## 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.**

**Solution:**

**Given:**

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.**

**Solution:**

**Given:**

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 C_{1} and C_{2}.

Join AC_{1 }and AC_{2}.

Then both ΔABC_{1} and ΔABC_{2 }are the required triangles.

**Verification:** In the triangle ABC_{1} if you measure the sides \(\overline{A B}\) and \(\overline{A C_1}\)

with a scale and the angle ∠ABC_{1} with a protractor, you will find that AB = c,

AC_{1} = b, and ∠ABC_{1} is equal to ∠B.

Similarly, in the triangle ABC_{2}, AB = c, AC% = b, and ∠ABC_{2} 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.**

**Solution:**

**Given:**

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.**

**Solution:**

**Given:**

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 \(\stackrel{\leftrightarrow}{P Q}\).

**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 \(\stackrel{\leftrightarrow}{P Q}\).

**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.

**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.

**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.