9 Green C Math Class

EXERCISE: 12.1

A. Very Short Questions

t A [General Sction] A B
M N
1. a. In trapezium ABCD, M and N are the mid points of AC C
and BD respectively. What AB, CD and MN are called? D

b. In the given figure, ABCD is a square. AC and BD are A O D
diagonals and AO = 5cm. Find the length of AC and BD. B O
C
P S
R
c. In the given figure, PQRS is a rectangle. If OS = 2x cm Q
and PR = 14cm, find x.

d. Inthegiven figure, A BCDisaparallelogram.If∠A=4xand A 4x D
∠B = 80°, find the value x and ∠D. B 80° C

P S

e. In the given parallelogram PQRS, ∠Q = 2x + 25° and 2x+25° 4x–15°

∠S = 4x – 15°. Find the value of ∠P . Q R

(2x – 3)cm
PS

f. In the given parallelogram PQRS, PS = (2x – 3)cm and Q R
QR = (x + 2) cm. Find the length of PS. (x + 2) cm

B. Short Questions

2. In each of the following figure, find the values of a and b.

E D C P A aD
a 120° 130° S a

a. b. a c. b 80°
C
A BQ R T
B

GREEN Mathematics Book-9 195

A

A D P S a
120° 37°
da e. a f. D F E
E b
B
C Q b 73° R 75° 35°
B
C

A 32° D A 40° a D Ab D

g. Fa h. b i. a 40°
b B 20°
B C 130° C
B A E F
D
E C

3. a. In a parallelogram ABCD, ∠ABC = 50°, E
find the measure of ∠CDE.

B 40° C
A
B
40°
b. In the given figure, ABCD is a rhombus. D
If ∠ADB = 40°, find the size of ∠CDE.

C

P QE
?
c. In the given figure, PQRS is a rhombus and 44°

∠SPR = 44°. Find the size of ∠PQR.

S R B
A C
d. In the adjoining figure, ABCD is a square. If ∠BFE = 100,
find the measure of ∠AED and ∠DAE. F
100°
e. In the adjoining figure, ABCD is a A D
rhombus. If ∠OBC = 30°. Find ∠OAD and B D E
∠BAD.
O C

f. In the adjoining figure, ABCD is a Aa D
parallelogram. Find ∠AEB. a
E
b

Bb C

g. In the adjoining figure PQRS is a square and OQR is an P S
equilateral triangle. Find the measure of ∠QPO and ∠PSO. O

QR

196 GREEN Mathematics Book-9

C. Long Questions A PD
4. a. Prove that diagonals of a rectangle are equal. O
QC
b. In the figure ABCD is a parallelogram P and Q are the mid- PC

points of AD and BC respectively. Prove that OP = OQ. B B
N
D
D
5. a. In the given figure, ABCD is a parallelogram and AP is the
MC
bisector of ∠BAD, prove that BC = DP. A A

A

b. In the figure AM and CN are bisectors of ∠A and ∠C

respectively. If ABCD is a parallelogram, prove that

AMCN is also a parallelogram. B

c. In the given figure, P and R are the mid points of AB and P Q

BC of ∆ABC respectively. If PQ||BC, prove that BP = RQ. BR C

6. a. In the figure, ABCD is a parallelogram. If AP = CQ, prove A D
that PBQD is a parallelogram. P
QC
B D

b. In the figure ABCD is a parallelogram. If AP = CQ, P
prove that ∠PDA = ∠QBC. A

7. a. In the figure, PQ = PR, MR = QN and QN||ML. BC
Prove that MN and QL bisect each other. Q

Q P

M

LR

N

b. In the figure, ABCD is a parallelogram. M and N are the A M D
mid- points of AD and BC respectively. Prove that: P Q

i. ANCM is a parallelogram B C
N
ii. BP = PQ = QD

c. ABCD is a square. If AP = BP prove that AP and BQ are A D
perpendicular to each other. B Q
C
P

GREEN Mathematics Book-9 197

12.3 Mid-point Theorems

Theorem - 14

Statement:

The straight line drawn through the mid-point of one side of a triangle and parallel to the
other side bisects the third side.

Experimental verification A

A

A

DE DE DE

B CB CB C

(i) (ii) (iii)

ABC is a triangle in which D is the mid-point of AB and DE||BC.

To verify: AE = EC

Measurement table:

Figure AE EC Remarks
i. AE = EC
ii.
iii.

Result : From the above figures, we found that AE and EC are equal.

Conclusion : The straight line drawn through the mid-point of one side of a triangle

and parallel to the other side bisects the third side.

Theoretical Proof A

Given: ABC is a triangle in which AX = XB and XY||BC. XY
To prove: AY = YC

Construction: Draw XB||YZ. So that YZ meets BC at Z. B C
Z

Statement Reasons
1. XB||YZ 1. By construction.
2. XBZY is a parm. 2. Opposite sides being parallel.

3. XB = YZ 3. Opposite sides of a parallelogram are equal.

4. XB = AX 4. Given
5. YZ = AX
6. ∠AXY = ∠ABC 5. Both are equal to XB.

6. Being pair of corresponding angles formed on
parallel lines.

198 GREEN Mathematics Book-9

7. ∠YZC = ∠ABC 7. Same as reason no. 6.
8. ∠AXY = ∠YZC 8. Both are equal to ∠ABC.
9. In DAXY and DYZC 9.

i. ∠AYX = ∠YCZ (A) i. Same as reason on. 6.

ii. ∠AXY = ∠YZC (A) iii. From statement no. 8.

iii. AX = YZ (S) iii. From statement no. 5.
10. DAXY ≅ ∆YZC 10. By A.A.S. theorem.
11. AY = YC 11. Corresponding sides of congruent triangles.

Therefore, the straight line drawn through the mid-point of one side of a triangle and
parallel to the other side bisects the third side.

Proved

Converse of Theorem - 14

Statement :

A line segment joining the mid-points of any two sides of a triangle is parallel to third side
and is equal to half of its length.

Experimental verification A

A

A

XY XY XY

B CB CB C

(i) (ii) (iii)

ABC is a triangle in which X and Y are the mid-points of the AB and AC respectively.

To verify: (i) XY = 1 BC
2
(ii) XY//BC, i.e. ∠AXY = ∠ABC or ∠AYX = ∠ACB.

Measurement table

Figure XY BC ∠ABC ∠AXY Remarks
i.
ii. XY = 1 BC
iii. 2
∠AXY = ∠ABC i.e. XY||BC
Result
: From the above figures, we found that XY = 1 BC and ∠ABC = ∠AXY
2
(XY||BC)

Conclusion : A line segment joining the mid-points of any two sides of a triangle is
parallel to third side and is equal to half of it.

GREEN Mathematics Book-9 199

Theoretical Proof A

Given : ABC is a triangle in which AX = XB and AY= YC. X YZ

To prove: (i) XY = 1 BC (ii) XY//BC BC
2
Construction: CZ is drawn such that it is parallel to BX and XY produced to meet at Z.

Statements Reasons

1. In DAXY and DYCZ 1.
i. ∠XYA = ∠ZYC (A) i. Vertically opposite angles.
ii. AY = AC (S) ii. Given.
iii. ∠XAY = ∠ZCY (A) iii. Pair of alternate angles formed on parallel lines.

2. DAXY ≅ DYCZ 2. BY A.S.A. axiom.

3. AX = CZ 3. Corresponding sides of congruent triangles.

4. AX = XB 4. Given.

5. XB = CZ 5. Both are equal to AX.

6. XB//CZ 6. By construction.

7. XZ = BC and XZ//BC 7. The line joining end points of same side of equal
and parallel lines are also equal and parallel.

8. XBCZ is a parallelogram 8. Opposite sides being parallel.

9. XY = YZ i.e. XY = 1 XZ 9. Same as reason on. 3.
2

10. XY = 1 BC 10. Being XZ = BC from statement 4.
2 11. From 7 and 10.

11. XY = 1 BC and XY//BC
2

Proved.

200 GREEN Mathematics Book-9

P

Worked Out EXAMPLES mB
2n
EXAMPLE 1 Look at the given figure carefully and A 136°
find the value of m and n. Q 36cm S

R

Solution : i. AB = 1 QR A and B are being the mid-points of PQ and PR
2 respectively.

or, m = 1 × 36cm
2

∴ m = 18cm

ii. AB//QR A and B are being the mid-points of PQ and PR
respectively.

iii. ∠ABR = ∠BRS Pair of alternate angle formed on the parallel
lines.

or, 2n = 136°

or, n = 136° ∴ n = 68°
2

Therefore, in the given figure, the values of m and n are 18cm and
68° respectively.

X

EXAMPLE 2 In ∆XYZ, AB||YZ and XY = XZ. Prove that AY = BZ. A B

Solution : Given : In the figure, AB||YZ and XY = XZ Y Z
To prove: AY = BZ
Proof:

Statements Reasons

1. XY = XZ 1. Given

2. ∠AYZ = ∠BZY 2. Base angles of an isosceles
triangle.

3. ∠XAB = ∠AYZ and 3. AB||YZ, corresponding angles.
∠XBA = ∠BZY

4. ∠XAB = ∠XBA 4. From statement 2 and 3.

5. XA = XB 5. From statement 4.

6. XY – XA = XZ – XB 6. Subtracting statement 5 from
statment 1.

7. ∴ AY = BZ 7. From statments 7.

Proved

GREEN Mathematics Book-9 201

EXAMPLE 3 In the given figure, PS bisects ∠QPR, P
Solution : T
QM⊥PS and QU = UR, then prove that
M
MU = 1 (PR – PQ). SU
2

Q R

Given : In DPQR.

∠QPM = ∠MPR, QU = UR and QM⊥PS

To prove: MU = 1 (PR – PQ)
2

Construction: QM is produced to meet PR at T.

Proof:

Statements Reasons
1. In DQPM and DTPM
1.

i. ∠QPM = ∠TPM i. Given

ii. PM = PM ii. Common sides of the both triangles

iii. ∠QMP = ∠TMP iii. QM being perpendicular on PS.

2. DQPM ≅ DTMP 2. By A.S.A. axiom.

3. QM = MT, PT = PQ 3. Correspond sides ≅ ∆

4. In DQTR 4.

i. M is mid-point of QT i. QM and MT being sides of
ii. U is mid-point of QR congruent triangles.

ii. Given.

5. MU = 1 TR 5. By using mid-points theorem. (M
2 and U being the mid-points of the
QT and QR respectively.)

6. TR = PR – PT 6. By whole part axiom.

7. TR = PR – PQ 7. From st. 5 and 6 since

8. MU = 1 (PR – PQ) 8. From statement 5 and 7.
2

Proved

P

EXAMPLE 4 In ∆PQR, M and N are the mid-points N
Solution : of the sides QR and PR respectively. X O

X is any points on PQ. Prove that MN M

bisects RX. Q R

Given : In ∆PQR, M and N are mid-points of sides QR and PR
respectively. MN intersects RX at 'O'

To prove: MN bisects RX i.e. RO = OX

202 GREEN Mathematics Book-9

Proof:

Statements Reasons
1. PN = NR 1. N is mid point of PR
2. QM = MR 2. M is mid points of RX
3 MN||PQ 3 By mid point theorem
4. PX||NO 4 ∴ PQ||MN
5. OX = OR 5. ∴ PN = NQ and PX||NO
6. MN bisects RX 6. From statement 5.

Proved

EXAMPLE 5 In the adjoining figure PQ//SR, PX = XS and P Q
Y
QY = YR. X R

Prove that: (i) XY//PQ. (ii) XY = 1 (PQ + SR). S
2

Solution : Given : PQRS is a trapezium in P Q
which PQ//SR, PX = XS X Y

and QY = YR. S R A

To prove: (i) XY//PQ (ii) XY = 1 (PQ + SR)
2
Construction: P and Y are joined and produce it to meet SR produced

at A.

Proof:

Statements Reasons
1. In DPQY and DRYA 1.

i. ∠PQY=∠YRA (A) i. PQ||AR and being alternate angle.

ii. QY = YR (S) ii. Given.

iii. ∠QYP=∠RYA (A) iii. Vertically opposite angles are equal.

2. \ DPQY ≅ DRYA 2. By A.S.A. axiom.

3. PQ = RA, PY = YA 3. Corresponding sides of congruent
triangles.

4. XY=1 SA and XY//SA 4. ∴ X and Y are mid- points of PS and PA
2 respectively of triangle PSA.

5. XY = 1 (SR + RA) and 5. By whole part axiom
2

XY//SR or PQ

6. XY = 12 (SR + PQ) 6. From statement 3 and 5.
and XY//PQ

Proved

GREEN Mathematics Book-9 203

EXAMPLE 6 In the adjoining figure, P, Q, R and S are mid- A C
PS
points AD, AB, BC and CD respectively. Q
D
Prove that : PQRS is a parallelogram. R

Solution : Given : P, Q, R and S are mid-point of AD, B C
AB , BC and CD respectively. A PS

To prove: PQRS is parallelogram. Q D
R
Construction: A and C are joined. B

Proof:

Statement Reasons
1. In DADC 1. P and S are mid-points of AD

PS = 1 AC and PS//AC and CD respectively.
2
2. Q and R mid-points of AB and
2. In DABC BC respectively on triangle
ABC.
QR = 1 AC and QR//AC
2

3. PS = QR and PS//QR 3. From statement (1) and (2)

4. PQ = SR and PQ//SR 4. The lines joining same sides
5. PQRS is parallelogram of equal and parallel lines are
equal and parallel.

5. Opposite sides are equal and
parallel.

X Proved

EXAMPLE 7 In ∆XYZ, XP = PR = RY; PQ||RS||YZ. P Q

Prove that. XQ = QS = SZ RO S

Y Z

Given : In ∆XYZ, XP = PR = RY and PQ||RS||YZ

To prove: XQ = QS = SZ

Solution : Construction: Joins QY which meets RS at O.

Proof:

Statement Reasons
1. In ∆XRS, XQ = QS 1. P is mid point of XR and PQ||RS.

2. In ∆PQY, QO = OY 2. R is mid point of PY and RO||PQ

3. In ∆QYZ, QS = SZ 3. Being QO = OY and OS||YZ
4. XQ = QS = SZ 4. From statement (1) and (3).

204 GREEN Mathematics Book-9 Proved

EXERCISE: 12.2

A. Very Short Questions A

1. a. In the figure, M and N are the mid-points of AC M

and BC respectively. What is the relation of AB B C
N
and MN? A

b. In the figure, P and Q are the mid-points of AB P Q
and AC respectively. If PQ = 5cm, find the length C

of BC. PB

c. In the figure if ER = 5cm, find PR. D E

QR D

d. In the figure P is the mid-point of DE and PQ||DF. P

Write the relation of EQ and QF.

EQ F

B. Short Questions

2. Find the values of x, y and z in the following figures.

a. A b. A c. A
z
y z
M 40° x
N D E M y xF
Q 60° x
RB y 60° C B 50° 110°
45° CD

A

d. z

E

F y x 20° C
50°

BD

3. Find the values of x, y and z in the following figures.

A L 10.6cm M A

a. b. c.

DE P xcm Q
5.7cm xcm z cm 8.4 cm

PQ ycm
4.7cm

BC B 10.2cm C
x cm N

3.6 cm X
PQ
d. e. 3cm 4.6cm
K xcm ycm M A xC
zy
L
YB Z
SR 3.8cm
9.6 cm

GREEN Mathematics Book-9 205

A

C. Long Questions M N
P
4. In the given figure, M and N are the mid-points of the B
sides AB and AC of ∆ABC respectively P is any point on D C
MN, prove that AP = PD. D
R

C

5. In the given figure ABCD is a square, the mid-points of S Q
the sides AB, BC, CD and AD are P, Q R and S respectively. A
Prove that PQRS is a square. B
P

P

6. In the given ∆PQR, K and M are the mid-point of the KE
sides PQ and QR respectively. If KE//QR, prove that QK
= ME. Q R
M

A

7. In the figure alongside, ABC is an equilateral triangle; D, D E
E and F are the middle points of the sides AB, AC and B C
BC respectively. Prove that DEF is also an equilateral
triangle. F

8. In the figure, AD//PQ//BC and DQ = CQ. Prove that : PB
A

(i) AD = 2QR R
(ii) 2PQ = AD + BC D

QC

9. In the given parallelogram ABCD, M and N are the mid- A M
points of the sides AB and CD respectively. Prove that B
(i) ANCM is a parallelogram
P

(ii) BP = PQ = QD Q C
D

N

10. In the given figure, PQ//SR. If N and X are the mid-point P Q
of QR and PR respectively, prove that M is the mid-point M N
of PS.
X

SR

206 GREEN Mathematics Book-9

C X D A P
Y S B
11. In the adjoining figure, AB//CD//XY and Q
BY = DY. Prove that AB = CD. R
R
AB
D
12. In given figure, P, Q, R and S are the mid-point of AB, AC, CD
B
and BD respectively. Prove that PQRS is a parallelogram.

P D C

13. In the given figure, PC bisects ∠QPR, QA⊥PC S
C
and B is mid-point of QR, prove that AB = 1 (PR A R
2 CB
– PQ).
Q

P

14. In the figure, A, B, C and D are the mid-points of PQ, A
QR, RS and SP respectively, prove that ABCD is a

parallelogram. Q

12.4 Pythagoras Theorem

Statement:

The square on the hypotenuse is equal to the sum of the squares of the other two sides.
N
L
Experimental verification
L

M

M

MN N L

In ∆LMN ∠M is a right angle and LN is the hypotenuse.

To verify: LN² = MN² + LM²

Measurement table

Figure LN² LM² MN² MN² + LM² Remarks
i. LN² = MN² + LM²
ii.
iii.

Result : From the above figure, we found that LN² = MN² + LM².

Conclusion : The square on the hypotenuse is equal to the sum of the squares on the
other two sides.

Converse of the Pythagoras theorem

Statement: If the square of one side of a triangle is equal to the sum of the squares of other
two sides then the angle opposite to the first side is a right angle.

GREEN Mathematics Book-9 207

Worked Out EXAMPLES B
13 cm
EXAMPLE 1 In the given figure, ABC is a right angled

triangle of AB = 5cm and AC = 13cm, find 4cm
6 cm
the perimeter of the ∆ABC. A
5cm
C

Solution : By the question,
In ∆ABC
∠B = 90°, AB = 5cm, AC = 13cm and BC = ?
By using the Pythagoras theorem
We can write,
AC² = AB² + BC²
or, (13cm)² = (5cm)² + (BC)²
or, 169cm² = 25cm² + BC²
or, 169cm² – 25cm² + BC²
or, 144cm² = BC²
\ BC = 12cm
Again,
Perimeter of the DABC = AB + BC + AC
= 5cm + 12cm + 13cm
= 30cm
Therefore, the perimeter of the DABC is 30cm.

EXAMPLE 2 Determine whether the following triangles are right angled or not?

a. A b. A

5cm 10 cm

B CB C
7cm 8 cm
Solution :
a. In ∆ABC, \ AB² = 16cm²
\ BC² = 49cm²
AB = 4cm \ AC² = 25cm²

BC = 7cm

AC = 5cm

208 GREEN Mathematics Book-9

Here, BC is the longest side of DABC and BC² ≠ AB² + AC² so, ABC
is not a right angled triangle.

b. In DABC,

AB = 6cm \ AB² = 36cm²

BC = 8cm \ BC² = 64cm²

AC = 10cm \ AC² = 100cm²

Here, AC is the longest side of DABC and AC² = AB² + BC² so, ABC
is a right angled triangle.

EXERCISE: 11.4

A. Very Short Questions

1. Determine whether the following are right angled triangles or not? A

A A c.
5cm 8 cm x6 cm
a. b. 6 cm
4 cm 3cm 13 cm
5.5cm
C
B C B 8 cm CB
4cm P 17cm 6 2 cm
d. f. O
A e.
5 cm 8cm R 20cm

12cm

B C Q 15cm M 16cm N
6 cm

2. In each of the triangles below, the angle of reference is coloured. Find perpendicular,
hypotenuse and base in each.

a. b. c. d. C e.

A E
P XY

BC B AF
QR X D

3. Calculate the value of x in the following figures. A

A A

a. x b. x

9cm B c. 13cm
12 cm
B C 10cm C
40cm
C B

GREEN Mathematics Book-9 209

B. Short Questions

4. Calculate the values of x, y and z in the following figures.

A

xP13 cm S Az
(x + 4)cm D

xy

BE
6 cm C 3 cm
a. (x + 7)cm b. c.
5 cm
y
8 cm
8 cm
C QR
B (x + 3) cm 7 cm

C. Long Questions A

5. a. In the figure, ABC is an isosceles right angled
triangle, prove that AC² = 2AB².

B C
A


b. In ∆ABC, AB = BC = CA = a. AD is a perpendicular drawn
from A to the side BC. Prove that :

i. AD = 3 a ii. Area of ∆ABC = 3 a2 B C
2 4 D

6. In the given figure, ABC is a triangle ∠A = 90°, AD⊥BC and AD = h. Prove that.

i. ha = bc ii. 1 = 1 + 1 A
h² c² b²
cb
h

BDC
a

Project Work:

1. Take a carton box and make piece of square whose sides are pythagorous triplet
and verify the relation of h2 = p2 + b2.

2. Make two equal parallelogram by using paper, verify the all properties of
parallelogram. Show to your teacher.

210 GREEN Mathematics Book-9

13

Constructions

Estimated Teaching Periods : 7

Pythagoras was a Greek philosopher who made important developments
in mathematics, astronomy, and the theory of music. The theorem now
known as Pythagoras's theorem was known to the Babylonians 1000 years
earlier but he may have been the first to prove it.

Contents

13.1 Construction of triangle
13.2 Construction of quadrilateral

Objectives

At the end of this unit, students will be able to:
introduce construction
able to construct triangle, quadrilateral, parallelogram, rectangle, rhombus, square,
kite, trapezium, etc.

Materials

Geometry box, geo board, chart paper, pencil, white paper, etc.
Triangular and quadrilateral shapes.

GREEN Mathematics Book-9 211

13.1 Construction of triangles

We are familiar with the word construction. In general, we think that construction means
the building the house or other physical structure. But in this chapter construction means
drawing geometrical figures according to the given measurements. Let us review some of
the steps to be followed in construction of geometrical figure like triangle, quadrilateral.
– At first draw a rough sketch of the required figure with help of given data.
– Now, mention the given measurements in the rough sketch.
– Finally, understand the rough sketch and construct actual figure.
Look at the following worked out example.

Worked Out EXAMPLES Rough

EXAMPLE 1 Construct an equilateral triangle ABC of side 4.5cm. A
4.5cm
A
C
Solution :

B

4.5cm 4.5cm

XB 4.5cm CY

Steps of construction:

Step - 1 Draw a line XY and cut BC = 4.5cm.

Step - 2 Taking B an iniial point, cut an arc of radius 4.5 cm up and
taking C as an initial point, cut an arc of radius of 4.5cm to
cut previous arc so that the two arcs intersect at A.

Step - 3 Join AB and AC.

Then, we get the equilateral ∆ABC of its side 4.5cm.

212 GREEN Mathematics Book-9

EXAMPLE 2 Construct an isosceles triage ABC in which base BC = 5.5 cm and
Solution : AB = AC = 4.2 cm.

Rough

AA

4.2cm

4.2cm B 5.5cm C

4.2cm

XB 5.5cm CY

Steps of construction:

Step - 1 Draw a line XY and cut BC = 5.5cm.

Step - 2 Taking B as an initial point, cut an arc of radius 4.2 cm and
C as an initial point, cut an arc of radius of 4.2cm so that two
arcs intersect at A.

Step - 3 Join AB and AC.
Then, we get an isosceles ∆ABC of the given measurement.

EXAMPLE 3 Construct a triangle ABC in which AB = 6cm, BC = 4cm and the
Solution : median CD = 3.5cm.

Rough

CC

4cm

3.5cm

AD B
6cm
Q

XA D 6cm Y
B

P

Steps of construction:

Step - 1 Draw a line XY and cut AB = 6cm.

Step - 2 Taking A as an initial point, cut an arc of greater than 3cm
above and below AB and taking B as an initial point cut
the previous arcs with the same length of as before up and
below. So, that we get the points P and Q. Join PQ to cut at D.

GREEN Mathematics Book-9 213

Step - 3 Taking B as an initial point, cut an arc of length 4cm and
taking D as an initial point cut an arc of length 3.5cm,
So, that it cuts the previous arc at C.

Step - 4 Join BC, CD and CA.

Then, we get the ∆ABC with the given measurements, where
∆ADC = ∆BDC, since median bisects the original triangle into
two equal parts.

EXAMPLE 4 Construct a triangle PQR, in which QR = 5cm, ∠PQR = 75° and
Solution : ∠QPR = 45°.

P Rough

P

45° 45°

Q 75° 5cm ?R

X 75° 60° Y
Q 5cm R
Steps of construction:

Step - 1 Draw a line XY and cut QR = 5cm.

Step - 2 Make 75° angle at point Q.

Step - 3 Find ∠QRP by 180° – (75° + 45°) = 60°.

Step - 4 Now, make 60° angle at point R.

Step - 5 Produce the arms of both angles to meet at P. Then we get
triangle PQR with the given measurements.

EXAMPLE 5 Construct a right angled triangle ABC, in which ∠ACB = 30°,
Solution :
AB = 4.5cm. Rough

C

30°

CA ? B
4.5cm

X 60° Y
214 GREEN Mathematics Book-9
4.5cm
AB

Steps of construction:
Step - 1 Draw a line XY and cut AB = 4.5cm.
Step - 2 Make 90° angle at point A.
Step - 3 Find ∠ABC by 180° – (90° + 30°) = 60°.
Step - 4 Draw 60° angle at point B.
Step - 5 Produce both the arms angles to meet at C. Then we get the

required triangle ABC.

EXERCISE: 13.1

1. a. Construct an equilateral triangle ABC having its side 4cm.
b. Construct an isosceles triangle PQR in which base PQ = 6.5cm and QR = PR =

4.8cm.
c. Construct a triangle ABC such that BC = AB = 6cm and the median BD = 4cm.
2. a. Construct an equilateral triangle PQR of a side 5.5cm.
b. Construct an equilateral triangle XYZ of a side 3.5cm.
3. a. Construct an isosceles triangle ABC in which base AB = 6.5cm and rest of sides

4.5cm each.
b. Construct an isosceles triangle PQR in which QR = 5cm, PQ = PR = 6cm.
4. a. Construct a triangle MNO with its median OP, where MN = 5.5cm, MO =

6.5cm and ON = 7cm.
b. Construct a triangle ABC in which median CD = 4.5cm, BC = 6cm and AB =

5.5cm.
5. a. Construct an isosceles ∆ABC with its base BC = 5cm ∠ABC = 50° = ∠ACB.
b. Construct a right angled triangle PQR, in which ∠PQR = right angle, PQ = 6cm

and QR = 7.5cm.
c. Construct a triangle MNO, in which MN = 6.5cm, ∠OMN = 75°, ∠MON = 60°.
d. Construct a scalene triangle XYZ, in which XY ≠ XZ by taking length of sides

by yourself.

GREEN Mathematics Book-9 215

13.2 Construction of quadrilaterals

EXAMPLE 1 Construct a parallelogram ABCD in which ∠B = 60°, AB = 5.6 cm
Solution :
and BC = 4.2 cm. Rough

D 5.6cm C DE

4.2cm

A 60° B

5.6cm

4.2cm 4.2cm

A 5.6cm 60°
X Y

B

Steps of construction:

Step - 1 Draw a line XY and cut AB = 5.6cm.

Step - 2 Draw an angle of 60° at B with base BA.

Step - 3 Taking B as an initial point cut an arc of length 4.2cm. So that
BC = 4.2cm. Similarly, by taking A as an initial point cut an
arc of length 4.2cm.

Step - 4 Taking C as an initial point with length of arc 5.6cm cut the
previous arc at D.

Step - 5 Join CD and DA.

Then, we get the parallelogram ABCD with the given measurements.

EXAMPLE 2 Construct a parallelogram ABCD in which ∠BAC = 30°, AB = 4.5cm.
Solution :
and diagonal BD = 4.8cm.

QD CP Rough

DC

M

A 30° B
4.5cm
M

XA 30° Y
4.5cm B

Steps of construction:
Step - 1 Draw a line XY and cut AB = 4.5cm.

216 GREEN Mathematics Book-9

Step - 2 Draw an angle of 30° at A with the base AB.
Step - 3 Taking B as an initial point cut an arc of length 2.4cm on AP

at M.
Step - 4 Join BM and produce it to Q.
Step - 5 Mark the point C such that AM = MC and mark the point D

such that BM = MD.
Step - 6 Join BC, CD and DA.
Then, we get the parallelogram ABCD with the given measurements.

EXAMPLE 3 Construct a square ABCD with a side 5cm.

Solution : DC Rough

DC

A 5cm B

X 5cm Y
A B

Steps of construction:
Step - 1 Draw a line XY and cut AB = 5cm.
Step - 2 Make 90° angle at point A.
Step - 3 Cut an arc of 5cm on the arm of 90° so that AD = 5cm.
Step - 4 Again cut another arc with length 5cm from point B and D

to intersect at C.
Step - 5 Join CD and BC. Then we get the required square.

EXAMPLE 4 Construct a square PQRS with its diagonal 6cm.

Solution : Y Rough
SR
SR

O P 5cm Q
6cm

P Q
X

Steps of construction:

Step - 1 Draw a line XY and cut PR = 6cm.

Step - 2 Draw bisector line of PR by taking more than 3cm arc.

Step - 3 Cut 3cm along OS and OQ.

Step - 4 Join SP, SR, PQ and QR. Then we get the required square.

GREEN Mathematics Book-9 217

EXAMPLE 5 Construct a rhombus with its PR = 6cm and diagonal QS = 7.5cm.

Solution : 3cm Rough
O 7.5cm
XQ SR
7.5cm

PQ

SY

Steps of construction:
Step - 1 Draw a line XY and cut QS = 7.5cm.
Step - 2 Draw a bisector line of QS.
Step - 3 Cut OP = OR = 3cm as shown in the figure.
Step - 4 Join QP, QR, PS and RS. Then we get the required rhombus.

EXAMPLE 6 Construct a parallelogram PQRS, in which PQ = 5cm, PR = 7cm and
QS = 6cm.

Solution : S

R

Rough

3cm SR

M 6cm 7cm

PQ

3.5cm

XP 5cm QY

Steps of construction:

Step - 1 Draw a line XY and cut PQ = 5cm.

Step - 2 Draw PM = 1 PR and QM = 1 QS with the given arcs.
2 2
Step - 3 Produce PM upto R and make PM = MR and QM upto S to
make QM = MS.

Step - 4 Join SR, QR and PS. Then, we get the required parallelogram.

218 GREEN Mathematics Book-9

EXAMPLE 7 Construct a parallelogram with its diagonals PR and QS respectively
Solution : of 7cm and 6cm. Also angle between the diagonals is 60°.

S

Rough

S 60°7cm R
P Q
60° Y
O 7cm R

XP

Q

Steps of construction:
Step - 1 Draw a line XY and cut PR = 7cm.
Step - 2 Draw a bisector of PR to get the mid point O of PR.
Step - 3 Make 60° angle at O and cut OS = OQ = 3cm from the point O.
Step - 4 Join PQ, QR, RS and PS. Then, we get the required parallel-

ogram.

EXAMPLE 8 Draw a rectangle with a diagonal 7cm and one of the adjacent sides
Solution : is 6cm.

D C Rough

DC

7cm

7cm A B

XA BY

Steps of construction:

Step - 1 Draw a line XY and cut AB = 6cm.

Step - 2 Make 90° angle at B.

Step - 3 Cut an arc of 7cm on the arm of 90° from A to cut at C.

Step - 4 With the points C and A draw intersecting point at D with
CD = 6cm and AD = BC.

Step - 5 Join DC and DA. Then, we get the required rectangle.

GREEN Mathematics Book-9 219

EXAMPLE 9 Construct a trapezium ABCD in which Rough
Solution : ∠ADC = 120°, AB = 6cm, AD = 4cm, BC = 5cm
and AB||DC. D C
120° 5cm
B
4cm

A 6cm

DC

4cm 120° 5cm

60°

XA 6cm BY

Steps of construction:

Step - 1 Draw a line XY and cut AB = 6cm.

Step - 2 Find ∠DAB = 60°. So, that ∠DAB + ∠ADC = 180°.

Step - 3 Make 60° angle at A and cut an arc of 4cm on the arm of 60°
at point D.

Step - 4 Make 120° angle at D as shown in figure.

Step - 5 Cut an arc of5cm from B on the arm of 120° angle to get the
point C.

Step - 6 Join BC and DC. Then, we get the required trapezium
ABCD.

EXERCISE: 13.2

1. Construct the square PQRS is which

a. PQ = 6.4cm. b. RS = 4.7cm

c. Diagonal PR = 5.9cm d. Diagonal QS = 6.1cm

2. a. Construct the rectangle ABCD is which AB = 6.1cm and BC = 4.7cm.

b. Construct the rectangle PQRS in which PQ = 6.5cm and QR = 4.1cm.

c. Construct the rectangle ABCD in which diagonals AC = BD = 6.6cm and angle
between them is 45°.

d. Construct the rectangle PQRS in which diagonals PR = QS = 7.2cm and angle
between them is 60°

3. a. Construct the rhombus ABCD in which AB = 5.3cm and ∠ABC = 30°.

b. Construct the rhombus PQRS in which diagonals PR = 7.2cm and QS = 5.6cm.

220 GREEN Mathematics Book-9

4. a. Construct the parallelogram ABCD in which AB = 4.6cm, BC = 3.2cm and
∠ABC = 60°.

b. Construct the parallelogram ABCD in which base AB = 5.6cm. diagonal BD =
7.2cm and ∠BAD = 30°.

c. Construct the parallelogram PQRS in which diagonals PR = 6.4cm, QS = 7.2cm
and they bisect each other making an angle of 60°.

5. a. Construct the trapezium ABCD in which AB = 5.4cm, BC = 6cm, ∠BAD = 60°,
∠BCD = 90° and AD//BC.

b. Construct the trapezium ABCD in which AB = 5.7cm, diagonal AC = 7.2cm,
∠BAC = 60°, CD = 5.4, and AB//DC.

c. Construct the trapezium ABCD in which AB = 4.6cm, diagonal AC = 7.4cm,
AD = BC = 5.4cm.

6. Construct PQRS quadrilateral under the following condition.
a. PQ = QR = 4.5cm, RS = SA = 5cm and ∠PQR = 75°
b. PQ = 4cm, QR = 3cm, PS = 4cm, QS = 6cm and RS = 4cm
c. RQ = 5cm, PQ = 3.6cm, PS = 4cm ∠Q = 75° and RS = 4.5cm.

Project Work:
1. Construct a triangle with sides of 3cm, 4cm and 7cm, if not discuss the condition.
2. Draw a quadrilateral with sides 3.5cm, 4.5cm, 4.5cm and 6cm with the help of

compass and ruler, if not discuss to construct quadrilateral.

GREEN Mathematics Book-9 221

14

Similarity

Estimated Teaching Periods : 5

Pythagoras was a Greek philosopher who made important developments
in mathematics, astronomy, and the theory of music. The theorem now
known as Pythagoras's theorem was known to the Babylonians 1000 years
earlier but he may have been the first to prove it.

Content

14.1 Introduction
14.2 Similar Triangles
14.3 Similar Polygons

Objectives

At the end of this unit, students will be able to :
know the difference of congruent triangles.
properties of smilarity of polygons.
use it in daily life.

Materials

Different types, models of triangles and quadrilaterals.
Chart of properties of different plane figure.

222 GREEN Mathematics Book-9

14.1 Introduction

Let us look at the following figures.

Figure - I Figure - II

In figure I, we have three right angled triangles of
different sizes. In figure II and we have three circle
of different radii and three hexagons of different
sizes, respectively. Do they have same shape? Are
they similar?

In the figure, all the triangles, circles and hexagons Figure - III
are similar though they have different sizes.

Thus, figures which have same shape but not necessarily same size are known as similar
figures.

14.2 Similar Triangle

Two or more triangles are said to be similar if A

a. the corresponding angles of two triangles are D
equal.

In the given triangles ABC and DEF if. CE F

B

∠A = ∠D, ∠B = ∠E and ∠C = ∠F, they are

similar.

∆ABC is similar to ∆DEF is written as ∆ABC ∼ ∆DEF.

b. the corresponding sides are proportional. The ratio of the area of two similar
triangles is equal to the ratio of the
In the given triangles ABC and DEF, squares of their corresponding sides

AB = 4 = 1 D AD
DE 8 2 10cm
A
BC 3 1 B 3cm C E B CE F
EF 6 2 4cm5cm 6cm
= = 8cm

and ADCF = 5 = 1 F ∆ABC = AB²
10 2 ∆DEF DE²

∴ ∆ABC ∼ ∆DEF. = BC² = AC²
EF² DF²

GREEN Mathematics Book-9 223

14.3 Similar Polygons

Similar polygons are combination of similar P U
Q T
triangles. The similar polygons have following A F
E R S
properties.
D
a. The corresponding angles are equal. B

In the figures, ∠A= ∠P, ∠B = ∠Q, ∠C = ∠R, C
∠D = ∠S, ∠E = ∠T and ∠F = ∠U.

b. The corresponding sides are proportional

In the figures AB = BC = CD = DE = EF = AF
PQ QR RS ST UT PU

c. The corresponding diagonals are proportional.

In the figure, AC = CF = CE
PR RU RT

d. The triangles separated by the corresponding diagonals are similar.

In the figure, ∆ABC ∼ ∆PQR, ∆ACF ∼ ∆PRU ∆CEF ∼ ∆PUT and ∆CDE ∼ ∆RST

All regular polygons of same number of sides such as equilateral triangles,
squares, rectangles etc. are similar.

All circles are similar.

Any two line segments are similar.

Theorem - 16

Two equiangular triangles are similar

Experimental verification: A A
P
A P

P

B CQ RB CQ RB CQ R

Figure - I Figure - II Figure - III

Construction: Three pairs of triangles ABC and PQR are drawn where ∠B = ∠Q and ∠C
= ∠R (we can draw any two corresponding angles equal). Here, the remaining angles ∠A
and ∠P are also equal.

Now, measure all the corresponding sides and calculate the ratio of respective sides and
tabulate the results.

224 GREEN Mathematics Book-9

Verification Table:

∆ABC ∆PQR Ratio of corresponding sides

Figure AB BC CA PQ QR PR AB BC CA Results
PQ QR PR

I AB = BC = CA
PQ QR PR

II AB = BC = CA
PQ QR PR

III AB = BC = CA
Conclusion: The equiangular triangles are similar. PQ QR PR

Theorem - 17

If any two corresponding sides of triangles are proportional and the angles included by
them are equal, the triangles are similar.

Experimental verification: A A
D
A D

D

B CE FB CE FB CE F

Figure - I Figure - II Figure - III AB
DE
Construction: Three pairs of triangle ABC and DEF are drawn in such a way that =
BC
EF and ∠B = ∠E in each pair.

Now, The remaining angles and ratio of remaining corresponding sides are measured and
the result tabulated in the table.

Verification Table:

Measurement Ratio of corresponding sides
of angles
Figure AC Results
∠B ∠E DF
AB DE AC DF

I AB = BC , ∠B = ∠E
DE EF

II AB = BC , ∠B = ∠E
DE EF

III AB = BC , ∠B = ∠E
DE EF

Conclusion: If any two corresponding sides of triangles are proportional and the angles
included by them are equal, the triangles are similar.

GREEN Mathematics Book-9 225

Theorem - 18

If two triangles have their corresponding sides proportional, then the triangles are similar.

Experimental verification: P P
A
P A

A

B CQ RB CQ RB CQ R
Figure - I Figure - II Figure - III

Construction: Three sets of triangles ABC and PQR are drawn such that AB = BC = AC .
PQ QR PR

Now, Measure all the corresponding angles of each of triangles and tabulate the results.

Verification Table:

Figure ∆ABC ∆PQR Results
I ∠A ∠B ∠C ∠P ∠Q ∠R ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R

∴∆ABC ∼ ∆PQR
II ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R

∴∆ABC ∼ ∆PQR
III ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R

∴∆ABC ∼ ∆PQR

Conclusion: If two triangles have their corresponding sides proportional, then the
triangles are similar.

Worked Out EXAMPLES

EXAMPLE 1 In the figure AB||DE. Show that ∆ABC ∼ ∆DEC. A
D
Also find the length of AB if BC = 12cm, EB = 4cm
E
Solution : and DE = 6cm. B C

In ∆ABC and ΔDEC,

i. ∠BAC = ∠EDC (A) [∵ Corresponding angles]

ii. ∠ABC = ∠DEC (A) [∵ Corresponding angles]

iii. ∠ACB = ∠DCE (A) [∵ Common angle]

∴ ∆ABC∼ ΔDEC, [They are equiangular]

226 GREEN Mathematics Book-9

Now, AB = BC [∵ Corresponding sides of similar triangles
or, DE EC are proportional]

AB = 12 [∵ EC = BC – BE =12 – 4 = 8]
6 8

or, AB = 6 × 12
8

∴ AB = 9cm

P

EXAMPLE 2 In the figure, ∠QPR = ∠PSQ. Prove that 4cm 10cm
ΔPQR ∼ΔPQS. Also find the length of QR.

Solution : In ∆PQR and ΔPQS,

i. ∠PQR = ∠PQS (A) [∵ Common angles] QR S
ii. ∠QPR = ∠PSQ (A) [∵ Given]
8cm

iii. ∠PRQ = ∠QPS (A) [∵ Remaining angles]

∴ ∆PQR ∼ ΔPQS, [∵ By AAA]

Now, QR = PQ [∵ Corresponding sides of similar triangles are proportional]
PQ QS

or, QR = 4
4 8

∴ QR = 2cm A

EXAMPLE 3 In the figure if ΔABE ≅ ΔACD, prove that D E
ΔADE ∼ΔABC. B C

Solution :
Given : ∆ABE ≅ ΔACD,

To prove : ∆ADE ∼ ΔABC

Statements Reasons
1. AB = AC, AE = AD 1. Corresponding sides of congruent

2. AB = AC or AB = AD triangles ABE and ACD.
AD AE AC AE 2. From statement 1.

3. ∠BAC = ∠DAE 3. Common angle
4. ∆ADE ∼ ΔABC 4. From statement 2 and 3; if two sides

of one triangle are proportional to the
corresponding sides of other and the
angles included by them are equal, the
triangles are similar.

Proved

GREEN Mathematics Book-9 227

EXAMPLE 4 In the figure ∠ACB = 90° and CD⊥AB. C

Prove that BC² = BD .
CA² AD

Solution : Given : In the figure, ∠ACB = 90°and CD⊥AB A B
D
BC² BD
To prove : CA² = AD

Proof:

Statements Reasons
1.
1. In ∆ACD and ∆ABC,
i. ∠ADC = ∠ACB i. Both are right angles
ii. ∠CAD = ∠BAC ii. Common angle
iii. ∠ACD = ∠ABC ii. Remaining angles of triangles
2. By A.A.A. axiom
2. ∆ACD ∼ Δ ABC 3. Corresponding sides of similar
triangles are proportional.
3. AC = AD ∴AC² = AD × AB 4. As reason 1.
AB AC
5. Corresponding sides congruent
4. Similarly triangles are proportional.

∆BCD ∼ΔABC 6. Dividing statement 5 by statement
5. BC AB 3

BD = BC ∴ BC² = BD × AB Proved

6. BC² = BD × AB
AC² AD × AB

∴ BC² = BD
AC² AD

EXAMPLE 5 In the figure, ABCD is a parallelogram A D
and M is the mid- point of BC. Prove C

that OD = 2OM.

Solution : Given : ABCD is a parallelogram and M is O
M
the mid- point of BC. B

To prove : OD = 2OM

Proof:

Statements Reasons
1. In ∆AOD and ∆COM, 1.

i. ∠OAD = ∠OCM i. Alternate angles
ii. ∠ADO = ∠CMO ii. Alternate angles
iii. ∠AOD = ∠COM iii. Vertically opposite angles

228 GREEN Mathematics Book-9

2. ∆AOD ∼ ΔCOM 2. By A.A.A. axiom
3. Corresponding sides of similar
3. OD AD
OM = CM triangles are proportional.
4. Opposite sides of a parallelogram
4. OD = BC
OM are equal and M is the mid - point
1 BC of BC.
2
Proved.
OD
or, OM = 2

∴ OD = 2OM

EXERCISE: 14.1

A. Very Short Questions

A D E
F
1. In the figures ΔABC and ΔDEF are similar triangles.
a. Name the corresponding sides of the triangles.

b. Which ratio from ∆DEF is equal to the ratio AB ? B C
BC

c. Write two ratios from the figures which is proportional to AB .
DE
d. Is DE = EF ?
AB BC

B. Short Questions

2. a. In the figure, AB = AC . Show that ∆ABC ∼ ΔADE. A
AD AE E
C
Which side of ΔADE is corresponding to BC of ΔABC? D
P
b. In the figure, ∠ABD = ∠ACB. Show that B AB
ΔABD ∼ ΔABC. Which angle of ΔABC is D
corresponding to ∠ADB of ΔABD? C

c. In the adjoining figure, MN||QR. Show the ΔPQR ∼ ΔPMN. M N
Also write the relation of corresponding sides. R

Q

GREEN Mathematics Book-9 229

d. In the adjoining figure, AB||CD. Show that ΔAOB ∼ ΔCOD. B D
O
A

A D 3cmE
5cm 4cm F
3. a. In the given triangles, ABC and DEF are
similar. Find the length of AC.

B C
C
P 6cm
y 5cm
b. In the adjoining figure, C
MN||QR and ∆PMN ∼
∆PQR. Find the value of x 2cmMN
and y. 4cm x

QR

8cm

A 5cm

c. In the figure, AB||CD. Find the length of AB.

O 7.5cm
2cm
B D
A
d. In the adjoining figure, ∠BAD = ∠ACB, BC = 8cm,
AC = 10cm and AB = 4cm. Find the length of BD. 4cm 10cm

B C
D
e. In the given triangle PQR, ∠QPR = 90° and MP⊥ 8cm

QR. Prove that ∆PQR ∼ΔPMR. Also find MP and P

MR if PQ = 3cm and PR = 4cm. A 3cm

4. a. In the figure, PQ||SR, TS = 2PT Q R
M
T S
and PQ = 5cm. Find the length P R

of AS . Q

D

S

E C 3.6cm R
6cm T
b. In the figure, ABCDE and PQRST are 6cm

similar polygons. Find the length of PR,

AD and TQ. AB PQ
4cm 2cm

230 GREEN Mathematics Book-9

C. Long Questions B
5. In the given figure, ∠ACO = ∠DBO. Prove that AC.OD = BD.OA. C

O

6. In the figure, MN||BC. Prove that AM.BC= AB.MN. AA D
X
MN

BC

7. In the figure, ∠WYZ = ∠WXY. Prove that WY² = WX. WZ. Z

Y W
P
8. In the figure, PQR is a right angled triangle in which N
∠PQR = 90° and MN⊥PR, Prove the QR.PN = PQ.MN. M
Q
R
A
9. In the ΔABC, ∠A = 90° and AD⊥BC. Prove that AB² +AC² = BC²

B D C
A

10. In the figure, AB⊥CE and ED⊥AC. Prove that AC = AB .
CE DE
D

C B E
A D
F E
11. In the figure, AD||BE, AB||CD and AE= 3AF. Prove that S
BE = 3BC. C R

B

12. In the figure, PQRS is a parallelogram. If OP = 2OR, prove that P

M is the mid-point of QR. O
M
Q

GREEN Mathematics Book-9 231

13. In the figure MQ||BP and QN||PC. Prove that MN||BC. M A N
B C
Q

P

E

14. In ∆DEF, ∠D = 90°, DG⊥EF. If EF = x, DF = Z, DE = P, GF = y P x
and GE = x – y, prove that x2 = p2 + z2. G

D Z F
A

15. In a right angled triangle ABC, AD⊥BC. Prove that

a. BC2 : BA2 = BC : BD B DC
b. BC2 : CA2 = BC : CD A

16. In ∆ABC, ∠A = 90° and PQRS is a square. Prove that PQ
SR2 = BS × RC.

BS R C
D
17. In the figure, ∆BFE ∼ ∆BCD and ∆EFC ∼ ∆ABC where A E z
FC z x y C

EF||DC. If AB = x, EF = y and CD = z, show that BF = x . F

B

232 GREEN Mathematics Book-9

15

Circle

Estimated Teaching Periods : 9

Pythagoras was a Greek philosopher who made important developments
in mathematics, astronomy, and the theory of music. The theorem now
known as Pythagoras's theorem was known to the Babylonians 1000 years
earlier but he may have been the first to prove it.

Contents

15.1 Introductions
15.2 Theorems

Objectives

At the end of this unit, students will be able to :
define different parts of the circle
verify experimentally and prove theorems theoretically
solve the problems related to circle
demonstrate different parts of circle

Materials

Models of circle, chart of circle with the labelling of its different parts
Instrument box, chart paper, geo-board, etc.

GREEN Mathematics Book-9 233

15.1 Introduction

A circle is the locus of a moving point which always moves an B A
equal distance from a fixed point. The fixed point is called the
centre of the circle. In other word, the circle is a plane figure OD
bounded by curved line in which all the points are on the curved E
line and are equidistant from the fixed point. In the figure O is the
centre of the circle. All the points A, B, C, D and E are equidistant C
from O. i.e. OA = OB = OC = OD = OE = radii

Circumference

The perimeter of a circle or total length of curved line is the
circumference of the circle.

O
Circumference

Radius (Plural : Radii) B A
D
The straight line segment which joins the centre and any point of O
the circumference is called radius. In the figure OA, OB, OC and C
OD are radii of the circle.

Radii of same circle are equal. i.e. OA = OB = OC = OD

Name Definition Figure

A line segment which joins any two points on the d
r Or
Diameter circumference and passes through the centre of circle A B
is called diameter. In the figure, AB is the diameter

which is double of radius. i.e. d = 2r

Chord The line segment joining any two points of the C D
circumference of a circle is called the chord of the A B
circle. In the figure, AB, CD and EF are the chords.
AB is also the diameter. Diameter is the longest chord. E OF

Secant A straight line which cuts the circle at two points is O B
called a secant. In the figure the line AB cuts the circle AM N
at two points M and N. Thus AB is a secant of the
circle.

234 GREEN Mathematics Book-9

Name Definition Figure

Tangent A straight line which touches the circle at a point OB
is called a tangent. The point where the tangent
Arc of a touches the circle is point of contact. In the figure P
circle AB is a tangent and P is the point of contact. A

Sector Arc is the part of the circumference of a circle. In major arc
the figure AB is the arc. The arc AB is written as AB
Segment A OB
Sector is the region bounded by two radii and the
Intersecting arc between them. E arc minor arc
circles
Concentric The region bounded by a chord and an arc is called A
circles segment of the circle. The segment containing the minor sector
minor arc is the minor segment.
OB
Two circles are said to be intersecting circles if
they intersect each other at two points. A and B are major sector
intersecting points. major segment

Two or more circles having the same centre are OB
called concentric circles. Circles A, B and C are
concentric circles. A minor segment
A

OP

B
C

B
A

Theorem - 18

The perpendicular drawn from the centre of a circle to a chord, bisects the chord.

Given : O is the centre of circle and OM is perpendicular to chord AB. O
To Prove : AM = BM

Construction : Join OA and OB AM B

S.N. Statements S.N. Reasons
1. In ∆ AOM and ∆ BOM 1.
i. Both are right angles; OM⊥AB
i. ∠AMO = ∠BMO (R) ii. Radii of the same circle.
ii. OA = OB (H) iii. Common side
iii. OM = OM (S) 2. By R.H.S. axiom
2. ∆AOM ≅ ∆BOM 3. Corresponding sides of congruent triangles.
3. AM = BM

∴ The perpendicular drawn from the centre to a chord bisects the chord. Proved.

GREEN Mathematics Book-9 235

Converse of Theorem - 18

A straight line joining the centre of circle and mid-point of a chord is O

perpendicular to the chord.
Given: O is the centre of circle and M is the mid-point of chord AB. A M B

To Prove : OM ⊥AB

Construction: Join OA and OB

Proof:

S.N. Statements S.N. Reasons
1. In ∆AOM and ∆BOM 1.
i. OA = OB (S) i. Radii of the same circle
2. ii. AM = BM (S) ii. Given
3. iii. OM = OM (S) iii. Common side
4. ∆AOM ≅ ∆BOM 2. By S.S.S. theorem
5. ∠AMO = ∠BMO 3. Corresponding angles of congruent triangles.
∠AMO + ∠BMO = 180° 4. Linear pair angles
∠AMO +∠AMO = 180° 5. From statements 3 and 4
or, 2 ∠AMO = 180°
∴ ∠AMO = 90°
i.e. OM⊥AB.

Proved.
∴ The line joining the centre and mid point of a chord is perpendicular to the chord.

Theorem - 19

The perpendicular bisector of the chord of a circle passes through the centre.
Experimental Verification:

AC AB AC
O

BO D

D BD

C

Construction : Using pencil, compass and ruler.

i. Three circles of different radii are drawn.

ii. Perpendicular bisectors of chords AB and CD are drawn which meet at O.

iii. In each figure O is joined with A, B, C and D.

236 GREEN Mathematics Book-9

Now, measures of OA, OB, OC, and OD are tabulated below.
Verification Table

Figure OA OB OC OD Results

I OA = OB = OC = OD which shows
that O is the centre of circle.

II OA = OB = OC = OD

III OA = OB = OC = OD

Conclusion : The perpendicular bisector of the chord of a circle passes through the centre.

Theorem - 20

Equal chords of a circle are equidistant from the centre of the circle.

Given: O is the centre of the circle, where chord AB = chord CD. OM and ON are drawn

perpendiculars to AB and CD respectively. AC

To Prove : OM = ON

Construction: Join OA and OC M ON
Proof: BD

S.N. Statements S.N. Reasons

1. AM = 1 AB and CN = 1 CD. 1. Perpendicular drawn from the centre to a
2 2 chord bisects it

2. AB = CD 2. Given

3. AM = CN 3. From statement 1 and 2

4. In ∆AMO and ∆CNO 4.

i. ∠AMO = ∠CNO (R) i. Both are right angles

ii. OA = OC (H) ii. Radii of the same circle

iii. AM = CN (S) iii. From statement 3.
5. ∆AMO ≅ ∆CNO
6. OM = ON 5. By R.H.S. theorem

6. Corresponding sides of congruent
triangles.

Proved.

∴Equal chords of a circle are equidistant from the centre of the circle.

GREEN Mathematics Book-9 237

Converse of Theorem - 20 A C

Chords which are equidistant from the centre of a circle are equal .

Given: Perpendiculars OM and ON are drawn to chords AB and CD M O N
BD
respectively from the centre O such that OM = ON. .

To Prove : AB = CD

Construction: OA and OC are joined.

Proof:

S.N. Statement S.N. Reasons
1. In ∆AMO and ∆CNO, 1.

i. ∠AMO = ∠CNO (R) i. Both are right angles

ii. OA = OC (H) ii. Radii of same circle.

iii. OM = ON (S) iii. Given

2. ∆AMO ≅ ∆CNO 2. By R.H.S. theorem

3. AM = CN 3. Corresponding sides of congruent triangles.

4. AM = 1 AB and CN = 1 CD 4. Perpendicular drawn from the centre
2 2 bisects the chord.

5. AB = CD 5. From statement 3 and 4

Proved.

Worked Out EXAMPLES

EXAMPLE 1 In the figure, O is the centre of the circle and O P
M B
OM⊥AB. If OP = 10cm and OM = 6cm, find the A
Solution : length of chord AB.

In the figure, OP = 10cm

∴ OA = 10 [∵ OA = OP, radii of same circle]

and OM = 6cm

In ∆AOM, AM = OA² – OM²
= 10² – 6²
= 100 – 36

= 64 = 8cm
Now, AB = 2 × AM q∵thPeecrepnetnrde itcoulaarchdorardwbnifsreocmtsq

= 2 × 8 cm the chord.
= 16 cm
238 GREEN Mathematics Book-9

EXAMPLE 2 In the figure, O is the centre of the circle, chord C O
Solution : AB = 6cm, chord CD = 12cm, OM⊥CD and A
ON⊥AB. Find the distance between the chords MD
if the radius is 3 5 cm. NB

Here, OA = OC = 3 5 cm

CM = 1 CD = 1 × 12cm = 6cm
2 2
1 1
AN = 2 AB = 2 × 6cm = 3cm

In right angle ∆AON

ON = OA² – AN²

= (3 5 )² – (3)²

= 45 – 9

= 36
= 6cm
In right angle ∆COM,

OM = OC² – CM²

= (3 5 )² – (6)²
= 45 – 36

= 9
= 3cm
Now,
MN = ON – OM
= (6 – 3) cm
= 3cm
∵ The distance between chord is 3cm.

EXAMPLE 3 In the figure, O is the centre of the circle and AO D
∆ABC ≅ ∆DBC. Prove that OM is the bisector of
∠AMD. PQ
M

BC

Solution : Given : In the figure O is the centre of the circle and ∆ ABC ≅ DDBC.
To Prove : OM is the bisector of ∠AMD. i.e. ∠AMO = ∠DMO.
Construction : OP ⊥AC and OQ⊥BD are drawn.

GREEN Mathematics Book-9 239

Prove:

S.N. Statements S.N. Reasons
1. AC = BD
1. Corresponding sides of
2. OP = OQ congruent traingles ABC and
BCD
3. In ∆OPM and ∆OQM
i. ∠OPM = ∠OQM (R) 2. Equal chords of a circle are
ii. OM = OM (H) equidistant from the centre of
iii. OP = OQ (S) circle.

3.

i. Both are right angles

ii. Common side

iii. From statement 2

4. ∆OPM ≅ ∆OQM 4. By R.H.S. theorem

5. ∠PMO = ∠QMO 5. Corresponding angles of
i.e. ∠AMO = ∠DMO congruent triangles

6. OM is the bisector of 6. From statement 5
∠AMD

Proved.

EXAMPLE 4 In the figure O is the centre of the circle and PM = QN. Prove that
∆POQ is an isosceles triangle.

Solution : Given : O is the centre of the circle and PM = QN.

To Prove : ∆POQ is an isosceles triangle. O

Construction : OM and ON are joined..

Proof: PM NQ

S.N. Statements S.N. Reasons
1. ∠OMN = ∠ONM
1. OM = ON; base angles of an
2. ∠OMP = ∠ONQ isosceles triangle OMN.

3. In ∆OMP and ∆ONQ, 2. ∠PMO + ∠OMN = 180°,
i. OM = ON (S) ∠ONQ + ∠ONM = 180°
ii. ∠OMP = ∠ONQ (A) ∠OMN = ∠ONM
iii. PM = QN (S)
3.

i. Radii of same circle

ii. From statement 2

iii. Given

4. ∆OMP ≅ ∆ONQ 4. By S.A.S. axiom

5. OP = OQ 5. Corresponding sides of
congruent triangles

6. ∆POQ is an isosceles 6. From statement 5
triangle

Proved.

240 GREEN Mathematics Book-9

EXERCISE 15.1

A. Very Short Questions AO D
1. a. In the figure if OM = ON, write the relation of AB and CD.
MN

CB
P

b. In the figure OQRS is a rectangle. If QS = 5cm, find the length R Q
O
of OP. TS

c. In the figure, OX⊥AB, write relation of AX and XB. O
A XB

d. In the figure, O is center of the circle. If AP = BP, write relation O
of OP and AB. PB

A

e. In the figure, O is centre of circle. If AB = CD, write relation of A P
O
of OX and OY. X
Y

2. a. In the figure, O is the centre of the circle and M is the mid- BQ
point of chord AB. A

OM

i. What type of triangle is OMB? B
ii. If OM = 3cm and MB = 4cm, find the area of ∆OMB. C

b. In the figure, O is centre of the circle. If OC = 10cm and O
MB
OM = 6cm, find the length of AB. A

B. Short Questions

3. a. In the figure, AB = 16cm, OA = 10cm and OM⊥AB. Find the O
MB
area of ∆AOB. A

b. In the figure, ABCD and OMCN are rectangles. If BC = 12cm A D

and AB = 8cm, find the area of rectangle OMCN. ON

B C
M 241

GREEN Mathematics Book-9

AM B
D
4. a. In the figure, AB||CD, OM⊥AB and ON ⊥CD. If AB = 16cm, O

CD = 12cm and radius of the circle is 10cm, find the length C
N
of MN.

b. In the adjoining figure, AB||CD and distance between them A O B
MN = 3cm. If AB = 12cm and CD = 6cm, find the radius of the C M D
circle. N

R

5. a. The radius of the given circle is 10cm. If chord PQ and RS O

are perpendicular to each other and PQ = RS = 16cm, find PM Q
the measure of OM.

b. In the figure, O is the centre of circle and ABC is an equilateral S
A

triangle. When AB = 16cm and radious of circle is 10cm, find O
OD. DC

B

C. Long Questions

6. In the figure, O is the centre of two circles and AB is a straight line. O

Prove that:

a. AC = BD b. AD = BC . AC DB

7. In the figure, OA is the diameter of smaller circle and radius of A D
C
bigger circle. Prove that AC = CD.
B
O

8. In the figure, P and Q are the centres of intersecting A CB
circles and AB||PQ. Prove that AB = 2PQ. PQ

9. In the figure, P and Q are the centres of intersecting

242 GREEN Mathematics Book-9

A

circles and AB is the common chord. Prove that PQ is P OQ
the perpendicular bisector of AB. B

10. In the given figure, O is the centre of the circle, OM⊥AB AC
and ON⊥CD . If OM = ON and QA = RC, prove that
OA = OC. QR

11. In the figure, O is the centre of the circle. M and N are MON
the mid-points of equal chords AB and CD respectively.
Prove that OMN is an isosceles triangle. BD

12. In the figure, O is the centre of the circle. If OM⊥AB, A C
ON⊥AC and OM = ON, prove that AB = AC. M N
BO
13. In the figure, AB = BC and O is the centre of the circle. P
Prove that OB is the bisector of ∠ABC.
A

P Q
M N
BO
C

A OC

14. In the figure, chords AB and CD intersect at M. If OM B
is the bisector of ∠BMD, prove that AB = CD. D

A
O

M

C
B

15. In the adjoining figure, RA = SB and O is the centre of O
the circle, prove that AM = BN.
R S B
16. In the figure, M and N are the mid-points of equal A N
chords AB and CD respectively and O is the center of
the circle. Prove that ∠MNA = ∠CMN. M

D

AM
O

N
C

B

GREEN Mathematics Book-9 243

A

17. In the figure, O is the centre of circle and equal chords MO D
AB and CD intersect at right angles at L. If M and N are
the mid-points of AB and CD respectively, prove that C LN
MLNO is a square. B

AD

18. In the figure, O is the centre of the circle and ABCD a B OC
rectangle. Prove that EB = CD.
E E
19. In the figure, O is the centre of the circle. If AB||CD O B
and CE is a straight line, prove that ∠BOD = ∠BOE.
A

CD

20. In the figure, O and Q are the centres of intersecting O
circles. If MN is the common chord, prove that M PN
MR = NR.
Q
21. In the adjoining figure, X and Y are centres of two
circles. If XY⊥AD, prove that R

i. AB = CD ii. AC = BD A
B
Project Work:
Copy the following figure and find the centre of circles. YX
C

D

a. b. c.

P QA

CD
RB

244 GREEN Mathematics Book-9