Class 7 Exam  >  Class 7 Notes  >  RD Sharma Solutions for Class 7 Mathematics  >  RD Sharma Solutions - Ex-17.1 , Constructions, Class 7, Math

Ex-17.1 , Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics PDF Download

Question 1:

Draw an ∠BAC of measure 50° such that AB = 5 cm and AC = 7 cm. Through C draw a line parallel to AB and through B draw a line parallel to AC, intersecting each other at D. Measure BD and CD.

Answer 1:

Steps of construction:

  1. Draw angle BAC =  50°° such that AB = 5 cm and AC = 7 cm.
  2. Cut an arc through C at an angle of 50°°.
  3. Draw a straight line passing through C and the arc. This line will be parallel to AB since ∠CAB=∠RCA=50°∠CAB=∠RCA=50°.
  4. Alternate angles are equal; therefore the line is parallel to AB.
  5. Again through B, cut an arc at an angle of 50°° and draw a line passing through B and this arc and say this intersects the line drawn parallel to AB at D.
  6. ∠SBA=∠BAC=50°∠SBA =∠BAC=50°, since they are alternate angles. Therefore BD∥AC.
  7. Also we can measure BD = 7 cm and CD = 5 cm.

Ex-17.1 , Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

 

Question 2:

Draw a line PQ. Draw another line parallel to PQ at a distance of 3 cm from it.

Answer 2:

1. Draw a line PQ.
2. Take any two points A and B on the line.
3. Construct ∠PBF=90° and ∠QAE=90°.
4. With A as centre and radius 3 cm cut AE at C.
5. With B as centre and radius 3 cm cut BF  at D.
6. Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.

Ex-17.1 , Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

 

Question 3:

Take any three non-collinear points A, B, C and draw ∆ABC. Through each vertex of the triangle, draw a line parallel to the opposite side.

Answer 3:

Steps of construction:

1. Mark three non collinear points A, B and C such that none of them lie on the same line.

2. Join AB, BC and CA to form triangle ABC.

 

Parallel line to AC

1. With A as centre, draw an arc cutting AC and AB at T and U, respectively.

2. With centre B and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at X.

3. With centre X and radius equal to TU, draw an arc cutting the arc drawn in the previous step at Y.

4. Join BY and produce in both directions to obtain the line parallel to AC.

 

Parallel line to AB

1. With B as centre, draw an arc cutting BC and BA at W and V, respectively.

2. With centre C and the same radius as in the previous step, draw an arc on the opposite side of BC to cut BC at P.

3. With centre P and radius equal to WV, draw an arc cutting the arc drawn in the previous step at Q.

4. Join CQ and produce in both directions to obtain the line parallel to AB.

 

Parallel line to BC

1. With B as centre, draw an arc cutting BC and BA at W and V, respectively (already drawn).

2. With centre A and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at R.

3. With centre R and radius equal to WV, draw an arc cutting the arc drawn in the previous step at S.

4. Join AS and produce in both directions to obtain the line parallel to BC.

 

Ex-17.1 , Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

 

Question 4:

Draw two parallel lines at a distance 5 cm apart.

Answer 4:

Steps of construction:

1. Draw a line PQ.
2. Take any two points A and B on the line.
3. Construct ∠PBF=90° and ∠QAE=90°.
4. With A as centre and radius 5 cm cut AE at C.
5. With B as centre and radius 5 cm cut BF  at D.
6. Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.

Ex-17.1 , Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

 

The document Ex-17.1 , Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
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FAQs on Ex-17.1 , Constructions, Class 7, Math RD Sharma Solutions - RD Sharma Solutions for Class 7 Mathematics

1. What are RD Sharma Solutions?
Ans. RD Sharma Solutions are a set of comprehensive and step-by-step solutions for the RD Sharma textbook for Mathematics. These solutions are designed to help students understand and solve the problems given in the textbook effectively.
2. What is the significance of Ex-17.1 in RD Sharma Solutions?
Ans. Ex-17.1 refers to Exercise 17.1 in the RD Sharma textbook for Mathematics. This exercise focuses on the topic of Constructions. It includes various problems related to constructing geometrical figures using a compass, ruler, and other tools.
3. What is the purpose of constructions in mathematics?
Ans. The purpose of constructions in mathematics is to learn how to construct accurate geometric figures using specific tools like a compass and ruler. It helps in understanding the properties and relationships of geometric shapes and enhances problem-solving skills.
4. How can RD Sharma Solutions for Class 7 Math help in exam preparation?
Ans. RD Sharma Solutions for Class 7 Math provide detailed explanations and solutions for all the problems given in the textbook. By referring to these solutions, students can understand the concepts better, practice different types of questions, and improve their problem-solving abilities. This, in turn, helps in preparing for exams effectively.
5. Are RD Sharma Solutions available online?
Ans. Yes, RD Sharma Solutions are available online. Many educational websites and platforms offer these solutions for free or for a nominal fee. Students can access these solutions in digital format, which makes it convenient for them to study and revise anytime, anywhere.
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