{"id":581528,"date":"2022-02-23T11:18:54","date_gmt":"2022-02-23T11:18:54","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=581528"},"modified":"2022-02-25T09:32:55","modified_gmt":"2022-02-25T09:32:55","slug":"maharashtra-board-solutions-class-9-maths-part-2-chapter-4-1-constructions-of-triangles","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-9-maths-part-2-chapter-4-1-constructions-of-triangles\/","title":{"rendered":"Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 4.1- Constructions of Triangles"},"content":{"rendered":"\n

Class 9: Maths Chapter 4.1 solutions. Complete Class 9 Maths Chapter 4.1 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 4.1- Constructions of Triangles<\/h2>\n\n\n\n

Maharashtra Board 9th Maths Chapter 4.1, Class 9 Maths Chapter 4.1 solutions<\/p>\n\n\n\n

Question 1.
Construct APQR, in which QR = 4.2 cm, m\u2220Q = 40\u00b0 and PQ + PR = 8.5 cm.
Solution:<\/strong>
\"Maharashtra
As shown in the rough figure draw seg QR = 4.2 cm
Draw a ray QT making an angle of 40\u00b0 with QR
Take a point S on ray QT, such that QS = 8.5 cm
Now, QP + PS = QS [Q-P-S]
\u2234 QP + PS = 8.5 cm \u2026\u2026.(i)
Also, PQ + PR = 8.5 cm \u2026\u2026(ii) [Given]
\u2234 QP + PS = PQ + PR [From (i) and (ii)]
\u2234 PS = PR
\u2234 Point P is on the perpendicular bisector of seg SR
\u2234 The point of intersection of ray QT and perpendicular bisector of seg SR is point P.<\/p>\n\n\n\n

Steps of construction:
i. Draw seg QR of length 4.2 cm.
ii. Djraw ray QT, such that \u2220RQT = 40\u00b0.
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, \u2206PQR is the required triangle.
\"Maharashtra<\/p>\n\n\n\n

Question 2.
Construct \u2206XYZ, in which YZ = 6 cm, XY + XZ = 9 cm, \u2220XYZ = 50\u00b0.
Solution:<\/strong>
\"Maharashtra
As shown in the rough figure draw seg YZ = 6 cm
Draw a ray YT making an angle of 50\u00b0 with YZ
Take a point W on ray YT, such that YW = 9 cm
Now, YX + XW = YW [Y-X-W]
\u2234 YX + XW = 9 cm \u2026.(i)
Also, XY + XZ = 9 cm \u2026.(ii) [Given]
\u2234 YX + XW = XY + XZ [From (i) and (ii) ]
\u2234 XW = XZ
\u2234 Point X is on the perpendicular bisector of seg WZ
\u2234 The point of intersection of ray YT and perpendicular bisector of seg WZ is j point X.<\/p>\n\n\n\n

Steps of construction:
i. Draw seg YZ of length 6 cm.
ii. Draw ray YT, such that \u2220ZYT = 50\u00b0.
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, \u2206XYZ is the required triangle.
\"Maharashtra<\/p>\n\n\n\n

Question 3.
Construct \u2206ABC, in which BC = 6.2 cm, \u2220ACB = 50\u00b0, AB + AC = 9.8 cm.
Solution:<\/strong>
\"Maharashtra
As shown in the rough figure draw seg CB = 6.2 cm
Draw a ray CT making an angle of 50\u00b0 with CB
Take a point D on ray CT, such that
CD = 9.8 cm
Now, CA + AD = CD [C-A-D]
\u2234 CA + AD = 9.8 cm \u2026\u2026.(i)
Also, AB + AC = 9.8 cm \u2026\u2026(ii) [Given]
\u2234 CA + AD = AB + AC [From (i) and (ii)]
\u2234 AD = AB
\u2234 Point A is on the perpendicular bisector of seg DB
\u2234 The point of intersection of ray CT and perpendicular bisector of seg DB is point A.<\/p>\n\n\n\n

Steps of construction:
i. Draw seg BC of length 6.2 cm.
ii. Draw ray CT, such that \u2220BCT = 50\u00b0.
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, \u2206ABC is the required triangle.
\"Maharashtra<\/p>\n\n\n\n

Question 4.
Construct \u2206ABC, in which BC = 3.2 cm, \u2220ACB = 45\u00b0 Solution:and perimeter of AABC is 10 cm.
Solution:<\/strong>
\"Maharashtra
Perimeter of \u2206ABC = AB + BC + AC
\u2234 10 = AB + 3.2 + AC
\u2234 AB + AC = 10 \u2013 3.2
\u2234 AB + AC = 6.8 cm
Now, In \u2206ABC
BC = 3.2 cm, \u2220ACB = 45\u00b0 and AB + AC = 6.8 cm \u2026.(i)
As shown in the rough figure draw j seg BC = 3.2 cm
Draw a ray CT making an angle of 45\u00b0 with CB
Take a point D on ray CT, such that
CD = 6.8 cm
Now, CA + AD = CD [C-A-D]
\u2234 CA + AD = 6.8 cm \u2026(ii)
Also, AB + AC = 6.8 cm \u2026.(iii) [From (i)]
\u2234 CA + AD = AB + AC [From (ii) and (iii)]
\u2234 AD = AB
\u2234 Point A is on the perpendicular bisector of seg DB
\u2234 The point of intersection of ray CT and perpendicular bisector of seg DB is point A.<\/p>\n\n\n\n

Steps of construction:
i. Draw seg BC of length 3.2 cm.
ii. Draw ray CT, such that \u2220BCT = 45\u00b0.
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, \u2206ABC is the required triangle.
\"Maharashtra<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 4.1- Constructions of Triangles<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions Class 9-Maths (Part 2): Chapter 4.1- Constructions of Triangles PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 9 Maths :<\/strong><\/h2>\n\n\n\n

Part 2<\/strong><\/p>\n\n\n\n