This GSEB Class 9 Maths Notes Chapter 11 Circles covers all the important topics and concepts as mentioned in the chapter.

## Circles Class 9 GSEB Notes

To draw precise figures, one must have a geometry box that contains the following:

- A graduated scale: On one side of the graduated scale centimetres and millimetres are marked off and on the. other side inches and their parts are marked off.
- A pair of set-squares: One set-square with angles 90Â°, 60Â° and 30Â° and other set-square with angles 90Â°, 45Â° and 45Â°.
- A pair of dividers (or a divides): A divider has two pointed ends which can be adjusted to compare different lengths.
- A pair of compasses (or a compass): A compass has provision of fitting a pencil at one end to draw desired circle.
- A protractor

The process of drawing a geometrical figure using only two instruments – an ungraduated ruler (straight edge) and a compass is called geometrical construction. In construction where measurements are also required a graduated scale as well as a protractor can be used.

In the first part of these chapter, we will study some basic constructions as listed below. We will use only a straight edge and a compass for these constructions.

- Construction of the bisector of a given angle.
- Construction of the perpendicular bisector of a given line segment.
- Construction of an angle of measure 60Â°.
- Construction of angles with measures in multiples of 15Â°.

We will also justify the validity of certain constructions.

Example 1:

To construct the bisector of a given angle.

Answer:

Given an angle ABC, we want to construct its bisector.

Steps of construction:

- Taking B as centre and any radius, draw an arc to intersect the rays BA and BC, say at E and D respectively [see figure (i )]
- Taking D and E as centres and with the radius more than DE, draw arcs to intersect each other, say at E
- Draw the ray BF [see figure (ii)].

This ray BF is the required bisector of the angle ABC.

Justification:

Join DF and EF.

In A BEF and A BDF,

BE = BD {Radii of the same arc)

EF = DF (Arcs of equal radii)

BF = BF (Common)

âˆ´ âˆ†BEF = âˆ†BDF (SSS rule)

âˆ´ âˆ EBF = âˆ DBF (CPCT)

âˆ´ âˆ ABF = âˆ CBF

Example 2:

To construct the perpendicular bisector of a given line segment.

Answer:

Given a line segment AB, we want to construct , its perpendicular bisector.

Steps of construction:

- Taking A and B as centres and radius more than AB, draw arcs on both sides of the line segment AB (to intersect each other).
- Let these arcs intersect each other at P and Q. Join PQ (see figure).
- Let PQ intersect AB at the point M. Then line PMQ is the required perpendicular bisector of AB.

Justification:

Join A and B to both P and Q to form AR AQ, BP and BQ.

In âˆ†PAQ and âˆ†PBQ,

AP = BP (Arcs of equal radii)

AQ = BQ (Arcs of equal radii)

PQ = PQ (Common)

âˆ´ âˆ†PAQ â‰… âˆ†PBQ (SSS rule)

âˆ´ âˆ APQ = âˆ BPQ (CPCT)

âˆ´ âˆ APM = âˆ BPM

Now in âˆ†PMA and âˆ†PMB,

AP = BP (As before)

PM = PM (Common)

âˆ APM = âˆ BPM (Proved above)

âˆ´ âˆ†PMA â‰… âˆ†PMB (SAS rule)

âˆ´ AM = BM and âˆ PMA = âˆ PMB (CPCT)

But, âˆ PMA + âˆ PMB = 180Â° (Linear pair)

âˆ´ âˆ PMA = âˆ PMB = \(\frac{180^{\circ}}{2}\) = 90Â°.

Therefore, PM, that is, PMQ is the perpendicular bisector of AB.

Example 3:

To construct an angle of 60Â° at the initial point of a given ray.

Answer:

Let us take a ray AB with initial point A [see figure (i)]. We want to construct a ray AC such that âˆ CAB = 60Â°.

Steps of construction:

1. Taking A as centre and some radius, draw an arc of a circle, which intersects AB, say at a point D.

2. Taking D as centre and with the same radius as before, draw an arc intersecting the previously drawn arc, say at a point E.

3. Draw the ray AC passing through E [see figure (ii)].

Then âˆ CAB is the required angle of 60Â°.

Justification:

Join DE.

Then, AE = AD = DE (By construction)

Therefore, âˆ†EAD is an equilateral triangle and the âˆ EAD, which is the same as âˆ CAB is equal to 60Â°.

**Some constructions of triangles:**

According to the rules of congruence viz. SAS, SSS, ASA (or AAS) and RHS rules if-

- two sides and the included angle is given,
- three sides are given,
- two angles and the included side (or anyone side) is given
- in a right triangle, the hypotenuse and one side is given, a unique triangle can be constructed.

Now, we will study some more constructions of triangle as mentioned below:

- Construction of a triangle, given its base, a base angle and the sum of other two sides.
- Construction of a triangle, given its base, a base angle and the difference of other two sides.
- Construction of a triangle, given its perimeter and its two base angles.

Example 1:

Construction 11.4:

To construct a triangle, given its base, a base angle and sum of other two sides.

Answer:

Given the base BC, a base angle, say âˆ B and the sum AB + AC of the other two sides of a triangle ABC, you are required to construct

it.

Steps of construction:

- Draw the base BC and at the point B make an angle, say XBC equal to the given angle.
- Cut a line segment BD equal to AB + AC from the ray BX.
- Join DC and make an angle DCY equal to âˆ BDC.
- Let CY intersect BX at A (see the figure).

Then, ABC is the required triangle.

Justification:

Base BC and Z B are drawn sis given.

Next in AACD,

âˆ ACD = âˆ ADC (By construction)

âˆ´ AC = AD

Then, AB = BD – AD = BD – AC

âˆ´ AB + AC = BD

Alternative method:

Follow the first two steps as above. Then draw perpendicular bisector PQ of CD to intersect BD at a point A (see the figure). Join AC. Then, ABC is the required triangle.

Note that A lies on the perpendicular bisector of CD, therefore AD = AC.

Remark: The construction of the triangle is not possible if the stun AB + AC = BC.

Example 2:

To construct a triangle given its base, a base angle and the difference of the other two sides.

Answer:

Given the base BC, a base angle, say Z B and the difference of other two sides AB – AC or AC – AB, you have to construct the triangle ABC.

Clearly there are following two cases:

Case 1: Let AB > AC, that is, AB – AC is given.

Steps of construction:

- Draw the base BC and at point B make an angle say XBC equal to the given angle.
- Cut the line segment BD equal to AB – AC from ray BX.
- Join DC and draw the perpendicular bisector, say PQ of DC.
- Let PQ intersect BX at a point A. Join < AC (see the figure).

Then, ABC is the required triangle.

Justification:

Base BC and âˆ B are drawn as given. The point A lies on the perpendicular bisector of DC.

âˆ´ AD = AC

BD = AB – AD = AB – AC

Case 2: Let AB < AC, that is, AC – AB is given.

Steps of construction:

- Same as in case 1.
- Cut line segment BD equal to AC – AB from the line BX extended on opposite; side of line segment BC.
- Join DC and draw the perpendicular bisector, say PQ of DC.
- Let PQ intersect BX at A. Join AC (see the figure).

Then, ABC is the required triangle.

We can justify the construction as in case 1.

Example 3:

Construction 11.6:

To construct a triangle, given its perimeter and its two base angles.

Answer:

Given the base angles, say Z B and Z C and BC + CA + AB, you have to construct the triangle ABC.

Steps of construction :

- Draw a line segment, say XY equal to BC + CA + AB.
- Make angles LXY equal to âˆ B and MYX equal to âˆ C.
- Bisect âˆ LXY and âˆ MYX. Let these bisectors intersect at a point A [see figure]

- Draw perpendicular bisectors PQ of AX and RS of AY.
- Let PQ intersect XY at B and RS intersect XY at C. Join AB and AC [see figure].

Justification:

B lies on the perpendicular bisector PQ of AX.

Therefore, XB = AB and similarly, CY = AC.

This gives, BC + CA + AB = BC + CY + XB = XY.

Again, âˆ BAX = âˆ AXB (As in âˆ†AXB, AB = XB) and âˆ ABC = âˆ BAX + âˆ AXB = 2âˆ AXB = âˆ LXY ,

Similarly, âˆ ACB = âˆ MYX as required.

Example 4:

Construct a triangle ABC, in which âˆ B = 60Â°, âˆ C = 45Â° and AB + BC + CA = 11 cm.

Answer:

Steps of construction:

- Draw a line segment PQ =11 cm. (PQ = AB + BC + CA).
- At P construct an angle of 60Â° and at Q, an angle of 45Â°.
- Bisect these angles. Let the bisectors of these angles intersect at a point A.
- Draw AP and AQ. Draw perpendicular bisectors DE of AP to intersect PQ at B and FG of AQ to intersect PQ at C.
- Join AB and AC (see the figure).

Then, ABC is the required triangle.