The chapter constructions come under the unit Geometry which discusses about drawing of shapes, angles or lines accurately. These constructions are done only by using a compass, straightedge and a pencil. In this chapter, you will learn pure form of geometric constructions of quadrilaterals where no numbers will be involved. For the convenience of the students, we have provided ICSE Class 8 Maths Selina Solutions Chapter 18 Constructions prepared by our in-house subject experts in a simple and understandable manner. By going through these solutions students will get an idea on how to solve the questions.

These Selina Solutions will also help them while preparing for their final exam. Students can access the pdf of ICSE Class 8 Maths Selina Solutions of Chapter 18 Constructions from the link provided below. Solve these questions before the exams to clear your doubts related to the chapter.

**Download ICSE Class 8 Maths Selina Solutions Chapter 18:-**Download Here

The exercise mentioned in Chapter 18 Constructions have a total of 15 questions and the detailed solutions of these questions are provided below.

### CHAPTER 18 – CONSTRUCTIONS

**Exercise**

**Question 1.Â **

Given below are the angles x and y.

Without measuring these angles, construct:

(i)Â âˆ ABC = x + y

(ii)Â âˆ ABC = 2x + y

(iii)Â âˆ ABC = x + 2y

**Solution:-**

**(i)Â StepsÂ ofÂ Construction:**

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

2. Taking B as centre, construct an arc of any suitable radius. With the same radius, construct arcs with the verticesÂ of given angles as centers. Consider these arcs to cut arc x at pointsÂ PÂ andÂ QÂ and arms of angleÂ yÂ at pointsÂ RÂ andÂ S.

3. From the arc, with B as centre,Â cut DEÂ = PQÂ arc ofÂ xÂ andÂ EF = RSÂ arc ofÂ y

4 Now join BF and produce up to pointÂ A.

Therefore, âˆ ABC = x + y

**(ii)Â StepsÂ ofÂ Construction:**

Continue the same steps as in part (i)

(i)Â Consider DEÂ = PQ = arc of x.

EF = PQ = arc of xÂ andÂ FG = RS = arc of y

Now joinÂ BGÂ and produce it up toÂ A.

Therefore, âˆ ABC=x+x+y=2x+y

**(iii)Â StepsÂ ofÂ Construction:**

Continue the same steps asÂ inÂ (ii)

Consider DE = PQ = arc of x. and EF = RS = arc of y and FG = RS = arc of y.

Now join BF and produce up to point A.

Thus âˆ ABC=x+y+y=x+2y

**QuestionÂ 2.Â **

Given below are the anglesÂ x,yÂ andÂ zÂ .

Without measuring these angles construct:

(i)Â âˆ ABC = x + y + z

(ii)Â âˆ ABC = 2x + y + z

(iii)Â âˆ ABC = x + 2y + z

**Solution:-**

(ii)Â StepsÂ ofÂ Construction:

1.Â Construct lineÂ segmentÂ BCÂ ofÂ anyÂ suitableÂ length.

2. Taking B as centre, construct an arc of any suitable radius. With the same radius, construct arcs with the verticesÂ of given angles as centers. Let these arcs cut arms of the angleÂ xÂ at the pointsÂ PÂ andÂ QÂ and arms ofÂ the angle y at pointsÂ RÂ andÂ SÂ and arms of the angleÂ zÂ at the pointsÂ LÂ andÂ M.

3. From the arc, with B as centre,Â cutÂ DE = PQ = arc of x, EF = RS = arc of yÂ andÂ FG = LM = arc of z

4. Now join BG and produce it up to A.

Therefore, âˆ ABC = x + y + z

(ii)Â âˆ ABC = 2x + y + z

(ii)Â Repeat asÂ inÂ partÂ (i)Â up toÂ stepÂ 2.

FromÂ theÂ arc,Â with B as centre,Â cut

DE=2PQ=2 arc of x

EF=RS=arc of y

FG=âˆ M=arc of z

Now join BG and produce it up to pointÂ A

Therefore, âˆ ABC=2x+y+z

(iii)Â âˆ ABC = x + 2y + z

(iii) Repeat as in (i) up to step 2

Cut arc DE = arc PQ = arc of x arc EF = 2arc RS = 2ARC OF Y arc FG = arc LM = arc of Z.

Now join BG and produce it up to A

Therefore, âˆ ABC=x+2y+z

**Question 3.Â **

Draw a line segmentÂ BC = 4cm. Construct angleÂ ABC = 60Â°.

**Solution:-**

**Steps of Construction:Â **

1. Construct a line segmentÂ BC = 4cm

2Â Taking BÂ as centre, construct an arc of any suitable radius which cutsÂ BCÂ at the pointÂ D.

3. Taking D as centre, and the same radius as in stepÂ 2,Â construct one more arc which cuts the previous arc at pointÂ E.

4. Now join BE and produce it to the pointÂ A.

Therefore, âˆ ABC = 60Â°

**Â QuestionÂ 4**.

Construct angleÂ ABC = 45Â°Â in whichÂ BC = 5cmÂ andÂ AB = 4.6cm.

**Solution:-**

**Â Steps of Construction:**

1. Construct a line segmentÂ BC = 5cm

2. With centre B, construct an arc of any suitable radius, which cutsÂ BCÂ at the pointÂ D.

3. Taking D as centre and the same radius, as taken in stepÂ 2,Â construct an arc which cuts the previous arc atÂ pointÂ E.

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

5. Taking E and F as centers and radii equal to more than half the distance between E atÂ F, construct arc whichÂ cut each other at point P.

6. Now join BP to meet EF at L and produce to pointÂ 0.Â ThenÂ âˆ OBC=90Â°

7. Construct BA, the bisector of angle OBC. [With D, L as centers and suitable radius construct two arc meetingÂ each other atÂ QÂ produced it to R]

=>âˆ ABC=45Â°[âˆ´BAÂ is bisector ofÂ âˆ OBCâˆ´âˆ ABC==45Â°]

8. From BR cut arc AB = 4.6 cm

**QuestionÂ 5.**

Construct angleÂ ABC = 90Â°.Â Draw BP, the bisector of angle ABC. State the measure of angle PBC.

**Solution:-**

1. Construct âˆ ABC=90Â°Â (as in Ques.Â 4)

2. Construct bisector ofÂ âˆ ABC

**Question 6.Â **

6. Draw angle ABC of any suitable measure.

(i) Draw BP, the bisector of angle ABC.

(ii) Draw BR, the bisector of angle PBC and draw BQ, the bisector of angle ABP.

(iii) Are the anglesÂ ABQ, QBP,Â PBRÂ andÂ RBCÂ equal?

(iv) Are the angles ABR and QBC equal?

**Solution:Â Â **

**Steps of Construction:Â **

1. Draw any angle ABC

2. Taking B as centre, construct an arc EF meeting BC at E and AB at F.

3. Taking E, FÂ as centers construct two arc of equal radii meeting each other at the pointÂ P

**Â Question 7.Â **

Draw a line segment AB of lengthÂ 5.3 cm. using two different methods bisect AB.

**Solution:-**

**Â Steps of Construction:Â **

1. Construct a line segmentÂ AB = 5.3cm

2. Taking A as centre and radius equal to more than half of AB, construct arcs on both sides of AB.

3. Taking BÂ as centre and with the same radius as taken in stepÂ 2, construct arcs on both the sides ofÂ AB.

4. Let the arcs intersect each other at pointsÂ PÂ andÂ Q.

5. Now joinÂ PÂ andÂ Q.

6. The line PQ cuts the given line segment AB at the point O.

Therefore, PQ is a bisector of AB such that

**SecondÂ Method**

**Â Steps of Construction:Â **

1. Construct the given line segmentÂ AB = 5.3 cm.

2. At the point A,Â draw âˆ PABÂ of any suitable measure. ThenÂ âˆ PAB = 60Â°Â constructÂ âˆ QBA = 60Â°

3. From AP, cut AR of any suitable length and from BQ; cut BS = AR.

4. Now join R and S

5. Let RS cut the given line segment AB at the point O.

Therefore, RS is a bisector of AB such that

**QuestionÂ 8.Â **

Draw a line segmentÂ PQ = 4.8 cm. Construct the perpendicular bisector ofÂ PQ.

**Solution:-**

Â Steps of Construction:

1. Construct a line segmentÂ PQ = 4.8cm.

2. Taking PÂ as centre and radius equal than half ofÂ PQ, construct arc on both the PQ.

3. Taking QÂ as centre and the same radius as taken in stepÂ 2, construct arcs on both sides ofÂ PQ.

4. Let the arcs intersect each other at pointÂ AÂ andÂ B

5. Now joinÂ AÂ andÂ B.

6. The line AB cuts the line segment PQ at the point O. Here OP = OQ and âˆ AOQ=90Â° Then the line AB is perpendicular bisector of PQ.

**Â Question 9.Â **

In each of the following, draw perpendicular through point P to the line segment AB.

Â

**Solution:-**

Â (i) Steps of Construction:

1. Taking P as centre, construct an arc of a suitable radius which cutsÂ ABÂ at pointsÂ C and D

2. Taking CÂ andÂ DÂ as centers, construct arcs of equal radii and let these arcs intersect each other at the pointÂ 0Â [The radius of these arcs must be more than half of CD and both the arcs must be drawn on the other side]

3. Now joinÂ PÂ andÂ Q

4. LetÂ PQÂ cutÂ ABÂ at the pointÂ O.

Therefore, OP is the required perpendicular clearly, âˆ AOP=âˆ BOP=90Â°

(ii) Steps of Construction:

1. Taking PÂ as centre, construct an arc of any suitable radius which cutsÂ ABÂ at pointsÂ CÂ andÂ D.

2. Taking CÂ andÂ DÂ as centers, construct arcs of equal radii which intersect each other at pointÂ A.

[This radius must be more than half of CD and let these arc intersect each other at the point 0]3. Now join P and O. Then OP is the required perpendicular.

âˆ OPA=âˆ OPB=90Â°

(iii) Steps of Construction:

1. Taking P as centre, construct an arc of any suitable radius which cutsÂ ABÂ at pointsÂ CÂ andÂ D

2. Taking CÂ andÂ DÂ as centre, construct arcs of equal radii

[The radius of these arcs must be more than half of CD and both the arcs must be drawn on the otherÂ side.]And let these arcs intersect each other at the point Q.

3. Now joinÂ QÂ andÂ P. LetÂ QPÂ cutÂ ABÂ at the pointÂ O.Â Then OP is the required perpendicular.

Clearly,Â âˆ AOP = âˆ BOP = 90Â°

**Question 10:**

DrawÂ aÂ lineÂ segmentÂ AÂ B=5.5Â cmMarkÂ aÂ pointÂ P,Â suchÂ thatÂ PÂ A=6Â cmÂ andÂ PÂ B=4.8Â cm. FromÂ theÂ pointÂ P drawÂ perpendicularÂ toÂ AB.

**Solution:-**

Â StepÂ ofÂ Construction:

1.Â Construct aÂ lineÂ segmentÂ AB=5.5Â cm

2.Â Taking AÂ asÂ centreÂ andÂ radiusÂ =6Â cmÂ construct anÂ arc.

3.Â Taking BÂ asÂ centreÂ andÂ radiusÂ =4.8Â cmÂ construct anotherÂ arc.

4.Â LetÂ theseÂ arcsÂ meetÂ eachÂ otherÂ atÂ theÂ pointÂ P.Â PA=6Â cm,Â PÂ B=4.8

5.Â Take PÂ asÂ centreÂ andÂ someÂ suitableÂ radiusÂ construct anÂ arcÂ meetingÂ ABÂ atÂ theÂ pointsÂ CÂ andÂ D.

6.Â Take CÂ asÂ centreÂ andÂ radiusÂ moreÂ thanÂ halfÂ ofÂ CDÂ construct anÂ arc.

7.Â Take DÂ asÂ centerÂ andÂ sameÂ radiusÂ asÂ inÂ stepÂ 6, construct anÂ arc.

8.Â LetÂ theseÂ arcsÂ meetÂ eachÂ otherÂ atÂ theÂ pointÂ Q.

9. Now joinÂ PQ.

10.Â TheÂ PQÂ meetÂ ABÂ atÂ pointÂ O.

ThenÂ POABÂ i.e;Â âˆ AOP=90Â° =âˆ POB

**QuestionÂ 11.**

Draw a line segmentÂ AB=6.2cm. Mark a pointÂ PÂ in AB such thatÂ BP=4cm. Through pointÂ PÂ drawÂ perpendicular to AB.

**Solution:Â **

Steps of Construction:

1. Construct a line segmentÂ AB=6.2cm

2. Cut offÂ BP=4cm

3. Take P as centre and some radius construct arc meeting AB at the pointsÂ C, D.

4. Take C, DÂ as centers and equal radii [each is more than half of CD] construct two arcs, meeting each otherÂ at the pointÂ 0.

5. Now join OP. Then OP is perpendicular forÂ AB.

**Â Question 12.Â **

Draw a lineÂ AB=6cm. Mark a pointÂ PÂ anywhere outside the line AB. Through the pointÂ P, construct a lineÂ parallel to AB.

**Solution:-**

Â Steps of construction:

1. Construct a lineÂ AB=6cm

2. Take any pointÂ QÂ on the lineÂ ABÂ and join it with the given pointÂ P.

3. At pointÂ P, draw âˆ CPQ=âˆ PQB

4. Produce CP up to any point D.

Therefore, CPD is the required parallel line.

**Question 13.**

Draw a line Mn=5. 8cm. Locate point A which is 4.5cm from M and 5cm from N. Through A draw a line parallel to line MN.

**Solution:-**

Â Steps of construction:

1. Construct a line MNÂ =5.8cm

2. Taking M as centre and radiusÂ =4.5cm, constructÂ an arc.

3. Taking NÂ as centre draw another arc of radiusÂ 5cm. These arcs intersect each other atÂ A.

4. Now join AM and AN.

5. At pointÂ A,Â construct âˆ DAN=âˆ ANM

6. Produce DA to any pointÂ C.

Therefore, CAD is the required parallel line.

**Â Question 14.Â **

Draw a straight lineÂ AB=6.5cm. Draw another line which is parallel toÂ ABÂ at a distance ofÂ 2.8cmÂ from it.

**Solution:-**

Â Steps of construction:

1. Construct a straight lineÂ AB=6.5cm

2. With A as centre, construct an arc of radiusÂ 2.8cm.

3. With B as centre, construct another arc of radiusÂ 2.8cm.

4. Construct a line CD which touches the two arcs drawn.

Therefore, CD is the required parallel line.

**Question 15.Â **

Construct an angleÂ PQR=80Â°.Â Draw a line parallel toÂ PQÂ at a distance ofÂ 3cmÂ from it and another lineÂ parallel toÂ QRÂ at a distance ofÂ 3.5cmÂ from it. Mark the point of intersection of these parallel lines asÂ A.

**Solution:-**

Â Steps of construction:

1.Â Construct âˆ PQR=80Â°

2. Taking PÂ as center construct an arc of radiusÂ 2cm.

3. Again with Q as centre, construct another arc of radiusÂ 2cm. ThenÂ BMÂ is a line which touches the twoÂ arcs. ThenÂ BMÂ is a line parallel toÂ PQ.

4. Taking QÂ as centre, construct an arc of radiusÂ 3.5cm.Â Taking RÂ as centre construct another arc of radiusÂ 3.5cm.Â Â Construct a line HC which touches these two arcs. Let these two parallel lines intersect at A.

### ICSE Class 8 Maths Selina Solutions Chapter 18 – Constructions

Chapter 18 of ICSE Class 8 Maths provides introduction of construction of quadrilaterals and explains the steps of constructing different types of quadrilaterals. The concepts included in this chapter is explained in a simple language for better understanding of the students. All the concepts taught in this chapter will be required when you will be promoted to the next class. Moreover, students can also access ICSE Class 8 Selina Solution for other subjects like Physics, Chemistry, Biology.

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