Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.5- Quadrilaterals

Class 9: Maths Chapter 5.5 solutions. Complete Class 9 Maths Chapter 5.5 Notes.

Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.5- Quadrilaterals

Maharashtra Board 9th Maths Chapter 5.5, Class 9 Maths Chapter 5.5 solutions

Question 1.
In the adjoining figure, points X, Y, Z are the midpoints of of ∆ABC respectively, cm. Find the lengths of side AB, side BC and side AC AB = 5 cm, AC = 9 cm and BC = 11c.m. Find the lengths of XY, YZ, XZ.

Solution:
i. AC = 9 cm [Given]
Points X and Y are the midpoints of sides AB and BC respectively. [Given]
∴ XY = 12 AC [Midpoint tfyeprem]
= 12 x 9 = 4.5 cm

ii. AB = 5 cm [Given]
Points Y and Z are the midpoints of sides BC and AC respectively. [Given]
∴ YZ = 12 AB [Midpoint theorem]
= 12 x 5 = 2.5 cm

iii. BC = 11 cm [Given]
Points X and Z are the midpoints of sides AB and AC respectively. [Given]
∴ XZ = 12 BC [Midpoint theorem]
= 12 x 11 = 5.5 cm
l(XY) = 4.5 cm, l(YZ) = 2.5 cm, l(XZ) = 5.5 cm

Question 2.
In the adjoining figure, □PQRS and □MNRL are rectangles. If point M is the midpoint of side PR, then prove that,
i. SL = LR
ii. LN = 12 SQ.

Given: □PQRS and □MNRL are rectangles. M is the midpoint of side PR.
Solution:
Toprove:
i. SL = LR
ii. LN = 12 (SQ)
Proof:
i. □PQRS and □MNRL are rectangles. [Given]
∴ ∠S = ∠L = 90° [Angles of rectangles]
∠S and ∠L form a pair of corresponding angles on sides SP and LM when SR is their transversal.

∴eg ML || seg PS …(i) [Corresponding angles test]
In ∆PRS,
Point M is the midpoint of PR and seg ML || seg PS. [Given] [From (i)]
∴ Point L is the midpoint of seg SR. ……(ii) [Converse of midpoint theorem]
∴ SL = LR

ii. Similarly for ∆PRQ, we can prove that,

Point N is the midpoint of seg QR. ….(iii)
In ∆RSQ,
Points L and N are the midpoints of seg SR and seg QR respectively. [From (ii) and (iii)]
∴ LN = 12SQ [Midpoint theorem]

Question 3.
In the adjoining figure, ∆ABC is an equilateral triangle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that ∆FED is an equilateral triangle.

Given: ∆ABC is an equilateral triangle.
Points F, D and E are midpoints of side AB, side BC, side AC respectively.
To prove: ∆FED is an equilateral triangle.
Solution:
Proof:
∆ABC is an equilateral triangle. [Given]
∴ AB = BC = AC ….(i) [Sides of an equilateral triangle]
Points F, D and E are midpoints of side AB and BC respectively.

∴ FD = 12AC …..(ii) [Midpoint theorem]
Points D and E are the midpoints of sides BC and AC respectively.

∴ DE = 12AB …..(iii) [Midpoint theorem]
Points F and E are the midpoints of sides AB and AC respectively.
∴ FE = 12BC
∴ FD = DE = FE [From (i), (ii), (iii) and (iv) ]
∴ ∆FED is an equilateral triangle.

Question 4.
In the adjoining figure, seg PD is a median of ∆PQR. Point T is the midpoint of seg PD. Produced QT intersects PR at M. Show that PMPR = 13. [Hint: Draw DN || QM]

Solution:
Given: seg PD is a median of ∆PQR. Point T is the midpoint of seg PD.
To Prove: PMPR = 13
Construction: Draw seg DN ||seg QM such that P-M-N and M-N-R.
Proof:
In ∆PDN,
Point T is the midpoint of seg PD and seg TM || seg DN [Given]
∴ Point M is the midpoint of seg PN. [Construction and Q-T-M]
∴ PM = MN [Converse of midpoint theorem]
In ∆QMR,
Point D is the midpoint of seg QR and seg DN || seg QM [Construction]

∴ Point N is the midpoint of seg MR. [Converse of midpoint theorem]
∴ RN = MN …..(ii)
∴ PM = MN = RN …..(iii) [From (i) and (ii)]
Now, PR = PM + MN + RN [ P-M-R-Q-T-M]
∴ PR = PM + PM + PM [From (iii) ]
∴ PR = 3PM
PMPR = 13

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Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.5- 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