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### Transcript of "Module5 dodong2"

1. 1. GRADE 9 MATHEMATICS QUARTER 3 Module 5 May 2014
2. 2. FIGURE, PICTURE, QUOTATION βn 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. 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. 4. Parallelograms Definition: A quadrilateral whose opposite sides are parallel. (Garces, et. Al., 2007)
5. 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. 6. Parallelograms The diagonal divides it into two congruent triangles. Proof: Given that β‘ABCD is a parallelogram. We have to prove that βABD β βCDB. By definition , π΄π΅ β₯ πΆπ· and π΅πΆ β₯ π΄π·, then we can say that β π·π΅πΆ β β π΅π·π΄ and β πΆπ·π΅ β β π΄π΅π· by Alternate Interior Angles theorem. This implies that βABD β βCDB by ASA congruence theorem. B A D C
7. 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·πΆπ΅. Because βABD β βCDB by ASA congruence theorem, then β π΅π΄π· β β π·πΆπ΅ by CPCTC. B A D C
8. 8. Parallelograms Consecutive angles are supplementary. Proof: Given that β‘ABCD is a parallelogram. We have to prove that β π΄π΅πΆ + β π΅πΆπ· = 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΅πΆ + 2β π΅πΆπ· = 360. thus the last equation is β π΄π΅πΆ + β π΅πΆπ· = 180
9. 9. Parallelograms Diagonals bisect each other. B A D C Proof: Given that β‘ABCD such that π΄π β πΆπ and π΅π β ππ·. We to prove that β‘ABCD is a parallelogram. By Vertical Angle Theorem , β π΄ππ· β β π΅ππΆ. This implies that βAPDββCPB by SASβ theo., and β π·π΄π β β π΅πΆπ by CPCTC. This means that π΄π· β₯ π΅πΆ. Similarly, π΄π΅ β₯ π·πΆ. Hence β‘ABCD is a parallelogram. QED P
10. 10. Parallelograms Opposite sides are congruent. Proof: Given that β‘ABCD such that π΄π΅ β πΆπ· and π΅πΆ β π΄π·. We to prove that β‘ABCD is a parallelogram. Since π΄π΅ β πΆπ· and π΅πΆ β π΄π·, then we can say that βABDββCDB by SSSβ theo. π΅π· is transversal line π΄π· and π΅πΆ, and β π·π΅πΆ β β π΄π·π΅ (CPCTC) then π΄π΅ β₯ π΅πΆ by the Converse of Alternate Interior Angles. Hence β‘ABCD is a parallelogram. QED B A D C
11. 11. Parallelograms The opposite sides are parallel and congruent. Proof: Given that β‘ABCD such that π΄π· β₯ π΅πΆ and π΄π· = π΅πΆ. We to prove that β‘ABCD is a parallelogram. The diagonal BD is a transversal of π΄π· and π΅πΆ. And by Alternate Interior angle theo., β π΄π·π΅ β β πΆπ΅π·. π΅π· is a common side between βADB and βCBD which are congruent by ASA. By CPCTC, AB=CD. Therefore, β‘ABCD is a parallelogram. B A D C
12. 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. 13. Parallelogram In the figure below, β‘AECG is a parallelogram. If π΅πΈ β πΊπ·, prove that β‘ABCD is also a parallelogram. A B C D E G
14. 14. Statement Reason 1. Draw π΄πΆ such that π΄πΆ intersect πΈπΊ at X 1. Construction 2. π΄πΆ is diagonal to β‘AECG and β‘ABCD 2. Def. of diagonal line 3. πΈπ β ππΊ 3. Prop. of parallelogram 4. π΅πΈ β πΊπ· 4. Given 5. π΅π β π΅πΈ + πΈπ; π·π β ππΊ + πΊπ· 5. Segment Addition Postulate 6. π΅π β π΅πΈ + πΈπ; π·π β πΈπ + π΅πΈ 6. Substitution 7. π΅π β π·π 7. Transitive PE 8. π΄π β ππΆ 8. Prop of Parallelogram (β‘AECG) 9. β΄ β‘ABCD is a parallelogram 9. Prop of Parallelogram (7 and 8) x
15. 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. 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΅πΆ β β πΆπ·π΄ β β π΅π΄π· β β π΅πΆπ· 2. Def. of rectangle 3. π΄π΅ β π·πΆ, π΅πΆ β π΄π· 3. Prop. Of rectangle 4. βABCβ βπΆπ·π΄ β βπ΄π΅π· ββDCB 4. SAS β Theorem 5. π΅π· β π΄πΆ 5. CPCTC A D C B Given: β‘ABCD is a rectangle Prove: π΅π· β π΄πΆ
17. 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 πΆ are congruent. And since β π΅ is opposite to β π·. Therefore parallelogram ABCD is a rectangle.
18. 18. Theorems on Rhombus 1. The diagonals of a rhombus are perpendicular and bisect each other. Given: π΄πΆ and π΅π· are diagonals intersect at P. π΄πΆ β₯ π΅π·, π΄π β ππΆ, π΅π β ππ·. Prove: β‘ABCD is a rhombus. A B C D P Statements Reasons 1. π΄πΆ β₯ π΅π· 1. Given 2. β π΄ππ΅ β β πΆππ· β β πΆππ΅ β β π΄ππ· 2. Def. of β₯ 3. π΄π β ππΆ, π΅π β ππ· 3. Given 4. βAPB β βπΆππ΅ β βπΆππ· ββAPD 4. SAS β Theorem 5. π΄π΅ β π΅πΆ β πΆπ· β π·π΄ 5. CPCTC 6. β΄ β‘ABCD is a rhombus 6. Prop of rhombus
19. 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 βABC, P and Q are midpoints of π΄π΅ and π΄πΆ, respectively. Prove: ππ β₯ π΅πΆ and PQ= 1 2 π΅πΆ. A P B Q C R Proof: Let R be a point in ππ such that ππ β ππ. Since Q is midpoint π΄πΆ, π΄π β ππ΅. And by Vertical Angle Theom., β π΄ππ β β πππΆ, so βAQPββCQR by SAS cong. Theo. which implies that β ππ΄π β β ππΆπ and π΄π β ππΆ by CPCTC. This means that β‘PBCR is parallelogram because π΄π΅ β₯ ππΆ (converse of AIA) and ππ΅ β ππΆ by transitivity PE. Therefore ππ β₯ π΅πΆ.
20. 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 βABC, P and Q are midpoints of π΄π΅ and π΄πΆ, respectively. Prove: ππ β₯ π΅πΆ and PQ= 1 2 π΅πΆ. A P B Q C R Since β‘PBCR is parallelogram, ππ β π΅πΆ. 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΅πΆ
21. 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. 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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