Chapter 17 Constructions

RD Sharma Solutions for Class 7 Chapter 17 help students understand the topic of Constructions clearly and. This chapter focuses on learning how to draw accurate geometrical figures using tools like a ruler, compass, and protractor. With RD Sharma Solutions for Class 7 Chapter 17 Constructions, students can practice step-by-step methods for drawing angles, bisecting lines, and constructing different shapes as per the given measurements.

The Class 7 Chapter 17 RD Sharma Solutions are designed to match the CBSE syllabus, making them a perfect study companion for exams. Each solution explains the method in an easy-to-follow manner so that students not only get the correct answer but also understand the logic behind every step. This approach helps in building a strong foundation in geometry. By using Class 7 Chapter 17 Constructions RD Sharma Solutions, students can develop accuracy, improve their problem-solving skills, and gain confidence in handling construction-based questions.

The solutions also include diagrams for better understanding, making learning more visual and interactive. One of the main advantages of studying from RD Sharma Solutions for Class 7 Chapter 17 Constructions is that they provide a variety of solved examples and practice questions. These cover all types of problems that might appear in the exam.

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Whether it’s constructing a triangle with given sides, drawing perpendicular bisectors, or creating angle bisectors, these solutions guide students through each step. In short, RD Sharma Solutions for Class 7 Chapter 17 make the topic of Constructions easy, interesting, and exam-ready. With consistent practice from these solutions, students can master the art of accurate constructions and score well in their mathematics exams.

Download RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions PDF Here

You can easily download RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions PDF here and study anytime, anywhere. This PDF includes all the step-by-step solutions for the chapter, making it easy to understand how to solve each construction problem. Whether you are learning to draw angles, bisectors, or triangles, the PDF gives you clear explanations along with neat diagrams. Having the Class 7 Chapter 17 Constructions RD Sharma Solutions in PDF form means you can revise even without the internet and keep it handy for quick reference before exams.

Access Answers to RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions

Question 1

Task: Draw △ABC with ∠A = 120°, AB = AC = 3 cm. Find ∠B and ∠C.

Construction Steps

1. Draw a 3 cm line segment and name its ends A and C (so AC = 3 cm).

2. At A, use a protractor to make a 120° angle. Draw a ray from A along this angle.

3. On this ray, mark point B so that AB = 3 cm (use a ruler or set the compass to 3 cm).

4. Join B to C to complete the triangle.

Result

AB = AC, so △ABC is isosceles with vertex at A. The remaining angles are equal: ∠B = ∠C = (180° − 120°) ÷ 2 = 30°.

Question 2

Task: Construct △ABC where ∠C = 90° and AC = BC = 4 cm.

Construction Steps

1. Draw a 4 cm segment and name it BC (so BC = 4 cm).

2. At C, draw a right angle (90°) and draw a ray along this angle.

3. On this ray, mark point A so that AC = 4 cm.

4. Join A to B. Triangle ABC is formed with a right angle at C and AC = BC.

Question 3

Task: Draw △ABC with BC = 4 cm, AB = 3 cm, and ∠B = 45°. Also draw a perpendicular from A to BC.

Construction Steps

1. Draw AB = 3 cm.

2. At B, make a 45° angle and draw a ray from B.

3. On this ray, locate point C so that BC = 4 cm (measure from B).

4. Join A to C to get △ABC.

5. Perpendicular from A to BC:

   1. With A as center, draw an arc that cuts BC at two points (say M and N).

   2. With M and N as centers and the same radius (more than half MN), draw two arcs above (or below) BC to meet at E.

   3. Join A to E. AE is perpendicular to BC and meets BC at D.

Question 4

Task: Draw △ABC with AB = 3 cm, BC = 4 cm, ∠B = 60°. Draw the angle bisectors of ∠A and ∠C; they meet at O. Find ∠COA.

Construction Steps

1. Draw BC = 4 cm.

2. At B, construct a 60° angle and draw a ray.

3. On that ray, mark A so that AB = 3 cm. Join A to C to complete △ABC.

4. Bisector of ∠A:

   1. With A as center, draw an arc to cut AB and AC at P and Q.

   2. With P and Q as centers (same radius), draw two arcs that meet at T.

   3. Join A to T. AT is the bisector of ∠A.

5. Bisector of ∠C:

   1. With C as center, draw an arc to cut CA and CB at M and N.

   2. With M and N as centers (same radius), draw arcs that meet at R.

   3. Join C to R. CR is the bisector of ∠C.

6. Let AT and CR meet at O, the incenter.

Result

On measuring ∠COA, you will get 120° (since it equals 90° + (∠B/2) in this configuration with ∠B = 60°).

Question 5

Task: Construct △ABC with BC = 4 cm, ∠B = 50°, and ∠C = 70°.

Construction Steps

1. Draw BC = 4 cm.

2. At B, make a 50° angle and draw ray BX.

3. At C, make a 70° angle on the same side of BC and draw ray CY.

4. Rays BX and CY meet at A. Join A to B and A to C. Triangle ABC is ready.

Question 6

Task: Construct △ABC with BC = 8 cm, ∠B = 50°, and ∠A = 50°.

Construction Steps

1. Draw BC = 8 cm.

2. At B, construct a 50° angle and draw ray BX.

3. Since ∠A = 50°, ∠C = 180° − 50° − 50° = 80°. At C, make an 80° angle and draw ray CY.

4. Rays BX and CY meet at A. Join A to B and A to C to complete the triangle.

Question 7

Task: Draw △PQR with ∠Q = 80°, ∠R = 55°, and QR = 4.5 cm. Also draw the perpendicular bisector of QR.

Construction Steps

1. Draw QR = 4.5 cm.

2. At Q, construct an 80° angle and draw a ray. At R, construct a 55° angle and draw a ray. Let the two rays meet at P.

3. Join P to Q and P to R to get △PQR.

4. Perpendicular bisector of QR:

   1. With Q as center and radius more than half of QR, draw arcs above and below QR.

   2. Repeat from R with the same radius to get two intersection points M and N.

   3. Join M and N. MN is the perpendicular bisector of QR.

Question 8

Task: Construct △ABC where AB = 6.4 cm, ∠A = 45°, and ∠B = 60°.

Construction Steps

1. Draw AB = 6.4 cm.

2. At A, make a 45° angle and draw ray AX.

3. At B, make a 60° angle on the same side of AB and draw ray BY.

4. Rays AX and BY meet at C. Join C to A and C to B. Triangle ABC is formed.

Question 9

Task: Draw a right triangle △PQR with hypotenuse 5 cm and one other side 4 cm.

Construction Steps

1. Draw QR = 4 cm.

2. At R, construct a right angle (90°) and draw ray RX.

3. With Q as center and radius 5 cm, draw an arc to cut RX at P.

4. Join P to Q. △PQR is a right triangle with hypotenuse PQ = 5 cm.

Question 10

Task: Construct a right triangle where the hypotenuse is 4 cm and one side is 2.5 cm.

Construction Steps

1. Draw QR = 2.5 cm.

2. At R, draw a 90° angle and ray RX.

3. With Q as center and radius 4 cm, draw an arc to cut RX at P.

4. Join P to Q to get the required right triangle △PQR.

Question 11

Task: Construct right triangle △ABC with hypotenuse 5.4 cm and one angle 30°.

Construction Steps

1. Draw BC = 5.4 cm (this will be the hypotenuse).

2. At C, construct an angle of 30° and draw ray CX.

3. At B, construct an angle of 60° (the other acute angle of a right triangle with 30°) and draw ray BY.

4. Rays CX and BY meet at A. Join A to B and A to C to complete △ABC.

Question 12 (Variant A)

Task: Construct △ABC where AC = 6 cm, ∠A = 90°, and ∠B = 60°.

Construction Steps

1. Draw AC = 6 cm.

2. At A, construct a 90° angle and draw ray AY.

3. Since ∠A = 90° and ∠B = 60°, ∠C = 180° − 90° − 60° = 30°. At C, construct a 30° angle and draw ray CX.

4. Rays AY and CX meet at B. Join B to A and B to C to form △ABC.

Question 12 (Variant B)

Task: Draw right triangle △ABC where AB = 5.8 cm, BC = 4.5 cm, and ∠C = 90°.

Construction Steps

1. Draw BC = 4.5 cm.

2. At C, construct a right angle (90°) and draw ray CX.

3. With B as center and radius 5.8 cm, draw an arc to cut CX at A.

4. Join A to B. △ABC is a right triangle at C with AB = 5.8 cm.

Question 13

Task: Draw right triangle △ABC with ∠C = 90°, AB = 5.2 cm, and BC = 4.6 cm.

Construction Steps

1. Draw BC = 4.6 cm.

2. At C, draw a right angle and extend a ray.

3. With B as center and radius 5.2 cm (the hypotenuse AB), draw an arc to cut the ray at A.

4. Join A to B. △ABC is complete with a right angle at C.

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