Math 10

b) Construct a parallelogram PQRS where PQ = 6.8 cm, QR= 5.9 cm, PR = 6.2 cm.
c) Also construct another parallelogram equal in area to the parallelogram PQRS,
d) whose one angle is 450.
e)
Construct a rhombus ABCD whose one side is 6.5 cm and one angle is 450.
f) Also construct another parallelogram equal in area to the rhombus ABCD,
whose one side is 7 cm.
2. a)
b) Construct a rectangle equal in area with parallelogram ABCD where,
c) BC = 5.2 cm, ∠ACB = 600 and AC = 5.6cm.

Construct a parallelogram PQRS where QR = 5.5 cm, diagonal PR = 6.4 cm,
diagonal QS = 6cm. Also construct another parallelogram equal in area with to
the parallelogram PQRS, whose one side is 5.9cm.

Construct a parallelogram ABCD where diagonal AC = 7cm, diagonal BD = 6cm
and angle between them is 600. Also construct a parallelogram equal in area with
to the parallelogram ABCD, whose one side is 6.8cm.

Construct a parallelogram PQRS where SR = 6.2 cm, QR = 5.6 cm, ∠PSR = 600.
Also construct a parallelogram SRTU where RT = 7cm.

Construct a parallelogram ABCD where AB = 5.4 cm, BC = 6cm, ∠ABC = 600.
Also construct a rectangle FBCE equal in area to the parallelogram ABCD.

Construct a parallelogram QXYP where ∠QXY = 300 equal in area to the
parallelogram WXYZ where WX = XY = WY = 6cm.

14.3 Construction of triangle equal in area to given parallelogram

Concept:

The area of a triangle is half of the area of parallelogram standing on the same base and
between the same parallel lines.

In other words, the area of a triangle is equal to the area of a parallelogram between the
same parallel lines if the base of the triangle is twice the base of the parallelogram.

Example: 1
Construct a rectangle whose length and breadth are respectively 5 cm and 3.5 cm. Then construct
a triangle which is equal to the area of the given rectangle, in which one of the sides is 8.5 cm.

A DQ Rough Sketch

A DQ

3.5cm 8.5cm 8.5cm
3.5cm
3.5cm

B 5cm C 5cm P 900 P

B 5cm C

Oasis School Mathematics-10 245

Steps:

(i) Draw a line BC = 5 cm.
(ii) Construct ∠B = 900 and cut AB = 3.5 cm.
(iii) Similarly, construct ∠C = 900 and cut CD = 3.5 cm.
(iv) Now, ABCD is a rectangle.
(v) Select the point P such that BC = CP i.e. BP = 2BC.
(vi) Cut an arc of 8.5 cm from B, to get Q.
(vii) Join BQ and QP.
∆QBP is the required triangle.

Analysis
Area of rectangle ABCD = Area of ∆QBP.
Base of ∆QBP = 2 (base of rectangle ABCD)

Exercise 14.2

1. a) Construct a parallelogram ABCD where AB = 6.5 cm, BC = 6 cm and ∠ABC = 600.
Also construct a triangle equal in area to the parallelogram ABCD, whose one
side is 7.6 cm.

b) Construct a parallelogram PQRS in which PQ = 5cm, diagonal PR = 6 cm and
QR = 5.6 cm. Also construct a triangle equal in area to the parallelogram PQRS,
whose one angle is 450.

c) Construct a right-angled triangle equal in area to a parallelogram whose two
diagonals are 6.8 cm and 7.4 cm, and the angle between the diagonals is 600.

2. a) Construct a parallelogram ABCD where BC = 6cm, AC = 6.4 cm, ∠ACB = 600. Also
construct a ∆FEC equal in area to the parallelogram ABCD, where FC = 6cm.

b) Construct a parallelogram PQRS, where PQ = 6.4 cm, QR = 5.6 cm and PR = 6cm.
Also construct a ∆SQT equal in area to the parallelogram PQRS.

c) Construct a parallelogram WXYZ, where XY = 5.8 cm, YZ = 6.5 cm,
∠XYZ = 1200. Also construct a ∆NXM equal in area to the parallelogram WXYZ
where ∠NXY = 450.


14.4 Construction of a triangle equal in area to given triangle

Concept:

Triangles on the same base and between the same parallel lines are equal in area.

246 Oasis School Mathematics-10

Example: 1
Construct a ∆ABC in which a = 6 cm, b = 5.5 cm and c = 7.5 cm. Construct a triangle equal to
∆ABC and having an angle of 60°.

PA Rough Sketch
A
PA

7.5cm 5.5cm7.5cm
5.5cm

600 C 600 6cm C
B B

6cm

Steps:
(i) Draw a line segment BC = 6 cm.
(ii) From B and C cut an arc length of 7.5 cm and 5.5 cm respectively to get point A.
(iii) Join AB and AC to get ABC.
(iv) From A draw a line parallel to BC.
(v) Draw an angle ∠PBC = 600 at B.
(vi) Join PC.
∆PBC is the required triangle.

Analysis
∆ABC and ∆PBC both lie on the same base BC and between the parallel lines PA and BC.

Exercise 14.3

1. (a) Construct a ∆ABC having AB = 6.2 cm, BC = 5.4 cm and AC = 6.5 cm. Also construct
a triangle equal in area to ABC, whose one angle is 600.

(b) Construct a ∆ABC in which a = 6.3 cm, ∠B = 300, ∠C = 600. Also construct a triangle
equal in area to a triangle, whose one side is 5.6 cm.

(c) Construct a ∆XYZ in which XY = 5.6cm, YZ = 6.2 cm, ∠XYZ = 600. Also construct
another triangle equal in area to XYZ, whose one side is 6.8 cm.

2. (a) Construct an equilateral triangle PQR having a side 6 cm. Also construct an
equivalent ∆SQR where SQ = 6.5 cm.



Oasis School Mathematics-10 247

(b) Construct a ∆ABC in which AB = 6 cm, BC = 6.8 cm, ∠ABC = 600. Also construct a
∆DBC where ∠DBC = 300.

(c) Construct a ∆GEF where ∠GEF = 900 equal in area to ∆DEF having
DE = EF = 6.8 cm and DF = 7.2 cm.



14.5 Construction of a parallelogram equal in area to given triangle

Concept:

The area of a triangle is half the area of a parallelogram standing on the same base and
between the same parallel lines.

Or, The area of a triangle is equal to the area of a parallelogram if they lie between the same
parallel lines and the base of the triangle is twice the base of the parallelogram.

Example: 1

Construct a triangle ABC in which a = 7.8 cm, b = 7.2 cm, c = 6.3 cm. Construct a parallelogram
equal in area to ∆ABC and having a side = 8.5 cm.

A P Q

7.2cm 8.5cm 8.5cm Rough Sketch

A PQ

6.3cm 7.2cm
6.3cm

8.5cm

B 7.8cmM C B MC

7.8cm

Steps:
(i) Draw a line segment BC = 7.8 cm.
(ii) Cut an arc of length AB = 6.3 cm and AC = 7.2 cm. respectively to get point A.
(iii) From A, draw a line parallel to BC.
(iv) Draw the perpendicular bisector of BC which meets BC at M.
(v) From M and C cut an arc 8.5 cm to get P and Q.
(vi) Join MP and CQ.
Hence, PMCQ is the required parallelogram.

Analysis
Area ∆ABC= area of parallelogram PMCQ. Base of ∆ABC = 2 (base of parallelogram PMCQ)

248 Oasis School Mathematics-10

Example: 2
Construct a triangle PQR in which PQ = 6.5 cm QR = 7.6 cm, PR = 6.9 cm. Construct a rectangle
equal in area to ∆PQR.

PN O

Rough Sketch

PN O

6.9cm 6.9cm
6.5cm
6.5cm

Q M R
R 7.6cm
Q 7.6cm M

Steps:
(i) Draw a line segment QR = 7.6 cm.
(ii) From Q and R cut an arc of length 6.5cm and 6.9 cm, to get point P.
(iii) Join PQ and PR.
(iv) Draw a perpendicular bisector of QR which meets QR at M.
(v) From P, draw a line parallel to QR.
(vi) Take an arc equal to MR and cut from N to get O.
(vii) Join OR.
Hence, MNOR is the required rectangle.

Exercise 14.4

1. (a) Construct a triangle ABC where AB = 5.7 cm, BC = 5.5 cm and ∠ABC = 600. Also
construct a parallelogram equal in area to ∆ABC whose one side is 6 cm.

(b) Construct a triangle XYZ where XY = YZ = 5.2cm and XZ = 5.9 cm. Also construct
a parallelogram equal in area to ∆XYZ whose one angle is 450.

(c) Construct a triangle PQR having PQ = 6.8 cm, ∠P = 600, ∠Q = 450. Also construct a
rectangle equal in area to given triangle.

2. (a) Construct a ∆ABC having AB = 6 cm, BC = 7cm and AC = 7.2 cm. Also construct a
parallelogram CFGH equal in area to the ∆ABC, where FG = 6.9 cm.

(b) Construct a right-angled isosceles triangle PQR where PQ = PR, QR = 6 cm and
∠QPR = 900. Also construct a parallelogram TSRU equal in area with ∆PQR where
RU = 5 cm.

(c) Construct a ∆ABC in which a = 6.5 cm, b = 7.5 cm, c = 6 cm. Reduce the triangle
ABC into an equivalent rectangle DEFC.

Oasis School Mathematics-10 249

14.6 Construction of a triangle equal in area to given quadrilateral

Concept: Triangles on the same base and between the same parallel lines are equal in area.

Example: 8
Construct a quadrilateral ABCD in which AB = 4.2 cm, BC = 4.8 cm, CD = 5.4 cm, DA = 5.8
cm and diagonal BD = 5.8 cm. Construct a triangle equal in area to the quadrilateral ABCD.

D

5.8cm Rough Sketch
A
D

5.8cm

A
5.4cm
4.2cm

5.4cm
4.2cm
5.8cm

5.8cm

E B 4.8cm C

E B 4.8cm C

Steps:
(i) Draw a line segment BC = 4.8 cm.
(ii) Cut an arc of 5.8 cm and an arc of 5.4 cm from B and C respectively to get the point of

intersection at D.
(iii) Cut an arc of length 4.2 cm and 5.8 cm respectively from B and D to get point A.
(iv) Join AD and AB, then ABCD is a quadrilateral.
(v) From A, draw AE//DB which meets CB produced at E.
(vii) Join DE.

∆DEC is the required triangle.

Analysis
∆EBD = ∆ABD ('Triangles on the same base BC and between the parallel lines AE and DB)
∆EBD + ∆DBC = ∆ABD + ∆DBC
∴ ∆DEC = quadrilateral ABCD

250 Oasis School Mathematics-10

Exercise 14.5

1. (a) Construct a quadrilateral ABCD where AB = 5.5 cm, BC = 5.1 cm, CD = 6cm,
AD = 4.9 cm and diagonal BD = 5.8 cm. Also construct a triangle equal in area
with quadrilateral ABCD.

(b) Construct a quadrilateral PQRS having PQ = 6 cm, PS = 6.4 cm, QR = 5.2 cm,
RS = 5.6 cm and ∠QPS = 600. Also construct a triangle equal in area with
quadrilateral PQRS.

(c) Construct a quadrilateral ABCD in which AB = 6.4 cm, BC = 5.5 cm, CD = 4.5
cm, ∠B = 1200 and ∠C = 600. Also construct a triangle equal in area to the given
quadrilateral.

2. (a) Construct a quadrilateral PQRS where PQ = QR = 5.2 cm PR = 5.5 cm,
PS = RS = 6.7 cm. Also construct a ∆PQT equal in area to the quadrilateral PQRS.

(b) Construct a quadrilateral ABCD having AB = 5.8 cm, AD = 5.2 cm, CD = 5.6 cm, BC=6
cm and ∠ADC = 600. Also construct a ∆BEC equal in area with quadrilateral ABCD.

(c) Construct a quadrilateral MNOP having MN = 5.6 cm, ON=6.2 cm,
OP = 5.8 cm, ∠MNO = 600, ∠PON = 1200. Also construct a ∆MNQ equal in area to
the quadrilateral MNOP.


14.7 Construction of a quadrilateral equal in area to given triangle

Concept: Triangles on the same base and between the same parallel lines are equal in area.

Example 1:
Construct a quadrilateral equal to the area of ∆ABC such that the sides of the triangle are
AB = 4 cm, BC = 4.5 cm and AC = 5 cm.

AQ Rough Sketch

AQ

D

D
5cm

5cm
4cm

4cm

B P 4.5cm C B P 4.5cm C

Oasis School Mathematics-10 251

Steps:
(i) Draw a line segment BC = 4.5 cm.
(ii) From B and C cut an arc of 4 cm and 5 cm respectively to get A.
(iii) Join AB and AC.
(iv) Choose any point P on BC and join AP.
(v) From C, CQ //PA.
(vi) Take any point D on CQ and join PD.
ABPD is the required quadrilateral.

Analysis
∆APD = ∆APC [Triangles on the same base and between the same parallel lines]
∆APD + ∆ABP = ∆ APC + ∆ABP
Quad. ABPD = ∆ABC.

Exercise 14.6

1. (a) Construct a ∆ABC in which ∠ABC = 600, AB = 6 cm, BC = 5.5 cm. Also construct a
quadrilateral equal in area to ∆ABC.

(b) Construct a ∆PQR in which PQ = 7 cm, QR = 6 cm and PR = 6.5 cm. Also construct
a quadrilateral equal in area to ∆PQR.

2. (a) Construct a ∆ABC in which AB = 7cm, ∠A = 600 and ∠B = 450. Also construct a
quadrilateral BDEC equal in area to ∆ABC where DE = 7.2 cm.

(b) Construct a ∆PQR in which PQ = 5.7 cm, QR = 5.6 cm. ∠Q = 600. Also construct a
quadrilateral PTSQ equal in area to ∆PQR where TS = 6cm.



Project Work

On a chart paper, draw a model of each type of construction mentioned above.

252 Oasis School Mathematics-10

Unit

15 Circle

15.1 Warm-up Activities

Discuss the following in your class and draw a conclusion.

• Draw a circle and show the following parts,

• Circumference • Radius • Diameter • Chord

• Arc • Semi circle • Sector • Segment

• If chord AB = 6cm, then find the length of AM and MB. What is the O

length of OB?

A M B

• If two chords are equidistant from the center, are they equal in

length?

• Draw a circle and show its different parts.

15.2 Circle

Basic definitions: S
OR
If a point moves on a plane in such a way that its distance from

a fixed point is always the same, the locus of the point is called a P

circle. It is denoted by . .

The fixed point which is equidistant from any point of the circle is Q
called the center of the circle. In the given figure, points P, Q, R and
S are equidistant from the point O, the center of the circle. Circumference

Circumference:

The total length of the rim of a circle is its circumference. In other words O

the perimeter of the circle is the circumference of the circle. Q

Radius:

A line segment which joins the center of a circle to any point of the Radius O
circumference is the radius. It is denoted by 'r'.
Q

Chord: Chord B

AO
A line segment which joins any two points of the circumference is a C
D
chord. AB and CD are chords.

Oasis School Mathematics-10 253

Diameter:

A chord which passes through the center of a circle is called the Diameter C

diameter, i.e., a diameter is the longest chord. AC is the diameter. A Chord D

Note: Diameter = 2 × Radius = 2r.

Arc: Major arc

Any part of the circumference of a circle is called an arc. It is denoted by C

' '. If the arc is less than half of the circumference, then it is called

minor arc and if it is more than half of the circumference, then it is a O

major arc.

In the given figure, AB is the minor arc and ACB the major arc. AB
Minor arc

Note: If a minor and a major arc together make up the whole circumference, then
they are called conjugate arcs.

Semi–circle: A

A diameter of a circle divides a circle into two equal parts, and each

part is called a semi–circle. D O B

In the figure, DAB and DCB are two semi–circles of the circle.

C

Sector:

A region enclosed by any two radii and an arc of a circle is major sector
called the sector of the circle. If the sector is less than a O
semi-circle, then it is a minor sector, and if it is more than a
semi-circle, then it is a major sector. PQ
minor sector

Segment: major
segment

A region enclosed by an arc and a chord of a circle is called the

segment of the circle. If the segment is less than half the circle then O

it is called a minor segment and if it is more than half the circle then A B
it is called a major segment.

minor segment

Concentric circles: OO
Two or more than two circles having the same centre are called
concentric circles, and each circle is called a family of circles. I
The portion between two circles is called the annulus. II
III annular region

(annulus)
254 Oasis School Mathematics-10

Line of centers:

The straight line joining the centers of two circles is called the O'
line of centers. In the adjoining figure, OO' is the line of centers. O

Intersecting circles:

When two circles intersect each other at two different points, they A

are said to be intersecting circles. The line which joins their points O O'
of intersection is their common chord.

In the given figure, two circles intersect at A and B, and AB is the B
common chord.

Concyclic points: A

B O
D
The points which lie on the circumference of the circle are called
concyclic points. C

In the figure, A, B, C and D are concyclic points.

Central angle: O

An angle at the center of a circle is central angle. The centre is Central
angle
the vertex for the angle. Here, AOB is the central angle.
A B

Some points to remember about a circle: P

1. A line cannot intersect a circle at more than two points.

2 The radii of the same circle or equal circles are equal.

3. Two circles are equal if their radii are equal.
4. The greatest chord of a circle is its diameter.

5. In general, the sector of a circle is referred to the area of this region.

Some important theorems related to the circle:

1. A perpendicular drawn from the center of a circle to a chord bisects the chord.

2. A line, passing through the center of a circle and bisecting a chord, is perpendicular to
the chord.

3. Equal chords of a circle are equidistant from the center.

4. Chords which are equidistant from the center of a circle are equal.

Note: The theorems mentioned above have already been explained in Grade IX.

Central angle and inscribed angles: O B
P
Any angle whose vertex is at the center of a circle is called a central angle Q
(i.e., angle at the center of a circle). In the figure alongside, ∠AOB standing A

on the APB and reflex ∠AOB standing on the arc AQB are the central angle
with the center at O.

Oasis School Mathematics-10 255

Any angle subtended by an arc or chord of a circle at its circumference (i.e. B
angles at the circumference of a circle) is called an inscribed angle. In the figure

alongside, ∠ABC is an inscribed angle subtended by the arc AC.

Theorem related to arcs and angle subtended by them A O

C

Theorem 15.1

In a circle, arcs that subtend equal angles at the centre are equal.

Experimental Verification:
• Draw two circles of different radii with centre O.
• Draw two radii OA and OC in each circle.
• Draw two equal angles AOB and COD at point O.

B

AO AO
C
BD
C D

To verify : AB = CD Fig. I Fig. II

Measure the lengths of AB and CD with the help of a thread and a ruler and tabulate them

as follows.

Figure AB CD Remarks
(i) ..... cm ...... cm AB = CD
(ii) AB = CD
..... cm ..... cm

Conclusion : Hence, if two arcs of a circle subtend equal angles at the center of the circle, they
are equal.

Note: Arcs that subtend equal angles at the circumference are equal.

Theorem 15.2

Equal arcs of a circle subtend equal angles at the center
Experimental Verification:
Draw two circles with centre O with different radii and also draw two equal arcs AB and CD
with the help of a pencil and a compass.

256 Oasis School Mathematics-10

B

AO AO

BD C

CD

Fig. I Fig. II

To verify : ∠AOB = ∠COD

Measure ∠AOB and ∠COD with the help of a protractor and tabulate them as follows:

Figure ∠AOB ∠COD Remarks

(i) ...... 0 ...... 0 ∠AOB = ∠COD

(ii) ...... 0 ...... 0 ∠AOB = ∠COD

Conclusion: Thus, equal arcs of a circle subtend equal angles at the center of the circle.

Note: Equal arcs subtend equal angles at the circumference.

Theorem 15.3

Equal chords of a circle form equal arcs in the circle.

Experimental Verification:

Draw two circles with centre O and having different lengths of radii. Also draw two equal
chords AB and CD in each circle.

AC

O D AB
B O

DC

Fig. I Fig. II

To verify : AB = CD

Measure the lengths of AB and CD with the help of a thread and a ruler and tabulate them

as follows:

Figure AB CD Remarks

(i) ..... cm ...... cm AB = CD

(ii) ..... cm ..... cm AB = CD

Conclusion: Equal chords of a circle form equal arcs on the circle.

Oasis School Mathematics-10 257

Theorem 15.4

Chords formed by the equal arcs of a circle are equal.

Experimental Verification

Draw two circles with centre O and different length of radii.
With the help of pencil and compass, draw two equal arcs AB and CD in each figure.

AC

O D AB
B O

DC

Fig. I Fig. II

To verify : AB = CD

Measure the lengths of AB and CD with the help of scale and tabulate them as follows:

Figure AB CD Remarks

(i) ..... cm ...... cm AB = CD

(ii) ..... cm ..... cm AB = CD

Conclusion: Thus, chords formed by the equal arcs of a circle are equal.

Theorem 15.5

If two chords are parallel, then the arc between them are equal.

Experimental Verification:

B

AB D
CD
A

Fig. I C
Fig. II

Draw two circles of different radii.

Draw two chords AB and CD, which are parallel to each other.

To verify : AC = BD

Measure the length of arc AC and arc BD with the help of a ruler and a thread and tabulate
as given below.

258 Oasis School Mathematics-10

Observation:

Figure Arc AC Arc BD Remarks

(i) ..... cm ...... cm Arc AC = Arc BD

(ii) ..... cm ..... cm Arc AC = Arc BD

Conclusion: Hence, if two chords are parallel, then the arcs between them are equal.

Theorems related to arc and central angles/inscribed angles

a. Central angle and corresponding arc:

If two radii of the same circle meet at the center and they make an angle O

at the center, it is called a central angle and the arc opposite to the central C
O
angle is called corresponding arc. A P B
B
In the adjoining figure,

∠AOB = central angle

AB = corresponding arc.

Write : AB ≗ ∠AOB
Read : Degree measure of AB is ∠AOB



Relations between central angle and corresponding arc
The degree measure of an arc is always equal to the central angle in degree.

i.e., ∠AOB ≗ AB

b. Inscribed angle and opposite arc:

If two chords of the same circle meet at a point on the circumference
of a circle and they make an angle, such an angle is called an inscribed
angle, and the arc opposite to it is called an opposite arc.

In the adjoining figure, A
∠ACB is an inscribed angle and AB is an opposite arc.

Relation between inscribed angle and opposite arc

O is the center of a circle, ∠ACB is an angle at the circumference, i.e., an inscribed angle
and APB is its opposite arc.

Half the degree measure of the arc APB is equal to the inscribed angle ACB,

i.e. 1 AB ≗ ∠ACB
2

Write : 1 AB ≗ ∠ACB
2

Read : Degree measure of 1 AB is ∠ACB
2

Oasis School Mathematics-10 259

Note: [Degree measure of an arc]

• minor arc APB ≗ ∠AOB C

• major arc ACB ≗ 3600 – ∠AOB

• semi circle ≗ 1800 O
A
• circumference ≗ 3600 B
≗ ∠ACB P
• 1 arc APB
2

Worked Out Examples

Example: 1

Find the value of x in the given figure. E F

Solution: 15 0 x

Given AB//CD A B
C D
then AC = BD [Arcs between parallel chords are equal]
B
∠AFC = ∠BED 800

[Inscribed angles standing on equal arcs] O
C
or x = 150.
D
Example: 2
In the given figure, if ∠ABC = 800, find the degree measure of arc ABC. O
Solution: C

Given ∠ABC = 800 A E

AC ≗ 2∠ABC

= 2 × 800

= 1600

Then degree measure of ABC = 3600 – 1600

= 2000

Example: 3

In the given figure, DE is the diameter, if BE = CE A
B
prove that ∠AED = 1 (∠ABC – ∠ACB).
2

Given : In the given figure, DE is the diameter and BE = CE
To prove
Proof : ∠AED = 1 (∠ABC – ∠ACB)
2

260 Oasis School Mathematics-10

S.N. Statements S.N. Reasons
1. DE being the diameter
1. DABE = DCE
2. AD + AB + BE = DC + CE 2. From (1)
3. BE = CE 3. Given

4. AD + AB = DC 4. From (1), (2) and (3)
5. 2AD + AB = DC + AD
6. 2 AD + AB = ADC 5. Adding AD on both sides
7. 2 AD = ADC – AB 6. From (5)
8. 2∠AED = ∠ABC–∠ACB
7. From (6)
9. ∠AED = 1 (∠ABC – ∠ACB)
2 8. From (6), using the relation of inscribed
angle with its opposite arc length

9. From (8)

Example: 4 Hence, proved.

In the given figure, PQ//RS and QT = RS, PR
prove that ∠PQR = ∠RQT.

Solution:

Given : In the given figure, PQ //RS and QT = RS Q T
To prove : ∠PQR = ∠RQT S
Proof

S.N. Statements S.N. Reasons
1. PQ//RS 1. Given
2. Arc between two parallel chords are equal
2. PR = QS
3. QT = RS 3. Given

4. QT = RS 4. Corresponding arcs of two equal chords
5. QS + ST = RT + ST 5. Using whole part axiom on statement 4
6. QS = RT 6. From statement 5
7. PR = RT 7. From statements 2 and 6
8. ∠PQR = ∠RQT 8. Inscribed angles on the equal arcs PR and RT.

Example: 5 Hence, proved.

In the given figure, ABCD is a square and ∆AEF is an equilateral AB
triangle, prove that DB//EF.
F
Solution: DC

Given : In the given figure, ABCD is a square and E

AEF is an equilateral triangle.

To prove : DB // EF

Proof

Oasis School Mathematics-10 261

S.N. Statements S.N. Reasons
1. Being the sides of a square
1. AB = AD 2. Being the sides of an equilateral triangle
3. Corresponding arcs of two equal chords
2. AF = AE
4. Same as statement 3
3. AB = AD
4. AF = AE 5. Subtracting statement 3 from 4
5. AF – AB = AE – AD
6. BF = DE 6. From statement 5
7. DB//EF
7. From statement 6, being arcs between two
chords equal

Example: 6 Hence, proved.
In the given figure, AD = BC then, prove that AC = BD.
Solution: AB

Given : In the given figure, AD = BC DC
To prove : AC = BD

Proof

S.N. Statements S.N. Reasons
1. AD = BC 1. Given
2. AD = BC 2. Corresponding arcs of two equal chords
3. AD + DC = BC + DC
4. ADC = BCD 3. Adding common arc DC on statement 2
5. AC = BD
4. From statement 4

5. From 4, corresponding chords of two equal arcs

Example: 7 Hence, proved.
In the given figure, two chords AB and CD are perpendicular to each
A

other, prove that AC + BD = AD + BC

Solution: C ED
Given B
: Two chords AB and CD are perpendicular each other at E.

To prove : AC + BD = AD + BC

Construction : Join AC and AD
Proof

S.N. Statements S.N. Reasons
1. ∠ACE = ∠CAE = ∠AED
1. Sum of two angles of triangle is equal to op-
2. ∠ADE + ∠DAE = ∠AEC posite exterior angle.

2. As statement 1.

3. ∠AED = ∠AEC 3. Both being 900.

4. ∠ACE + ∠CAE = ∠ADE + ∠DAE 4. From statement 1, 2 and 3.

5. AC + BC = AC + BD 5. From statement 4, using the relation of
inscribed angle with corresponding arc.

Hence, proved.

262 Oasis School Mathematics-10

Exercise 15.1 A
BC
1. (a) In the given figure, write the relation between
∠BAC and arc BC.

(b) In the given figure, write the relation between O
∠BOC and arc BC. BC

(c) In the given figure, AB//CD. Which two arcs are AB
equal?

(d) In the given figure, PR = QS , write the relation C D
between chord PQ and RS. Q
P
R S

P A

(e) In the given figure, BC = QR , which two angles R
are equal? B

CQ

(f) In the given figure, O is the center of the circle and O
∠AOB = ∠BOC. Which two arcs are equal. C

2. Find the value of x in the given figures. A
B
(a) (b) Q P

D 200 x

O C
x B

600 Q
R
AB

(c) E (d) Q (e) 20 0 A
A Gx B P
x B x
D
C 450 H D 350 S C
B E
R


3. In the given figure ∠BAC = 60°. A 600 C
Find the degree measure of arc BC and arc BAC.

Oasis School Mathematics-10 263

4. (a) In the adjoining figure, AB is the DC B
diameter with center O. If arc BC = arc A
CD, prove that AD//OC.
O

DC

(b) In the adjoining figure, AB is the diameter. If AO B

AD//OC, prove that BC = DC . Q
B
A B
D
5. (a) In the given circle, ∠WAN = ∠MBZ.

Prove that WZ||MN. W Z
N
M

P

(b) In the given figure, if AB||CD prove that ∠APC = ∠BQD. A
C

6. (a) In the given figure, if AC= BD FO E
prove that ∠AEC = ∠DFB.
A B
C D

(b) In the given figure, ∠AEC = ∠BFD. FO E
Prove that AC= BD.
A B
7. (a) C D

In the adjoining figure, POQ and ROT are two PO T
diameters of the circle with center at O. If Q is Q
the mid-point of arc TQS and ∠QOR is an obtuse, R
prove that PQ//RS. T S

(b) In the given figure, O is the center of the circle. P OQ

Prove that: TQ = QS . RS

8. (a) In the given figure, chords MN and RS of the circle

intersect externally at X. M N
R X
Prove that ∠MXR ≡ 1 ( MR – NS ).
2 S

264 Oasis School Mathematics-10

(b) In the given figure, chords MN and RS meet at X. MS

Prove that ∠MXR ≡ 1 ( MR + NS ). E X
2 N

9. In the given figure, O is the center of the circle. A O B R
If ∠AOC = 2 ∠BED, prove that AB//CD. C D
P

10. In the given figure, chord PQ is perpendicular to chord RS at M. M
S
Prove that: PQ + QS = PS + RQ . R
Q
11. In the given figure, PT is the bisector of U P
PQ Q R
T
∠QPS and UR is the bisector of ∠SRQ. S R
Prove that UT is a diameter of the circle. T S

12. In the given figure, chord PS and QR intersect at T,
prove that: (i) PQ = RS, (ii) PS = QR.

P

13. In the given figure, PQ//RS and ∠PQR = ∠RQT, Q R
prove that QT = RS. T

14. In the given figure, AB//CD then, A B S
prove that AD = BC. C D
A
15. In the given figure, AB = AC and AD = AE, then B C
prove that: BC//DE. D E

16. In the given figure, If AQ = PB, prove that A B
(i) AP = BQ, (ii) AB //PQ.

PQ C

17. In the given figure, AOB is a diameter of the circle. AO B
AOB bisects ∠CAD prove that AC = AD.

Answer D
1. Consult your teacher 2. (a) 600 (b) 200 (c) 450 (d) 350 (e) 200 3. 1200, 2400

Oasis School Mathematics-10 265

Theorem 15.5

The angle at the center of a circle is twice the angle at its circumference standing on the
same arc.

Experimental Verification:

Draw two circles of different radii, with centre O and also draw an angle BOC at the centre
and ∠BAC at the circumference standing on the same arc BC.

A
AC

OO

B C
Fig. I
B
To verify : ∠BOC = 2∠BAC Fig. II

Observation: Remarks
∠BOC = 2∠BAC
Figure ∠BAC ∠BOC
...... 0
I ..... 0

II ..... 0 ..... 0 ∠BOC = 2∠BAC
Conclusion : Hence, the central angle is twice the angle at its circumference standing on the
same arc.

Theoretical Proof:

Given : O is the center of the circle. Central angle BOC and inscribed

angle BAC are standing on the same arc BC. A

To prove : ∠BOC = 2∠BAC O
Construction : Join AO and produce it to K. B KC

Proof

S.N. Statements S.N. Reasons

1. In AOB, AO = OB 1. Radii of same circle

2. ∠OAB = ∠OBA 2. Base angles of an isosceles triangle

3. ∠BOK = ∠OAB + ∠OBA 3. An exterior angle is equal to sum of its
opposite interior angles.

4. ∴ ∠BOK = 2∠OAB 4. From (2) and (1)

266 Oasis School Mathematics-10

5. Similarly, in ∆AOC, 5. Same as above
∠COK = 2∠CAO

6. ∠BOK + ∠COK = 2(∠OAB +∠CAO) 6. Adding (4) and (5)

7. ∠BOC = 2∠BAC 7. From (6) whole part axiom

Hence, proved.

Hence, the angle at the center of a circle is twice the angle at the circumference standing on
the same arc.

Alternative method

Given : O is the center of the circle. ∠BOC and ∠BAC A

are the central angle and inscribed angle standing O
on the same arc BC.

To prove : ∠BOC = 2∠BAC C
Proof
B

S.N. Statements S.N. Reasons
1. Relation of central angle with arc length
1. ∠BOC ≗ BC 2. Relation of inscribed angle with arc length

2. ∠BAC ≗ 1 BC
2

3. ∠BOC = 2∠BAC 3. From (1) and (2)

Hence, proved.

Corollary: The angle in a semi-circle is 900 (a right angle).

Experimental Verification:

Draw two circles of different radii with center 'O' and in each circle draw ∠BAC on the semi-circle
taking BC as the diameter.

AA

BO C B OC

Fig. I Fig. II

To verify: ∠BAC = 900

Now, measure the angle BAC with the help of a protractor and tabulate them as follows;

Figure ∠BAC Remarks

I 900 ∠BAC = 900

II 900 ∠BAC = 900

Conclusion: Hence, the angle on a semi-circle is a right angle.

Oasis School Mathematics-10 267

Theoretical Proof : A

Given : O is the center of a circle and ∠BAC is an angle the

circumference of the circle BO C
To prove : ∠BAC = 900

Proof :

S.N. Statements S.N. Reasons
1. ∠BAC = 12∠BOC
1. Relation between inscribed angle and cen-
tral angle

2. ∠BOC = 1800 2. BOC is a straight angle
3. From (1) and (2)
∠BAC = 1 × 1800 = 900
3. 2

Hence, the angle on the semi-circle is a right angle. Hence, proved.

Theorem 15.6

Angles on the same segment of a circle are equal.

Or,

Inscribed angles standing on the same arc are equal.

Experimental Verification:

Draw two circles of different radii with center O and also draw angles BAC and BDC on the

same segment BC. A

A

D

O

B D BO
C
C
Fig. I
Fig. II

To Verify : ∠BAC = ∠BDC

Observation:

Figure ∠BAC ∠BDC Remarks

I ....° ....° ∠BAC = ∠BDC

II ....° ....° ∠BAC = ∠BDC

Conclusion : Thus angles on the same segment of a circle are equal.

268 Oasis School Mathematics-10

Theoretical Proof: AD
O
Given : O is the center of the circle. ∠BAC and ∠BDC are the
BC
inscribed angles on the same segment BC.

To Prove : ∠BAC = ∠BDC

Construction : Join B and O, C and O.

Proof :

S.N. Statements S.N. Reasons

1. ∠BAC = 1 ∠BOC 1. Relation between inscribed angle and cen-
2 tral angle

∠BDC = 1 ∠BOC 2. Same as (1)
2
2.

3. ∴ ∠BAC = ∠BDC 3. From (1) and (2)

Hence, proved.

Hence, angles on the same segment of a circle are equal.

Theorem 15.7

The opposite angles of a cyclic quadrilateral are supplementary.

Experimental Verification:

Draw two circles of different radii with center O and also draw a cyclic quadrilateral ABCD in each circle.

B

A A D
O O
C

D Fig. I BC

To verify: (ii) Fig. II

(i) ∠A + ∠C = 1800 ∠B + ∠D = 1800

Observation:

Figure ∠A ∠B ∠C ∠D Remarks
I ......0 ......0 ......0 ......0 ∠A + ∠C = 1800
II ......0 ......0 ......0 ......0 ∠B + ∠D = 1800

Conclusion: Hence, the opposite angles of a cyclic quadrilateral are supplementary.

Theoretical Proof: A

Given : O is the center of the circle. ABCD is a cyclic quadrilateral. O y0

To prove : (i) ∠A + ∠C = 1800 (ii) ∠B +∠D = 1800 B x D

Construction: Join B and O, D and O. Let the obtuse ∠BOD be x and reflex

∠BOD be y. C

Oasis School Mathematics-10 269

Proof :

S.N. Statements S.N. Reasons

1. ∠BAD = 1 x 1. Relation between the central angle and
2 inscribed angle standing on the same arc

2. ∠BCD = 1 y 2. Same as statement 1
2

3. ∠BAC + ∠BCD = 1 (x + y) 3. Adding statements 1 and 2
2

4. x + y = 3600 4. Angle in one complete rotation is 3600
5. From 3 and 4
5. ∴ ∠BAC + ∠BCD = 1 × 3600
2

= 1800

i.e. ∠A + ∠C = 1800

6. Similarly, ∠B + ∠D = 1800 6. By joining A and O, C and O and following

similar process

Hence, the opposite angles of a cyclic quadrilateral are supplementary. Hence, proved.

Note: If a pair of opposite angles of a quadrilateral are supplementary, then its vertices
are concyclic.

Corollary

If a side of a cyclic quadrilateral is produced, the exterior angle so formed is equal to the
opposite interior angle.

Experimental Verification:

A D
A O

OD C E

B E B Fig. II
C

Fig. I

Draw two circles of different radii taking O as the center.

Draw cyclic quadrilateral ABCD in each figure.
Produce side BC to E.
To verify : ∠DCE = ∠BAD

270 Oasis School Mathematics-10

Observation: ∠BAD ∠DCE Remarks
......0 ......0 ∠DCE = ∠BAD
Figure
(i)

(ii) ......0 ......0 ∠DCE = ∠BAD

Conclusion: Hence, if a side of a cyclic quadrilateral is produced, an exterior angle so formed
is equal to the opposite interior angle.

Theoretical Proof:

If a side of a cyclic quadrilateral is produced, the exterior angle so formed is equal to the
opposite interior angles.

Given : O is the center of the circle and the side BC of the cyclic A D
quadrilateral ABCD is produced to E. O
To prove
Proof: : ∠DCE = ∠BAD B C E

S.N. Statements S.N. Reasons
1. ∠BAD +∠BCD = 1800 1. Opposite angles of a cyclic quadrilateral

are supplementary

2. ∠BCD + ∠DCE = 1800 2. Sum of linear pair is 1800

3. ∠BAD + ∠BCD = ∠BCD + ∠DCE 3. From statements 1 and 2

4. ∴ ∠BAD = ∠DCE 4. From 3

Hence, proved

Hence, if a side of a cyclic quadrilateral is produced, the exterior angle so formed is equal to

opposite interior angles.

Worked Out Examples

Example: 1
Find the angles marked by the letters x0, y0 and z0 in the following figures:

Solution: C

∠ADB = ∠ACB = 300 (Angles on the same segment) D O y0

∴ y = 300 300 x0 B

or, 2∠ADB = ∠AOB (Relation between inscribed angle A
and central angle standing on the same
arc.)

or, 2 × 300 = ∠AOB

∴ ∠AOB = 600

∴ x = 600

Oasis School Mathematics-10 271

Example: 2 A
Find the value of x in the given figure.
Solution: O C
Construction: Join A and O
i. ∠BAO = ∠ABO = 400 (Base angles of isosceles ∠AOB as AO = OB) 400 300

x

B

ii. ∠OAC = ∠OCA = 300 (Base angles of isosceles ∠AOB as AO = OC) A

iii. ∴ ∠BAO + ∠OAC = ∠BAC = 400 + 300 = 700 (whole part axiom) O

∴ Obtuse ∠BOC = 2 ∠BAC (Relation between inscribed angle and central 400 300
x
angle standing on the same arc) C
B

x = 2 × 70

x = 1400

∴ x = 1400
Example: 3

In the given figure, BC is a diameter and AB = AC, find the value of x. A

Solution: Here,

∠BAC = 900 (Angle at the semi-circle) O C

B x0

∠ABC = ∠ACB (Being AB=AC) D

In ∆ABC,

∠BAC + ∠ABC + ∠ACB = 1800

or, 900 + ∠ACB + ∠ACB = 1800 (Base angles of an isosceles triangle, AB = AC)

or, 2∠ACB = 1800 – 900

∴ ∠ACB = 900 = 450
2

iii. ∠ADB = ∠ACB = 450 (Angles on the same segment)

∴ x = 450
Example: 4

In the given figure, 'O' is the centre of the circle, BP⊥ AC, find the value of x.

Solution: Here, C
Px
2∠CAB = ∠BOC (Relation between the inscribed
or, 2 × 600 600 O
angle and central angle standing on A

the same arc)

= ∠BOC B

∴ ∠BOC = 1200

In ∆CPO, BP⊥ AC ∠CPO + ∠OCP = ∠BOC (Sum of two angles of a triangle

is equal to the opposite exterior angle)

272 Oasis School Mathematics-10

900 + ∠OCP = 1200
∴ ∠OCP = 1200 – 900
= 300
∴ x = 300

Example: 5

In the given figure, ABCD is a cyclic quadrilateral. BC is produced to E, find the value of

x and y.

A D
Solution: x0

Here, ∠BCD +∠DCE = 1800 y0 700 E
y + 700 = 1800 (Sum of linear pair)
∴ y = 1800 – 700 = 1100 BC

Again, x + y = 1800 (Opposite angles of a cyclic quadrilateral)

x + 1100 = 1800

∴ x = 1800 – 1100 = 700

Example: 6

(a) In the given figure, O is the center of a circle in which ∠AOC = 1600 and D
∠DAB = 400. Find the value of x.
x

Solution: O C
B
Here, 1600
400
Obtuse ∠AOC + Ref. ∠AOC = 3600 (Angle in one complete rotation)

A

or, 1600 + Ref. ∠AOC = 3600

or, Ref. ∠AOC = 3600 – 1600 = 2000

∴ ∠ABC = 1 × Ref. ∠AOC (Relation between inscribed angle and
2 central angle standing on the same arc)

= 1 × 2000
2

= 1000

Again, In ∆ABD

∠BAD + ∠ABD +∠ADB = 1800 (Sum of angles of a triangle)

400 + 1000 + x = 1800

∴ x = 1800 – 1400 = 400

Oasis School Mathematics-10 273

Example: 7 C B
In the adjoining figure, O is the center of the circle. Calculate the D
value of x.
1150 x
Solution: AO
Construction: Join B and C,

i. ∠ADC + ∠ABC = 1800 (Opposite angles of the cyclic quadrilateral)

or, 1150 + ∠ABC = 1800

or, ∠ABC = 1800 – 1150

= 650

ii. In ∆BOC,

∠OBC = ∠OCB = 65° (OB = OC, radii of the same circle)

And ∠BOC + ∠OBC + ∠OCB = 180° (Sum of angles of the triangle)

or, ∠BOC + 65° + 65° = 180°
or, ∠BOC = 180° – 130°
∴ x = 50°

Example: 8 A OB
In the given figure, AB is a diameter and ∠AEB = 500. Find the
value of ∠COD.

Solution: CD
Given AOB is a diameter 500
∠AEB = 500 E

Join AD [Angle on the semi-circle]
Here,
∠ADB = 900
In ∆ADE,

∠AED + ∠ DAE = ∠ADB

[Sum of two angles of a triangle is equal to
opposite exterior angles]

or, 500 + ∠DAE = 900

or, ∠DAE = 900 – 500

or, ∠DAE = 400
Again,

∠COD = 2∠DAE (Central angle is twice the inscribed angle standing on the same arc.]

or, ∠COD = 2 × 400

= 800

274 Oasis School Mathematics-10

Example: 9 P Q
Points P, Q, R, S and T are on the circumference of a circle, such that

PS ⊥ PQ and RT ⊥ PR. Prove that PQ = ST.

Solution: P, Q, R are on the circumference of a circle, such that S T
Given : PS ⊥ PQ and RT ⊥ PR. R
PQ = ST
Join Q and T
To prove :
Construction :

Proof :

S.N. Statements S.N. Reasons
1. ∠PRT = ∠PST = 900 1. Angles on the same segment and ∠PRT = 900

2. ∠PST + ∠PQT = 1800 2. Opposite angles of the cyclic quadrilateral

3. ∴ ∠PQT = 900 3. From 1 and 2

4. ∠SPQ + ∠QTS = 1800 4. Opposite angles of the cyclic quadrilateral.

5. ∴ ∠QTS = 900 5. ∠SPQ = 900 and from 1st no. (4)

6. ∴ PQTS is a rectangle 6. Each angle of quadrilateral being 900

7. PQ = ST 7. Opposite sides of the rectangle
Hence, proved

Example: 10 Y
In the given figure, O is the center of a circle.
Prove that: ∠XOZ = 2(∠XZY + ∠YXZ). XZ
O
Solution:
Given : O is the center of the circle.
To prove : ∠XOZ = 2(∠XZY + ∠YXZ)

Proof :

S.N. Statements S.N. Reasons

1. ∠XOY = 2∠XZY 1. Relation between inscribed angle and
central angle standing on the same arc

2. ∠YOZ = 2 ∠YXZ 2. Same as 1.

3. ∠XOY + ∠YOZ = 2(∠XZY+∠YXZ) 3. Adding 1 and 2

4. ∠XOZ = 2(∠XZY + ∠YXZ) 4. From 3, whole part axiom
Hence, proved.

Oasis School Mathematics-10 275

Example: 11

Prove that the quadrilateral formed by an angular bisector of a cyclic D P C
quadrilateral ABCD is also cyclic. S

Solution: Q
Given
A R B

: ABCD is a cyclic quadrilateral. AP, BP, CR and DR are

the bisectors of ∠A, ∠B, ∠C and ∠D respectively.

To prove : PQRS is a cyclic quadrilateral.

Proof

S.N. Statements S.N. Reasons

1. ∠APB +∠PAB + ∠PBA = 1800 1. Sum of angles of a triangle

2. ∠CRD +∠RCD + ∠RDC = 1800 2. Same as 1

3. ∴ ∠APB + ∠PAB + ∠PBA + ∠CRD + 3. Adding 1 and 2
∠RCD + ∠RDC = 3600

4. ∠APB + ∠CRD + 4. AP, BP, CR and DR are the bisectors
of ∠A, ∠B, ∠C and ∠D
1 (∠A+∠B+∠C+∠D) = 3600
2

5. ∠APB + ∠CRD = 3600 – 1 × 3600 5. Sum of the angles of the
2 quadrilateral

= 1800

6. ∴ PQRS is a cyclic quadrilateral 6. Opposite angles of the
quadrilateral being supplementary

Hence, proved

Example: 12 A B
In the given figure, chords AB and CD are produced to meet at E. O
Prove that: ∠AOC – ∠BOD = 2∠AEC. E
C D
Solution:

Given : O is the center of the circle chord and AB and CD are
A
produced to meet at E.

To prove : ∠AOC – ∠BOD = 2∠AEC B
O
Construction : Join AD
E
Proof : D

C

S.N. Statements S.N. Reasons

1. ∠AOC = 2∠ADC 1. Central angle is twice the inscribed angle
standing on the same arc

276 Oasis School Mathematics-10

2. ∠BOD = 2∠BAD 2. As statement 1

3. ∠AOC – ∠BOD = 2(∠ADC – ∠BAD) 3. Subtracting statement 2 from 1

4. ∠ADC = ∠BAD + ∠AED 4. Sum of two angles of a triangle is equal
to the opposite exterior angle.

5. ∠AED = ∠ADC – ∠BAD 5. From 4

6. ∠AOC – ∠BOD = 2 ∠AED 6. From 3 and 5

Hence, proved.

Example: 13 AC
In the given figure, AB is a diameter of the circle with centre O.
If AD = DE, prove that OE // BD. EO
D
Solution: : In the given figure, AE = DE.
Given B

To prove : OE // BD

Construction : Join OD

Proof

S.N. Statements S.N. Reasons

1. In ∆ADE and ∆DOE 1.
(i) AE = DE (S) (i) Given
(ii) OA = OD (S) (ii) Radii of same circle
(iii) OE = OE (S) (iii) Common side

2. ∠AOE ≅ ∠DOE 2. By SSS axiom

3. ∠AOE = ∠DOE 3. Corresponding angles of congruent triangles

4. ∠AOE = 12∠AOD 4. From statement 3
5. ∠ABD = 12∠AOD
5. Inscribed angle is half of the centre angle
standing on the same arc

6. ∠AOE = ∠ABD 6. From statements 4 and 5

7. OE // BD 7. From statement 6, being corresponding
angles equal

Hence, proved.

Oasis School Mathematics-10 277

Exercise 15.2

1. (a) What is the relation between the inscribed angle and central angle standing on the
same arc?

(b) What is the relation between the inscribed angles standing on the same arc?

(c) What is the value of an angle on a semi-circle?

(d) What is the relation between the opposite angles of a cyclic quadrilateral?

2. (a) In the given figure, write the relation between ∠BAC A
and ∠BOC. O

BC

A
D

(b) In the given figure, write two pairs of angles

which are equal.

B
C

CB

(c) In the given figure, what is the value of

∠ACB? Why? O

A
A

(d) In the given figure, write the relation between D
∠A and ∠C as well as ∠B and ∠D. C

B
P

(e) In the given figure, write the angle which is equal T
S
to ∠PST. Q

R

3. Find the sizes of unknown angles in the following circles with center O.

(a) A (b) 2300 C (c) A

x0 A O xy

x O

1000 C B 30 z 20

B BC

278 Oasis School Mathematics-10

(d) S (e) D B

yo R A (f)

x 300 x0 1200 x
C O O
PQ
Z 1400
Qy S
(g) (h) P
R
O x
(i) B
1000 Q y S
1400 x
A xC O
R
B y
350 C
(j) (k) A A

D (l) A
x
C O
B
3y y0 C OD
2 O x0
A
B
B Dx C 550
(o) E
(m) A E E
A x0 B
x0 (n) A
O
500

B O D x Z
C O
500 250 C 230 D
y0 x0
YD

M N

(p) AD (q) (r) A E

B E A B x0

1200 x 920 100 E OD
400 B 350 C
O Dx F
C

C

(s) (t) (u) T
P 1100
R P

2300 R

O O 400 x0 S QS

x T R (x–20)0
U
QP Q

(v) A (w) A

B xD B x0 C
250 C
200 1200

D O

Oasis School Mathematics-10 279

4. (a) In the given figure, PQ = 6 cm, QR = 8 cm. Q

Find the length of OQ. 6cm 8cm

P OR

(b) In the given figure, AC = 6cm, OC = 5 cm. A O B
Find the length of BC.
6cm 5cm AQ
5. (a) O
C

B

In the given figure, O is the center of the circle, PQ ⊥AB.
If ∠BAP = 700, find ∠ABO.

C P
OD
(b) In the given figure, O is the center of the circle.

A

AD⊥BC. If ∠OCD = 300, find ∠ABC.

B

6. (a) OABC is a parallelogram in the given figure, O
where O is the centre of the given circle. Find x and y.
A x0 y0
B C

P S

(b) In the given figure, PQRS is a parallelogram. Find the

value of ∠SPQ. A Q R

7. (a) In the given figure, O is the center O B A
of the circle. Find ∠ADB. C 300 O

D C
A
(b) In the given figure, O is the center of the circle. C is the
D
mid-point of BD and ∠ADB = 500. Find ∠ABD. D B

(c) In the given figure, ∆ABC is an isosceles triangle

with AB = AC and ∠ABC = 500. Find ∠BDC and ∠BEC. B C
E

(d) A chord of a circle is equal to its radius. Prove that the angle subtended by the
chord at a point on the circumference of the circle is either 1500 or 300.

280 Oasis School Mathematics-10

D

8. In the given figure, O is the center of the circle, AB is the diameter C

and DO⊥AB. Prove that ∠AMC = ∠ODA.

D A B
O

9. In the figure alongside DM⊥EF, FN⊥ED. N

Prove that ∠EMN = ∠EDF. M

E

MF

10. (a) In the given cyclic quadrilateral ABCD, DC is the A D
B E
bisector of ∠BDE. Prove that AC = BC.
C

(b) In the given cyclic quadrilateral ABCD, AC = BC. AD
Prove that DC is the bisector of ∠BDE. E

C
B

11. ABCD is a parallelogram. The circle through A, B and C ED C
intersects CD produced at E. Prove that AE = AD. A B

12. In the figure alongside, O is the center of the circle. Then prove AOB

that 2∠APC = ∠AOC + ∠BOD. P

BA C D

13. In the given figure, O is the center of the circle. M OC N
Prove that ∠AOB = ∠ACB, if AC = CD. D

PMS

14. In the given figure, two circles intersect at M and N. PQ R Q

and RS pass through M. R, P, S and Q are joined to N. Prove

that ∠PNR = ∠SNQ. N

15. In the given figure, if AB = AC, A B
prove that BE = DC. D

O

EC

Oasis School Mathematics-10 281

16. In the given figure, AD and BC are chords perpendicular to each A C
other, and O is the center of the circle.
O
Prove that ∠AOC + ∠BOD = 180°. B

D

P A Q
B R
17. Two circles intersect each other at points A and B. PAQ and SBR
are two straight lines. Prove that PS||QR.

S
A

18. In the given figure, PQ//BC, prove that AP = AQ. P Q

BC

19. In the given diagram, M and N are the centers of two circles, which AB

intersect at A and C. MA and MC are produced to meet the other M N
circle at B and D. Prove that AB = CD. CD

20. In the figure, AP is the radius of the bigger circle and diameter of C
Q
the smaller circle. Prove that AQ = QC. A
P

B

A

21. In the given figure, P and Q are mid–points of arcs AB and AC Px yQ
respectively. Prove that AX = AY.
BC


22. In the given figure, if AB//DC and AD = BC, A B
C
show that ABCD is a cyclic quadrilateral.
AD
D
BC
23. In the figure alongside, BD = DP and ∠ABD = ∠DBC. P
Prove that AB = CP.

282 Oasis School Mathematics-10

AD

24. (a) In the given figure, AM is the bisector of ∠BAC. E F
Prove that EF//BC. B C

(b) In the given figure, EF // BC. A D M
Prove that AM is the bisector of ∠BAC. E
F
B C

MA

R Q
P
25. In ∆ABC, P, Q and R are mid-points of BC, AC and AB respectively.

If AW⊥BC, prove that W, P, Q and R are concyclic points. BW C

26. (a) In the given figure, AB is a diameter, and OE//BD, AC
prove that AE = ED. DB

(b) In the given figure, AB is a diameter and AE = ED, AC
prove that OE // BD.
DB

(d) x=600, y=300
Answer (h) x=400, y=700
(l) x=1100
1. Consult your teacher 2. Consult your teacher (p) x = 800
(t) x = 700
3. (a) x = 500 (b) x= 650 (c) x=300, y=200, z=1000 6. (a) x=600 , y=1200

(e) x = 300 (f) x = 700, y=1100 (g) x=1350

(i) x = 550, y=1100 (j) x = 540, y=360 (k) x = 600

(m) x = 500, y=550 (n) x = 400 (o) x = 440

(q) x = 980 (r) x = 1250 (s) x = 500

(u) x = 900 (v) x = 1150 (w) x = 400

4. (a) 5 cm (b) 8cm 5. (a) 500 (b) 600

(b) 900 7. (a) 600 (b) 500 (c) 800 and 1000

Project Work

DWo iytohutkhneowhe!lp of a thin string, prepare a circle on a chart paper. With the help of a
Thtoeogtehompiectrkymofaakpelatnheefignuerceeissskanroywannags lEeuscalindedanvGereiofmyetthrye. above theorems.

Oasis School Mathematics-10 283

15.3 Tangent and Secant

Some important terms related to tangents: P QR
• Secant: A secant is a line which intersects a circle distinctly

through two different points. In the figure alongside, AB X O Z
A D
is a secant which intersects the circle through the points Y B
C and D. PQR and XYZ are the other two secants of the C

circle.

• Tangent: We can draw a line which touches a circle at a T O
single point. In the given figure, S is a point through SN
which TN passes. Here TN is called a tangent of the circle.
Y P
A tangent is a straight line which touches a circle at a O B
single point. The common point is called the point of
contact. In the given figure, S is the point of contact of the AS
circle and the tangent.

* Angle in alternate segments: In the figure, PS is a chord

drawn from the point of contacts. So ∠PYS is an angle
on the alternate segment with respect to the ∠PSB.
Similarly, ∠YPS is the angle on the alternate segment
with respective to ∠ASY.

Some important properties of tangent:

1. The radius of a circle is perpendicular to the tangent O
at the point of contact, TS

In the given figure, ∠OST = ∠OSN = 90°. A N
O
i.e. OS ⊥ TN. B P
P
2. The lengths of the tangents drawn from any external R Q
O N
point of a circle to it are always equal. In the given
figure, AP = BP. TS



3. The angle made by a chord drawn from the point of
contact with the tangent is equal to the angle on the
alternate segment. In the figure,

∠PST = ∠PQS

∠PRS = ∠PSN, ∠RST = ∠RPS.

Note : Properties 2 and 3 are not included in the curriculum prescribed by CDC.

284 Oasis School Mathematics-10

Theorem 15.8

"Radius of a circle is perpendicular to the tangent at the point of contact."

Experimental Verification:

Draw two circles with different radii (r >3 cm) with center at O.
Draw a tangent AB through C and radius OC.

OO

AB A CB
C

Fig. (i) Fig. (ii)

To verify : OC ⊥ AB (i.e., ∠OCA = ∠OCB = 900)

Observation :

Figure ∠OCA ∠OCB Remarks
(i) 900 900 OC ⊥ AB

(ii) 900 900 OC ⊥ AB

Conclusion : The radius of a circle is perpendicular to the tangent at the point of contact.

Theoretical proof.:

Given : OC is the radius drawn from the point of contact C of O

the tangent AB.

To prove : OC ⊥ AB. A P' P C Q Q' B

Construction : Let P, P', Q, Q' be different points on AB, all
joined to O.

Proof

S.N. Statements S.N. Reasons
1. OP, OP', OQ, OQ' all are > OC 1. Since P, P', Q, Q' are points out of the

circle

2. OC is the shortest distance of O from 2. From statement (1)
AB

3. ∴ OC ⊥ AB 3. From statement (2)

Hence, proved.

Oasis School Mathematics-10 285

Theorem 15.9

* "The length of the tangents drawn from any external point of a circle on it are equal."

Experimental Verification

Draw two circles with centre O and different lengths of radii.
From any external point P, draw two tangents PA and PB in each figure.

O B A
A O
P
B
P
Fig. (ii)
Fig. (i)

To verify : AP = BP

Observation:

Figure AP BP Remarks
I ............... ............... AP = BP

II ................ ................. AP = BP

Conclusion: Hence, the lengths of the tangents drawn from any external point of a circle on it are equal.

T heoretical Proof:

Given : O is the center of the circle in which AP and BP are two tangents drawn from

any external point P. A

To prove : AP = BP PO
Construction : Join OB, OA and OP

Proof B

S.N. Statements S.N. Reasons

1. In ∆AOP & BOP 1.
(i) ∠OAP = ∠OBP = (900) (R) (i) Radius is perpendicular to the tangent at
(ii) OP = OP (H) the point of contact
(iii) OA = OB (S) (ii) Common sides
(iii) Being radii of the same circle

2. ∴∆AOP ≅ ∆BOP 2. By R.H.S. axiom
3. ∴ AP = BP 3. Corresponding sides of congruent triangles

Hence, proved.

286 Oasis School Mathematics-10

Theorem 15.10

"The angle formed by a tangent to a circle and a chord drawn from the point of contact are
equal to respective angles in the alternate segments."

(Alternate segments theorem)

Experimental Verification: BT

C

BO OA

N

A C N
T Fig. (ii)

Fig. (i)

• Draw two circles of different radii taking 'O' as the center.
• Draw a tangent TAN at point A.
• Draw two secants AB and AC and join BC.
Now, the angle on the alternate segment of ∠BAT is ∠ACB and the angle on the alternate
segment of ∠CAN is ∠ABC.

To verify:

(i) ∠BAT = ∠ACB (ii) ∠CAN = ∠ABC

Observations

Figure ∠BAT ∠ACB ∠CAN ∠ABC Remarks
I
............ ............ ........... .......... ∠BAT = ∠ACB

∠CAN = ∠ABC
II ............ ............ ............ .......... ∠BAT = ∠ACB

∠CAN = ∠ABC

Conclusion: Hence, the angle made by a chord drawn from the point of contact with the
tangent is equal to the angle of the alternate segment.

Theoretical Proof:

Given: A D
(i) TN is a tangent of the circle with center at O. O
(ii) S is the point of contact through which chord SB is drawn. B
S C
(iii) ∠BSN, ∠BAS and ∠AST, ∠ABS are angles on alternate segments.
N
To prove : (i) ∠AST = ∠ABS (ii) ∠BSN = ∠BAS T
Construction : Join diameter SD and join BD.

Oasis School Mathematics-10 287

Proof:

S.N. Statements S.N. Reasons

1. ∠DBS = 90° 1. Angle of semi circle

2. ∠BDS + ∠BSD = 90° 2. Remaining angles of ∆BDS

3. ∠BSN + ∠BSD = 90° 3. Since DS ⊥ TN

4. ∴ ∠BDS +∠BSD =∠BSN + ∠BSD 4. From statements (2) & (3)

5. ∴ ∠BDS = ∠BSN 5. Remaining parts

6. But, ∠BDS = ∠BAS 6. Angles of the same segments

7. ∴ ∠BSN = ∠BAS 7. From statements 5 & 6

8. Similarly, ∠AST = ∠ABS 8. Proof as above.

Hence, proved.

Note : Theorem 15.9 and 15.10 are not in the curriculum prescribed by CDC.

Worked Out Examples

E xample: 1 O 20cm P
Calculate the value of x in the given figure. 16cm
x
Solution:
AS

Here, OS⊥AP (radius is perpendicular to the tangent at the point of contact)

∆OSP is a right angled triangle

∴ (OS)2 = (OP)2 − (SP)2
OS = (OP)2 − (SP)2

= (20cm)2 − (16cm)2

= 400cm2 − 256cm2

= 144cm2 = 12 cm
x = 12 cm

Example 2:

Find the value of y in the given figure.

Solution: Q O

Here, OA⊥PN (radius is perpendicular to the tangent at the point of P y 7cm N
contact) 24cm
A

288 Oasis School Mathematics-10

Also, OA = OQ = 7 cm (radius)

Now, applying the Pythagoras Theorem in ∆OAP

OP2 = OA2 + PA2
= (7)2 + (24)2
= 49 + 576
625
OP2 =

∴ OP = 25 cm.

Now, y = PQ = OP – OQ = (25 – 7) cm = 18 cm

Example 3: O
Find the value of x from the given figure.
Solution: Qx B
In the given figure,
T A 300 N

∠OAN = 900 [Radius is perpendicular to the tangent at the point of contact]

or, ∠OAB + ∠BAN = 900

or, 300 + ∠OAB = 900

or, ∠OAB = 900 – 300

or, ∠OAB = 600

Again,

OA = OB [Radii of same circle]

∠OAB = ∠OBA [Base angles of an isosceles triangle]

In ∆OAB,

or, ∠OAB + ∠OBA + ∠AOB = 1800

or, 600 + 600 + ∠AOB = 1800

or, 1200 + ∠AOB = 1800

or, ∠AOB = 1800 – 1200 = 600

Again , ∠ACB = 1 ∠AOB [An inscribed angle is half of the central
2 angle standing on the same arc]


or, x = 1 × 600
2

= 300

Oasis School Mathematics-10 289

Exercise 15.3

1. (a) What is the relation between the radius of a circle and tangent at the point of
contact?

(b) In the given figure, 'O' is the center of the circle, what is the OB
relation between OM and AMB?
M
A

(c) In the given figure, write the relation among OA, TA and OT. O

TA N

C
B

(d) In the given figure, write the relation among AB, OB and OA. O

A

2. (a) In the given figure, OA = 15cm, AB = 12cm, find the length of S O
AB. A
BC
(b) In the given figure, PQR is a tangent at Q. If PQ = 40cm, and R
OP = 41cm, find the length of OS.
OQ
(c) In the given figure, OQ = 7cm, and MN = 24 cm, find the
length of ON. S
P

P

MO

Q
N

(d) In the given figure, AB = 16cm, AD = 8cm and OD = x cm, find O

the value of x. D
A
B C

(e) In the given figure, NZ = 9cm and YZ = 15cm, find the length

of its radius. O
N

XY Z

290 Oasis School Mathematics-10

3. Find the value of x from the given figure.

a) P c) O

b) 250 T x

O O x0 R A

550

X x0 Y Z N

Q

d) O B R f) R

x e) xO

T A 500 N O Q 750 F

150 x DE

P S

4. Find the values of x, y and z from the given figures.

a) 2200 A b) A c) O

O C Ax B

Tx O1200

B x0 B 500
P
T
d) P e) T P f) D

x0 z0 A

A B R Oy0 x0 x0

1060 y0 750 O T

O NQ S y0 240 B

C

A N A

g) z0 h) O S i) 400 T
250
O O
M Cx
B x0 y0
yx B
600 P R
S Q
T

j) A k) B E

600 OC l)

RQ 400 F

x DO

xy

O 400 N A 600 B C
y
T
PC A

B P



5. (a) In the given figure, QS is a tangent where R is the M O
point of contact. If ∠MRQ = 33°, find the measure of Q 330 R
the ∠MQR. S

Oasis School Mathematics-10 291

(b) In the given figure, RN is the tangent to M
the circle and N the point of contact. If O
∠PRN = 28°, calculate the value of the ∠RMN.
P
280 N
R

P

6. (a) In the given figure, O is the the centre of circle, TS R
tangent, Q point of contact. If PR//OS, prove that O
∆PQR~OSQ.
TQ S

(b) In the figure, TAN is the tangent and AB is the B
diameter of the circle with center at O. Prove that:
(i) ∆ABN ~ ∆ACN. O
C
(ii) AN2 = BN.CN.
T AN

Answer
1. Consult your teacher 2. (a) 9cm (b) 9 cm (c) 25 cm (d) 12 cm
(e) 8cm 3. (a) 350 (b) 650 (c) 450 (d) 400 (e) 750 (f) 1500 4. (a) x = 400 (b) x = 600 (c) x = 250

(d) x = 320, y = 1480 (e) x = 1500, z = 750, y=150 (f) x = 330, y=570 (g) x = 600, y = 600, z = 300 (h) x = 650, y = 250
(i) x = 700 (j) x=1200, y=600 (k) x = 950 (l) x = 700, y = 350 5. (a) 570 (b) 310

Note
Problems using Alternate segment theorem are not in the syllabus prescribed by CDC.
These questions are given for further practice.

Project Work

On a chart paper, prepare a circle with the help of a thin string, paste toothpick and verify
the above theorem.

Do you know !
Thales' Theorem, named after Thales of Miletus states that if A, B and C are points on a
circle where the line AC is a diameter of the circle, then the angle ABC is a right angle.

292 Oasis School Mathematics-10

Miscellaneous Exercise A C
P
Area of Triangle
1. In the adjoining figure, AD is one of the medians of BD
D
∆ABC and P is a point on AD.
Prove that: (i) ∆BDP = ∆CDP (ii) ∆BAP = ∆ACP E C
GA BF
2. In the given figures ABCDE is a pentagon.
EG//DAand CF//DB. Show that the area of the pentagon
ABCDE = ∆DGF.

PQ

3. In the given trapezium PQRS, PQ||SR, M and N are MN
the mid-points of the diagonals PR and QS respectively.
Prove that ∆PNR = ∆QMS. S R
A B
4. MN is the median of the trapezium ABCD. If AB// DC, M
prove that 2∆BCM = trapezium ABCD. N
D
5. In the given figure, ABC is an equilateral triangle. From C
the point O, OE, OF and OG are drawn perpendicular A
to BC, AC and AB respectively.
GO F
Prove that: AD = OE + OF + OG . B ED C

6. In the given figure, D, E, F and G are the mid-points on A
BC, AD, BE and CF respectively.
F E C
Prove that: ABC = 8∆EFG. B G
D
7. In the quadrilateral ABCD, AO = OC. Prove that, A D C
2∆ABD = quadrilateral ABCD.
O
8. In the given figure, F, G and H are the mid-points on B
AE, BE and CE respectively.
A
Prove that: ∆ABC = 4∆FGH.
F H
GE
B C

Oasis School Mathematics-10 293

9. In the given figure, ABCD, ADEF and BCEF are three E
parallelograms. Prove that: parallelogram ABCD = F
parallelogram ADEF + parallelogram BCEF. DC

10. In the given figure, AB//DC, AD // BE, AH//DE, A B
DG // EH. Prove that: parallelogram ABCD = A B
parallelogram DEHG. G
D C
Circle
1. In the given figure, O is the center of the circle. C H
E
and D are two points on the chord. If AC = BD and
AM = BN , prove that ∠ACM = ∠BDN. A CD B

O
MN

2. In the given figure, O is the centre of the circle in which D C
AO B
triangle OCE is an isosceles triangle. E

Prove that BC = 1 AD .
3
A
3. In the adjoining figure, ABCD is a cyclic quadrilateral,

the side CD is produced to the point E where BC = DE. B

If CA bisects ∠BCD, prove that ∆ACE is an isosceles C DE
AD
triangle.
P
4. Chords AB and CD of a circle intersect at a right angle NM
at P. M is the mid-point on BD and the the MP produced
meets AC at N. Prove that PN⊥AC. C B A
F E
5. In the given figure, chords AD, BE and CF intersect at A
G. Prove that ∠BGC =∠BAC + ∠FDE. O B G
D C
6. In the given figure, two chords AB and CD intersect C E
at E, prove that: ∠AOC + ∠BOD = 2∠BED. B D

7. In the given figure, BE = AD, AB = CD, prove that: A E
AEBD is a parallelogram.
B

DC

294 Oasis School Mathematics-10