Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.4- Quadrilaterals

Class 9: Maths Chapter 5.4 solutions. Complete Class 9 Maths Chapter 5.4 Notes.

Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.4- Quadrilaterals

Maharashtra Board 9th Maths Chapter 5.4, Class 9 Maths Chapter 5.4 solutions

Question 1.
In □IJKL, side IJ || side KL, ∠I = 108° and ∠K = 53°, then find the measures of ∠J and ∠L.
Solution:

i. ∠I = 108° [Given]
side IJ || side KL and side IL is their transveral. [Given]
∴ ∠I + ∠L = 180° [Interior angles]
∴ 108° + ∠L = 180°
∴ ∠L = 180° – 108° = 72°

ii. ∠K = 53° [Given]
side IJ || side KL and side JK is their transveral. [Given]
∴ ∠J + ∠K = 180° [Interior angles]
∴ ∠J + 53° = 180°
∴ ∠J= 180°- 53° = 127°
∴ ∠L = 72°, ∠J = 127°

Question 2.
In □ABCD, side BC || side AD, side AB ≅ side DC. If ∠A = 72°, then find the measures of ∠B and ∠D.
Construction: Draw seg BP ⊥ side AD, A – P – D, seg CQ ⊥ side AD, A – Q – D.
Solution:

i. ∠A = 72° [Given]
In □ABCD, side BC || side AD and side AB is their transversal. [Given]
∴ ∠A + ∠B = 180° [Interior angles]
∴ 72° +∠B = 180°
∴ ∠B = 180° – 72° = 108°

ii. In ∆BPA and ∆CQD,
∠BPA ≅ ∠CQD [Each angle is of measure 90°]
Hypotenuse AB ≅ Hypotenuse DC [Given]
seg BP ≅ seg CQ [Perpendicular distance between two parallel lines]
∴ ∆BPA ≅ ∆CQD [Hypotenuse side test]
∴ ∠BAP ≅ ∠CDQ [c. a. c. t.]
∴ ∠A = ∠D
∴ ∠D = 72°
∴ ∠B = 108°, ∠D = 72°

Question 3.
In □ABCD, side BC < side AD, side BC || side AD and if side BA ≅ side CD, then prove that ∠ABC = ∠DCB.

Given: side BC < side AD, side BC || side AD, side BA = side CD
To prove: ∠ABC ≅ ∠DCB
Construction: Draw seg BP ⊥ side AD, A – P – D
seg CQ ⊥ side AD, A – Q – D
Solution:
Proof:
In ∆BPA and ∆CQD,
∠BPA ≅ ∠CQB [Each angle is of measure 90°]
Hypotenuse BA ≅ Hypotenuse CD [Given]
seg BP ≅ seg CQ [Perpendicular distance between two parallel lines]
∴ ∆BPA ≅ ∆CQD [Hypotenuse side test]
∴ ∠BAP ≅ ∠CDQ [c. a. c. t.]
∴ ∠A = ∠D ….(i)
Now, side BC || side AD and side AB is their transversal. [Given]
∴ ∠A + ∠B = 180°…..(ii) [Interior angles]
Also, side BC || side AD and side CD is their transversal. [Given]
∴ ∠C + ∠D = 180° …..(iii) [Interior angles]
∴ ∠A + ∠B = ∠C + ∠D [From (ii) and (iii)]
∴ ∠A + ∠B = ∠C + ∠A [From (i)]
∴ ∠B = ∠C
∴ ∠ABC ≅ ∠DCB

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Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.4- Quadrilaterals

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Chapterwise Maharashtra Board Solutions Class 9 Maths :

Part 2

• Chapter 1.1- Basic Concepts in Geometry
• Chapter 1.2- Basic Concepts in Geometry
• Chapter 1.3- Basic Concepts in Geometry
• Chapter 2.1- Parallel Lines
• Chapter 2.2- Parallel Lines
• Chapter 3.1- Triangles
• Chapter 3.2- Triangles
• Chapter 3.3- Triangles
• Chapter 3.4- Triangles
• Chapter 3.5- Triangles
• Chapter 4.1- Constructions of Triangles
• Chapter 4.2- Constructions of Triangles
• Chapter 4.3- Constructions of Triangles
• Chapter 5.1- Quadrilaterals
• Chapter 5.2- Quadrilaterals
• Chapter 5.3- Quadrilaterals
• Chapter 5.4- Quadrilaterals
• Chapter 5.5- Quadrilaterals
• Chapter 6.1- Circle
• Chapter 6.2- Circle
• Chapter 6.3- Circle
• Chapter 7.1- Co-ordinate Geometry
• Chapter 7.2- Co-ordinate Geometry
• Chapter 8.1- Trigonometry
• Chapter 8.2- Trigonometry
• Chapter 9.1- Surface Area and Volume
• Chapter 9.2- Surface Area and Volume
• Chapter 9.3- Surface Area and Volume