Construction Class 9 Free Notes Maths Chapter 11 | Edugrown

CBSE Class 9 Maths Notes Chapter 11 Construction Pdf free download is part of Class 9 Maths Notes for Quick Revision. Here we have given NCERT Class 9 Maths Notes Chapter 11 Construction. According to new CBSE Exam Pattern, MCQ Questions for Class 9 Maths Carries 20 Marks.

Class 9 Construction Notes

NCERT Class 9 Notes becomes a vital resource for all the students to self-study from NCERT textbooks carefully. That is why we have arranged every important points given in the chapter and simplified them so you can easily understand every point clearly. Through these notes a student can boost their preparation and assessment of understood concepts. It is quite easy to retain the answers once you are fully aware of the concept thus notes can be beneficial for you. As you already know the importance of NCERT textbooks for Class 9 thus these NCERT Notes is prove very useful during the preparation of exams. 

Quick Revision Notes of Ch-11 Construction Class 9th Maths.

Introduction to Constructions

Constructions tell us about the steps of drawing of perfect geometrical figures like triangle, circle, polygons etc by using geometrical tools with the given measurements.

Geometry Box

To draw the geometrical figures we need some geometrical instruments which we can find in the geometry box. Some instruments are:

1. Graduated scale

This is the scale to make the straight lines. Its one side is marked with centimetres and millimetres and the other side is marked with inches.

2. A pair of set squares

It is a set of two squares. One with 90°, 60° and 30° angles and other with angles 90°, 45° and 45°.

3. Divider

It helps in measuring the length.

4. Compass

It is used to draw the circles and angles.

5. Protector

It is used to mark and measure the angles.

Geometrical Construction

In the Geometrical construction, we use only two instruments – A non graduated ruler also called a Straight Edge and a compass for drawing a geometrical figure. For measurements, we may use a graduated scale and protractor also.

Construction of Angle Bisector

1. Steps for construction of Angle Bisector for a given angle.
Given: ∠POQ. Construction: To construct the bisector of ∠POQ. Steps of Construction:  

1. With O as centre and any suitable radius, draw  an arc to meet OP at R and OQ at S.
2. With R as centre and any suitable radius (not  necessarily) equal to radius of step 1 (but > 1/2 RS),  draw an arc. Also, with S as centre and same  radius, draw another arc to meet the previous arc  at T.
3. Join OT and produce it, then OT is the required  bisector of ∠POQ.  

2. Construction of Important Angles with some measurement. 
(i) To construct an angle of 60°
Steps of Construction:

1. Draw any line OP.
2. With O as centre and any suitable radius, draw  an arc to meet OP at R.
3. With R as centre and same radius (as in step 2),  draw an arc to meet the previous arc at S.
4. Join OS and produce it to Q, then ∠POQ = 60°. 

(ii) To construct an angle of 30°
Steps of Construction:

1. Construct ∠POQ = 60° (as above).
2. Bisect ∠POQ (as in construction I). Let OT be the bisector of ∠POQ, then ∠POT = 30°.

(iii) To construct an angle of 120°
Steps of Construction:

1. Draw any line segment OP.
2. With O as centre and any suitable radius, draw  an arc to meet OP at R.
3. With R as centre and same radius (as in step 2), draw an arc to meet the previous arc at T. With T as  centre and same radius, draw another arc to cut  the first arc at S.
4. Join OS and produce it to Q, then ∠POQ = 120°

(iv) To construct an angle of 90°Steps of Construction: 

1. Construct ∠POQ = 60°.
2. Construct ∠POV = 120°. 
3. Bisect ∠QOV. 
4. Let OU be the bisector of ∠QOV, then ∠POU = 90°. 

(v) To construct an angle of 45°
Steps of Construction:

1. Construct ∠AOP = 90°
2. Bisect ∠AOP.
3. Let OQ be the bisector of ∠AOP, then ∠AOQ = 45°. 

3. Construction of Perpendicular Bisector
Given: Any line segment AB.Construction: To construct a perpendicular bisector of line segment AB.
Steps of Construction:

1. Draw a line segment AB.
2. Taking A and B as the centres and radius of more than half the length of AB, draw arcs on both  sides of AB.
3. Let these arcs intersect each other at points M and N.  4. Join the points of intersection M and N. Thus, MN is the required perpendicular bisector of AB.

4. Construction of a triangle, given its base, difference of the other two sides and one base angle.
Construction: Construct a triangle with base of length 7·5 cm, the difference of the other two sides is 2·5 cm, and  one base angle of 45°.Given: In DABC, base BC = 7·5 cm, the difference of the other two  sides, AB – AC or AC – AB = 2·5 cm and one base angle is 45°.Required: To construct the DABC. 
Case (i) AB – AC = 2·5 cm. 

Steps of Construction: 

1. Draw BC = 7·5 cm.
2. At B, construct ∠CBX = 45°.
3. From BX, cut off BD = 2·5 cm. 
4. Join CD. 
5. Draw the perpendicular bisector RS of CD intersecting  BX at a point A.
6. Join AC. Then DABC is the required triangle.

Case (ii) AC – AB = 2·5 cm. 

Steps of Construction:

1. Draw BC = 7·5 cm. 
2. At B, construct ∠CBX = 45° and produce XB to  form a line XBX’.
3. From BX’, cut off BD’ = 2·5 cm.
4. Join CD’. 
5. Draw the perpendicular bisector RS of CD’  intersecting BX at a point A.
6. Join AC. Then DABC is the required triangle.

5. Construction of a triangle of given perimeter and base angles.
Construction: Construct a triangle with perimeter  11·8 cm and base angles 60° and 45°.Given: In DABC, AB + BC + CA = 11·8 cm, ∠B = 60° and ∠C = 45°.Construction: To construct the DABC.

Steps of Construction: 

1. Draw DE = 11·8 cm. 
2. At D, construct ∠EDP = 1/2 of 60° = 30° and at E, construct ∠DEQ = 1/2 of 45° = 22½°. 
3. Let DP and EQ meet at A.  4. Draw a perpendicular bisector of AD to meet DE at B.
4. Draw a perpendicular bisector of AE to meet DE at C.
5. Join AB and AC. Then DABC is the required triangle.