Constructions Class 9 Notes CBSE Maths Chapter 11 (Free PDF Download)

Introduction:

In geometrical constructions, the process of drawing is done only with the help of an ungraduated ruler known as straight edge and a compass. Sometimes when the measurements are also required, a graduated scale and a protractor may also be used.

Basic constructions:

To construct the bisector of a given angle – Let us have an $\angle ABC$ and bisect it. The steps that needed to be followed are as follows;

1. Draw an arc of any radius, taking $B$ as the centre, that intersects the rays $BA$ and $BC$ at $E$ and $D$ respectively.

2. Now, draw arcs taking $E$ and $D$ as centers taking radius more than $\dfrac{1}{2}DE$ such that they intersect at $F$.

3. Now, draw a ray joining $B$ and $F$ and extending it. Thus, $BF$ is the required angle bisector.

To construct the perpendicular bisector of a given line segment – Let us have a line segment $\overline{AB}$, for which we will draw a perpendicular bisector. The steps to be followed are as follows;

1. Taking $A$ and $B$ as centres, draw arcs on both sides of the line segment with radius more than $\dfrac{1}{2}AB$. The arcs drawn should intersect each other.

2. Let us say that the arcs intersect at points $P$ and $Q$ respectively. Join $P$ and $Q$.

3. Let’s say that the line $\overleftrightarrow{PQ}$ intersects the segment $\overline{AB}$ at point $M$. Then the line $PMQ$ is the required perpendicular bisector.

To construct an angle of ${{60}^{\circ }}$ at the initial point of a given ray – Take a ray $AB$ having initial point $A$ from where we have to draw another ray $AC$ such that $\angle BAC={{60}^{\circ }}$. The steps to be followed are as follows;

1. Take $A$ as centre and draw an arc that cuts the ray $AB$ at some point $D$. Now, taking $A$ and $D$ as centres, and $AD$ as radius, draw two arcs that cut each other, say at point $E$.

2. Join $AE$ and extend it till point $C$ on the ray $AC$. Hence, $\angle CAB={{60}^{\circ }}$.

Some Construction of Triangles:

Measurements of at the least $3$ elements of a triangle are required for the development of a triangle. But all of the combos of $3$ elements aren't enough for the purpose. For example, it isn't viable to assemble a completely unique triangle whilst the measurements of aspects and a perspective that isn't protected in among the given aspects are given. A triangle may be built when (i) the base, one base attitude and the sum of the opposite aspects are given (ii) the base, a base attitude and the distinction among the opposite aspects are given (iii) perimeter and base angles are given.

To construct a triangle, given its base, a base angle and sum of other two sides – Let us suppose, we have to draw $\Delta ABC$, whose base $BC$ and $\angle B$ is given. And the sum $AB+AC$ is also given. The required triangle will be constructed in the steps as follows;

1. First draw the base $BC$ and then make $\angle CBX$ same as the angle given to us at point $B$.

2. Now, mark a point $D$ on the ray $BX$ such that $BD=AB+AC$.

3. Then Join $DC$. Construct $\angle DCY$ such that it is equal to $\angle BDC$.

4. Let us say that the ray $CY$ intersects $BD$ at point $A$. Then $\Delta ABC$ is the required triangle.

To construct a triangle given its base, a base angle and the difference of the other two sides – Let us say, we have to construct $\Delta ABC$, of which the base $BC$, one base angle $\angle B$ and the difference of other two sides either $AB-AC$ or $AC-AB$ are given. This may have two cases as follows;

Case (i): When $AB>AC$

1. First draw the base $BC$ and then make $\angle CBX$ same as the angle given to us at point $B$.

2. Now, mark a point $D$ on the ray $BX$ such that $BD=AB-AC$.

3. Draw the perpendicular bisector of line $CD$, say $PQ$ when extended cut the ray $BX$ at $A$. Hence, $\Delta ABC$ has been constructed.

Case (ii): When $AB<AC$

1. First draw the base $BC$ and then make $\angle CBX$ same as the angle given to us at point $B$.

2. Now, mark a point $D$ on the ray $BX$ extended on opposite side of $BX$ such that $BD=AC-AB$.

3. Draw the perpendicular bisector of line $CD$, say $PQ$ when extended cut the ray $BX$ at $A$. Hence, $\Delta ABC$ has been constructed.

To construct a triangle, given its perimeter and its two base angles – Construct $\Delta ABC$ when the base angles $\angle B$ and $\angle C$ and the perimeter $AB+BC+CA$ are given. The steps need to be followed are as follows;

1. Draw $XY$ that is equal to $AB+BC+CA$.

2. Then make $\angle LXY=\angle B$ and $\angle MYX=\angle C$.

3. Bisect the angles $\angle LXY$ and $\angle MYX$ such that the bisectors meet at point $A$.

4. Now, construct perpendicular bisectors $PQ$ and $RS$ of $AX$ and $AY$ respectively.

5. Suppose $PQ$ intersect $XY$ at $B$ and $RS$ intersect $XY$ at $C$. Then, join $AB$ and $AC$. Hence, we get the required $\Delta ABC$.