Class 7 Exam  >  Class 7 Notes  >  RD Sharma Solutions for Class 7 Mathematics  >  RD Sharma Solutions - 17.2, Constructions, Class 7, Math

17.2, Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics PDF Download

QUESTION 1:

Draw ∆ ABC in which AB = 5.5 cm, BC = 6 cm and CA = 7 cm. Also, draw perpendicular bisector of side BC.

ANSWER 1:

Steps of construction:

  1. Draw a line segment AB of length 5.5 cm.
  2. From B, cut an arc of radius 6 cm.
  3. With centre A, draw an arc of radius 7 cm intersecting the previously drawn arc at say, C.
  4. Join AC and BC to obtain the desired triangle.
  5. With centre B and radius more than 1/2BC, draw two arcs on both sides of BC.
  6. With centre C and the same radius as in the previous step, draw two arcs intersecting the arcs drawn in the previous step at X and Y.
  7. Join XY to get the perpendicular bisector of BC.

17.2, Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics


QUESTION 2:

Draw ∆ PQR in which PQ = 3 cm, QR = 4 cm and RP = 5 cm. Also, draw the bisector of ∠Q.

ANSWER 2:

Steps of construction:

  1. Draw a line segment PQ of length 3 cm.
  2. With Q as centre and radius 4 cm, draw an arc.
  3. With P as centre and radius 5 cm, draw an arc intersecting the previously drawn arc at R.
  4. Join PR and QR to obtain the required triangle.
  5. From Q, cut arcs of equal radius intersecting PQ and QR at M and N, respectively.
  6. From M and N, cut arcs of equal radius intersecting at point S.
  7. Join QS and extend to produce the angle bisector of angle PQR.
  8. Verify that angle PQS and angle SQR are equal to 45° each.

17.2, Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

 

QUESTION 3:

Draw an equilateral triangle one of whose sides is of length 7 cm.

ANSWER 3:

Steps of construction:

  1. Draw a line segment AB of length 7 cm.
  2. With centre A, draw an arc of radius 7 cm.
  3. With centre B, draw an arc of radius 7 cm intersecting the previously drawn arc at C.
  4. Join AC and BC to get the required triangle.

17.2, Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

 

QUESTION 4:

Draw a triangle whose sides are of lengths 4 cm, 5 cm and 7 cm. Draw the perpendicular bisector of the largest side.

ANSWER 4:

Steps of construction:

  1. Draw a line segment PR of length 7 cm.
  2. With centre P, draw an arc of radius 5 cm.
  3. With centre R, draw an arc of radius 4 cm intersecting the previously drawn arc at Q.
  4. Join PQ and QR to obtain the required triangle.
  5. From P, draw arcs with radius more than half of PR on either sides.
  6. With the same radius as in the previous step, draw arcs from R on either sides of PR intersecting the arcs drawn in the previous step at M and N.
  7. MN is the required perpendicular bisector of the largest side.

17.2, Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics


QUESTION 5:

Draw a triangle ABC with AB = 6 cm, BC = 7 cm and CA = 8 cm. Using ruler and compass alone, draw (i) the bisector AD of ∠A and (ii) perpendicular AL from A on BC. Measure LAD.

ANSWER 5:

Steps of construction:

  1. Draw a line segment BC of length 7 cm.
  2. With centre B, draw an arc of radius 6 cm.
  3. With centre C, draw an arc of radius 8 cm intersecting the previously drawn arc at A.
  4. Join AC and BC to get the required triangle.

 

Angle bisector steps:

1. From A, cut arcs of equal radius intersecting AB and AC at E and F, respectively.

2. From E and F, cut arcs of equal radius intersecting at point H.

3. Join AH and extend to produce the angle bisector of angle A, meeting line BC at D.

 

Perpendicular from Point A to line BC steps:

1. From A, cut arcs of equal radius intersecting BC at P and Q, respectively (Extend BC to draw these arcs).

2. From P and Q, cut arcs of equal radius intersecting at M.

3. Join AM cutting BC at L. 

4. AL is the perpendicular to the line BC.

5. Angle LAD is 15o.

17.2, Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

 


QUESTION 6:

Draw ∆ DEF such that DEDF = 4 cm and EF = 6 cm. Measure ∠E and ∠F.

ANSWER 6:

Steps of construction:

  1. Draw a line segment EF of length 6 cm.
  2. With E as centre, draw an arc of radius 4 cm.
  3. With F as centre, draw an arc of radius 4 cm intersecting the previous arc at D.
  4. Join DE and DF to get the desired triangle.DF, .

By measuring we get,∠E=∠F=40°∠E=∠F=40°

17.2, Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics


QUESTION 7:

Draw any triangle ABC. Bisect side AB at D. Through D, draw a line parallel to BC, meeting AC in E. Measure AE and EC.

ANSWER 7:

We first draw a triangle ABC with each side = 6 cm.
 

Steps to bisect line AB:

1. Draw an arc from A on either side of line AB.

2. With the same radius as in the previous step, draw an arc from B on either side of AB intersecting the arcs drawn in the previous step at P and Q.

3. Join PQ cutting AB at D. PQ is the perpendicular bisector of AB.

 

Parallel line to BC:

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

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

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

4. Join XD and extend it to intersect AC at E.
5. DE is the required parallel line.

17.2, Constructions, Class 7, Math RD Sharma Solutions | RD Sharma Solutions for Class 7 Mathematics

The document 17.2, 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 17.2, 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 solutions for the RD Sharma textbook, which is widely used by students studying mathematics in India. These solutions provide step-by-step explanations and solutions to the various problems and exercises presented in the textbook.
2. What is the significance of Class 7 RD Sharma Solutions?
Ans. Class 7 RD Sharma Solutions are significant as they help students understand and master the concepts of mathematics taught in the textbook. These solutions provide clarity and guidance on solving problems, thereby enhancing the students' problem-solving skills and preparing them for higher-level mathematics.
3. How can RD Sharma Solutions benefit Class 7 students?
Ans. RD Sharma Solutions can benefit Class 7 students in several ways. They provide a structured approach to solving mathematical problems, helping students develop logical thinking and analytical skills. These solutions also offer additional practice exercises and examples, allowing students to reinforce their understanding of concepts and improve their performance in exams.
4. Are RD Sharma Solutions available online?
Ans. Yes, RD Sharma Solutions are available online. Many websites and educational platforms provide free or paid access to these solutions. Students can access them anytime and anywhere, making it convenient for self-study and revision purposes.
5. Can RD Sharma Solutions help in exam preparation?
Ans. Yes, RD Sharma Solutions can be extremely helpful in exam preparation. By practicing the solved examples and exercises provided in these solutions, students can gain confidence and improve their problem-solving speed. Additionally, the step-by-step explanations help students understand the concepts thoroughly, enabling them to tackle similar problems in exams effectively.
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