X Mat Ch 2 Full Notes (E)

2

Circles

Prepared by
Cecilia Joseph
St. John De Britto’s, A.I.H.S,
Fortkochi

Circles 1

Online Class – X − 13 19 / 07 / 2021
To view class
2 . Circles – Class 1

Previous Knowledge:

Basics of circle

• Fixing the pointed end of the compass at one spot and then
drawing the compass around we get a circle.
This fixed point is called the centre of the circle.

• The distance from the centre to the circle is Centre Radius (r)
called the radius(r) of a circle.
Radii of a circle are equal .

• Diameter(d) of a circle is the line joining Diameter (d)
two points on the circle through its centre.
Diameter is two times the radius .
d = 2r

Radius is half the diameter

r= d
2

• The line joining any 2 points on a circle
is a chord.

• We can draw many chords in a circle.
• Diameter is the longest chord of a circle.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 2

• An arc is a part of a circle. Arc of
a circle

• The angle formed by the line joining the Central angle
endpoints of an arc to its centre is known
as the central angle of that arc.



• The length of an arc of any circle depends on its central angle
and the radius of that circle .

600 900 1500
2500

•When the central angle is 180°, the arc 1800
will be a semicircle (half of a circle).

Sector

• A sector is said to be a part of a circle
made by the arc and its two radii.




Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 3

• Area of a circle = πr2 πd
• Perimeter of a circle = 2πr = πd πr2

where ‘ r ’ is the radius of the circle , r
π ≈ 3.14

Activity 1
Draw a right triangle of hypotenuse 5 centimetres .
Method 1
Draw a line 5 centimetres long. Draw a triangle by drawing
two angles at the ends of this line such that the sum of the
angles is 90° .

We can draw so many right triangles by changing the angles at
the ends of the line such that their sum is 900 .
Some other pair of angles whose sum is 90° are
(20° ,70° ), (45° ,45° ),(15° ,75° ),(35° ,55° ) etc

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 4

Method 2
We can also use a set square to draw this.
Draw 5cm long line .
Then place a set square with the
right angle on top and edges
passing through the ends of the line.

By adjusting the set square we can draw so many right
triangles. We can draw at the bottom of the line also .

Drawing several such triangles and looking at the third
vertices we can see all of them lie on a circle.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 5

Previous Knowledge

• If two sides of a triangle are equal, then the
angles opposite to these sides are also equal.

• If two angles of a triangle are equal, then the
sides opposite to the equal angles are equal.

• A triangle with two sides equal, is called
an isosceles triangle.

•Triangles with two angles equal are also
isosceles.

Activity 2 A P
In the figure , AB is a diameter and P OB
is a point on the circle .
Prove that the angle at P
is 90°.

Ans) P

Join P to the centre O of the circle. Line PO splits
∠P into two and △ APB into two triangles.

Let ∠APO = x° & ∠ BPO = y° O B

OA = OB = OP ( Radii of a circle are equal )

A

△ AOP is an isosceles triangle (OA = OP )
So ∠ APO = ∠ A = x°

△ BOP is an isosceles triangle (OB = OP )
So ∠ BPO = ∠ B = y°

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 6

In triangle APB
∠A + ∠B + ∠ APB = 180°

( Sum of the angles of a triangle is 180° )

x° + y° + ( x° + y° ) = 180°

2 x° + 2 y° = 180°

2( x° + y°) = 180°
180 °
x° + y° = 2 = 90°

If we join the ends of a diameter of a circle to a point
on the circle , we get a right angle.
or
Angle in a semicircle is right.

Assignment
T. B Page 42

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 1

Online Class – X − 14 22 / 07 / 2021
To view class
2 . Circles – Class 2

Assignment Answer

Ans)

Observe the picture carefully, we can see that the right angled

vertex of the set square is on the circle.

The length of the perpendicular sides are 3cm & 6 cm.

When we remove the set square from the circle, we get a figure

as given below . C

When we join the ends A ,B we get a A B

chord AB .
We know ∠C = 90°.
Since ∠C = 90° , AB is the diameter of the

circle.
Consider ∆ ACB, it is a right triangle .

By Pythagoras theorem
AB2 = AC2 + BC2
= 32 + 62 = 9 + 36 = 45

AB = √45 = 6.71

Diameter (d) = 6 .71 cm

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 2

So, Radius (r) = 6.71 = 3.36 cm π = 3.14
2

Area of circle = πr2 = 3.14 × 3.36 × 3.36 = 35.45 cm2

Perimeter of circle = πd = 3.14 × 6.71 = 21.07 cm.

Previous Knowledge

• In any isosceles triangle the perpendicular
from the point joining equal sides to the
opposite side bisects the angle at this point
and the side opposite.

• The perpendicular from the centre of a
circle to a chord bisects the chord.
The line joining the centre of a circle
and the midpoint of a chord is
perpendicular to the chord.

• A line which divides two sides of a x
triangle in the same ratio is parallel to 2x
the third side.
The length of the line joining the
midpoints of two sides of a triangle is
half the length of the third side.

• Outer angle at any vertex of a triangle x°

is the sum of the inner angles at other y° x°+y°
two vertices.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 3

Activity 1
Draw a 5cm long line .Draw a circle through the ends of this
line . Mark some points inside the circle. Join these points to
the ends of the diameter. Measure these angles .What is the
peculiarity of these angles ?
Ans)
We can see that all of these angles formed
inside the circle are greater than 90° .

Proof :

AB is the diameter of the circle and

P is a point inside the circle.

Join AP and BP.

Extend AP to meet the circle at Q. Join QB. Q

Since angle in a semicircle is 90°, P
∠PQB = 90°

Consider ∆ BQP.

∠APB is an outer angle at P of ∆ BQP. A B

So,

∠APB = ∠PQB + ∠QBP

= 90° + ∠QBP

which means ∠APB is greater than 90° ( ∠APB > 90°)

If we join the ends of the > 90°
diameter of a circle to a
point inside the circle
gives an angle greater than 90°

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 4

Activity 2
Draw a 5 cm long line . Draw a circle through the ends of this
line . Mark some points outside the circle. Join these points
to the ends of the diameter. Measure these angles .What is the
peculiarity of these angles ?
Ans)
We can see that all of these angles formed
outside the circle are less than 90° .

Proof

AB is the diameter of the circle and

P is a point outside the circle.

Join AP and BP. P
B
Q is a point on the circle where AP Q

meet the circle. Join BQ.

Since angle in a semicircle is 90°,

∠AQB = 90°

Consider the ∆ BQP. A

∠AQB is an outer angle at Q of ∆ BQP.

So,

∠AQB = ∠QPB + ∠PBQ

90° = ∠QPB + ∠PBQ

which means ∠QPB is less than 90° ( ∠QPB < 90°)

If we join the ends of the < 90°
diameter of a circle to a
point outside the circle
gives an angle less than 90°

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 5

Findings

The angle formed by joining the ends of diameter of a

circle to a point

a) inside the circle is greater than 90° (>90°) < 90°

b) on the circle is 90° 90°
c) outside the circle is less than 90° (< 90°) > 90°

If the lines drawn from the ends of a diameter
of a circle are perpendicular to each other,

then they meet on the circle

Assignment
T.B Page 42

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 1

Online Class – X − 15 23 / 07 / 2021
To view class
2 . Circles – Class 3

Assignment Answers

Ans)
Angle given in the first triangle =110°
As 110° > 90° the top corner will be inside the circle .
Angle given in the second triangle = 90°
So the top corner will be on the circle.
Angle given in the the third triangle = 70°
As 70° < 90° the top corner will be outside the circle .

Ans)
We Know that Sum of angles of a quadrilateral is 360°.
So , ∠ A + ∠ B + ∠ C + ∠D = 360°
Given ∠A = 105° , ∠ B = 55° , ∠C = 110°

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 2

105° + 55° + 110° + ∠D = 360°
∴ ∠D = 360° − (105° + 55° + 110°)

= 360° − 270°

= 90°

Drawing diagonal AC and taking it

as diameter of a circle.
As, ∠D = 90° , D will be on the circle.
∠B = 55° ( 55°< 90° )
∴ B will be outside the circle

Drawing diagonal BD and taking it

as diameter of a circle.
∠A = 105°, ∠C = 110°

As both angles are greater than 90°, they lie inside the circle.

T.B Page 42

Ans)
In this figure, A is the point where
the two circles meet.
Join CD & BE .
∠ ADC = 90° & ∠AEB = 90°
( Angle in a semicircle is 90° )
We know a perpendicular drawn
from the centre of a chord bisects
the chord .
Since C is the centre and CD is perpendicular to the chord AE

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 3

D is the midpoint of AE.
That is chord AE is bisected by the small circle.
Any chord drawn from A to the larger circle is cut by the small
circle.
So any chord of the larger circle through the point where the
circles meet is bisected by the smaller circle.

Ans)

i)Given , AQ is the diameter of the big circle &

AP is the diameter of the small circle.

Join BP, BA & BQ .
∠ ABP = ∠ ABQ = 90°

(angle in a semicircle is 90° )

Since

∠ABP + ∠ ABQ = 180° 90° 90°
(Linear Pair),

P, B, Q fall on a straight line.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 4

ii)Let ‘ X ’ & ‘ Y ’ be the centres of the small circle & the big
circle respectively.
Consider ∆PAQ,
X is the midpoint of side AP and Y is the midpoint of side AQ.
Join XY.
We know , the line joining the midpoints of any two sides of a
triangle is parallel to the third side and is half of the third
side.

∴ XY is parallel to PQ and XY is half of PQ or PQ = 2XY

Ans)
Here, ∆ABC is an isosceles triangle,

AB = AC
AD is the perpendicular
drawn from the vertex
joining the equal sides .
We know,
The perpendicular drawn from
the vertex joining equal sides of
an isosceles triangle pass trough the
midpoint of the third side.

So, D is the midpoint of BC .

Since∠ADB = 90° & ∠ADC = 90° , If we draw a circles with AB
and AC as diameter, that circles will pass through D.

That is, the two circles drawn on the two equal sides of an
isosceles triangle as diameters pass through the mid point of
the third side.

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 5

Ans) O
i) In the figure, ABCD is a rhombus.

In a rhombus;
• all sides are equal,
• opposite angles are equal
• diagonals are perpendicular

bisectors.
So AB = BC = CD = DA
Draw diagonals AC & BD
‘O’ is the midpoint of AC and BD.

∆ABD and ∆ CBD are isosceles triangles.
We know that the two circles drawn on the two equal sides of
an isosceles triangle as diameters pass through the mid point
of the third side.

∴ if we draw circles with diameters AB, AD, BC and CD ,
that circles pass through the common point ‘O’ .

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 6

ii) In the quadrilateral ABCD, O
adjacent sides are equal.
That is AB = AD and CB = CD.
Draw diagonals AC and BD.
∆ABD and ∆ CBD are isosceles
triangles.

We know that the two circles drawn on the two equal sides
of an isosceles triangle as diameters pass through the mid
point of the third side.
∴ If we draw circles with diameters AB and AD ,
that circles pass through ‘O’ . If we draw circles with
diameters CB and CD; that circles pass through ‘O’ .

Ans) 1
2
Area of △ ABC = ×b×h C

= 1 × 2r × h II
2
I
= rh A h

Area of semicircle drawn = πr2 B
with AB as diameter 2

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 7

Area of semicircle drawn = π r 2
with AC as diameter 1

2

Area of semicircle drawn π r 2
with BC as diameter = 2

Area of shaded part I & II = 2
=
Area of semicircle Area of
△ ABC
drawn with AB as −

diameter
πr2
2 − rh

Area of blue crescent − Area of red crescent

= Area of semicircle + Area of semicircle Area of
drawn with AC as shaded part
drawn with BC as −

diameter diameter I & II

=π r 2 + π r 2 − πr2 − rh
1 2 2

2 2

= π r 2 + π r 2 − πr2 + rh
1 2 2

2 2

= π ( r12 + r22 ) − πr2 + rh Consider △ ABC,
2 2 By Pythagoras theorem

= π r2 − πr2 + rh (2 r) 2 = (2 r1)2 + (2 r2)2
2 2 4r 2 = 4 r12 + 4 r22
r 2 = r12 + r22
= rh

= Area of △ ABC

Cecilia Joseph, St. John De Britto’s A . I. H. S, Fortkochi

Circles 1

Online Class – X − 16 27 / 07 / 2021
To view class
2 . Circles – Class 4

Chord, angle and arc

A diameter of a circle divides it into
two equal parts .
Joining the ends of the diameter to a
point in any part gives a right angle.

If we draw a chord other than the Larger part
diameter then also the circle is divided Smaller part
into two parts, one part is larger
and other part is smaller.

Activity1
 Draw a circle with radius 5 cm & draw a chord which is

not a diameter.
 Mark 3 points on the larger part of the circle .
 Mark 3 points on the smaller part of the circle.

Join these points to the ends of the chord.
 Measure the angles .

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 2

We can observe that,
When we draw a non diametrical
chord in a circle , that chord divides
the circle into two parts.

All of the angles in the larger part
are equal .
All of the angles in the smaller part
are equal .

Relation between the central angle of a non diametrical
chord and the angles formed by joining the ends of the
chord to the points on the larger parts of the circle .

Draw a circle centred at O . Draw chord AB . Mark a point P
on the larger part of the circle . Join the ends of the chord to
the point P .The following situations may arise .

Case 1 Case 2 Case 3

Case 1 (The lines AP & BP may on the either sides of the centre )
Case 2 (The line AP may passes through the centre )
Case 3 ( The lines AP and BP may on the same side of the centre )

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 3

Case 1
When the lines AP & BP are on either sides of the centre.

Consider the following figure

Draw the radii OA, OB, OP .
OA = OB = OP (Radii of same

circle)
Let ∠ APO = x° , ∠ BPO = y°
∠P = x° + y°
c° Consider ∆AOP
∆AOP is an isosceles triangle
∠APO = ∠PAO = x°
∠AOP =180° − 2x°

Consider ∆BOP
∆BOP is an isosceles triangle
∠BPO = ∠PBO = y°
∠BOP =180° − 2y°

We know that the sum of the angles around a point = 360°
So ∠ AOP + ∠ BOP + ∠ AOB = 360°

Let ∠ AOB = c°
∴ 180° − 2x° + 180°− 2y° + c° = 360°

360° − 2x° − 2y° + c° = 360° < P = < APO +< BPO
< P = x° + y°
c° = 2x° + 2y°

c° = 2 ( x° + y° )

c° = 2 ∠P
c°
∴ ∠P = 2

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 4

Case 2
When the line AP passes through the centre of the circle

Consider the following figure
Let ∠APB = xo

Draw OB
Consider ∆ OPB

OB = OP (Radii of same circle)

∆ OPB is an isosceles triangle.

∠OPB = ∠OBP = xo

∠POB = 180o − 2xo

180o− 2xo + co = 180o ( Linear pair )

co = 2xo

co = 2∠P
c°
∴ ∠P = 2

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 5

Case 3
When the lines AP and BP are on the same side of the centre.

Draw the lines OA , OB and OP.

OA = OB = OP ( Radii of same

circle )
Let ∠APO = xo & ∠BPO = y°
∠APB = ∠BPO − ∠APO

= y° − xo

∆ AOP is an isosceles triangle ( OA = OP)
Since OA = OP , ∠ OAP = ∠ OPA = xo

∠ AOP = 1800 − ( xo + xo ) = 1800 − 2xo
∆ BOP is an isosceles triangle ( OB = OP)
Since OB = OP , ∠ OBP = ∠ OPB = yo

∠ BOP = 1800 − ( yo + yo ) = 1800 – 2yo

Let ∠AOB = co

co = 180° − 2 xo − ( 180° − 2 yo )

= 180° − 2 xo − 180° + 2 yo

= 2 yo − 2 xo

= 2 ( yo − xo )

co = 2 × ∠APB
c°
∠APB = 2

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 6

From the above 3 cases we can arrive at a conclusion as
follows .

Conclusion

If we joining the ends of a non diametrical chord to
any point on the larger part of the circle ,

we get an angle which is half the size of the angle ,
we get by joining them to the centre of the circle .

Assignment
What is the relation among the angles formed by joining the
ends of a non diametrical chord to the points on the larger
and smaller parts of the circle ?

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 1

Online Class – X − 17 29 / 07 / 2021

2 . Circles – Class 5

To view class

Relation between the central angle of a non diametrical chord
and the angles formed by joining the ends of the chord to the
points on the smaller part of the circle .

Consider the given figure

Draw OA, OB, OQ
Let ∠AQO = x° & ∠BQO = y°

∠Q = x° + y°

Consider ∆ AOQ,

∆AOQ is an isosceles triangle (OA = OQ)

∠AQO =∠QAO = x°

∠AOQ =180° − ( x° + x°)

=180° − 2x°

Consider ∆ BOQ,

∆BOQ is an isosceles triangle (OB = OQ)

∠BQO =∠QBO = y°

∠BOQ =180° − ( y° + y°)

=180° − 2y°

∠AOB = ∠AOQ + ∠BOQ

Let ∠AOB = c°

So, c° = 180° − 2x°+ 180° − 2y°

= 360° − 2( x° + y° )

= 360° − 2∠Q

2∠Q = 360° − c°

∠Q = 360 °−c ° = 180° − c°
2 2

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 2

Looking at the angles in the two parts of the circle and the
angle at the centre together we have,

Any chord which is not a diameter splits the
circle into unequal parts.

The angle got by joining any point on the larger part
to the ends of the chord is

half the angle got by joining the centre of the circle
to these ends.

The angle got by joining any point on the smaller part

to the ends of the chord is
half the angle at the centre subtracted from 180 o .

Q) If the chord AB makes an angle 140° at the centre of the

circle find ∠APB & ∠ AQB.

Ans) ∠ AOB
∠APB = 2

= 140 °
2

= 70°

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 3

∠AQB = 180° − ∠ AOB
2
140 °
= 180° − 2

= 180° − 70°

= 110°

Putting the results obtained in terms of arcs and their central
angles

Any two points on a circle divide it
into two arcs.
Each of these two arcs can be called the
alternate arc or complementary arc
of the other.
In the figure the two arcs are
arc APB and arc AQB.
Alternate arc or complementary arc of arc APB is arc AQB
Alternate arc or complementary arc of arc AQB is arc APB

Central angle of an arc is the angle made by
the arc at the centre of the circle .
Let central angle of arc AQB = c°
and central angle of arc APB = d°

We know angle around a point is 360°

So, c° + d° = 360°

d° = 360° − c°
d° 360°−c °
2 = 2

d° = 180° − c°
2 2

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 4

Conclusion

The angle made by any arc of a circle
on the alternate arc is

half the angle made at the centre

Note:

Sum of the angles on an arc = ∠P +∠ Q
and its alternate arc c° c°
= 2 + 180° − 2

= 180°

Sum of the angles on an arc and its alternate arc is 180 o .

Pairs of angles of sum 180 o are usually called
supplementary angles.
Conclusion

All angles made by an arc on the alternate arc are equal
and

a pair of angles on an arc and its alternate arc are
supplementary

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 5
B
In the figure, AB is the diameter of the circle. P

Arc APB and arc AQB are semicircles.

Central angle of a semicircle is 180°.

∠P is the angle made by the arc AQB at its A

alternate arc APB and

∠Q is the angle made by the arc APB at its

alternate arc AQB

Q

∠P = 180 ° = 90°
2
180 °
∠Q = 2 = 90°

Angles in semicircle are right or 90°

Assignment
T.B Page 53

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 1

Online Class – X − 18 02 / 08 / 2021

2 . Circles – Class 6

To view class

Assignment Answers

T.B Page 53

Q1)

a) Join OA

Since OA = OB & OC = OA ,
∆ OAB & ∆ OAC are isosceles triangles

Given ∠ OBA = 20° , Given ∠ OCA = 30°

∴ ∠OAB = 20° ∴∠OAC = 30°

∠BAC = ∠OAB + ∠OAC

= 20° + 30° = 50°
∠BOC = 2 × 50° = 100°

Since OB = OC , ∆ OBC is an isosceles triangle.
180 °−100° 80 °
∴ ∠OBC = ∠OCB = 2 = 2 = 40°

Angles of ∆ ABC are ∠A = 50° , ∠B = 60°, ∠C = 70°
Angles of ∆ OBC are ∠OBC = 40°, ∠OCB = 40°, ∠BOC = 100°

b)

∆OAC is an isosceles triangle.

Given ∠ OAC= 40°

∴ ∠OCA = 40°

∠AOC= 180° − 80° = 100°
100 °
∴ ∠ABC= = 2 = 50°

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 2

Join OB, ∆ OBC is an isosceles triangle

So ∠ OCB = ∠OBC = 30°

∴ ∠OBA = 50° – 30° = 20°

∆ OBA is an isosceles triangle , So ∠ OAB= 20°

Angles of ∆ ABC are ∠ A = 60°, ∠B = 50°, ∠C = 70°

Angles of ∆ OBC are ∠OBC = 30°, ∠OCA = 30°, ∠BOC = 180° – 60°

=120°

c) Given ∠BOC = 70°

∆ OBC is an isosceles triangle
180 °−70 °
∠ OBC = ∠ OCB = 2

= 110 ° = 55°
2

Angles of ∆OBC are ∠OBC = 55° ,

∠BOC = 70°, ∠OCB = 55°

Since ∠AOC = 40° , ∠ABC = 40 ° = 20°
2
70 °
Since ∠BOC = 70° , ∠BAC = 2 = 35°

∠ACB = 180° − (20° + 35°) = 180° − 55° = 125°

Angles of ∆ABC are 125°, 20°, 35°

T B Page 53
Q2)

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles In a clock’s face 3
O
Ans)
a)

60 minute = 3600
360 °
1 minute = 60 ° = 6°

5 minute = 30°

∠ BOC = 4 × 30° = 120°
120 °
∴∠A= 2 = 60°

∠ AOC = 5 × 30° = 150°
150 °
∴∠ B = 2 = 75°

∠ AOB = 3 × 30° = 90°
90 °
∴∠C= 2 = 45°

b)

We can make 4 equilateral triangles by joining the numbers

on the clock

(1 , 5, 9), (2, 6, 10), (3, 7, 11), (4, 8, 12)

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 4

Assignment
Q) In the figure O is the centre of the

circle and ABC is an equilateral
triangle.
Find ∠BAC and ∠ABO.

T.B Page 54

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 1

Online Class – X − 19 04 / 08 / 2021

2 . Circles – Class 7

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Angles formed by an arc on the circle are
1. Central angle of an arc
2. Angle on the alternate arc
3. Angle within the arc

Assignment Answer

T.B Page 54

Q5)

Ans) Join OC
Let ∠ABC = x0 and ∠OAC = y0

Then ∠AOC = 2x0
and ∠OCA = y0

In ∆AOC,

2x0 + y0 + y0 = 1800

2x0 + 2y0 = 1800

2 ( x0 + y0 ) = 1800

∴ x0 + y0 = 180 ° = 900
2

So, ∠OAC + ∠ABC = 900

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 2

Construction 1 1
2
a) Draw an angle of size 22 0

Ans)

Steps

Step1. Draw circle of radius 3cm ( Any radius can be taken)

Step2. Draw radius OA ( Double of 22 1 0 is 450 )
Step3. Measure 450 angle at O 2

Step4. Draw OB

Step5. Extend AO to meet the circle at C
1
Step6. Join CB , we get ∠BCO as 22 2 0

b) Draw an angle of size 11 1 0.
4

Ans)

Steps

Step1. Draw circle of radius 3cm ( Any radius can be taken)

Step2. Draw radius OA (Double of 22 1 0 is 450 )
Step3. Measure 450 angle at O 2

Step4. Draw OB

Step5. Extend AO to meet the circle at C

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 3

Step6. Join CB , we get ∠BCO as 22 1 0
2

Step7. With C as centre draw circle of radius 3cm .

Step8. Extend OC to meet the circle at D
1
Step9. Join DE , we get ∠EDC as 11 4 0

Construction 2

a) Draw a triangle of circum radius 3 centimetres and two of

the angles 50 o and 60 o .

Ans)

Steps
Step1. Draw circle of radius 3cm

Step2. Draw radius OC

Step3. Measure1000 angle at O ( Double of 500 is 1000 )

Step4. Draw OB

Step5. Measure 1200 angle at O ( Double of 600 is 1200 )

Step6. Draw OA

Step7. Join AB, BC, AC

Step8.We get ∆ ABC with ∠A = 500 , ∠C = 600

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 4

b) Draw a triangle of circumradius 3 centimetres and two of the
1 1
angles 57 2 0 and 62 2 0.

Ans)

Double of 57 1 o is 1150
2

Double of 62 1 o is 1250
2

T B Page 53
Q3)

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 5

Ans)
i) Given all angles on one part 80°.
So the central angle of arc AB is double of 800 = 1600

Draw a circle .

A Draw arc AB with central
angle 160°.

B All angles on one part of arc AB
will be 80° .

ii) Given all angles on one part 110°
If angle on an arc is 110° , then
angle on its alternate arc is 180° − 110° = 70°
Central angle of arc AB is 2 × 70° = 140°

A Draw a circle .
Draw arc AB with central
angle 140°.

B All angles on one part of the arc
AB will be 70° and
All angles on the other part will
be 110°.

iii) Given all angles on one part is half of all the angles on the
other part.
So, if angle on one part is x° ,then angle on other part is 2x°.
We have x° + 2x° = 180°
3x° = 180°

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 6

x° = 180 ° = 60°
3

Angle on one part is 60° .

So central angle of arc AB is 2 × 60° = 120°

If angle on an arc is 60° , then

angle on its alternate arc is 120° .

A

Draw a circle .
Draw arc AB with central
angle 120°.
B All angles on one part of the arc
will be 60° and
All angles on the other part will
be 120°.

iv) Given all angles on one part is one and a half times the

angles on the other part.

So , if angle on one part is x° , 1 3
2 2
then angle on the other part is 1 x° = x°

We have x° + 3 x° = 180°
2
5x
2 = 180°

5x° =180° × 2 = 360°
360 °
x° = 5 = 72°

Angle on one part is 72° .

So central angle of arc is 2 × 72° = 144°

If angle on an arc is 72° , then

angle on its alternate arc is 180°− 72° = 108°

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 7
A
Draw a circle .
Draw arc AB with central
angle 144°.
B All angles on one part of the
arc will be 72° and
All angles on the other part
will be 108°.

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 1

Online Class – X − 20 06/ 08 / 2021
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Circle and Quadrilateral
Finding the relation between the angles of a quadrilateral
if the four vertices are on a circle

Consider quadrilateral ABCD.
Draw diagonals AC & BD

Diagonals AC & BD are chords of the circle .

Since a pair of angles on an arc and its alternate arc are

supplementary , & ∠A +∠C =180°
we get , ∠B +∠D =180°

Conclusion:

If all the four vertices of a

quadrilateral are on a circle, then its

opposite angles are supplementary.
∠B +∠D =180°
∠A +∠C =180°

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 2

To check ,if the opposite angles of a quadrilateral are
supplementary, then all its vertices are on a circle

Consider quadrilateral ABCD.

We can draw a circle passing through three of the vertices
A, B, C .

Then fourth vertex D may be
(i) Outside the circle
(ii) Inside the circle
(iii) On the circle

(i) (ii) (iii)

Case (i) When the fourth vertex D is outside the circle
Let CD intersect the circle at E.
Join AE
Consider quadrilateral ABCE
∠B + ∠AEC =180°. . . . . . . . . . 1
Consider △ AED
∠AEC = ∠EAD + ∠D
So, ∠D < ∠AEC . . . . . . . . . . . 2

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 3

From 1 & 2 ∠B + ∠D < 180°

Case (ii) When the fourth vertex D is inside the circle
Extend CD to meet the circle at E

Join AE
Consider quadrilateral ABCE
∠B + ∠E = 180° . . . . . . . . 1

Consider △ AED
∠ADC = ∠E + ∠EAD
So, ∠ADC > ∠E . . . . . . . . 2

From 1 & 2 ∠B + ∠ADC > 180°

i.e, ∠B + ∠D > 180°

Case (iii)

From case(i) & case(ii) we have seen,

When the fourth vertex D is outside the circle then
∠B + ∠D <180°

When the fourth vertex D is inside the circle then
∠B + ∠D >180°

So,
If ∠B + ∠D = 180°

then , ∠D must be on the circle

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 4

If the opposite angles of a quadrilateral are supplementary,
we can draw a circle passing through all four of its vertices.

Conclusion

If vertex D of quadrilateral ABCD is ,

(i) (ii) (iii)
Outside the circle Inside the circle On the circle

drawn through the drawn through the drawn through the

other three veritces, other three veritces, other three veritces,

then then then

∠B + ∠D < 180° ∠B + ∠D > 180° ∠B + ∠D = 180°

If all four vertices of a quadrilateral are on a
circle,

then its opposite angles are supplementary,

If the opposite angles of a quadrilateral are
supplementary,

then all its vertices are on a circle .

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 5

If the opposite angles of a quadrilateral
are supplementary, we can draw a circle
passing through all four of its vertices.

This quadrilateral can be called as a
Cyclic Quadrilateral

Cyclic quadrilaterals are those quadrilaterals with
opposite angles supplementary.

* All rectangles are cyclic quadrilaterals

Assigmnent
Q) ABCD is an isosceles trapezium. Check whether it is a

cyclic quadrilateral.

DC

AB

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

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Assignment Answer
Since ABCD is an isosceles trapezium
AD = BC and also
AB is parallel to DC .

We have to prove ∠ A+ ∠ C = 180° & ∠ B + ∠ D =180°
Since ABCD is an isosceles trapezium,
∠ A = ∠ B . . . . . . . . (1)
Since AB // DC
∠ A + ∠ D = 180° . . . . . . . . (2) ( Co-interior angles are
supplementary)
From (1) & (2) we have
∠ B + ∠ D = 180°
Also, ∠ A + ∠ C = 180°
Since the opposite angles are supplementary , ABCD is cyclic.

T. B Page 59
Q1)

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 2

Ans) Given , ∠BDC = 50° All angles made by an arc on the
So , ∠BAC = 50° alternate arc are equal

Given , ∠ACD = 30°
So, ∠ABD = 30°

Given , ∠ CBD = 45°
So, ∠CAD = 45°

Consider △ ABC,
∠ACB = 180° − ( 50° + 75° ) = 180° − 125° = 55°

So ∠ ADB = 55°

Consider △ AOD , ∠ AOD = 180° − ( 55° + 45° ) = 180° – 100° = 80°
∠ DOC = 180° − 80° = 100° (Linear pair)

∠ AOD = ∠ BOC = 80° ( opposite angles )
∠ DOC = ∠ AOB = 100° ( opposite angles )

∴ Angles of the quadrilateral are Angles between diagonals are

∠ A = 45° + 50° = 95° ∠ AOD = ∠ BOC = 80°

∠ B = 30° + 45° = 75° ∠ DOC = ∠ AOB = 100°

∠ C = 55° + 30° = 85°

∠ D = 50° + 55° = 105°

Assignment
Q1) In the figure PQRS is an isosceles trapezium

and QR is extended to X .
If < SRX = 100o ,
find all angles of PQRS ?

Q2) Prove that any non- isosceles trapezium is not cyclic ?

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

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Cyclic quadrilaterals are those quadrilaterals with opposite
angles supplementary.

Quadrilaterals which are always cyclic are
(i) Square
(i) Rectangle
(iii) Isosceles Trapezium

Assignment Answer
(Q) Prove that any non- isosceles trapezium is not cyclic ?
Ans)

Given ABCD is a non- isosceles trapezium
So ∠A ≠ ∠B . . . . . . . . 1

Since ABCD is a trapezium, AB // CD.

So ∠A +∠D = 180° . . . . . 2 ( Co-interior angles are

supplementary)

From 1 & 2 we have ∠B +∠D ≠ 180°

Since the opposite angles are not supplementary ,

ABCD is not cyclic.

That is , any non- isosceles trapezium is not cyclic .

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 2

T .B page 59
Q2) Prove that any outer angle of a cyclic quadrilateral is equal

to the inner angle at the opposite vertex.
Ans)

In the figure ∠CBE is an outer angle at a vertex of
cyclic quadrilateral ABCD .
We have to prove ∠CBE =∠ADC

Since ABCD is cyclic
∠ABC + ∠ADC = 180° . . . . . . . . .1

Also
∠ABC +∠CBE = 180° . . . . . . . . 2

(Linear pair)

From 1 & 2 we have
∠ABC + ∠ADC = ∠ABC +∠CBE
∴ ∠ADC = ∠CBE

That is any outer angle of a cyclic quadrilateral is equal

to the inner angle at the opposite vertex .

Any outer angle of a cyclic quadrilateral
is equal to

the inner angle at the opposite vertex.
∠ADC = ∠CBE
∠ABC = ∠GDA
∠DAB = ∠DCF
∠HAB = ∠DCB

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi