (rectangle, rhombus, square)
5. Proves the Midline Theorem.
6. Proves theorems on trapezoids and kites.
7. Solves problems involving parallelograms, trapezoids
and kites.
4.
Parallelograms
Definition:
A quadrilateral whose opposite sides are parallel.
(Garces, et. Al., 2007)
5.
Properties of Parallelograms
1. The diagonal divides it into two congruent triangles.
2. Opposite angles are congruent.
3. Two consecutive angles are supplementary.
4. The diagonals bisect each other.
5. Opposite sides are congruent.
6. The opposites sides are parallel and congruent.
6.
Parallelograms
The diagonal divides it into
two congruent triangles.
Proof:
Given that β‘ABCD is a parallelogram. We have to prove
that βABD β βCDB.
By definition , π΄π΅ β₯ πΆπ· and π΅πΆ β₯ π΄π·, then we can say that
β π·π΅πΆ β β π΅π·π΄ and β πΆπ·π΅ β β π΄π΅π· by Alternate Interior
Angles theorem. This implies that βABD β βCDB by ASA
congruence theorem.
B
A D
C
7.
Parallelograms
Opposite angles are congruent.
Proof:
Given that β‘ABCD is a parallelogram. We have to prove
that β π΅π΄π· β β π·πΆπ΅.
Because βABD β βCDB by ASA congruence theorem, then
β π΅π΄π· β β π·πΆπ΅ by CPCTC.
B
A D
C
8.
Parallelograms
Consecutive angles are supplementary.
Proof:
Given that β‘ABCD is a parallelogram. We have to prove
that β π΄π΅πΆ + β π΅πΆπ· = 180.
Since the sum of interior angles of quadrilaterals is 3600
and the opposite angles of parallelogram are congruent,
then we can say that 2β π΄π΅πΆ + 2β π΅πΆπ· = 360. thus the last
equation is β π΄π΅πΆ + β π΅πΆπ· = 180
9.
Parallelograms
Diagonals bisect each other.
B
A D
C
Proof:
Given that β‘ABCD such that π΄π β πΆπ and π΅π β ππ·. We
to prove that β‘ABCD is a parallelogram.
By Vertical Angle Theorem , β π΄ππ· β β π΅ππΆ. This implies
that βAPDβ βCPB by SASβ theo., and β π·π΄π β β π΅πΆπ by
CPCTC. This means that π΄π· β₯ π΅πΆ. Similarly, π΄π΅ β₯ π·πΆ.
Hence β‘ABCD is a parallelogram. QED
P
10.
Parallelograms
Opposite sides are congruent.
Proof:
Given that β‘ABCD such that π΄π΅ β πΆπ· and π΅πΆ β π΄π·. We
to prove that β‘ABCD is a parallelogram.
Since π΄π΅ β πΆπ· and π΅πΆ β π΄π·, then we can say that
βABDβ βCDB by SSSβ theo. π΅π· is transversal line π΄π· and
π΅πΆ, and β π·π΅πΆ β β π΄π·π΅ (CPCTC) then π΄π΅ β₯ π΅πΆ by the
Converse of Alternate Interior Angles. Hence β‘ABCD is a
parallelogram. QED
B
A D
C
11.
Parallelograms
The opposite sides are parallel
and congruent.
Proof:
Given that β‘ABCD such that π΄π· β₯ π΅πΆ and π΄π· = π΅πΆ. We
to prove that β‘ABCD is a parallelogram.
The diagonal BD is a transversal of π΄π· and π΅πΆ. And by
Alternate Interior angle theo., β π΄π·π΅ β β πΆπ΅π·. π΅π· is a
common side between βADB and βCBD which are
congruent by ASA. By CPCTC, AB=CD. Therefore, β‘ABCD
is a parallelogram.
B
A D
C
12.
Parallelogram
Find the values of x and y in the parallelogram ABCD.
B
A
x0
C
y0
400
600
Answer: y = 1200 and x = 200
13.
Parallelogram
In the figure below, β‘AECG is a parallelogram. If π΅πΈ β
πΊπ·, prove that β‘ABCD is also a parallelogram.
A
B C
D
E
G
14.
Statement Reason
1. Draw π΄πΆ such that π΄πΆ intersect πΈπΊ at X 1. Construction
2. π΄πΆ is diagonal to β‘AECG and β‘ABCD 2. Def. of diagonal line
3. πΈπ β ππΊ 3. Prop. of parallelogram
4. π΅πΈ β πΊπ· 4. Given
5. π΅π β π΅πΈ + πΈπ; π·π β ππΊ + πΊπ· 5. Segment Addition Postulate
6. π΅π β π΅πΈ + πΈπ; π·π β πΈπ + π΅πΈ 6. Substitution
7. π΅π β π·π 7. Transitive PE
8. π΄π β ππΆ 8. Prop of Parallelogram (β‘AECG)
9. β΄ β‘ABCD is a parallelogram 9. Prop of Parallelogram (7 and 8)
x
15.
Kinds of Parallelogram
Rectangle β All angles are congruent
Rhombus β All sides are congruent
Square β All angles and sides are congruent
16.
Theorems on Rectangle
1. In a rectangle, the two diagonals are congruent.
Statements Reasons
1. β‘ABCD is rectangle 1. Given
2. β π΄π΅πΆ β β πΆπ·π΄ β β π΅π΄π· β β π΅πΆπ· 2. Def. of rectangle
3. π΄π΅ β π·πΆ, π΅πΆ β π΄π· 3. Prop. Of rectangle
4. βABCβ βπΆπ·π΄ β βπ΄π΅π· β βDCB 4. SAS β Theorem
5. π΅π· β π΄πΆ 5. CPCTC
A D
C
B
Given: β‘ABCD is a rectangle
Prove: π΅π· β π΄πΆ
17.
Theorems on Rectangle
2. If one angle of a parallelogram is right, then it is a
rectangle.
Proof:
Given that β π΄ of parallelogram ABCD is right, we need to prove that the
parallelogram is rectangle.
Since the consecutive angles of a parallelogram are supplementary, we can say β π΄
and β π΅ 900 since they are supplementary. From the properties of parallelogram
that the opposite angles are congruent, then β π΄ and β πΆ are congruent. And since
β π΅ is opposite to β π·. Therefore parallelogram ABCD is a rectangle.
18.
Theorems on Rhombus
1. The diagonals of a rhombus are perpendicular and
bisect each other.
Given: π΄πΆ and π΅π· are diagonals intersect at P.
π΄πΆ β₯ π΅π·, π΄π β ππΆ, π΅π β ππ·.
Prove: β‘ABCD is a rhombus.
A
B C
D
P
Statements Reasons
1. π΄πΆ β₯ π΅π· 1. Given
2. β π΄ππ΅ β β πΆππ· β β πΆππ΅ β β π΄ππ· 2. Def. of β₯
3. π΄π β ππΆ, π΅π β ππ· 3. Given
4. βAPB β βπΆππ΅ β βπΆππ· β βAPD 4. SAS β Theorem
5. π΄π΅ β π΅πΆ β πΆπ· β π·π΄ 5. CPCTC
6. β΄ β‘ABCD is a rhombus 6. Prop of rhombus
19.
Midline Theorem
The segment that joins the midpoints of two sides of a
triangle is parallel to the third side and half as long.
Given: In βABC, P and Q are midpoints of
π΄π΅ and π΄πΆ, respectively.
Prove: ππ β₯ π΅πΆ and PQ=
1
2
π΅πΆ.
A
P
B
Q
C
R
Proof: Let R be a point in ππ such that ππ β ππ .
Since Q is midpoint π΄πΆ, π΄π β ππ΅. And by Vertical Angle
Theom., β π΄ππ β β π ππΆ, so βAQPβ βCQR by SAS cong. Theo.
which implies that β ππ΄π β β ππΆπ and π΄π β π πΆ by CPCTC.
This means that β‘PBCR is parallelogram because
π΄π΅ β₯ π πΆ (converse of AIA) and ππ΅ β π πΆ by transitivity PE.
Therefore ππ β₯ π΅πΆ.
20.
Midline Theorem
The segment that joins the midpoints of two sides of a
triangle is parallel to the third side and half as long.
Given: In βABC, P and Q are midpoints of
π΄π΅ and π΄πΆ, respectively.
Prove: ππ β₯ π΅πΆ and PQ=
1
2
π΅πΆ.
A
P
B
Q
C
R
Since β‘PBCR is parallelogram, ππ β π΅πΆ. We can say that
PR=PQ+QR by Segment Addition postulate. Since PQ=QR
and PR=BC, then BC=2PQ. Simplify, the results will be
PQ=
1
2
π΅πΆ
21.
Theorems on Trapezoids
1. The median of a trapezoid is half of the sum of its
bases.
2. The base angles of an isosceles trapezoid are
congruent.
3. Opposite angles of an isosceles trapezoid are
supplementary.
4. The diagonals of an isosceles trapezoid are
congruent.
22.
Theorems on kite
1. The diagonals are perpendicular to each other.
2. The area is half of the product of its diagonals.
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