The chapter constructions come under the unit Geometry which discusses about Geometric Constructions like 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 student, 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 Solution 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 Constructions

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

**Question 1.Â **

Given below are the angles x and y.

Without measuring these angles, construct:

(i)Â âˆ ABC = x + y

**Solution:-**

**(i)Â StepsÂ ofÂ Construction: **

1. Draw a 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 verticesÂ of given angles as centers. Let these arcs cut arc x at pointsÂ PÂ andÂ QÂ and arms of angleÂ yÂ at pointsÂ RÂ andÂ S.

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

4Â Join BF and produce up to pointÂ A.Â Â ThusÂ âˆ ABC = x + y

(ii)Â âˆ ABC = 2x + y

**(ii)Â StepsÂ ofÂ Construction: **

Proceed in exactly the same way as in part

(i)Â Â Takes DEÂ = PQ = arc of x.

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

JoinÂ BGÂ and produce it up toÂ A.

ThusÂ âˆ ABC=x+x+y=2x+y)

(iii)Â âˆ ABC=x+2y

**(ii)Â StepsÂ ofÂ Construction: **

ProceedÂ inÂ exactlyÂ theÂ sameÂ wayÂ asÂ inÂ (ii)

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

4. 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

**Solution:-**

(ii)Â StepsÂ ofÂ Construction:

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

2. With B as centre, draw an arc of any suitable radius. With the same radius, draw 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 centreÂ B,Â cutÂ DE = PQ = arc of x, EF = RS = arc of yÂ andÂ FG = LM = arc of z

4. Join BG and produce it up to A.Â ThenÂ âˆ ABC = x + y + z

(ii)Â âˆ ABC = 2x + y + z

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

3.Â FromÂ theÂ arc,Â withÂ centreÂ B,Â cut

DE=2PQ=2 arc of x

EF=RS=arc of y

FG=âˆ M=arc of z

4. Join BG and produce it up to pointÂ A

Thenâˆ ABC=2x+y+z

(iii)Â âˆ ABC = x + 2y + z

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

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

4. Join BG and produce it up to A

5. Then âˆ ABC=x+2y+z

**Question 3.Â **

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

**Solution:-**

**Steps of Construction:Â **

1. Draw a line segmentÂ BC = 4cm

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

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

4. Join BE and produce it to the pointÂ A.Â ThusÂ 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. Draw a line segmentÂ BC = 5cm

2. Taking B as centre, draw an arc of any suitable radius, which cutsÂ BCÂ at the pointÂ D.

3. With D as centre and the same radius, as taken in stepÂ 2,Â draw an arc which cuts the previous arc atÂ pointÂ 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 centers and radii equal to more than half the distance between E atÂ F, draw arc whichÂ cut each other at point P.

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

7. Draw BA, the bisector of angle OBC. [With D, L as centers and suitable radius draw 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. Draw âˆ ABC=90Â°Â (as in Ques.Â 4)

2. Draw bisector ofÂ âˆ ABC

Then \(\angle P B C=\frac{1}{2}\left(90^{\circ}\right)=45^{\circ} ) \)

**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. Construct any angle ABC

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

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

4. JoinÂ BP. ThenÂ BPÂ is the bisector of \(\angle A B C=\angle A B P=\angle P B C=\frac{1}{2} \angle A B C\)

5. Similarly draw BR, the bisector ofÂ âˆ PBCÂ and draw BQ as the bisector ofÂ âˆ ABP [With the same methodÂ as in stepsÂ 2,3

6. ThenÂ âˆ ABQ = âˆ QBP = âˆ PBR = âˆ RBO

7. \(\angle A B R=\frac{3}{4} \angle A B C \text { and } \angle Q B C=\frac{3}{4} \angle A B C=\angle A B R=\angle O B C\)

**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. Draw a line segmentÂ AB = 5.3cm

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

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

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

5. JoinÂ PÂ andÂ Q.

6. The line PQ cuts the given line segment AB at the point O.Â Â Thus, PQ is a bisector of AB such thatÂ O \(A=O B=\frac{1}{2} A B\)

**SecondÂ Method **

**Â Steps of Construction:Â **

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

2. AtÂ A,Â constructÂ 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. Join R and S

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

Thus RS is a bisector of AB such that \(OA =O B=\frac{1}{2} A B\)

**QuestionÂ 8.Â **

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

**Solution:-**

Â Steps of Construction:

1. Draw a line segmentÂ PQ = 4.8cm.

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

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

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

5.Â 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:

- .p

**Solution:-**

Â (i) Steps of Construction:

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

2. WithÂ CÂ andÂ DÂ as centers, draw 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. JoinÂ PÂ andÂ Q

4. LetÂ PQÂ cutÂ ABÂ at the pointÂ O.

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

(ii)

(ii) Steps of Construction:

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

2. WithÂ CÂ andÂ DÂ as centers, draw 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. Join P and O. Then OP is the required perpendicular.

âˆ OPA=âˆ OPB=90Â°

(iii)

(iii) Steps of Construction:

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

2. WithÂ CÂ andÂ DÂ as centre, draw 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 pointÂ O.

3. 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 DrawÂ aÂ lineÂ segmentÂ AB=5.5Â cm

2.Â WithÂ AÂ asÂ centreÂ andÂ radiusÂ =6Â cmÂ drawÂ anÂ arc.

3.Â WithÂ BÂ asÂ centreÂ andÂ radiusÂ =4.8Â cmÂ drawÂ 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 WithÂ PÂ asÂ centreÂ andÂ someÂ suitableÂ radiusÂ drawÂ anÂ arcÂ meetingÂ ABÂ atÂ theÂ pointsÂ CÂ andÂ D.

6.Â WithÂ CÂ asÂ centreÂ andÂ radiusÂ moreÂ thanÂ halfÂ ofÂ CDÂ drawÂ anÂ arc.

7.Â WithÂ DÂ asÂ centerÂ andÂ sameÂ radiusÂ asÂ inÂ stepÂ 6, drawÂ anÂ arc.

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

9.Â JoinÂ PQ.

10.Â TheÂ PQÂ meetÂ ABÂ atÂ pointÂ O.

ThenÂ POABÂ i.e;Â âˆ AOP=90Â° =âˆ POB

**Â **

**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. Draw a line segmentÂ AB=6.2cm

2. Cut offÂ BP=4cm

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

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

5. 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. Draw 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, constructÂ âˆ CPQ=âˆ PQB

4. Produce CP up to any point D.

Thus, 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. Draw a line MNÂ =5.8cm

2. With M as centre and radiusÂ =4.5cm,

Draw an arc.

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

4. Join AM and AN.

5. At pointÂ A,Â drawÂ âˆ DAN=âˆ ANM

6. Produce DA to any pointÂ C.

Thus 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. Draw a straight lineÂ AB=6.5cm

2. Taking point A as centre, draw an arc of radiusÂ 2.8cm.

3. Taking B as centre, drawn another arc of radiusÂ 2.8cm.

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

Thus, 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 DrawÂ âˆ PQR=80Â°

2. With PÂ as center draw an arc of radiusÂ 2cm.

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

4. WithÂ QÂ as centre, draw an arc of radiusÂ 3.5cm.Â WithÂ RÂ as centre draw another arc of radiusÂ 3.5cm.Â Â Draw 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

Construction chapter of ICSE Class 8 Maths provides basic 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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