RD Sharma Solutions For Class 7 Maths Exercise 17.3 Chapter 17 Constructions

RD Sharma Class 7 Solutions Chapter 17 Ex 17.3 PDF Free Download

Get the detailed RD Sharma Solutions for Class 7 Maths Exercise 17.3 Chapter 17 Constructions which is available here. From the exam point of view, subject experts have prepared the solutions in a step by step format in accordance with CBSE syllabus. Students gain knowledge and increase their confidence by following the steps, when practised regularly. This exercise explains SAS triangle constructions and steps involved in constructing the triangles. To know more about constructions students can download RD Sharma Solutions for Class 7 Maths.

Download the PDF of RD Sharma Solutions For Class 7 Chapter 17 – Constructions Exercise 17.3

 

rd sharma class 7 maths solution ch 17 ex 3
rd sharma class 7 maths solution ch 17 ex 3
rd sharma class 7 maths solution ch 17 ex 3
rd sharma class 7 maths solution ch 17 ex 3
rd sharma class 7 maths solution ch 17 ex 3

 

Access other exercises of RD Sharma Solutions For Class 7 Chapter 17 – Constructions

Exercise 17.1 Solutions

Exercise 17.2 Solutions

Exercise 17.4 Solutions

Exercise 17.5 Solutions

Access answers to Maths RD Sharma Solutions For Class 7 Chapter 17 – Constructions Exercise 17.3

1. Draw △ABC in which AB = 3 cm, BC = 5 cm and ∠Q = 70o.

Solution:

RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions Image 12

Steps of construction:

1. Draw a line segment AB of length 3 cm.

2. Draw ∠XBA=70o.

3. Cut an arc on BX at a distance of 5 cm at C.

4. Join AC to get the required triangle.

2. Draw △ABC in which ∠A=70o. AB = 4 cm and AC= 6 cm. Measure BC.

Solution:

RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions Image 13

Steps of construction:

1. Draw a line segment AC of length 6 cm.

2. Draw ∠XAC=70o.

3. Cut an arc on AX at a distance of 4 cm at B.

4. Join BC to get the desired triangle.

5. We see that BC = 6 cm.

3. Draw an isosceles triangle in which each of the equal sides is of length 3 cm and the angle between them is 45o.

Steps of construction:

RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions Image 14

1. Draw a line segment PQ of length 3 cm.

2. Draw ∠QPX=45o.

3. Cut an arc on PX at a distance of 3 cm at R.

4. Join QR to get the required triangle.

4. Draw △ABC in which ∠A = 120o, AB = AC = 3 cm. Measure ∠B and ∠C.

Solution:

RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions Image 15

Steps of construction:

1. Draw a line segment AC of length 3 cm.

2. Draw ∠XAC = 120o.

3. Cut an arc on AX at a distance of 3 cm at B.

4. Join BC to get the required triangle.

5. By measuring, we get ∠B = ∠C = 30o.

5. Draw △ABC in which ∠C = 90o and AC = BC = 4 cm.

Solution:

RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions Image 16

Steps of construction:

1. Draw a line segment BC of length 4 cm.

2. At C, draw ∠BCY=90°.

3. Cut an arc on CY at a distance of 4 cm at A.

4. Join AB. ABC is the required triangle.

6. Draw a triangle ABC in which BC = 4 cm, AB = 3 cm and ∠B = 45o. Also, draw a perpendicular from A on BC.

Solution:

RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions Image 17

Steps of construction:

1. Draw a line segment AB of length 3 cm.

2. Draw an angle of 45o and cut an arc at this angle at a radius of 4 cm at C.

3. Join AC to get the required triangle.

4. With A as center, draw intersecting arcs at M and N.

5. With center M and radius more than half of MN, cut an arc on the opposite side of A.

6. With N as center and radius the same as in the previous step, cut an arc intersecting the previous arc at E.

7. Join AE, it meets BC at D, then AE is the required perpendicular.

7. Draw a triangle ABC with AB = 3 cm, BC = 4 cm and ∠B = 60o. Also, draw the bisector of angles C and A of the triangle, meeting in a point O. Measure ∠COA.

Solution:

RD Sharma Solutions for Class 7 Maths Chapter 17 Constructions Image 18

Steps of construction:

1. Draw a line segment BC = 4 cm.

2. Draw ∠CBX = 60o.

3. Draw an arc on BX at a radius of 3 cm cutting BX at A.

4. Join AC to get the required triangle.

Angle bisector for angle A:

5. With A as center, cut arcs of the same radius cutting AB and AC at P and Q, respectively.

6. From P and Q cut arcs of same radius intersecting at R.

7. Join AR to get the angle bisector of angle A.

Angle bisector for angle C:

8. With A as center, cut arcs of the same radius cutting CB and CA at M and N, respectively.

9. From M and N, cut arcs of the same radius intersecting at T

10. Join CT to get the angle bisector of angle C.

11. Mark the point of intersection of CT and AR as O.

12. Angle ∠COA = 120o.


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