Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 4.1- Constructions of Triangles
Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 4.1- Constructions of Triangles

Class 9: Maths Chapter 4.1 solutions. Complete Class 9 Maths Chapter 4.1 Notes.

Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 4.1- Constructions of Triangles

Maharashtra Board 9th Maths Chapter 4.1, Class 9 Maths Chapter 4.1 solutions

Question 1.
Construct APQR, in which QR = 4.2 cm, m∠Q = 40° and PQ + PR = 8.5 cm.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 1
As shown in the rough figure draw seg QR = 4.2 cm
Draw a ray QT making an angle of 40° with QR
Take a point S on ray QT, such that QS = 8.5 cm
Now, QP + PS = QS [Q-P-S]
∴ QP + PS = 8.5 cm …….(i)
Also, PQ + PR = 8.5 cm ……(ii) [Given]
∴ QP + PS = PQ + PR [From (i) and (ii)]
∴ PS = PR
∴ Point P is on the perpendicular bisector of seg SR
∴ The point of intersection of ray QT and perpendicular bisector of seg SR is point P.

Steps of construction:
i. Draw seg QR of length 4.2 cm.
ii. Djraw ray QT, such that ∠RQT = 40°.
iii. Mark point S on ray QT such that l(QS) = 8.5 cm.
iv. Join points R and S.
v. Draw perpendicular bisector of seg RS intersecting ray QT.
Name the point as P.
vi. Join the points P and R.
Hence, ∆PQR is the required triangle.
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 2

Question 2.
Construct ∆XYZ, in which YZ = 6 cm, XY + XZ = 9 cm, ∠XYZ = 50°.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 3
As shown in the rough figure draw seg YZ = 6 cm
Draw a ray YT making an angle of 50° with YZ
Take a point W on ray YT, such that YW = 9 cm
Now, YX + XW = YW [Y-X-W]
∴ YX + XW = 9 cm ….(i)
Also, XY + XZ = 9 cm ….(ii) [Given]
∴ YX + XW = XY + XZ [From (i) and (ii) ]
∴ XW = XZ
∴ Point X is on the perpendicular bisector of seg WZ
∴ The point of intersection of ray YT and perpendicular bisector of seg WZ is j point X.

Steps of construction:
i. Draw seg YZ of length 6 cm.
ii. Draw ray YT, such that ∠ZYT = 50°.
iii. Mark point W on ray YT such that l(YW) = 9 cm.
iv. Join points W and Z.
v. Draw perpendicular bisector of seg WZ intersecting ray YT. Name the point as X.
vi. Join the points X and Z.
Hence, ∆XYZ is the required triangle.
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 4

Question 3.
Construct ∆ABC, in which BC = 6.2 cm, ∠ACB = 50°, AB + AC = 9.8 cm.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 5
As shown in the rough figure draw seg CB = 6.2 cm
Draw a ray CT making an angle of 50° with CB
Take a point D on ray CT, such that
CD = 9.8 cm
Now, CA + AD = CD [C-A-D]
∴ CA + AD = 9.8 cm …….(i)
Also, AB + AC = 9.8 cm ……(ii) [Given]
∴ CA + AD = AB + AC [From (i) and (ii)]
∴ AD = AB
∴ Point A is on the perpendicular bisector of seg DB
∴ The point of intersection of ray CT and perpendicular bisector of seg DB is point A.

Steps of construction:
i. Draw seg BC of length 6.2 cm.
ii. Draw ray CT, such that ∠BCT = 50°.
iii. Mark point D on ray CT such that l(CD) = 9.8 cm.
iv. Join points D and B.
v. Draw perpendicular bisector of seg DB intersecting ray CT. Name the point as A.
vi. Join the points A and B.
Hence, ∆ABC is the required triangle.
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 6

Question 4.
Construct ∆ABC, in which BC = 3.2 cm, ∠ACB = 45° Solution:and perimeter of AABC is 10 cm.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 7
Perimeter of ∆ABC = AB + BC + AC
∴ 10 = AB + 3.2 + AC
∴ AB + AC = 10 – 3.2
∴ AB + AC = 6.8 cm
Now, In ∆ABC
BC = 3.2 cm, ∠ACB = 45° and AB + AC = 6.8 cm ….(i)
As shown in the rough figure draw j seg BC = 3.2 cm
Draw a ray CT making an angle of 45° with CB
Take a point D on ray CT, such that
CD = 6.8 cm
Now, CA + AD = CD [C-A-D]
∴ CA + AD = 6.8 cm …(ii)
Also, AB + AC = 6.8 cm ….(iii) [From (i)]
∴ CA + AD = AB + AC [From (ii) and (iii)]
∴ AD = AB
∴ Point A is on the perpendicular bisector of seg DB
∴ The point of intersection of ray CT and perpendicular bisector of seg DB is point A.

Steps of construction:
i. Draw seg BC of length 3.2 cm.
ii. Draw ray CT, such that ∠BCT = 45°.
iii. Mark point D on ray CT such l(CD) = 6.8 cm. that
iv. Join points D and B.
V. Draw perpendicular bisector of seg DB intersecting ray CT. Name the point as A.
vi. Join the points A and B.
Hence, ∆ABC is the required triangle.
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 8

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Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 4.1- Constructions of Triangles

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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.