Module5 dodong2
- 1. GRADE 9 MATHEMATICS QUARTER 3 Module 5 May 2014
- 2. FIGURE, PICTURE, QUOTATION ‘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. 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 □ABCD is a parallelogram. We have to prove that ∆ABD ≅ ∆CDB. By definition , 𝐴𝐵 ∥ 𝐶𝐷 and 𝐵𝐶 ∥ 𝐴𝐷, then we can say that ∠𝐷𝐵𝐶 ≅ ∠𝐵𝐷𝐴 and ∠𝐶𝐷𝐵 ≅ ∠𝐴𝐵𝐷 by Alternate Interior Angles theorem. This implies that ∆ABD ≅ ∆CDB by ASA congruence theorem. B A D C
- 7. Parallelograms Opposite angles are congruent. Proof: Given that □ABCD is a parallelogram. We have to prove that ∠𝐵𝐴𝐷 ≅ ∠𝐷𝐶𝐵. Because ∆ABD ≅ ∆CDB by ASA congruence theorem, then ∠𝐵𝐴𝐷 ≅ ∠𝐷𝐶𝐵 by CPCTC. B A D C
- 8. Parallelograms Consecutive angles are supplementary. Proof: Given that □ABCD is a parallelogram. We have to prove that ∠𝐴𝐵𝐶 + ∠𝐵𝐶𝐷 = 180. Since the sum of interior angles of quadrilaterals is 3600 and the opposite angles of parallelogram are congruent, then we can say that 2∠𝐴𝐵𝐶 + 2∠𝐵𝐶𝐷 = 360. thus the last equation is ∠𝐴𝐵𝐶 + ∠𝐵𝐶𝐷 = 180
- 9. Parallelograms Diagonals bisect each other. B A D C Proof: Given that □ABCD such that 𝐴𝑃 ≅ 𝐶𝑃 and 𝐵𝑃 ≅ 𝑃𝐷. We to prove that □ABCD is a parallelogram. By Vertical Angle Theorem , ∠𝐴𝑃𝐷 ≅ ∠𝐵𝑃𝐶. This implies that ∆APD≅∆CPB by SAS≅ theo., and ∠𝐷𝐴𝑃 ≅ ∠𝐵𝐶𝑃 by CPCTC. This means that 𝐴𝐷 ∥ 𝐵𝐶. Similarly, 𝐴𝐵 ∥ 𝐷𝐶. Hence □ABCD is a parallelogram. QED P
- 10. Parallelograms Opposite sides are congruent. Proof: Given that □ABCD such that 𝐴𝐵 ≅ 𝐶𝐷 and 𝐵𝐶 ≅ 𝐴𝐷. We to prove that □ABCD is a parallelogram. Since 𝐴𝐵 ≅ 𝐶𝐷 and 𝐵𝐶 ≅ 𝐴𝐷, then we can say that ∆ABD≅∆CDB by SSS≅ theo. 𝐵𝐷 is transversal line 𝐴𝐷 and 𝐵𝐶, and ∠𝐷𝐵𝐶 ≅ ∠𝐴𝐷𝐵 (CPCTC) then 𝐴𝐵 ∥ 𝐵𝐶 by the Converse of Alternate Interior Angles. Hence □ABCD is a parallelogram. QED B A D C
- 11. Parallelograms The opposite sides are parallel and congruent. Proof: Given that □ABCD such that 𝐴𝐷 ∥ 𝐵𝐶 and 𝐴𝐷 = 𝐵𝐶. We to prove that □ABCD is a parallelogram. The diagonal BD is a transversal of 𝐴𝐷 and 𝐵𝐶. And by Alternate Interior angle theo., ∠𝐴𝐷𝐵 ≅ ∠𝐶𝐵𝐷. 𝐵𝐷 is a common side between ∆ADB and ∆CBD which are congruent by ASA. By CPCTC, AB=CD. Therefore, □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, □AECG is a parallelogram. If 𝐵𝐸 ≅ 𝐺𝐷, prove that □ABCD is also a parallelogram. A B C D E G
- 14. Statement Reason 1. Draw 𝐴𝐶 such that 𝐴𝐶 intersect 𝐸𝐺 at X 1. Construction 2. 𝐴𝐶 is diagonal to □AECG and □ABCD 2. Def. of diagonal line 3. 𝐸𝑋 ≅ 𝑋𝐺 3. Prop. of parallelogram 4. 𝐵𝐸 ≅ 𝐺𝐷 4. Given 5. 𝐵𝑋 ≅ 𝐵𝐸 + 𝐸𝑋; 𝐷𝑋 ≅ 𝑋𝐺 + 𝐺𝐷 5. Segment Addition Postulate 6. 𝐵𝑋 ≅ 𝐵𝐸 + 𝐸𝑋; 𝐷𝑋 ≅ 𝐸𝑋 + 𝐵𝐸 6. Substitution 7. 𝐵𝑋 ≅ 𝐷𝑋 7. Transitive PE 8. 𝐴𝑋 ≅ 𝑋𝐶 8. Prop of Parallelogram (□AECG) 9. ∴ □ABCD is a parallelogram 9. Prop of Parallelogram (7 and 8) x
- 15. Kinds of Parallelogram Rectangle – All angles are congruent Rhombus – All sides are congruent Square – All angles and sides are congruent
- 16. Theorems on Rectangle 1. In a rectangle, the two diagonals are congruent. Statements Reasons 1. □ABCD is rectangle 1. Given 2. ∠𝐴𝐵𝐶 ≅ ∠𝐶𝐷𝐴 ≅ ∠𝐵𝐴𝐷 ≅ ∠𝐵𝐶𝐷 2. Def. of rectangle 3. 𝐴𝐵 ≅ 𝐷𝐶, 𝐵𝐶 ≅ 𝐴𝐷 3. Prop. Of rectangle 4. ∆ABC≅ ∆𝐶𝐷𝐴 ≅ ∆𝐴𝐵𝐷 ≅∆DCB 4. SAS ≅ Theorem 5. 𝐵𝐷 ≅ 𝐴𝐶 5. CPCTC A D C B Given: □ABCD is a rectangle Prove: 𝐵𝐷 ≅ 𝐴𝐶
- 17. Theorems on Rectangle 2. If one angle of a parallelogram is right, then it is a rectangle. Proof: Given that ∠𝐴 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 ∠𝐴 and ∠𝐵 900 since they are supplementary. From the properties of parallelogram that the opposite angles are congruent, then ∠𝐴 and ∠𝐶 are congruent. And since ∠𝐵 is opposite to ∠𝐷. Therefore parallelogram ABCD is a rectangle.
- 18. Theorems on Rhombus 1. The diagonals of a rhombus are perpendicular and bisect each other. Given: 𝐴𝐶 and 𝐵𝐷 are diagonals intersect at P. 𝐴𝐶 ⊥ 𝐵𝐷, 𝐴𝑃 ≅ 𝑃𝐶, 𝐵𝑃 ≅ 𝑃𝐷. Prove: □ABCD is a rhombus. A B C D P Statements Reasons 1. 𝐴𝐶 ⊥ 𝐵𝐷 1. Given 2. ∠𝐴𝑃𝐵 ≅ ∠𝐶𝑃𝐷 ≅ ∠𝐶𝑃𝐵 ≅ ∠𝐴𝑃𝐷 2. Def. of ⊥ 3. 𝐴𝑃 ≅ 𝑃𝐶, 𝐵𝑃 ≅ 𝑃𝐷 3. Given 4. ∆APB ≅ ∆𝐶𝑃𝐵 ≅ ∆𝐶𝑃𝐷 ≅∆APD 4. SAS ≅ Theorem 5. 𝐴𝐵 ≅ 𝐵𝐶 ≅ 𝐶𝐷 ≅ 𝐷𝐴 5. CPCTC 6. ∴ □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 ∆ABC, P and Q are midpoints of 𝐴𝐵 and 𝐴𝐶, respectively. Prove: 𝑃𝑄 ∥ 𝐵𝐶 and PQ= 1 2 𝐵𝐶. A P B Q C R Proof: Let R be a point in 𝑃𝑄 such that 𝑃𝑄 ≅ 𝑄𝑅. Since Q is midpoint 𝐴𝐶, 𝐴𝑃 ≅ 𝑃𝐵. And by Vertical Angle Theom., ∠𝐴𝑄𝑃 ≅ ∠𝑅𝑄𝐶, so ∆AQP≅∆CQR by SAS cong. Theo. which implies that ∠𝑃𝐴𝑄 ≅ ∠𝑄𝐶𝑅 and 𝐴𝑃 ≅ 𝑅𝐶 by CPCTC. This means that □PBCR is parallelogram because 𝐴𝐵 ∥ 𝑅𝐶 (converse of AIA) and 𝑃𝐵 ≅ 𝑅𝐶 by transitivity PE. Therefore 𝑃𝑄 ∥ 𝐵𝐶.
- 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 ∆ABC, P and Q are midpoints of 𝐴𝐵 and 𝐴𝐶, respectively. Prove: 𝑃𝑄 ∥ 𝐵𝐶 and PQ= 1 2 𝐵𝐶. A P B Q C R Since □PBCR is parallelogram, 𝑃𝑅 ≅ 𝐵𝐶. 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 𝐵𝐶
- 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.