math

Proof Statements S.N Reasons
S.N
1. In AMN and CDN 1.
(i)
(ii) MAN = DCN (A) (i) Being alternate angles as, BA // CD
(iii)
AN = CN (S) (ii) Given
2.(i)
ANM = CND (A) (iii) Vertically opposite angles.
(ii)
(iii) AMN  CDN By A.S.A.
(iv)
(v) AM = CD 2.(i) Being corresponding sides of the
3.(i) congruent triangles.

(ii) AM = BM (ii) Given

(iii) BM = CD (iii) From statements 2. (i) and (ii)

BM //CD (iv) By construction

MD = BC and MD//BC (v) Being, BM = CD and BM//CD.

BC // MN 3.(i) Same line segment (MD//BC)

MN = 1 MD (ii) MN = DN (corresponding sides of the
2 congruent triangles.

 MN = 1 BC (iii) MD = BC {from statements 2. (v) and
2 3. (iii)}

Proved.

Worked Out Examples A

Example 1: In the adjoining figure, AD = BD and AE = CE. If AX D YE
Solution: = 7cm and ABC = 52o, find ADE and length of XY.

Example 2: Here, [Given] B XC
(i) AD = BD and AE = CE,

DE//BC [ AD = BD and AE = CE]

(ii) ABC = ADE = 52o [Corresponding angles, as DE //BC.]

Again, In ABX

AD = BD [Given]

DY // BX [ DE // BC]

AY = XY

Thus, XY = 1 AX
2

 XY = 1 × 7cm = 3.5cm A 3.2cm D
2

In the figure alongside, ABCD is a trapezium in M PN
which AD//BC. If AM = BM, MN//BC, AD = 3.2cm
and BC = 5.8cm, find the length of MP and NP.

B 5.8cm C

194 | Mathematics - 9 Geometry

Solution: Here,

1. In ABC, [Given]
AM = BM, MP//BC

 AP = CP [ AM = BM and MP//BC]
[AM = BM and AP = CP]
MP = 1 BC
2

= 1 × 5.8cm = 2.9cm
2

2. In ACD, [AP = CP and AD//PN]
i. DN = CN

ii. PN = 1 AD [AP = CP and DN = CN]
2

= 1 × 3.2cm = 1.6cm
2

Example 3: In the adjoining figure, P, Q, R and S are mid-points of AB, BC, CD and
Solution:
AD respectively. Prove that PQRS is a parallelogram. A

Given: P, Q, R and S are mid-points of AB, BC, CD and AD PS
To prove: respectively, which are joined in order. C

PQRS is a parallelogram

Construction: Join A and C. Q R
Proof B D

S.N Statements S.N Reasons

1. In ABC 1.

i. BP = AP, BQ = CQ i. Given

ii. PQ //AC, PQ = 1 AC ii. Being BP = AP and BQ = CQ [from
2 (i)]

2. In ADC, 2.

i. DS = AS, DR = CR i. Given

ii. SR//AC, SR = 1 AC ii. Being DS = AS, DR = CR [from (i) of
2 2]

3.i. PQ = SR, PQ//SR 3.i. From (ii) of 1 and (ii) of 2.

ii. PS = QR, PS//QR ii. PQ = SR, PQ//SR [from (i) of 3]

4. PQRS is a parallelogram 4. Opposite sides being equal and
parallel.

Proved.

Geometry Mathematics - 9 |195

Example 4: In the adjoining figure, AD is an angle bisector of BAC. BEAD and

BF = CF. Prove that EF//AC and EF = 1 (AC – AB).
2

Solution:

Given: In ABC, AD is an angle bisector of BAC. i.e. BAD = CAD.

A

BE  AD; BF = CF; E and F are joined.

To prove: EF//AC and EF = 1 (AC – AB) G
Construction: 2 E

Produce BE to meet AC at G.

Proof B DF C

S.N Statements S.N Reasons

1. In ABE and AGE 1.

(i) BAE = GAE (A) (i) AD being an angle bisector of BAC.

(ii) AE = AE (S) (ii) Common side

(iii) AEB = AEG (A) (iii) Being right angles

(iv)  ABE  AGE (iv) By A.S.A.

(v) BE = GE (v) Corresponding sides of the congruent
triangles.

2. In BCG 2.

(i) BE = GE (i) From statements 1. (iv)

(ii) BF = CF (ii) Given

(iii)  EF // GC and EF = 1 GC (iii) From statements 2. (i) and (ii)
2

3.(i) EF//AC 3.(i) Same line segment

(ii) EF = 1 (AC – AG) (ii) CG = 1 (AC – AG)
2 2

(iii) EF = 1 (AC – AB) (iii) AB = AG (corresponding sides of
2 congruent triangles.

Proved.

Exercise 5.4

Group 'A'

1. Find the value of x & y in the following figures.

(a) A (b) P
x
81o

Py Q S X y 55o Y
x

50o C 50o R
B Q

196 | Mathematics - 9 Geometry

(c) B M yA (d) A
x 42o 26o
D E x
N
E 48o F
80o B y
C
124o
CD

2. Find the value of x & y in the following figures.

(a) P (b) A

3.4cm

M T D 2.2cm 2.6cm E
x 5.2cm
Sy Q
R
QN Bx PyC

(c) A 3.4cm D (d)

P x y Q
R

B 7.2cm C

3. (a) In the given figure, AD = BD, AE = CE.
Show that AP = PQ.

(b) In the given figure, BD = AD and BF = FG. A G
Prove that BE = CE. C
D
(c) In the given trapezium ABCD, if F
AD//BC, AE = CE and DN = CN, prove
that AM = BM. BE

Geometry AD
M EN
BC

Mathematics - 9 |197

AD

(d) In the adjoining figure, AD//PQ//BC. If P RQ
AP = BP, prove that DQ = CQ. BC

Group 'B' A HD
1. In the figure alongside, ABCD is a quadrilateral. E, F, G and H
EG
are the mid-points of AB, BC, CD and AD respectively. Prove
that EFGH is a parallelogram. BF C

2. In the adjoining figure, P, Q, R and S are the mid-points of AB, APB
BC, CD and AD respectively. Prove that PQRS is a
parallelogram. QS

3. In the given figure, A, B, C and D are the mid-points of EH, C R D
EF, FG and HG respectively. Prove that AD = BC and AB = H CF
DC.
D

AG

B

E

4. In the given figure, PQRS is a quadrilateral in which A, C, B P D
and D are the mid-points of PQ, QR, RS and PS respectively. If A
AB and CD intersect at O, prove that AO = BO and CO = DO. S
QO
C B Q

R

5. In the figure alongside, PQRS is a parallelogram in which M PM
B
and N are the mid points of PQ and SR respectively. PN and
A
MR intersect the diagonal QS at A and B respectively. Prove

that:

(a) PNRM is a parallelogram S NR

(b) AS = AB = BQ. AD

6. In the adjoining figure, ABCD is a trapezium where AD // BC. If P RQ
AD // PQ and DQ = CQ, prove that:
(a) QR // BC BC
(b) 2PQ = AD + BC.
Geometry
198 | Mathematics - 9

7. In the given figure, PQRS is a trapezium in which PS // QR. If P S
MN is a median, prove that: M N
Q R
(a) MN // QR

(b) MN = 1 (PS + QR).
2

8. In the given figure, AD//MN//BC and BM = 12AB. D A
Prove that AD = BC. M N

B C
P

9. In the adjoining figure, PS is an angle bisector of QPR. PS MA
Q SN
and QA interest at M at right angle. If MN = 1 (PR – PQ), R
2

prove that QN = RN.

AM D

10. In the figure alongside, ABCD is a parallelogram. Median MN P
and diagonal BD intersect at P. Prove that: B NC

(a) MBND is a parallelogram

(b) PN = 1 CD.
2

11. In the given figure, ABCD is a quadrilateral in which P, Q, R A PB
and S are the mid-points of respective parts. Prove that PQRS QS
is a parallelogram. DR C
D
12. In the given figure, M is a mid-point of BC. CA is produced to N
A
D. If BN is perpendicular to AN where AN is an angle bisector
BMC
of BAD, prove that MN = 1 (AB + AC).
2 AED

13. E is the mid-point of side AD of a parallelogram ABCD. F is B MC
any point of diagonal BD such that 4DF = BD. EF if produced N
to meet CD at G. Prove that CG = DG.

Geometry Mathematics - 9 |199

14. In the given figure, ABCD is a parallelogram in which E is A C
mid-point of AD and AM//EC. AM and DC and produced to D
meet at N. Prove that:
(a) 2AB = DN P
(b) 2CE = AN.
F
15. In the figure alongside, P is mid-point of AB and DF.
Similarly, Q is mid-point of BC and EF. Prove that AC = DE. BQ
E

16. In the adjoining figure, ABCD is a trapezium in which AD
AD//BC. M and N are mid points of diagonals BD and AC MN
respectively. Prove that:

(a) MN//AD BA C

(b) MN = 1 (BC – AD).
2
E
17. In the figure alongside, D is mid-point of BC. If 2CG = AC and
BA//DE, prove that 3DF = FG. F

BD C

5.4 Similarity G

Look at the figures alongside, these figures are similar (i)
figures. The figures having the same shape but same or
different sizes are called similar figures; as shown in (ii) (iii)
figures (i), (ii) and (iii). A D

Similarly, geometric figures having the same shape but BC F
different sizes are similar figures. In the figure (i) E
alongside, the triangles, the squares and the circles are
similar to each other separately because these figures r1 r2
are of same shape but different sizes. A picture and its X Y
photocopy obtained by enlarging or reducing in size are
the similar pictures.

(ii) (iii)

Similar Triangles A D

In the adjoining figures (i) and (ii), where A = D = 60°, 60° 60°
B = E = 40°, C = F = 80° in figure (i) and both
triangles are equilateral triangles in figure (ii), these pair B 40° 80° C E 40° 80°
of two triangles have same shape but different sizes. (i)
F

200 | Mathematics - 9 Geometry

They are similar triangles. Symbolically, we denote, LR
ABC DEF and LMN  RST.

M NS T
(ii)

Geometrically, two or more triangles are said to be similar triangle, under the following conditions.

(i) When all angles of one triangle are respectively equal to the A D
corresponding angles of another triangle, the triangles are
said to be similar triangles. In the adjoining triangles,  F

A = D, B = E and C = F. B CE

 ABC DEF
Note: When the triangles are similar, the corresponding sides are proportional.

(ii) When the corresponding sides of two triangles are P X
proportional, the triangles are also similar. Z

i.e. PQ = QR = PR  PQR XYZ. Q
XY YZ XZ RY

Note: When the triangles are similar, the corresponding angles are equal.

(iii) When any two corresponding sides are proportional and the angles included by them are equal,

the triangles are similar. LR

 i.e. LM = LN and MLN = SRT 
RS RT

 LMN RST ML NS T
OR
R

 LM = MN and LMN = RST 
RS ST
NS T
M
 LMN RST

OR L R

 MN = LN and LNM = RTS 
ST RT

 LMN RST M NS T

Note:

(i) Two congruent triangles are always similar but two similar triangles are not necessarily to be
congruent.

(ii) According to the present curriculum, these conditions are not necessary to prove theoretically or
practically (Experimentally).

Geometry Mathematics - 9 |201

Similar Polygons

The properties and conditions for similar polygons are as follows. A D PS
R
(i) If the angles of one polygon are respectively 

equal to the angles of the other polygon, the CQ
(i)
polygons are similar. B

For examples:

In figures (i), A = P, B = Q, C = R PT D H G
and D = S, then, polygon ABCD  polygon Q SE
PQRS.
R F
Similarly, in figure (ii), P =D, Q = E, (ii)
R = F, S = G and T = H, then
pentagon PQRST  pentagon DEFGH.

If the polygons are similar, then their corresponding sides are proportional

i.e. AB = BC = CD = AD [from (i)]
PQ QR RS PS

and PQ = QR = RS = ST = PT [from (ii)]
DE EF FG GH DH

(ii) If the sides of one polygon are proportional to the corresponding sides of the other polygon, the

polygons are similar. In the adjoining figure, AD

AB BC CD AD  PS
PQ QR RS PS
= = =

Then, polygon ABCD  polygon PQRS B CQ R

(iii) The corresponding diagonals of the similar A D (i) P S
polygons are proportional to their

corresponding sides. i.e. polygon ABCD 
polygon PQRS CQ R

Then, AB = AC = BD = ………… B
PQ PR QS

(iv) Similar polygons can be divided into the same P T H
number of similar triangles. Q D G

Then, PQT  DEH, QRS  EFG and 
TQS  HEG
SE

RF

202 | Mathematics - 9 Geometry

(v) The areas of similar polygons are proportional to A FP U
squares of their corresponding sides. T

For example, B
EQ

If polygon ABCDEF  polygon PQRSTU RS
D
area of polygon ABCDEF AB2 BC2 C
area of polygon PQRSTU PQ2 QR2
Then, = = = ……

Worked Out Examples

Example 1: In the adjoining figure, AB//CD, AD and BC A B
Solution: intersect at P. If AB = 4.5cm, BC = 4cm and P
`PC= 1cm, find the size of CD.
Example 2:
1. In ABD and CDP CD
ABP = DCP (A) [Being alternate angles as AB//CD.]

BAP = CDP (A) [Being alternate angles as AB//CD.]

APB = CPD (A) [Remaining angles.]

 BAP  CDP [By A.A.A]

2. (i) AB = BP [Corresponding sides of the similar triangles.]
CD CP

or, 4.5cm = BC - CP
CD CP

or, 4.5cm = (4 - 1)
CD 1

or, 3CD = 4.5cm

 CD = 1.5cm

In the given figure, if ABC = CAD, prove that: A

Solution: (i) ABC  ACD DC
Given: (ii) AC2 = BC . CD
To prove:
B
In ABC, AD meets BC at D such that ABC = CAD.
(i) ABC  ACD
(ii) AC2 = BC . CD

Geometry Mathematics - 9 |203

Proof

S.N Statements S.N Reasons

1. In ABC and ACD 1.

(i) ABC = CAD (A) (i) Given

(ii) ACB = ACD (A) (ii) Common angle

(iii) BAC = ADC (A) (iii) Remaining angles of triangles

 ABC  DAC By A.A.A

2. (i) AC = BC 2. (i) Corresponding side of the similar
CD AC triangles.

(ii) AC2 = BC . CD (ii) From statement 2. (i)

Proved

Example 3: In the given figure, ABCD is a A D

parallelogram in which E is the mid-point

of BC. Diagonal AC and DE intersect at M. M

Prove that 2CM = AM

Solution: B EC

Given (i) ABCD is a parallelogram

(ii) E is mid-point of BC i.e BE = CE.

(iii) AC and DE are intersecting at M.

To prove: 2CM = AM

Proof

S.N Statements S.N Reasons

1. In ADM and CEM 1.

(i) ADM = CEM (A) (i) Being alternate angles as AD//EC

(ii) MAD = MCE (A) (ii) Being alternate angles as AD//EC

(iii) AMD = CME (A) (iii) Remaining angles of triangles

 ADM  CEM By A.A.A

2.(i) AD = AM 2.(i) Corresponding sides of the similar
CE CM triangles.

(ii) 2CE = AM (ii) AD = BC = 2CE
CE CM

(iii) 2CM = AM (iii) From statement 2. (iii)

Example 4: In the given figures, polygon ABCD  Proved
polygon PQRS. If AB = 3cm, PQ = 2cm, PR =
4cm, CD = 2.25cm, SR = 1.5cm, AD = 7.5cm, D
S

C
R

find the size of PS. BP

Solution: A Q

1. ABC  PQR Similar polygons can be divided into same
number of similar triangles.
2. AB = AC
PQ PR [ABC  PQR]

204 | Mathematics - 9 Geometry

or,32ccmm = AC [Given values]
4cm
Similar polygons can be divided into same
AC = 6cm. number of similar triangles.
[ACD  PRS]
3. ACD  PRS
[Given values and AC = 6cm]
AC = AD
PR PS

or,64ccmm = 7.5cm
PS

or,PS = 30cm
6cm

PS = 5cm.

Exercise 5.5

Group 'A' P
1 Find the values of x and y from the following figures.
y 7.5cm
(a) A (b) 2cm T
S
D 1.5cm

x

By E 1cm C Q 3cm R

(c) D (d) P
y M
x 6cm
M 5cm N Q y N 3cm R
x

E 7.5cm F

2. Find the values of x and y with statements and reasons in the following figures.

(a) A (b) P

S
yx

B x D 8cm C Qy R
Geometry
Mathematics - 9 |205

(c) D (d)
yx A

xy

E 6cm GF B 4cm D 12cm C

9cm

A

3. (a) In the given figure, DE//BC. If BC = 6cm, AE = 2.25cm, y 2.5cm
DE = 4cm and BD = 3cm, find the values of x and y.
D 4cm E
3cm x

B 6cm C
A

(b) In the adjoining figure, BAD = ACB. If BC =
9cm and BD = 4cm, find the length of AB.

BD C
A
(c) In the adjoining figure, AB//FC and 2EF = DE. If
CF = 2.5cm, find AD.

D EF

(d) In the given figure, AB//PQ//DC. If AB = 4xcm, B C
CD = 3cm, AP = 5cm and CP = 3cm, find PQ. A
D
P

4. (a) In the given figure, AB//CD. If 2DP = AP, AB = 14cm B QC
and BP = 8cm, find CP and CD. AB

(b) In the given figure, BAC = BDE, find the values of P
CD and DE, if AC = 15cm, AB = 10cm, BD = 5cm and CD
AE = 2cm.
A
206 | Mathematics - 9 E

B DC
Geometry

(c) In the given figure, ABC is a right angled triangle where A
B = 90°, BDAC. If AC = 16cm, CD = 9cm, find the
length of BC. D
CB
(d) In the adjoining figure, ABCD is a parallelogram. If
3EF = CF and BC = 12cm, find the length of DE. EA D
F C

B

D S
AP
5. In the given figure, polygons ABCD and PQRS are
similar. If BC = 6cm, CD = 5cm, BD = 8cm, AB = 4cm
AD = 5cm and QR = 9cm. find PS, QS and RS.

B CQ R
PS
Group 'B'

1. In the adjoining figure, PS//QR and PQS = QRS
,prove that:

(a) QS2 = PS.QR

(b) PQ.QR = QS.RS Q R
A
(c) PQ.QS = PS.RS

2. In the given figure, ABC = CAD, prove that:
(a) AC2 = BC.AD

(b) AB.AC = BC.BD

(iii) BD = AB BD C
CD AC
P
3. In the adjoining figure, PQR is a right angled triangle,
in which Q = 90° and STPR. Prove that: T
S
(a) PR.PT = PS.PQ
RQ
(b) PR = PS PT
QR ST A

(c) PT = PQ Q
ST QR

4. In the figure alongside, PQRS is a parallelogram. If
2PT = QT, prove that 3AP = AR.

Geometry SR
Mathematics - 9 |207

5. In the adjoining figure, ABCD is a parallelogram. Prove P D
that AD.PQ = PB.QR AQ R

6. In the given figure, ABCD is a parallelogram. If 3CF = B C
A D

BC, prove that EF = 1 AF. B
4
E

7. In the adjoining figure, ABC is a right angled triangle C F
in which B = 90° and BDAC. Prove that A
(a) ABC  ABD and AB2 = AC.AD
(b) ABC  BCD and BC2 = AC.CD D
C
(c) ABD  BCD and AB.CD = BC.BD P

B

8. In the given figure, QPR = PSQ = 90°. Prove that:
(a) PS2 = QS.RS

(b) PQ = PR
QS PS

(c) PQ = PR QS R
PS RS A D
C
9. In the adjoining figure, AB//PQ//DC. Prove that P
C
1 = 1 + 1 BQ
PQ AB DC
A
10. In the given figures, BDE = ACB and ABD = D
CBE. Prove that:
E
(a) ABC  BDE B
(b) AC.BD = BC.DE

208 | Mathematics - 9 Geometry

5.5 Pythagoras Theorem AP



In the figures alongside, ABC and PQR are right angled hp hb
triangles where B and R are right angles. The sides AC

and PQ are opposite to the right angles and are called 

hypotenuses. C b BQ p R

(i) (ii)

In figure (i), the sides AB and BC are the perpendiculars (p) and bases (b) respectively but in figure

(ii) the side PR and QR are the bases (b) and perpendiculars (p) respectively. The side opposite to the

reference angle is called perpendicular and remaining side is base.

Pythagoras was a Greek Mathematician who lived around 500 BC and discovered the significant facts
about the right angled triangles known as Pythagoras Theorem.

Theorem - 15

In a right angled triangle, square of the hypotenuse is equal to the sum of the squares of remaining
two sides. Verify experimentally.

OR

Area of square formed on hypotenuse of the right angled triangle is equal to the sum of the squares
formed on two remaining sides. Verify experimentally.

Experimental Verification

Experiment: Draw three right angled triangles ABC right angled at B in different sizes and shapes
with the help of pencil compasses and ruler.

A

C
B

C

A (i) B A B (iii) C
(ii)

To verify: AB2 + BC2 = AC2

Verification: Measure the lengths of AB, BC and AC in each figure and tabulate.

Fig AB AB2 BC BC2 AC AC2 AB2 + BC2 Results
i AB2 + BC2 = AC2
Ii AB2 + BC2 = AC2
iii AB2 + BC2 = AC2

Conclusion: The above experiment shows that in a right angled triangle, square of the hypotenuse
is equal to the sum of squares of two remaining sides.

Geometry Mathematics - 9 |209

Theoretical proof B

Given: ABC is a right angled triangle in which B = 90°
To prove: AC2 = AB2 + BC2

Construction: Draw BD  AC

Proof AD C
Reasons
S.N Statements S.N
1. 1. Both are right angles (90°)
(i) In ABC and ABD (i) Common angle
(ii) (ii) Remaining angles of right angled
(iii) ABC = ADB (A) (iii) triangle.
By A.A.A fact.
BAC = BAD (A) (iv) Corresponding sides of the similar
triangles.
ACB = ABD (A) (v) From statements 1. (iv)
2. (i) Same as above facts and reasons.
 ABC  ADB Corresponding sides of the similar
(ii) triangles.
(iv) AB = AC Form statement 2. (ii)
AD AB (iii) Form statements 1. (v) and 2. (iii)
(v) 3. (i) From statement 3 (i)
2. (i) AB2 = AC.AD (ii) Whole part axiom.
(iii)
(ii) Similarly in ABC  BDC

(iii) BC = AC
3.(i) CD BC
(ii)
(iii) BC2 = AC.CD
AB2 + BC2 = AC.AD + AC.CD
AB2 + BC2 = AC(AD + CD)
AB2 + BC2 = AC.AC

AC2 = AB2 + BC2

Proved

Pythagorean Triplets (Triples or Triads)

These are the triple of positive integers a, b and c that can represent legs and hypotenuse of a right
angled triangle such that a2+b2=c2. For example 3, 4, 5; 6, 8,10; 5, 12, 13 etc.

Pythagoras himself devised the given formula to calculate the Pythagorean triples

2n+1, 2n(n + 1), 2n(n + 1) + 1 where n = 1,2,3, ……………

n a b c Triples a2 + b2 = c2
2n + 1 2n(n + 1) 2n(n + 1) + 1 a, b, c
2 × 1 (1 + 1) = 4 2 × 1(1 + 1) + 1 = 5 32 + 42 = 52
1 2×1+1=3 3, 4, 5 52 + 122 = 132
5, 12, 13 72 + 242 = 252
2 2 × 2 + 1 = 5 2 × 2(2 + 1) = 12 2 × 2(2 + 1) + 1 = 13 7, 24, 25

3 2 × 3 + 1 = 7 2 × 3(3 + 1) = 24 2 × 3(3 + 1) + 1 =25

To obtain Pythagorean triplets associated to a number we can devise the formula n, n2 - k2 , n2 + k2
2k 2k

where k is factors of n not exceeding n, k<n and k is odd or even according as n is odd or even.

 n is any positive integer.

 k is prime factors of n, their powers or products.

 If the factor is only power of 2, say 2p, maximum value of k is 2p-1.

210 | Mathematics - 9 Geometry

 If the factor is power of other primes like 3p, minimum value of k is 30 = 1 and maximum value
is 3p-1.

 In combination (products), power of 2 rises maximum to 2p-1 and power of other prime rise
maximum to 2p.

n2 - k2 n2 + k2 Triples associated to n
2k 2k n2 - k2 n2 + k2
nk (n, 2k , 2k )
4 5
31 3 5 3,4,5
12 13
42 8 10 4,3,5
24 25
51 15 17 5,12,13
6 10
62 40 41 6, 8, 10
12 15
71 7, 24, 25

8 2 8, 15, 17
4
8, 6, 10

9 1 9, 40, 41
3
9, 12, 15

Example: To find all the possible triplets associated to 24.

nk n2 - k2 n2 + k2 Associated triplets
2k 2k
24, 143, 145
2 143 145 24, 70, 74
24, 45, 51
4 70 74 24, 32, 40
24, 18, 30
6 45 51 24, 10, 26
24, 7, 25
24 8 32 40

12 18 30

16 10 26

18 7 25

Note: Any multiples of a Pythagorean triplet is also a triplet e.g. if 3, 4, 5 is a triplet then 2(3, 4, 5) i.e. 6,
8, 10 is also triplet similarly 3(3, 4, 5) i.e. 9, 12, 15 is also triplet

Worked Out Examples

Example 1: Find the value of x from the given figure. Ax
Solution: In right angled ABC, B 4cm C
AC2 = AB2 + BC2
D
= (3)2 + (4)2 = 9 + 16 = 25

AC = 5 cm

Again, In right angled ACD,
AD2 = AC2 + CD2 = (5)2 + (12)2 = 25 + 144 = 169

AD = x = 13cm.

Geometry Mathematics - 9 |211

Example 2: Find the values of x, y and z from the given figure.
Solution:
In the right angled QRS. P

QR2 = QS2 + SR2 ... (i) S
= x2 + 25

In the right angled triangle PQS. y
PQ2 = PS2 + QS2 x

or, y2 = x2 + 16 ... (ii) Qz R

Again, in the right angled triangle PQR

PR2 = PQ2 + QR2

or, 81 = y2 + QR2 ... (iii) [ PR = PS + RS = 4 + 5 = 9]

Now, setting the relations of (i) and (ii) in (iii)

81 = x2 + 16 + x2 + 25

or, 81 = 2x2 + 41

or, 81 - 41 = 2x2

or, x2 = 40 = 20
2

x = 2 5 cm.
Now, y2 = x2 + 16 = 20 + 16
y2 = 36

 y = 6cm
And QR = z2 = x2 + 25
or, z2 = 20 + 25
or, z2 = 45

 z = 3 5 cm

Example 3: In the adjoining figure, ABC is a right angled triangle where B = 90°. D
Solution: is any point on AB. Prove that AB2 + CD2 = AC2 + BD2.

In the right angled ABC A

BC2 = AC2 - AB2 -------- (i)

Again, in the right angled DBC, D
BC2 = CD2 - BD2 -------- (ii)

From relation (i) and (ii) BC
 AC2 - AB2 = CD2 - BD2

 AC2 + BD2 = CD2 + AB2 Proved

212 | Mathematics - 9 Geometry

Exercise 5.6

1. (a) Verify and select the Pythagorean triplets from the given.

(i) 3, 4, 5 (ii) 4, 5, 6 (iii) 5, 12, 13

(iv) 9, 12, 15 (v) 8, 15, 17 (vi) 7, 24, 25

(b) (2n + 1), 2n(n + 1) and 2n(n + 1) + 1 are sides of triangle. Substituting the values of
n = 1, 2, 3, 4, 5, verify that all triangles are right angled triangles.

(c) Find the Pythagorean triplets associated with

(i) 7 (ii) 11

(iii) 21 (iv) 15

2. Check whether the following are right angled triangles.

(a) A (b) P

B 13cm C Q 6cm R
(c) (d) G
A

H 17.5cm I

E 2.1cm F

3. Calculate the unknown length of the sides of following triangles:

(a) A (b) P

xy
S

B 9.6cm C Q xR
(c) D (d) A y

Ax B xD
B 4cm C C
Geometry
Mathematics - 9 |213

4. Find the length of all sides of the given figures. E
AD
(a) A F (b)
(d) A B 24cm C
E
D

B 8cm C

(c) D
A

E B 9cm C

B 8cm C D 7cm E

5. If the sides of a triangle are a2 + b2, a2 - b2 and 2ab, prove that the triangle is a right angled

triangle where a  b.

6. For what value of x, the triangle having sides x, x + 7, and x + 8 is a right angled triangle?

7. For what value of y, the triangle having sides y, y - 1 and y - 18 is a right angled triangle?
AE

8. In the given figure, ABC is a right angled triangle in which D C
AB = 4cm, BC = 3cm. Prove that ABD + BCF = ACE

B

9. In the given figure semicircles are formed in the sides of the 4 cm F
A
right angled ABC. Prove that Area of semicircle on
AC = Area of the circle on AB + Area of semi Circle on 5 cm
AB + Area of Semi Circle on BC.

B 3 cm C
A

10. In the adjoining figure, the diagonals of the quadrilateral
intersect at P at right angle.

Prove that:

(a) AB2 - BC2 = AP2 - CP2 B P
(b) AD2 + CP2 = CD2 + AP2 D

C

214 | Mathematics - 9 Geometry

11. In the given figure, PR and QS are diagonals of the rhombus. P S
Prove that 4PQ2 = QS2 + PR2

O

Q R

5.6 Construction of Quadrilaterals

A. Construction of square X
DC
When a side of a square is given
1. Construct a square ABCD in which AB = 6cm.

Steps of construction 6cm
(i) Draw a line segment AB = 6cm

(ii) At A, construct BAX = 90° 90o B
(iii) From centre A, cut an arc on AX at D with radius AD = 6cm
A 6cm

(iv) Similarly, from centre D and B, take arcs with radius DC = BC = 6cm which intersect
each other at C.

(v) Join B, C and D, C.

Thus, ABCD is the required square.

When the length of diagonal is given

2. Construct a square ABCD in which diagonal AC = 6.4cm. X

Steps of construction B

(i) Draw the diagonal AC = 6.4cm.

(ii) Draw a perpendicular bisector XY on AC which cuts A P C
AC at P.

(iii) From centre P, take PB = PB = PD = 3.2cm (half of
diagonal AC) along XY.

(iv) Join A, B; B, C; C, D; and A, D. D
Thus, ABCD is the required square. Y

Geometry Mathematics - 9 |215

B. Construction of rectangle

When the adjacent sides of a rectangle are given

1. Construct a rectangle ABCD in which AB = 7.4cm and BC = X

5.2cm. D C

Steps of construction

(i) Draw a line segment AB = 7.4cm.

(ii) At A, construct BAX = 90°.

(iii) Along AX take AD = 2.5cm. 90o

(iv) From B and D take arcs with radius BC = 2.5cm and A 7.4cm B

DC = 7.4cm to intersect at C.

(v) Join B, C and D, C.

Thus, ABCD is the required rectangle.

When the diagonals and angle between them are given

2. Construct a rectangle ABCD in which diagonal AC = 6.8cm and angle between diagonal AC

and BD is 60°. X

Steps of construction B

(i) Draw AC = 6cm and draw its perpendicular bisector to
find its mid-point P.

(ii) At P, construct APX = 60° and produces XP up to A PC
Y.
D
(iii) From P, take PB = 3cm on PX and PD = 3cm on PY Y
(half of diagonal AC or BD)

(iv) Join A, B; A, D; B, C and C, D.

Thus, ABCD is the required rectangle.

C. Construction of rhombus

When the length of diagonals are given

1. Construct a rhombus ABCD in which AC = 7.2cm and BD = 5.6cm. X
B
Steps of construction
P
(i) Draw diagonal AC = 7.2cm.

(ii) Draw a perpendicular bisector XY of AC which cuts A C
AC at P.

(iii) From centre P, take PB and PD both equal to 2.8cm (half D
of diagonal BD). Y

216 | Mathematics - 9 Geometry

(iv) Join A, B; B, C; C, D and A, D.
(v) Thus, ABCD is the required rhombus.
When a side and angle made by two adjacent sides are given

2. Construct a rhombus ABCD in which AB = 5.5cm and ABC = 60° X

Steps of construction D 5.5cm C

(i) Draw line segment AB = 5.5cm.

(ii) At B, construct ABX = 60°. 5.5cm
(iii) Along BX take BC = 5.5cm.
(iv) Centre A and C, take arcs with radius AD = CD = 5.5cm

5.5cm, which intersect each other at D. 60o
(v) Join A, D and C, D.
A 5.5cm B

Thus, ABCD is the required rhombus.

D. Construction of parallelograms

When two adjacent sides and angle between them are given.

1. Construct a parallelogram ABCD in which AB = 6.7cm , BC = 4.8cm and ABC = 45°.

Steps of construction D X
C

(i) Draw a line segment AB = 6.7cm. 4.8cm
(ii) At B, construct ABX = 45°.

(iii) Along BX take BC = 4.8cm.

(iv) From A and C, take arcs with radius AD = 4.8cm and A 6.7cm 45o
CD = 6.7cm which intersect at D.
B

(v) Join A, D and C, D.

Thus, ABCD is the required parallelogram.

When base, diagonal and angle made by the diagonal with base are given

2. Construct a parallelogram ABCD in which AB = 5.1cm, ABD = 30° and BD = 7.4cm.

Steps of construction XD C
(i) Draw a line segment AB = 5.1cm.

(ii) At B, construct ABX = 30°.

(iii) Along BX, take BD = 7.4cm and join A and D.

(iv) From centre D, cut an arc with radius DC = 5.1cm and A 30o B
from centre B, cut another arc with centre BC equal to 5.1cm
AD intersecting at C.

(v) Join D, C and B, C.

Thus, ABCD is the required parallelogram.

Geometry Mathematics - 9 |217

When the length of two diagonal and base are given

3. Construct a parallelogram ABCD in which AB = 5.7cm, AC = 6.4cm and BD = 8.2cm.

Steps of construction D C
A O
(i) Draw a line segment AB = 5.7cm.
5.7cm B
(ii) From A and B, cut arcs with radius AO = 3.2cm
(half of AC) and with radius BO = 4.1cm
(half of BD), which are interest at O.

(iii) Produce AO to C making AO = CO and BO to D
making BO = DO.

(iv) Join A, D; B, C and D, C.

Hence, ABCD is the required parallelogram.

E. Construction of trapezium

When two adjacent sides and base angles are given

1. Construct a trapezium ABCD in which AB = 4.7cm, BC = 4.5cm, ABC = 120° and
BAD = 135°.

Steps of construction
(i) Draw a line segment AB = 4.7cm.

(ii) Construct ABX = 120° at B and BAY = 135° at A. X
C
(iii) From B take BC = 4.5cm along BX. YD
(iv) From the point C, draw CD parallel

to AB which cut AY at D 135o 120o 4.5cm
[taking help of alternate angles]

Thus, ABCD is required trapezium. A 4.7cm B

F. Construction of quadrilaterals

When four sides and a diagonal are given

1. Construct a quadrilateral ABCD in which AB = 5.2cm, BC = 4.7cm, CD = 6.1cm, AD = 4.1cm
and BD = 5.5cm.

Steps of construction D 6.1cm C
4.7cm
(i) Draw a line segment AB = 5.2cm. 4.1cm 5.5cm
A
(ii) From the centre A and B cut arcs with radius AD 5.2cm B
= 4.1cm and BD = 5.5cm which intersect at D
and join A, D and B, D.

(iii) Similarly, from D and B take arcs with radius DC =
6.1cm and BC = 4.7cm which intersect at C.

(iv) Join D, C and B, C.

Thus, ABCD is the required quadrilateral.

218 | Mathematics - 9 Geometry

When four sides and one angle are given

2. Construct a quadrilateral ABCD in which AB = BC = 5.1cm, CD = AD = 4.5cm and
BAD = 60°.

Steps of construction X 4.5cm C
(i) Draw a line segment AB = 5.1cm. D

(ii) At A construct BAX = 60°. 4.5cm 5.1cm

(iii) From A cut an arc on AX at D with radius
4.5cm.

(iv) From D and B cut arcs with radius DC = 4.5cm 60o 5.1cm B
and BC = 5.1cm which intersect each other at C. A

(v) Join D, C and B, C.

Thus, ABCD is the required quadrilatral.

Exercise 5.7

1. Construct a square PQRS in which,

(a) PQ = 5.4cm (b) QR = 5.7cm

(c) RS = 6.1cm (d) PS = 4.9cm

2. Construct a square ABCD in which,

(a) Diagonal AC = 6.2cm (b) Diagonal BD = 5.8cm

(c) Length of diagonal = 6.4cm (d) length of diagonal = 5.6cm

3. Construct a rectangle ABCD in which,

(a) AB = 7.4cm and BC = 5.1cm

(b) CD = 6.4cm and AD = 5.2cm

(c) Length of diagonal = 7.4cm and diagonals making an angle 45° to each other.

(d) AC = BD = 6.8cm and diagonal making an angle of 30°.

4. Construct a rectangle PQRS in which,

(a) PQ = 5.8cm and diagonal PR = 6.1cm

(b) PQ = 7.2cm and diagonal QS = 8.1cm

(c) PQ = 5.2cm and RPQ = 60°

(d) PQ = 8.1cm and PQS = 30°
5. Construct a rhombus ABCD in which,

(a) AB = 4.8cm, BAD = 75° (b) AB = 5.4cm, BAC = 30°

(c) AB = 5.7cm, ABD = 60° (c) AB = 5.2cm, ABC = 120°
6. Construct a rhombus PQRS in which,

(a) Diagonals PR = 6.6cm and QS = 8.2cm

(b) Diagonals PR = 7.8cm and QS = 5.6cm

Geometry Mathematics - 9 |219

(c) PQ = 5.7cm and PR = 7.6cm
(d) PQ = 6.1cm and QS = 5.6cm
7. Construct a parallelogram WXYZ in which,
(a) WX = 7.1cm, XWY = 30° and XY = 4.9cm
(b) WX = 6.9cm, WXZ = 60° and WZ = 7.2cm
(c) WX = 7.4cm, XWZ = 75° and WZ = 4.7cm
(d) WX = 6.2cm, WXY = 120° and XY = 4.6cm
8. Construct a parallelogram ABCD in which,
(a) AB = 5.3cm, diagonals AC = 6.2cm and BD = 7.8cm
(b) AB = 6.2cm, AC = 8.4cm and BD = 5.6cm
(c) Diagonals AC = 5.6cm, BD = 6.4cm and they make an angle of 30°
(d) AC = 6.6cm, BD = 8.8cm and they make an angle of 45°
9. Construct a trapezium ABCD in which,
(a) AB = 7.2cm, BC = 4.3cm, ABC = 60° and CD = 5.1cm where AB//CD
(b) AB = 7.2cm, ABC = 45°, BAD = 60° and AD = 4.1cm where AB//CD
(c) AB = 5.3cm, ABC = 120°, BAD = 135° and BD = 8.9cm where AB//CD
(d) AB = 6.1cm, BC = 4.2cm, BAC = 30°, AD = 4.9cm where AB//CD.
10. Construct a trapezium PQRS in which,
(a) PQ = 5.7cm, diagonal PR = 6.6cm, QR = 4.3cm, PS = 4.9cm, where PQ//SR
(b) PQ = 7.3cm, PQS = 45°, QS = 5.1cm, SR = 4.2cm where PQ//SR
(c) PQ = 7.6cm, QPS = 60°, PS = 4.4cm, diagonal PR = 6.7cm where PQ//SR
(d) PQ = 4.3cm, QPR = 30°, PQR = 135°, diagonal QS = 6.3cm where PQ//SR
11. Construct a quadrilateral ABCD in which,
(a) AB = 5.3cm, BC = 6.1cm, CD = 4.2cm, AD = 5.9cm and BAD = 75°
(b) AB = BC = 5.6cm, CD = AD = 4.9cm and ABC = 60°
(c) AB = 6.1cm, BC = 5.2cm, CD = 4.5cm, AD = 5.5cm and BD = 4.7cm
(d) AB = 4.6cm, BC = 6.2cm, CD = 5.5cm, AD = 5.1cm and AC = 5.6cm

5.7 Circle

Introduction

Circle and centre of the circle B

In the adjoining figure, ABC is a circle. A circle is a closed plane figure

formed by a curved line, all of whose points are equidistance from a fixed O
point in the same plane. The fixed point is called centre of the circle. In the

given figure, O is the centre of the circle. The circle ABC is represented as A C
ABC.

220 | Mathematics - 9 Geometry

Circumference A
B
The perimeter of the circle is known as its circumference. In other words,
the total length of the boundary of a circle is known as the circumference of
the circle. In the adjoining figure, ABCA is the circumference of the circle.

Radius C
C
In the adjoining figure, O is the centre of the circle. A, B, C, D and E are the
points on the circumference of the circle. So OA is the radius of the circle. BD
Similarly, OB, OC, OD and OE are also radii (plural form of radius) of the
circle. The distance between the centre and any point on the O
circumference of the circle is called the radius of the circle. And it is E
denoted by r or R. Radii of the same circle are equal.
A
Diameter
d
In the adjoining figure, O is the centre of the circle. AOB is a diameter. The A rOr A

length of diameter of a circle is twice its radius. As shown in figure, a line

segment passing through the centre of a circle and having its end points

on circumference of the circle is called the diameter. And it is denoted by

d. i.e. 2r = d or r = d . The diameter bisects the circle. P
2

Semi Circle AO B

In the figure alongside, O is the centre and AOB or AB is a diameter of the Q
circle. APB or AQB are semi circles. A diameter divides the circle into two
equal parts and each part is called semi circle. Its denoted by .

Chords AB

In the adjoining figure, AB is a chord. Similarly CD and MN are also chords MO N
of the circle. The line joining any two points on the circumference of the C
circle is called chord. A diameter is also a chord and it is the longest chord
of the circle. In the given figure, MN is a diameter which is the longest chord D
of the circle. M

AB

Arc

In the adjoining figure, AB is a chord. The chord AB divides the N
circumference of the circle in two parts which are AMB and ANB. So AMB
and ANB are arcs of the circle. An arc is the part of the circumference of a

circle and it is denoted by the symbol . There are two types of arcs,

where is a minor arc and ANB is a major arc. A

Segment M ON

In the adjoining figure, AB is a chord. and ANB are minor and major B
Mathematics - 9 |221
arcs of the ANBM. There are two regions, one is shaded and other

unshaded. These both regions are called circle segments. The area enclosed

Geometry

by arc and chord of a circle is called circle segment. In the given figure, N
shaded part or AMB is minor circle segment and unshaded part or ANB is
major circle segment.

Sector

In the adjoining figure, O is the centre and OA and OB are radii of a circle. O
There are two regions, one is shaded and another unshaded. Both regions are B
enclosed by two radii and corresponding arcs. These regions are called
sectors. In short, the area enclosed between any two radii of a circle and AM
the corresponding arc is called sector of the circle. In the figure, OAMB is
a minor and OANB major sectors. P H

Concentric Circles DA E Q

In the figure alongside, ABC, DEF, GHI and PQR have same centre G O
O. These circles have different radii but the same centre. These types of CB
circles are called concentric circles. If two or more circles with different FI
radii and have the same centre, they are called concentric circles.

Intersecting Circles: AR

In the adjoining figure, AMB and ANB intersect at A and B. If M XY N
two circles intersect each other at two points, they are called B
intersecting circles. The intersecting circles have a common chord.
AB is a common chord as shown in the figure.

Theorem - 16

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

Given: i. O is the centre of a circle.

To prove: ii. PQ is a chord O
Construction: iii. ORPQ PRQ
Proof: PR = PQ
Join O, P and O, Q.

S.N Statements S.N Reasons

1. In OPR and OQR 1.

i. ORP = ORQ (R) i. Both of them are 90° (ORPQ) 

ii. OP = OQ (H) ii. Radii of thesame circle.

iii. OR = OR (S) iii. Common side

OPR  OQR By R.H.S.

2. PR = QR 2. Corresponding sides of the congruent
triangles

Proved

222 | Mathematics - 9 Geometry

Converse of theorem - 16

A line joining the centre of a circle and the mid-point of a chord is perpendicular to the chord.
(chord is not a diameter)

Given: i. O is the centre of a circle. O
ii. M is a mid-point of chord PQ. P MQ
To prove: iii. O and M are joined.
Construction: OMP = OMQ = 90°
Proof Join O, P and O, Q.

S.N Statements S.N Reasons

1. In OPM and OQM 1.

i. OP = OQ (S) i. Radii of the same circle.

ii. PM = QM (S) ii. Given (M is mid-point of PQ)

iii. OM = OM (S) iii. Common side

OPM  OQM By S.S.S.

2. OMP = OMQ 2. Corresponding angles of congruent
triangles.

3. OMPQ 3. From statement 3 (adjacent angles of the
linear pair equal)

Proved

Theorem - 17 MR

Equal chords of a circle are equidistant from the centre. AB
O
Given: i. O is the centre of a circle.
NS
ii. MN and RS are two equal chords. i.e (MN = RS)

iii. OAMN and OBRS

To prove: OA = OB

Construction: Join OM and OR.

Proof

S.N Statements S.N Reasons
1. (i) Given
1. (i) MN = RS
(ii) Being OAMN and OBRS.
(ii) MA = 1 MN and RB = 1 RS
2 2

(iii) MA = RB (iii) From statement 1. (i) and (ii)

2. In OAM and OBR 2.

OA and OB are perpendicular to MN

(i) OAM = OBR (R) (i) and RS respectively. (being right

angle)

(ii) OM = OR (H) (ii) Radii of the same circle.

(iii) MA = RB (S) (iii) From statement 1. (iii)

 OAM  OBR By R.H.S

3. OA = OB 3. Corresponding sides of the congruent
triangles.

Proved

Geometry Mathematics - 9 |223

Converse of theorem - 17

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

Given: i. O is the centre of a circle. AC
ii. AB and CD are two chords.

To prove: iii. OXAB, OYCD XY
iv. OX = OY O
AB = CD

Construction: Join A and C with centre O. BD
Proof

S.N Statements S.N Reasons
1.
1. In OAX and OCY
(i) Being OXAB and OYCD (given)
(i) OXA = OYC (R) (ii) Radii of the same circle
(iii) Given
(ii) OA = OC (H)
By R.H.S.
(iii) OX = OY (S)
2. (i) Corresponding sides of the congruent
 OAX  OCY triangles.

2. (i) AX = CY (ii) Multiplication axiom.
(iii) OXAB so OX bisects AB, at X
(ii) 2AX = 2CY
(iii) AB = CD similarly, OYCD so OY bisects CD
at Y (2AX = AB, 2CY = CD).

Proved

Theorem 18

Perpendicular bisector of a chord of a circle passes through the centre of the circle.

Given: O is the centre of the circle in which AB is a
chord. M be the mid-point of AB and MP is the
perpendicular bisector of AB.

To prove: MP passes through centre O.

Construction: Join mid-point M and centre O

Proof: Statements S.N Reasons
1. Given
S.N MP  AB 2. Line joining the centre of the circle and
1. MO  AB
2. the mid point of the chord is the
perpendicular to the chord.
3. OMB = PMB = 90 3. From statements (1) and (2)
4. MO and MP are name line 4. From the given point on the line there
passes only one line perpendicular to
5. MP passes through centre the given line.
Q.E.D. 5. 5. From statement (4)

224 | Mathematics - 9 Proved

Geometry

Worked Out Examples

Example 1: In the adjoining figure, find the length of chord O B
Solution: AB which is at a distance of 2.8cm from the AM
centre O of the circle of radius (OA) = 10cm. B
Here, D

O is the centre of a circle. OMAB,
OA (r) = 10cm
OM = 2.8cm
Length of AB = ?

In right angled OMA,

AM = (OA)2 - (OM)2

= (10)2 - (2.8)2

= 100 - 7.84

= 92.16 = 9.6cm
 AB = 2AM

= 2 × 9.6 cm = 19.2cm

Example 2: In the given figure, O is the centre of a circle. If A M
Solution: O
AB//CD, OMAB, ONCD, AB = 16cm, CD = 12cm
and the diameter of the circle is 20cm, find the N
length of MN.

Here, O is the centre of the circle. C
AB//CD, AB = 16cm, CD = 12cm,

OMAB, ONCD

Join B and D with centre O.

Now, OB = OD = 1 × 20cm = 10cm [radius = 1 of diameter]
2 2

BM = 1 AB DN = 1 CD
2 2

= 1 × 16cm = 1 × 12cm
2 2

= 8cm = 6cm

In the right angled triangle OMB.

MO = (OB)2 - (BM)2

= (10cm)2 - (8cm)2

= 100 - 64 = 36 = 6cm
Again, in the right angled OND
NO = (OD)2 - (DN)2

Geometry Mathematics - 9 |225

= (10cm)2 - (6cm)2

= 100 - 36

= 64 = 8cm
Then,
MN = MO + NO

= 6cm + 8cm = 14cm

Example 3: In the given figure, M and N are the A P CQ B

Solution: centres of two intersecting circles. If
Given:
To prove: ACB//MN, prove that MN = 1 AB. MN
2 D

i. M and N are the centres of two intersecting circles.

ii. ACB//MN

MN = 1 AB.
2

Construction: Draw perpendiculars MP and NQ on AB.
Proof

S.N Statements S.N Reasons
1. (i) Being PQ//MN and MPQ = 90°
1. (i) PMN = 90° (ii) Being PQ//MN and NQP = 90°

(ii) QNM = 90° Being all angles right angles.
2. (i) Opposite sides of the rectangle.
PMNQ is a rectangle
(ii) Being MPAC and NQBC.
2. (i) PQ = MN
(iii) From statement 2. (ii)
(ii) PC = 1 AC and QC = 21BC
2

(iii) PC + QC = 1 (AC + BC)
2

(iv) PQ = 1 AB (iv) Whole part axiom.
2 (v) From statement 2. (i)

(v) MN = 1 AB
2

D Proved

Example 4: In the adjoining figure, O is the centre of a A

circle. AB and CD are two chords, which

intersect at P. If OP is angle bisector of

BPD, prove that: (i) AB = CD, (ii) AP = CP. CP O

Solution:

Given: i. O is the centre of a circle.
ii. AB and CD intersect at P.
B

iii. OP is an angle bisector of BPD. i.e OPB = OPD.

226 | Mathematics - 9 Geometry

To prove: i. AB = CD AN D
Construction: ii. AP = CP P O
Draw OMAB and ONCD.
CM

Proof B
S.N
1. Statements S.N Reasons
In OPM and OPN
(i) OMP = ONP (A) 1.
(ii)
(iii) (i) Being OMAB and ONCD. (by
construction)
2. (i)
OPM = OPN (A) (ii) Given OPB = OPD
OP = OP (S)
(iii) Common side
OPM  OPN
OM = ON By A.A.S.

2. (i) Corresponding sides of the congruent
triangles.

(ii) AB = CD (ii) Chords being equidistant from the
centre of a circle.
(iii) PM = PN
(iii) Corresponding sides of the congruent
AM = CN triangles.
(iv)
(iv) AM = 1 AB and CN = 1 CD, where
2 2

AB = CD.

(v) AM – PM = CN – PN (v) From statements 2. (iii) and (iv)
(vi) AP = CP
(vi) From statement (v) (Remaining
facts)

Proved

Exercise 5.8

Group 'A' O

1. (a) In the given figure, O is the centre of a circle and OXAB. A XB
If AB = 4cm, OX = 4.8cm, find the length of the diameter. N P

(b) In the given figure alongside, O is the centre of a circle. PQ M
is a chord and OMPQ. If OM = 4.8cm and ON = 6cm, O
find the length of PQ.
Q

Geometry Mathematics - 9 |227

(c) In the adjoining figure, O is the centre and AB is a diameter C
of a circle. Chords AB and CD intersect at E such that
CE = DE = 12cm. If BE = 6cm, find the length of OC. A O EB
D
(d) In the given figure, O is the centre of a circle. OR and PQ P
intersect at S at right angle. If PQ = 6cm and SR = 3cm,
find the length of radius of the circle. O SR
Q
2. (a) In the given figure, O is the centre of a circle. If AB//CD,
AB = 8cm, CD = 6cm and diameter of the circle is 10cm, O
find the distance between AB and CD. AB

CD

(b) In the adjoining figure, O is the centre of a circle in which PQ
chords PQ and RS are parallel. If PQ = 18cm, RS = 24cm O
and radius of the circle is 15cm, find the distance between
PQ and RS. RS

(c) In the figure alongside, O is the centre of a circle. MN and MN
RS are equal and parallel chords. If the radius of the circle O
is 10cm and the distance between two chords is 12cm, find
the length of the chord MN. RS

(d) In the given figure, O is the centre of a circle. AC is a C
diameter and AB is a chord. If OMAB, AC = 10cm and O
AB = 9.6cm, find the length of OM and BC.
AM B

3. (a) In a circle of radius 2.5cm, AB and CD are two parallel chords of length 2.4cm and 1
cm respectively. Find the distance between the chords, if they lie (i) on the same side of
the centre (ii) on the opposite side of the centre.

(b) PQ and RS are two parallel chords which lie on the opposite sides of the centre of the
circle. If these chords are 14cm apart, PQ = 12cm and RS = 16cm, find the radius of the
circle.

(c) AB and CD are two parallel chords which lie on the same side of the centre of the
circle. If these chord are 1 cm apart, AB = 8cm and CD = 6cm, find the radius of the
circle.

228 | Mathematics - 9 Geometry

(d) MN and RS are two parallel chords of a circle such that MN = 18cm, RS = 24cm. If the

chords are on the opposite sides of the centre and the distance between two chords is

21cm, find the diameter of the circle. PR

Group 'B' MN
X
1. In the adjoining figure, PQ and RS are two equal chords of the circle
with centre X. If XMPQ and XNRS, show that MX = NX. QS

2. In the given figure, P is the centre of the circle. PA and PB are MR
perpendiculars on the chords MN and RS respectively. If PA = P
PB, prove that MN = RS.
AB

NS

3. In the adjoining figure, O is the centre of the circle. A chord PQ O
intersects two concentric circles at the points P, M, N and Q as P M NQ
shown in the figure. Prove that PM = QN.
D
4. In the given figure, O is the centre of the circle. Two equal chords
AB and CD intersect at P. Prove that OP is an angle bisector of A
BPD. PO

5. In the adjoining figure, O is the centre of a circle. Chords MN and C
RS intersect at P. If OP is an angle bisector of SPN, prove that:
(i) MN = RS, (ii) MP = RP and (iii) NP = SP B
MR
6. Two equal chords AB and CD meet at an external point E.
Prove that AE = CE and BE = DE. P

7. In the figure alongside, two chords PQ and RS of a circle with O
centre O are produced to meet at T. If PT = RT and QT = ST, SN
prove that OT is an angle bisector of PTR.
A
8. In the adjoining figure, O is the centre of a circle. Two equal B
chords AB and CD intersect at P. If M and N are mid-points of AB E
and CD respectively, prove that PMN is an isosceles triangle. D

Geometry C

P Q
O T

R S

D

AN
PO

CM

B

Mathematics - 9 |229

M B
A
9. In the figure alongside, O is the centre of a circle. If PB = PD,
P O
prove that OM = ON.

C D
N

B

10. In the given figure, O is the centre of circle. If OB is the bisector O
of ABC, prove that AB = BC. AC

AM B
D
11. In the given circle, O is the centre of a circle. OMAB, ONCD O P

and OM = ON. Prove that AP = CP. CN

12. In the adjoining figure, O is the centre of a circle. Two equal A D
chords intersect at right angle at G. If E and F are mid-points of
AB and CD respectively, prove that OEGF is a square. EO
CG F

B

13. In the figure alongside, P and Q are centres of two intersecting P Q
circles. If ACB//PQ, prove that 2PQ = ACB. CB
A
A
N
14. In the given figure, M and N are centres of two intersecting circles M
which interest at A and B. Prove that the line joining the centres of
the circle is the perpendicular bisector of the common chord AB.

15. In the adjoining figure, two circles with centres A and B are B
intersecting at C and D. If EF//CD, prove that EG = FH and CE
EH = FG.
G
16. In the given figure, X and Y are centres of two intersecting equal AB
circles. If XY = AB, prove that AXBY is a square.
 H
DF
A

X PY

B

230 | Mathematics - 9 Geometry

Unit Test

Time: 40 minutes F.M.- 24

Group-A A B xD
1. From the given figure, find the value of x.

2. Write the converse statement of "If two sides of a C
triangle are equal, the angles opposite to them are A
equal."
40°
Group B

3. Find the value of the angles denoted by x and y. 3x y 2x
C B
D
D
A

4. In the adjoining figure, ABCD is a square and AEC
is an equilateral triangle. Find the size of EAB.

B C
P
E

5. In the given figure, PRQ = QPS. If QR = 9 cm Q
and QS = 4 cm, find the length of PQ.
S
Group C R

6. Prove that equal chords of a cirlce are equidistance AP B
from the centure
S
7. In the given figure, ABCD is a quadrailateral. P, Q, R DQ
and S are mid poins of the sides AB, BC, CD and AD
respectively. Prove that PQRS is a parallelogram. R C
A D
8. Construct a parallelogram ABCD inb which
diagonals AC = 6 cm and BD = 8 cm and angle
between the diagonals = 60°. Give steps of
construction.

Group D

1 1 D1C. P
PQ AB
9. In the given figure, AB//DC//PQ. Prove that: = +

 BQ C

Geometry Mathematics - 9 |231

Answers ____________________________________________________________

Exercise 5.1

1. (a) 16° (b) 17° (c) 34° (d) 18° (e) 32°, 44°
(f) 44°, 54°, 82° (g) 12°, 120° (h) 22.5°, 45° (i) 48°, 42°, 48°
(b) 90° (c) 154° (d) 76°, 52° 52°
2. (a) 78° (b) 36° (c) 120° (d) 12°, 36°
3. (a) 127°

Exercise 5.2

1. (a) 30° (b) 112°, 34° (c) 44° (d) 18°, 54° (e) 30°
(f) 34° (g) 116°, 72° (h) 72°, 36° (i) 64°
(b) 44°, 78° (c) 26°, 52° (d) 22.5°, 22.5°
2. (a) 34°, 44° (f) 22°, 22°
(e) 45, 25° (b) 67.5° (c) 83° (d) 50°

3. (a) 72°

Exercise 5.3

A. 1. (a) 108°, 72° (b) 106°, 106° (c) 30°, 120° (d) 45°, 45°
(e) 74°, 106° (f) 45°, 54° (g) 34°, 101° (h) 116°, 52°
(i) 76°, 31° (j) 103°, 122° (k) 38°, 83° (l) 112°
(b) 90° (c) 64° (d) 34.5°
2. (a) 79° (b) 68° (c) 120° (d) 107°
3. (a) 15°

Exercise 5.4

1. (a) 49°, 130° (b) 75°, 130° (c) 42°, 32° (d) 76°, 124°
2. (a) 1.7cm, 2.6cm (b) 4.4cm, 5.2cm (c) 3.6cm, 1.7cm (d) 7.4cm, 8.4cm

Exercise 5.5 (b) 2.5cm, 3cm (c) 2.25cm, 6cm (d) 2.4cm, 2cm
(b) 4.5cm, 6cm (c) 3.6cm, 2.88cm (d) 8cm, 8 3cm
1. (a) 5cm, 1cm (b) 6cm (c) 5cm (d) 1.25cm
2. (a) 2cm, 3cm (b) 5cm, 7.5cm (c) 12cm (d) 16cm
3. (a) 1.25cm, 7cm
4. (a) 4cm, 7cm
5. 7.5cm, 12cm, 7.5cm

Exercise 5.6

1. (a) Show to your teacher (b) Show to your teacher

(c) (i) 7, 24, 25 (ii) 11, 60, 61 (iii) 21, 20, 29; 21,72, 75; 21, 220, 221 (iv) 15, 8, 17; 15, 20, 25; 15, 36, 39; 15, 112, 113

2. Show to your teacher

3. (a) 10.4 cm (b) 5cm, 13cm (c) 5cm, 12cm (d) 12cm, 15cm

4. (a) AC = 10 cm, AD = 5 5 cm, DE = 4.36 cm, AF = 12.65 cm

(b) AC = 25 cm, AD = 26.93 cm, AE = 30.8 cm

(c) AE = CE = 5cm, DE = 12 cm, BE = 6.24 cm

(d) AB = BD = 12 cm, AE = 25 cm

6. 5 cm 7. 25 cm

Exercise 5.7

Show to your Subject Teacher

Exercise 5.8

1. (a) 10.4cm (b) 7.2cm (c) 15cm (d) 3cm
2. (a) 1cm (b) 21cm (c) 16cm (d) 1.4cm, 2.8cm
3. (a) 1.7cm, 3.1cm (b) 10cm (c) 5cm (d) 15cm



232 | Mathematics - 9 Geometry

6Chapter

Trigonometry

Objectives:

At the end of this chapter, the
students will be able to:
 find the value of trigonometric

ratios of different standard
angles like 0°, 30°, 45°, 60°, 90°
 solve the different identities
related to trigonometric ratios.
 solve the triangle using
trigonometric ratios.

Teaching Materials:

Chart paper, flash card, chart of
trigonometrical formula and value
chart of standard angles.

Historical fact

Traditionally, trigonometric functions were considered with respect to arcs of a circle. Georg Joachim
Rhaeticus (1514 – 1576), a German mathematician was the first to define trigonometric functions as
the ratio of the sides of a right angled triangle. Thus, all six trigonometric functions came into full use
and here trigonometry came of age and started modern trigonometry.

6.1 Trigonometric Ratios C

Let’s consider a right angled triangle ABC with B = 90°. Among the other two A Base (b) B
acute angles, one is considered on reference angle in terms of which the ratios
are defined. The side opposite to the right angle is called hypotenuse (h), the side
opposite to the reference angle is perpendicular (p) and the remaining side is
called base (b).

Using three sides, six ratios can be formed as C

(i) CB = p (ii) AC = h
AC h CB p

AB b AC h hp
AC h AB b
(iii) = (iv) =

(v) CB = p (vi) AB = b A bB
AB b CB p

Until this step, we are simply dealing with pure geometry. Further, we relate these ratios with an angle
of the right angled triangle which is an important turn in mathematics from where trigonometry starts.

Now, taking C as reference angle, we define the above six ratios of sides trigonometrically as

(i) CB = p = sine of A or Sin A and AC = h = Cosecant of A or Cosec A
AC h CB p

(ii) AB = b = Cosine of A or Cos A and AC = b = Secant of A or Sec A
AC h AB h

(iii) CB = p = Tangent of A or Tan A and AB = b = Cotangent of A or Cot A
AB b CB p

Relations between the trigonometric ratios

A. Identities related to receiprocal

1. Sin A and Cosec A being reciprocals

Sin A . Cosec A = p × h = 1.
h p

 Sin A . Cosec A = 1
Similarly Cos A. Sec A = 1; Tan A . Cot A = 1

234 | Mathematics - 9 Trigonometry

B. Identities related to quotient

p b

1. Sin  = p = b = tan  2. Cos  = b = p = cot 
h h sec  h h cosec 

b p

p

3. Tan A = p = h = SinA 4. Cot A = CosA
b b CosA SinA

h

h h

5. Sec  = h = p = cosec  6. Cosec  = h = b = Sec 
b b cot  p p tan 

p b

C. Pythagorus Identities
1. p2 + b2 = h2

or, ph2 + hb2 = 1

or, Sin2 + Cos2 = 1
 Sin2 = 1 – Cos2

 Sin = 1 – Cos2
2. h2 – p2 = b2

or, hb2 + bp2 = 1

or, sec2 – tan2 = 1

3. h2 – b2 = p2

or, Ph2 + pb2 = 1

or, cosec2 – cot2 = 1

4. Sin A = p and Cos A = b
h h

 Sin2 A = p2 and Cos2 A = b2
h2 h2

Adding these two relations, we get

Sin2 A + Cos2 A = p2 + b2 = p2 + b2 = h2 =1 [p2 + b2 = h2, by Pythagoras theorem]
h2 h2 h2 h2

 Sin2 A + Cos2 A = 1

Trigonometry Mathematics - 9 |235

Similarly, we get

Sec2 A - Tan2 A = 1 and Cosec2 A - Cot2 A = 1
Note:
(i) Reference angles are also represented by Latin alphabets ,, ,  etc.
(ii) Being related to right angled triangle, Pythagoras theorem, h2 = p2 + b2 is very useful in

trigonometry.

Worked Out Examples

Example 1: Write the trigonometric ratios with respect to the given reference angle
Solution: of the given right angled triangle.

Example 2: Here,
Solution:
In the right angled triangle ABC,  B = 90° A
Reference angle  ACB = 

 h = AC, p = AB, b = BC

Now, Sin = p = AB
h AC
C
b BC B
h AC
Cos = =

Tan = p = AB
b BC

Cosec = h = AC
p AB

Sec = h = AC
b BC

And Cot = b = BC
p AB

In the right angled triangle PQR,  PQR = P
90°, PQ = 8cm, QR = 15cm. Find all the
trigonometric ratios of the angle R = 

Here,

In right angled triangle PQR, Q 15cm R
 PQR = 90°, PQ = 8cm, QR = 15cm.

 PR = PQ2 + QR2 [By Pythagoras theorem]

= (8cm)2 + (15cm)2

= 64cm2 + 225cm2

= 289cm2 = 17cm.

Now with respect to R = 

236 | Mathematics - 9 Trigonometry

Sin = p = PQ = 8cm = 8
h PR 17cm 17

Cos = b = QR = 15cm = 15
h PR 17cm 17

Tan = p = PQ = 8cm = 8
b QR 15cm 15

Cosec = h = PR = 17cm = 17
p PQ 8cm 8

Sec = h = PR = 17cm = 17
b QR 15cm 15

And Cot = b = QR = 15cm = 15
p PQ 8cm 8

Example 3: Using the table of values of trigonometric ratios, find the value of the
Solution:
side x of the given right angled triangle. A

Here, 35°
In right angled triangle ABC, ABC = 90°,

BAC = 35° and AB = 6.8cm, BC = x = ?

Now, with respect to A = 35°

Tan = BC C xB
AB

or, Tan 35° = x
6.8cm

or, 0.70 = x [Using table, tan 35° = 0.70]
6.8cm

 x = 4.76cm

Example 4: Using the table of values of trigonometric ratios, find the angle  in the
Solution: given right angled triangle.

Here,

In the rt.  MNO, OM = 51cm, ON = 45cm, MNO = 90°, MON =  = ? M

Now, with respect to angle 

Cos = ON
OM

or, Cos = 45cm O  45cm N
51cm

or, Cos = 0.88.

or, Cos = Cos 28°

 = 28°

Trigonometry Mathematics - 9 |237

Example 5: Express Cos in terms tan.
Solution: We know,

Cos = 1 = 1 [ Sec = 1 + tan2 ]
Sec 1 + tan2 A

Alternately: 1 + x2
x
Let tan = x B 1 C

 p = x, b = 1 Trigonometry

h = p2 + b2

= 12 + x2 = 1 + x2

Now, Cos = b
h

= 1
1 + x2

Cos = 1
1 + tan2

Example 6: If Sin = 11 , find tan.
Solution: 61

Here,

Sin = 11
61

Now, we have

Tan = Sin = Sin
Cos 1 - Sin2

Or, tan = 11 = 11 = 11
61 61 60
60
1 - 61112 61

Alternatively:

Here, Sin = 11
61

If P = 11x then, h = 61x

b = h2 - p2 = (61x)2 - (11x)2

= 3721x2 - 121x2

= 3600x2 = 60x

Now tan = p = 11x
b 60x

tan = 11
60

238 | Mathematics - 9

Example 7: Prove the given trigonometric identity: (1 + tan2) cos2 = 1.
Solution:
Here,

L.H.S = (1 + tan2) cos2

= 1 + sin2  . cos2
cos2

= cos2 + sin2 × cos2
cos2

= 1 × cos2 [ cos2 + sin2 = 1]
cos2

= 1 R.H.S proved.

Exercise 6.1

1. Write the basic trigonometric ratios with respect to the given angle as the ratios of the sides of

the triangle. B (b) L
(a) A



C N M
(c) (d) D

D

B C B C

2. In each of the following right angled triangles, find the ratios of the given angle.

(a) A (b) 4.8cm E
D

5cm

BC F

Trigonometry Mathematics - 9 |239

X
(c) M (d)

Z

L  2.4cm N 
Y

3. Use the table of values of trigonometric ratios of the given angles and find the lengths of the
unknown sides.
(a) A (b) P

10 c1m0 ? ??
B
C 30° R 60° Q
? 7.2cm

(c) X (d) A

? D 60°

?

Y ? 45° Z C? B

4. Find the indicated angles in each of the following right angled triangles. (Use table for the

value of the ratios)

(a) A (b) P

6cm

B C Q  R
(c) (d) 8cm

P Q P

b 25.98 cm D

30cm

R Q R

5. (a) Express tan and cos in terms of sin.

(b) Express sin in terms of tan.

(c) Express tanα in terms of cosα.

(d) Express sin in terms of cos.

6. (a) If sin = 6 , find the value of cos.
10

240 | Mathematics - 9 Trigonometry