ICSE Class 8 Maths Chapter 24 Constructions PDF Download

Chapter 24

Constructions

Basic Constructions Using Ruler and Compass

Construction 1

To construct an angle equal to a given angle.

Given. Any ∠AOB and a point P.

Required. To construct an angle at P equal to ∠AOB.

Steps of construction

1. Through P draw a ray PQ.

2. With O as centre and any (suitable) radius, draw an arc to meet ray OA at C and ray OB at D.

3. Taking P as centre and same radius (as in step 2), draw an arc to meet PQ at R.

4. Measure the segment CD with compass.

5. With R as centre and radius equal to CD, draw an arc to meet the previous arc at S.

6. Join PS and produce it to form a ray QT, then ∠QPT = ∠AOB.

Construction 2

To bisect a given angle.

Given. Any ∠AOB.

Required. To bisect ∠AOB.

Steps of construction

1. With O as centre any (suitable) radius, draw an arc to meet ray OA at C and ray OB at D.

2. With C as centre and any suitable radius (not necessarily equal to radius of step 1 but greater than \(\frac{1}{2}\) CD), draw an arc. Also, with D as centre and same radius draw another arc to meet the previous arc at E.

3. Join OE and produce it to form a ray, then ray OE is the required bisector of ∠AOB.

Construction 3

To construct angles of 60°, 30°, 120°, 90° and 45°

(i) To construct an angle of 60°

Steps of construction

1. Draw any ray OA.

2. With O as centre and any suitable radius, draw an arc to meet ray OA at C.

3. With C as centre and same radius (as in step 2), draw an arc to meet the previous arc at D.

4. Join OD and produce it to form ray OB, then ∠AOB = 60°.

(ii) To construct an angle of 30°

Steps of construction

1. Construct ∠AOB = 60° (as above).

2. Bisect ∠AOB (as in construction 2). Let ray OE be the bisector of ∠AOB, then ∠AOE = 30°.

(iii) To construct an angle of 120°

Steps of construction

1. Draw any ray OA.

2. With O as centre and any suitable radius, draw an arc to meet ray OA at C.

3. With C as centre and same radius (as in step 2), draw an arc to meet the previous arc at D. With D as centre and same radius, draw another arc to cut the first arc at E.

4. Join OE and produce it to form ray OB, then ∠AOB = 120°.

(iv) To construct an angle of 90°

Steps of construction

1. Construct ∠AOB = 60° (as in construction 3 (i)).

2. Construct ∠AOF = 120° (as above).

3. Bisect ∠BOF (as in construction 2). Let ray OP be the bisector of ∠BOF, then ∠AOP = 90°.

(v) To construct an angle of 45°

Steps of construction

1. Construct ∠AOP = 90° (as above).

2. Bisect ∠AOP (as in construction 2). Let ray OQ be the bisector of ∠AOP, then ∠AOQ = 45°.

Construction 4

(i) To bisect a given line segment.

Given. Any line segment AB.

Required. To bisect line segment AB.

Steps of construction

1. At A, construct any suitable angle PAB.

2. At B, construct ∠ABQ = ∠PAB on the other side of the line AB.

3. With A as centre and any suitable radius, draw an arc to meet AP at C.

4. From BQ, cut off BD = AC.

5. Draw a line passing through points C and D to meet AB at M, then the line CD is a bisector of the line segment AB and M is mid-point of AB.

(ii) To divide a given line segment in a number of equal parts.

Example. Divide a line segment AB of length 7-5 cm into 5 equal parts.

Steps of construction

1. Draw line segment AB = 7-5 cm.

2. At A, construct any suitable angle XAB.

3. At B, construct ∠YBA = ∠XAB on the other side of the line AB.

4. From AX, cut off 5 equal distances at the points C, D, E, F and G such that AC = CD = DE = EF = FG.

5. With the same radius, cut off 5 equal distances along BY at the points H, I, J, K and L such that BH = HI = IJ = JK = KL.

6. Join AL, CK, DJ, EI, FH and GB. Let CK, DJ, EI and FH meet the line segment AB at the points M, N, O and P respectively. Then, M, N, O and P are the points of division of AB such that AM = MN = NO = OP = PB.

Construction 5

To draw a perpendicular bisector of a line segment.

Given. Any line segment AB.

Required. To draw a perpendicular bisector of line segment AB.

Steps of construction

1. With A as centre any suitable radius \(\left( > \frac{1}{2} \text{AB} \right)\), draw two arcs, one on each side of AB.

2. With B as centre and same radius (as in step 1), draw two more arcs, one on each side of AB cutting the previous arcs at P and Q.

3. Draw a line passing through points P and Q to meet the line AB at M, then the line PQ bisects AB at M and is perpendicular to AB. Thus, the line PQ is the required perpendicular bisector of AB.

Construction 6

To draw a perpendicular to a line at a point on the line.

Given. A line AB and a point P on it.

Required. To draw a perpendicular to AB at the point P.

Steps of construction

1. With P as centre any suitable radius, draw an arc to cut the line AB at the points C and D.

2. With C and D as centres, draw two arcs of equal radius \(\left( > \frac{1}{2} \text{CD} \right)\) cutting each other at Q.

3. Draw a line passing through points P and Q, then QP is the required line perpendicular to the line AB at the point P.

Construction 7

To draw a perpendicular to a line from a point outside the line.

Given. A line AB and a point P outside AB.

Required. To draw a perpendicular to AB from the point P.

Steps of construction

1. With P as centre and any suitable radius, draw an arc to cut the line AB at points C and D.

2. With C and D as centres, draw two arcs of equal radius \(\left( > \frac{1}{2} \text{CD} \right)\) cutting each other at Q on the other side of AB.

3. Draw a line through points P and Q to meet the line AB at N, then segment PN is the required perpendicular from the point P to the line AB.

Construction 8

To draw a line parallel to a given line through a given point.

Given. Any line AB and a point P outside AB.

Required. To draw a line parallel to AB and passing through the point P.

Steps of construction

1. Take any point Q on AB. Join P and Q.

2. With Q as centre and any suitable radius, draw an arc to meet AB at C and QP at D.

3. With P as centre and same radius (as in step 2), draw an arc to meet PQ at E.

4. With E as centre and radius equal to CD, draw an arc to cut the previous arc at F.

5. Draw a line through points P and F, then PF is the required line parallel to the line AB and passing through P.

Note. In the above construction, we have drawn ∠QPF = ∠PQB. So, alternate angles are equal and hence FP ∥ AB.

Exercise 24.1

1. Construct an angle of 45° and bisect it. Measure each part by protractor.

2. By using ruler and compass, construct an angle of :

(i) 15° (ii) 75° (iii) 150° (iv) 135°.

3. Draw a line segment of 5-3 cm and draw its perpendicular bisector.

4. Draw a line segment PQ = 4-9 cm. Draw a perpendicular to it

(i) from a point A outside PQ (ii) at a point A on PQ

5. Draw any triangle ABC. Through A, draw a line parallel to BC.

6. Draw a line AB = 5-7 cm. Using ruler and compass, construct ∠CAB = 30° and ∠CBA = 45°. From C, draw altitude to AB.

7. Construct an angle of 135° and bisect it. Measure any one part by protractor and see how accurate you are.

8. Draw line segment PQ = 5-8 cm. Construct ∠RQP = 60° and ∠PQR = 45°. Through R, draw a line parallel to PQ.

9. Draw a line segment of length 6-4 cm and divide it into three equal parts.

10. Draw a line segment AB of length 7 cm and divide it into the ratio 2 : 3. [Hint. Divide AB into (2 + 3) i.e. 5 equal parts. Mark a point M on AB after 2 parts from A, then AM : MB = 2 : 3.]

Teacher's Note

Geometric constructions help us understand how to create precise shapes and angles, which is fundamental in engineering, architecture, and design fields where accuracy is essential.

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