ICSE Class 8 Maths Chapter 28 Constructions Using Ruler Compasses only PDF Download

Chapter 28

Constructions

(Using ruler and compasses only)

28.1 Construction Of An Angle

1. To Construct An Angle Equal To Given Angle

Let the given angle be ∠ABC as shown alongside and we have to construct another angle (say, ∠DEF) equal to ∠ABC.

Steps:

1. Draw a line segment EF of any suitable size.

2. With B as centre, draw an arc of any suitable radius which cuts AB at point P and BC at point Q.

3. With E as centre and the same radius as taken in step 2, draw an arc which cuts EF at point R.

4. With R as centre and radius equal to PQ, draw an arc which cuts the previous arc at point S.

5. Join E and S, and produce upto point D.

∠DEF so obtained is equal to ∠ABC.

2. To Draw The Bisector Of A Given Angle

Let the given angle be ∠ABC whose bisector is to be drawn.

Steps:

1. With B as centre, draw an arc of any suitable radius which cuts AB at point D and BC at point E.

2. Taking D and E as centres, draw arcs of equal radii and let these arcs cut each other at point O.

The radii of these arcs must be more than half the distance between points D and E.

3. Join BO and produce upto point P.

∴ BP is the required bisector of ∠ABC.

Thus, ∠ABP = ∠PBC = \(\frac{1}{2}\) ∠ABC.

3. Construction Of Angles Of 60°, 30°, 90° And 45°

1. Construction Of Angle Of 60°:

Steps:

1. Draw a line segment BC of any suitable length.

2. With B as centre, draw an arc of any suitable radius which cuts BC at point D.

3. With D as centre and the same radius as taken in step 2, draw one more arc which cuts the previous arc at point E.

4. Join BE and produce upto any point A.

∴ ∠ABC so obtained is of 60° i.e. ∠ABC = 60°.

2. Construction Of Angle Of 30°:

Steps:

1. Draw angle ABC = 60°.

2. Draw BP, the bisector of ∠ABC.

∴ ∠PBC = \(\frac{1}{2}\) ∠ABC = \(\frac{1}{2}\) × 60° = 30°

3. Construction Of Angle Of 90°:

Steps:

1. Draw a line segment BC of any suitable length.

2. Taking B as centre, draw an arc of any suitable radius, which cuts BC at point D.

3. With D as centre and the same radius, as taken in step 2, draw an arc which cuts previous arc at point E.

4. With E as centre and the same radius, draw one more arc which cuts the first arc at point F.

5. With E and F as centres and radii equal to more than half the distance between E and F, draw arcs which cut each other at point P.

6. Join BP and produce upto any point A.

∴ ∠ABC so obtained is of 90° i.e. ∠ABC = 90°.

Since, ∠ABC = 90° => AB and BC are perpendicular to each other.

4. Construction Of Angle Of 45°:

Steps:

1. Draw a line segment BC of any suitable length.

2. Construct angle OBC = 90°.

3. Draw BA, the bisector of angle OBC.

∴ ∠ABC so obtained is the angle of 45°.

Since, BA is bisector of angle OBC, ∠ABC = ∠ABO = \(\frac{90°}{2}\) = 45°.

Example 1:

Given below are the two angles x and y. Construct an angle ABC such that:

(i) ∠ABC = x + y (ii) ∠ABC = 2x + y

Solution:

(i)

Steps:

As shown above:

1. Draw line segment BC of any suitable length.

2. With B as centre, draw an arc of any suitable radius. With the same radius, draw arcs with the vertices of given angles as centres. Let these arcs cut arms of the angle x at points P and Q, and arms of the angle y at points R and S.

3. From the arc, with centre B, cut DE = PQ = x and EF = SR = y.

4. Join BF and produce upto point A.

Thus, ∠ABC = x + y

(ii)

Proceed in exactly the same way as in part (i) taking DE = PQ = x, EF = PQ = x and FG = RS = y.

Thus, ∠ABC = x + x + y = 2x + y

Teacher's Note

Angle construction is fundamental in architecture and engineering, where precise angles are needed for building structures, creating blueprints, and designing mechanical components that must fit together perfectly.

This is a preview of the first 3 pages. To get the complete book, click below.