Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions -Ex-17.3, Constructions, Class 9, Maths

Ex-17.3, Constructions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q1. Construct a △ABC in which BC=3.6cm, AB+AC=4.8cm and ∠B=600.

 

Ex-17.3, Constructions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Steps of Construction:

  1. Construct a line segment BC of 3.6cm.
  2. At the point B, draw ∠XBC=600.
  3. Keeping B as center and radius 4.8cm draw an arc which intersects XB at D.
  4. Join DC.
  5. Draw the perpendicular bisector of DC which intersects DB at A.
  6. Join AC.

Hence △ABC is the required triangle.

 

Q2. Construct a △ABC in which AB+AC=5.6cm ,BC=4.5cm and ∠B=450.

Ex-17.3, Constructions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Steps of Construction:

  1. Construct a line segment BC of 4.5cm.
  2. At the point B, draw ∠XBC=450.
  3. Keeping B as centre and radius 5.6cm draw an arc which intersects XB at D.
  4. Join DC.
  5. Draw the perpendicular bisector of DC which intersects DB at A.
  6. Join AC.

Hence △ABC is the required triangle

  

Q3. Construct a △ABC in which BC=3.4cm, AB-AC=1.5cm and ∠B=450.

Ex-17.3, Constructions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Steps of Construction:

  1. Construct a line segment BC of 3.4cm.
  2. At the point B, draw ∠XBC=450.
  3. Keeping B as centre and radius 1.5cm draw an arc which intersects XB at D.
  4. Join DC.
  5. Draw the perpendicular bisector of DC which intersects DB at A.
  6. Join AC.

Hence △ABC is the required triangle.

 

Q4. Using rulers and compasses only, construct a △ABC, given base BC=7cm, ∠ABC=600 AB+AC=12cm.

Ex-17.3, Constructions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Steps of Construction:

  1. Construct a line segment BC of 7cm
  2. At the point B, draw ∠XBC=600.
  3. Keeping B as center and radius 12cm draw an arc which intersects XB at D
  4. Join DC
  5. Draw the perpendicular bisector of DC which intersects DB at A
  6. Join AC

Hence △ABC is the required triangle.

 

Q5. Construct a triangle whose perimeter is 6.4cm, and angles at the base are 600 450.

Ex-17.3, Constructions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Steps of Construction:

  1. Draw a line segment XY of 6.4cm.
  2. Draw ∠DXY=600 and ∠EYX=450.
  3. Draw the angle bisectors of ∠DXY and ∠EYX which intersect each other at A.
  4. Draw the perpendicular bisector of AX and AY which intersect XY at B and C respectively.
  5. Join AB and AC.

Hence △ABC is the required triangle.

 

Q6. Using rulers and compasses only, construct a △ABC from the following data:

AB+BC+CA=12cm, ∠B=45and ∠C=600

Ex-17.3, Constructions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Steps of Construction:

  1. Draw a line segment XY of 12cm.
  2. Draw ∠DXY=450 and ∠EYX=600.
  3. Draw the angle bisectors of ∠DXY and ∠EYX which intersect each other at A.
  4. Draw the perpendicular bisector of AX and AY which intersect XY at B and C respectively.
  5. Join AB and AC.

Hence △ABC is the required triangle.

 

Q7. Construct a right-angled triangle whose perimeter is equal to 10cm and one acute angle equal to 600.

Ex-17.3, Constructions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Steps of Construction:

  1. Draw a line segment XY of 10cm.
  2. Draw ∠DXY=900 and ∠EYX=600.
  3. Draw the angle bisectors of ∠DXY and ∠EYX which intersect each other at A.
  4. Draw the perpendicular bisector of AX and AY which intersect XY at B and C respectively.
  5. Join AB and AC.

Hence △ABC is the required triangle.

 

Q8. Construct a triangle ABC such that BC= 6cm, AB= 6cm and median AD=4cm.

Ex-17.3, Constructions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Steps of construction:

  1. Draw a line segment BC of 6cm.
  2. Take mid-point O of side BC.
  3. With center B and D and radii 6cm and 4cm, draw two arcs which intersect each other at A.
  4. Join AB, AD and AC.

Hence △ABC is the required triangle.

 

Q9. Construct a right-angled triangle ABC whose base BC is 6cm and the sum of the hypotenuse AC and other side AB is 10cm.

Ex-17.3, Constructions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Steps of Construction:

  1. Construct a line segment BC of 6cm.
  2. At the point B, draw ∠XBC=900.
  3. Keeping B as center and radius 10cm draw an arc which intersects XB at D.
  4. Join DC.
  5. Draw the perpendicular bisector of DC which intersects DB at A.
  6. Join AC.

Hence △ABC is the required triangle.

 

Q10. Construct a triangle XYZ in which ∠Y=300, ∠Z=900 and XY+YZ+ZX=11cm.

Ex-17.3, Constructions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Steps of construction:

  1. Draw a line segment AB of 11cm.
  2. Draw ∠DAB=300 and ∠FBA=900.
  3. Draw the angle bisectors of ∠DAB and ∠EBA which intersect each other at X.
  4. Draw the perpendicular bisector of XA and XB which intersect AB at Y and Z respectively
  5. Join XY and XZ.

Hence △XYZ is the required triangle.

The document Ex-17.3, Constructions, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on Ex-17.3, Constructions, Class 9, Maths RD Sharma Solutions - RD Sharma Solutions for Class 9 Mathematics

1. What are RD Sharma Solutions?
Ans. RD Sharma Solutions are a series of textbooks and study materials designed to help students understand and solve mathematical problems. These solutions are specifically created for the CBSE board and follow the latest syllabus. They provide step-by-step explanations and solutions to the exercises given in the RD Sharma textbook, making it easier for students to grasp the concepts and improve their problem-solving skills.
2. How can RD Sharma Solutions help in my Class 9 Maths exam preparation?
Ans. RD Sharma Solutions can greatly assist in your Class 9 Maths exam preparation. These solutions are comprehensive and cover all the topics and chapters included in the syllabus. By referring to RD Sharma Solutions, you can gain a deeper understanding of the concepts, practice a wide range of problems, and strengthen your problem-solving abilities. This will not only help you in scoring well in the exams but also build a strong foundation for higher-level mathematics.
3. Are RD Sharma Solutions available for free online?
Ans. No, RD Sharma Solutions are not available for free online. They are copyrighted material and can only be accessed through authorized sources. However, there are various websites and online platforms that offer RD Sharma Solutions at a nominal cost or as part of their study packages. It is advisable to purchase these solutions from trusted sources to ensure their accuracy and reliability.
4. Can I solely rely on RD Sharma Solutions for my Class 9 Maths exam preparation?
Ans. While RD Sharma Solutions are an excellent resource for Class 9 Maths exam preparation, it is recommended to use them in conjunction with other study materials. RD Sharma Solutions provide detailed explanations and solutions, but it is essential to practice solving a variety of problems from different sources to enhance your problem-solving skills. Additionally, referring to your textbook, attending classes, and seeking clarification from your teacher will further strengthen your understanding of the subject.
5. Are RD Sharma Solutions only suitable for CBSE board students?
Ans. RD Sharma Solutions are primarily designed for CBSE board students as they closely follow the CBSE syllabus. However, these solutions can also be beneficial for students studying in other boards or preparing for competitive exams. The concepts and problem-solving techniques discussed in RD Sharma Solutions are universally applicable and can help students develop a strong foundation in mathematics, irrespective of the board they belong to.
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