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- 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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