Application of the Properties of Parallelogram
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The document provides an overview of the properties of parallelograms, including the relationships between angles and sides, such as congruence and supplementary angles. It includes examples and solutions to determine unknown side lengths and angle measures, emphasizing the congruence of opposite sides and angles and the bisecting nature of diagonals. Additionally, it demonstrates how to solve for variables using algebraic equations derived from the properties of parallelograms.
Application of the Properties of Parallelogram
- 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.
- 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.