Maharashtra Board for Class 9, Maths Chapter 5.2

Class 9: Maths Chapter 5.2 solutions. Complete Class 9 Maths Chapter 5.2 Notes.

Contents

1 Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.2- Quadrilaterals
2 Download PDF
3 Chapterwise Maharashtra Board Solutions Class 9 Maths :
4 FAQs
5 About Maharashtra State Board (MSBSHSE)

Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.2- Quadrilaterals

Maharashtra Board 9th Maths Chapter 5.2, Class 9 Maths Chapter 5.2 solutions

Question 1.
In the adjoining figure, □ABCD is a parallelogram, P and Q are midpoints of sides AB and DC respectively, then prove □APCQ is a parallelogram.

Given: □ABCD is a parallelogram. P and Q are the midpoints of sides AB and DC respectively.
To prove: □APCQ is a parallelogram.
Solution:
Proof:
AP = 12 AB …..(i) [P is the midpoint of side AB]
QC = 12 DC ….(ii) [Q is the midpoint of side CD]
□ABCD is a parallelogram. [Given]
∴ AB = DC [Opposite sides of a parallelogram]
∴ 12 AB = 12 DC [Multiplying both sides by 12]
∴ AP = QC ….(iii) [From (i) and (ii)]
Also, AB || DC [Opposite angles of a parallelogram]
i.e. AP || QC ….(iv) [A – P – B, D – Q – C]
From (iii) and (iv),
□APCQ is a parallelogram. [A quadrilateral is a parallelogram if its opposite sides is parallel and congruent]

Question 2.
Using opposite angles test for parallelogram, prove that every rectangle is a parallelogram.

Given:
□ABCD is a rectangle.
To prove: Rectangle ABCD is a parallelogram.
Solution:
Proof:
□ABCD is a rectangle.
∴ ∠A ≅ ∠C = 90° [Given]
∠B ≅ ∠D = 90° [Angles of a rectangle]
∴ Rectangle ABCD is a parallelogram. [A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent]

Question 3.
In the adjoining figure, G is the point of concurrence of medians of ADEF. Take point H on ray DG such that D-G-H and DG = GH, then prove that □GEHF is a parallelogram.

Given: Point G (centroid) is the point of concurrence of the medians of ADEF.
DG = GH
To prove: □GEHF is a parallelogram.
Solution:
Proof:

Let ray DH intersect seg EF at point I such that E-I-F.
∴ seg DI is the median of ∆DEF.
∴ El = FI ……(i)
Point G is the centroid of ∆DEF.
∴ DGGI = 21 [Centroid divides each median in the ratio 2:1]
∴ DG = 2(GI)
∴ GH = 2(GI) [DG = GH]
∴ GI + HI = 2(GI) [G-I-H]
∴ HI = 2(GI) – GI
∴ HI = GI ….(ii)
From (i) and (ii),
□GEHF is a parallelogram [A quadrilateral is a parallelogram, if its diagonals bisect each other]

Question 4.
Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.

Given: □ABCD is a parallelogram.
Rays AS, BQ, CQ and DS bisect ∠A, ∠B, ∠C and ∠D respectively.
To prove: □PQRS is a rectangle.
Solution:
Proof:
∠BAS = ∠DAS = x° …(i) [ray AS bisects ∠A]
∠ABQ = ∠CBQ =y° ….(ii) [ray BQ bisects ∠B]
∠BCQ = ∠DCQ = u° …..(iii) [ray CQ bisects ∠C]
∠ADS = ∠CDS = v° ….(iv) [ray DS bisects ∠D]
□ABCD is a parallelogram. [Given]
∴ ∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary]
∴ ∠BAS + ∠DAS + ∠ABQ + ∠CBQ = 180° [Angle addition property]
∴ x°+x°+ v° + v° = 180 [From (i) and (ii)]
∴ 2x° + 2v° =180
∴ x + y = 90° ……(v) [Dividing both sides by 2]
Also, ∠A + ∠D= 180° [Adjacent angles of a parallelogram are supplementary]
∴ ∠BAS + ∠DAS + ADS + ∠CDS = 180° [Angle addition property]
∴ x° + x° + v° + v° = 180°
∴ 2x° + 2v° = 180°
∴ x° + v° = 90° …..(vi) [Dividing both sides by 2]
In ∆ARB,
∠RAB + ∠RBA + ∠ARB = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ x° + y° + ∠SRQ = 180° [A – S – R, B – Q – R]
∴ 90° + ∠SRQ = 180° [From (v)]
∴ ∠SRQ = 180°- 90° = 90° …..(vi)
Similarly, we can prove
∠SPQ = 90° …(viii)
In ∆ASD,
∠ASD + ∠SAD + ∠SDA = 180° [Sum of the measures of angles a triangle is 180°]
∴ ∠ASD + x° + v° = 180° [From (vi)]
∴ ∠ASD + 90° = 180°
∴∠ASD = 180°- 90° = 90°
∴ ∠PSR = ∠ASD [Vertically opposite angles]
∴ ∠PSR = 90° …..(ix)
Similarly we can prove
∠PQR = 90° ..(x)
∴ In □PQRS,
∠SRQ = ∠SPQ = ∠PSR = ∠PQR = 90° [From (vii), (viii), (ix), (x)]
∴ □PQRS is a rectangle. [Each angle is of measure 90°]

Question 5.
In the adjoining figure, if points P, Q, R, S are on the sides of parallelogram such that AP = BQ = CR = DS, then prove that □PQRS is a parallelogram.

Given: □ABCD is a parallelogram.
AP = BQ = CR = DS
To prove: □PQRS is a parallelogram.
Solution:
Proof:
□ABCD is a parallelogram. [Given]
∴ ∠B = ∠D ….(i) [Opposite angles of a parallelogram]
Also, AB = CD [Opposite sides of a parallelogram]
∴ AP + BP = DR + CR [A-P-B, D-R-C]
∴ AP + BP = DR + AP [AP = CR]
∴ BP = DR ….(ii)
In APBQ and ARDS,
seg BP ≅ seg DR [From (ii)]
∠PBQ ≅ ∠RDS [From (i)]
seg BQ ≅ seg DS [Given]
∴ ∆PBQ ≅ ∆RDS [SAS test]
∴ seg PQ ≅ seg RS …..(iii) [c.s.c.t]
Similarly, we can prove that
∆PAS ≅ ∆RCQ
∴ seg PS ≅ seg RQ ….(iv) [c.s.c.t]
From (iii) and (iv),
□PQRS is a parallelogram. [A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent]

Question 1.
Points D and E are the midpoints of side AB and side AC of ∆ABC respectively. Point F is on ray ED such that ED = DF. Prove that □AFBE is a parallelogram. For this example write ‘given’ and ‘to prove’ and complete the proof. (Text book pg. no. 66)

Given: D and E are the midpoints of side AB and side AC respectively.
ED = DF
To prove: □AFBE is a parallelogram.
Solution:
Proof:
seg AB and seg EF are the diagonals of □AFBE.
seg AD ≅ seg DB [Given]
seg DE ≅ seg DF [Given]
∴ Diagonals of □AFBE bisect each other.
∴ □AFBE is a parallelogram. [ By test of parallelogram]

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Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 5.2- Quadrilaterals

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Chapterwise Maharashtra Board Solutions Class 9 Maths :

Part 2

FAQs

Where do I get the Maharashtra State Board Books PDF For free download?

You can download the Maharashtra State Board Books from the eBalbharti official website, i.e. cart.ebalbharati.in or from this article.

How to Download Maharashtra State Board Books?

Students can get the Maharashtra Books for primary, secondary, and senior secondary classes from here. You can view or download the Maharashtra State Board Books from this page or from the official website for free of cost. Students can follow the detailed steps below to visit the official website and download the e-books for all subjects or a specific subject in different mediums.
Step 1: Visit the official website ebalbharati.in
Step 2: On the top of the screen, select “Download PDF textbooks”
Step 3: From the “Classes” section, select your class.
Step 4: From “Medium”, select the medium suitable to you.
Step 5: All Maharashtra board books for your class will now be displayed on the right side.
Step 6: Click on the “Download” option to download the PDF book.

Who developed the Maharashtra State board books?

As of now, the MSCERT and Balbharti are responsible for the syllabus and textbooks of Classes 1 to 8, while Classes 9 and 10 are under the Maharashtra State Board of Secondary and Higher Secondary Education (MSBSHSE).

How many state boards are there in Maharashtra?

The Maharashtra State Board of Secondary & Higher Secondary Education, conducts the HSC and SSC Examinations in the state of Maharashtra through its nine Divisional Boards located at Pune, Mumbai, Aurangabad, Nasik, Kolhapur, Amravati, Latur, Nagpur and Ratnagiri.

About Maharashtra State Board (MSBSHSE)

The Maharashtra State Board of Secondary and Higher Secondary Education or MSBSHSE (Marathi: महाराष्ट्र राज्य माध्यमिक आणि उच्च माध्यमिक शिक्षण मंडळ), is an autonomous and statutory body established in 1965. The board was amended in the year 1977 under the provisions of the Maharashtra Act No. 41 of 1965.

The Maharashtra State Board of Secondary & Higher Secondary Education (MSBSHSE), Pune is an independent body of the Maharashtra Government. There are more than 1.4 million students that appear in the examination every year. The Maha State Board conducts the board examination twice a year. This board conducts the examination for SSC and HSC.

The Maharashtra government established the Maharashtra State Bureau of Textbook Production and Curriculum Research, also commonly referred to as Ebalbharati, in 1967 to take up the responsibility of providing quality textbooks to students from all classes studying under the Maharashtra State Board. MSBHSE prepares and updates the curriculum to provide holistic development for students. It is designed to tackle the difficulty in understanding the concepts with simple language with simple illustrations. Every year around 10 lakh students are enrolled in schools that are affiliated with the Maharashtra State Board.