WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions

Geometry Chapter 3 Constructions

Constructions Introduction

You are familiar with geometrical constructions from class VI. With the help of several instruments found in the geometrical box, you have already constructed a number of geometrical figures under different conditions in classes VI and VII. In this chapter, our aim is to discuss mainly some problems and to find out their probable applications.

Read And Learn More WBBSE Solutions For Class 8 Maths

Geometry Chapter 3 Constructions  Some Questions

 

Question 1

Construct a triangle when its two angles and the side opposite to one of them are given.

 

WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions 1

 

Solution:

Given:

 

WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions 1

 

Let the two angles of a triangle be a and p. The length of the side opposite to the angle β is m.

It is required to construct the triangle.

Maths Solutions Class 8 Wbbse

Construction: Draw any straight line \(\overrightarrow{A X}\) From \(\overrightarrow{A X}\) cut off AB equal to the length m.

Draw ∠CAB and ∠DBX equal to the angle a at the endpoints A and B

of the line segment \(\overline{A B}\).

Draw ∠DBE equal to p at point B of the straight

line \(\overrightarrow{B D}\) on that side of \(\overrightarrow{B D}\) in which \(\overrightarrow{A C}\).

lies. Let BE and AC intersect each other at point F.

Then AFAB is the required triangle.

Proof: According to the construction, ∠BAF = ∠XBD

But they are corresponding angles.                   [Here \(\overline{A B}\) has intersected \(\overrightarrow{A B}\) and \(\overrightarrow{B D}\)]

Hence, \(\overrightarrow{A C}\) || \(\overrightarrow{B D}\).

Again, \(\overline{B F}\) is the transversal of the parallel

straight lines \(\overrightarrow{A C}\) and \(\overrightarrow{B D}\).

∠AFB = alternate ∠FBD.

In the ΔFAB, ∠FAB = α

∠AFB = β

and the length of the side opposite to β is

\(\overline{A B}\) = m.

Hence, ΔFAB is the required triangle (Proved).

Maths Solutions Class 8 Wbbse

Question 2

Construct a triangle when the lengths of its two sides and the angle opposite to one of them are given.

 

WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions 2

 

Solution:

Given:

 

WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions 2

 

Let b and c be the lengths of the two sides of a triangle and the angle opposite to the side b be ∠B.

It is required to construct the triangle.

Construction: Draw any straight line

\(\overrightarrow{B D}\). Now at point B of the straight

line, \(\overrightarrow{B D}\) constructs an angle ∠DBE equal to the given angle B.

Now from BE, cut off \(\overline{B A}\) equal to the length of the given side c.

With A as the center and radius equal to the length of the given side b construct an arc of a circle. Let this arc

intersects \(\overrightarrow{B D}\) at C1 and C2.

Join AC1 and AC2.

Then both ΔABC1 and ΔABC2 are the required triangles.

Verification: In the triangle ABC1 if you measure the sides \(\overline{A B}\) and \(\overline{A C_1}\)

with a scale and the angle ∠ABC1 with a protractor, you will find that AB = c,

AC1 = b, and ∠ABC1 is equal to ∠B.

Similarly, in the triangle ABC2, AB = c, AC% = b, and ∠ABC2 are equal to ∠B.

Maths Solutions Class 8 Wbbse

Question 3

Construct a triangle when the length of its one side and the angles adjacent to it are given.

 

WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions 3

 

Solution:

Given:

 

WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions 3

 

Let a be the length of one side of a triangle and ∠B and ∠C be the angles adjacent to side a.

It is required to construct the triangle.

Construction: Draw any straight-line BD. Now from BD cut off BC equal to the length of the side a.

Now at point B on BC draw an angle ∠CBE equal to the given angle B. Also at point C on BC

draw an angle ∠BCF equal to the given angle C. Let BE and CF intersect each other at A.

Thus, AABC is the required triangle.

Verification: Measuring the side BC with a scale and the angles ∠ABC and ∠ACB with a protractor you will find that BC = a, ∠ABC, and ∠ACB are equal to the angles B and C respectively.

 

Question 4

The lengths of the two sides of a triangle and the opposite angle of one side are given. To construct the triangle.

 

WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions 4

 

Solution:

Given:

 

WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions 4

 

Let, the two sides of a triangle be a and b and the angle opposite to the side b be a.

It is required to construct the triangle.

Construction: Draw any straight line \(\overline{A X}\). From \(\overline{A X}\) cut off a portion \(\overline{A B}\)  equal to a.

At a point, A draw an

angle ∠XAY equal to a. From \(\overline{A Y}\) cut off \(\overline{B C}\) = b.

Then AABC is the required triangle.

Verification: By measuring with a scale it is found that \(\overline{A B}\) = a and \(\overline{B C}\) = b, and by measuring with a protractor it is found that, ∠BCA = α.

Ganit Prabha Class 8 Solution

Question 5

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

Solution:

Let \(\stackrel{\leftrightarrow}{P Q}\) be a given straight line and R be a given point outside it.

It is required to draw a straight line

through R parallel to \(\stackrel{\leftrightarrow}{P Q}\).

 

WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions 5

 

Construction: A point S is taken on

the straight line\(\stackrel{\leftrightarrow}{P Q}\). SR is joined.

An angle, ∠ARS is constructed at the point R on \(\overline{S R}\) which is equal to ∠RSQ and is on the opposite side of \(\overline{R S}\) in

which ∠RSQ lies. AR is produced upto the point B.

Then the straight line \(\stackrel{\leftrightarrow}{A B}\) is the required straight line which is parallel

to \(\stackrel{\leftrightarrow}{P Q}\) and passes through R.

Proof: Here is the line segment \(\overline{S R}\) intersects \(\stackrel{\leftrightarrow}{P Q}\) at S and \(\stackrel{\leftrightarrow}{A B}\) at R.

Also by construction,

∠ARS = ∠RSQ.

But they are alternate angles.

Hence, \(\stackrel{\leftrightarrow}{A B}\) || \(\stackrel{\leftrightarrow}{P Q}\).

Alternative Method:

Ganit Prabha Class 8 Solution

Let \(\stackrel{\leftrightarrow}{P Q}\) be a given straight line and R be a given point outside it.

It is required to draw a straight line

through R parallel to \(\stackrel{\leftrightarrow}{P Q}\).

 

WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions 5.1

 

Construction: A point S is taken on

the straight line \(\stackrel{\leftrightarrow}{P Q}\). \(\overline{S R}\) is joined and it is produced to T.

An angle, ∠TRB is constructed at the

point R on \(\overline{R T}\) which is equal to ∠RSQ

and is on the same side as \(\overline{R S}\) in which

∠RSQ lies. \(\overleftarrow{B R}\) is produced upto the point A.

Then the straight line \(\stackrel{\leftrightarrow}{A B}\) is the required straight line which is parallel

to \(\stackrel{\leftrightarrow}{P Q}\)  and passes through R.

Proof:  Here, \(\overline{S T}\) intersects \(\stackrel{\leftrightarrow}{P Q}\) at S and \(\stackrel{\leftrightarrow}{A B}\) at R.

Also by construction, ∠TRB – ∠RSQ. But they are corresponding angles.

Hence, \(\stackrel{\leftrightarrow}{A B}\) || \(\stackrel{\leftrightarrow}{P Q}\).

Ganit Prabha Class 8 Solution

Question 6

To divide a line segment into three equal parts.

Solution:

Given:

Let, \(\overline{A B}\) be a segment.

 

WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions 6

 

Construction: At point A draw any angle ∠BAC. Then ∠ABD is drawn equal to the measure of ∠BAC on the opposite side of \(\overline{A B}\) in which ∠BAC lies.

Cut off two line segments \(\overline{A E}\) and \(\overline{E F}\) of equal length from \(\overline{A C}\).

Again from \(\overline{B D}\) cut off \(\overline{B G}\) and \(\overline{G H}\) equal to the previous length.

Join \(\overline{E H}\) and \(\overline{F G}\).

They intersect the line segment \(\overline{A B}\) at points S and T.

In this way, the line segment \(\overline{A B}\) is divided at S and T into three equal parts.

It is required to divide the line segment \(\overline{A B}\) into three equal parts.

 

Question 7

To divide a line segment into five equal parts.

Solution:

Let \(\overline{A B}\) be a given line segment.

It is required to divide \(\overline{A B}\) into five equal parts.

 

WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions 7

 

Construction: A straight line \(\overline{A C}\) is drawn at point A making an angle with \(\overline{A B}\).

From \(\overline{A C}\), the line segments \(\overline{A D}\), \(\overline{D E}\), \(\overline{E F}\), \(\overline{F G}\) and \(\overline{G H}\) of equal length are cut off one after another.

\(\overline{B H}\) is joined.

Now, at points, D, E, F, and G angles are drawn making it equal to ∠AHB on the same side of \(\overline{A C}\) in which ∠AHB lies.

Let the sides of the angles intersect \(\overline{A B}\) at I, J, K, and L respectively.

Thus the line segment \(\overline{A B}\) is divided at the points I, J, K, and L into five equal parts.

Alternative method :

Let \(\overline{P Q}\) be a given line segment.

It is required to divide \(\overline{P Q}\) into five equal parts.

Construction: Any angle ∠QPR is drawn at point P.

Then, ∠PQS is drawn equal to the measure of ∠QPR on the opposite side of \(\overline{P Q}\) in which ∠QPR lies. Four line segments of equal lengths \(\overline{P A}\), \(\overline{A B}\), \(\overline{B C}\)

and \(\overline{C D}\) are cut off from \(\overline{P R}\) one after another.

Also, four line segments of former equal lengths \(\overline{Q D_1}\), \overline{D_1 C_1}, \overline{C_1 B_1}, and \overline{B_1 A_1} are cut off from QS one after another.

\(\overline{A A_1}\), \(\overline{B B_1}\) , \(\overline{C C_1}\), and \(\overline{D D_1}\) are joined.

They intersect the line segment \(\overline{P Q}\)at E, F, G, and H respectively.

Thus the line segment \(\overline{P Q}\) is divided at points E, F, G, and H into five equal parts.

 

WBBSE Solutions For Class 8 Maths Geometry Chapter 3 Constructions 7.1

 

 

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