Vedanta Excel in Mathematics Teachers' Manual -9 Final

4. In 'DBC and 'ABC 4.
i. BD = AC (S) i. From (3)
ii. ‘DBC = ‘ACB (A) ii. Given
iii. BC = BC (S) iii. Common side
5. 'DBC # 'ABC
6. CD = AB 5. By S.A.S. axiom

6. Corresponding sides of congruent triangle are
equal

44. In the figure alongside, PQRS is a square. U is the mid - point of V Proved
PQ, ‘RUT = 90°, TU and RQ are produced to meet at V. Prove
that : TR = PT PQ PT S
U
Solution:
QR
Given: PQRS is a square, U is the mid - point of PQ and ‘RUT = 90°

To prove: TR = PT PQ

Proof

Statements Reasons
1. In 'PUT and 'UVQ
i. ‘TPU = ‘UQV (A) 1.
ii. PU = UQ (S) i. Both are right angles
iii. ‘PUT = ‘VUQ (A) ii. U is the mid - point of PQ
2. 'PUT # 'UVQ iii. Vertically opposite angles
3. PT = VQ and UT = UV 2. By A.S.A. axiom
3. Corresponding sides of congruent triangle are
4. In 'TUR and 'UVR
i. UT = UV (S) equal
ii. ‘TUR = ‘VUR (A) 4.
iii. UR = UR (S) i. From (3)
5. 'TUR # 'UVR ii. Both are right angles
6. TR = VR iii. Common sides
7. VR = VQ QR = PT PQ 5. By S.A.S. axiom
8. TR = PT PQ 6. Corresponding sides of congruent triangles
7. Whole part axiom, VQ = PT and QR = PQ
8. From (6) and (7)

45. In the adjoining figure, AC = BC, ‘PCA = ‘QCB and ‘PBA = P Proved
‘QAB. Prove that 'OPQ is an isosceles triangle.
A Q
Solution: O
CB
Given: AC = BC, ‘PCA = ‘QCB and ‘PBA = ‘QAB

To prove: 'OPQ is an isosceles triangle

Proof

Statements Reasons
1. ‘PCA = ‘QCB 1. Given
2. ‘PCA ‘PCQ = ‘QCB ‘PCQ 2. Adding ‘PCQ on both sides of (1)
3. ‘ACQ ‘PCB 3. From (2), by whole part axiom

97 Vedanta Excel in Mathematics Teachers' Manual - 9

4. In 'QAC and 'PCB 4.
i. ‘QAC = ‘PBC (A) i. Given
ii. AC = BC (S) ii. Given
iii. ‘ACQ = ‘PCB (A) iii. From (3)
5. 'QAC # 'PCB
6. ‘AQC = ‘BPC 5. By A.S.A. axiom

7. CQ = CP 6. Corresponding angles of congruent
triangle
8. ‘CQP = ‘CPQ
9. CQP ‘AQC = ‘CPQ ‘BPQ 7. Corresponding sides of congruent
triangle
i.e, ‘OQP = ‘OPQ
10. 'OPQ is an isosceles tringle 8. From (7)

9. Subtracting (6) from (8)

10. Base angle are equal

Proved

Mid - point theorems P

1. In 'PQR, X and Y are mid - points of PQ and PR respectively.

If ‘P ‘Q = 50° and ‘P ‘R = 150°, find size of ‘PXY and XY
‘PYX.

Solution: R
In 'PQR, X and Y are the mid - point of PQ and PR respectively. ? Q

XY // QR.

i) ‘R = 180° (‘P ‘Q) = 180 ° 50° = 130°

? ‘PYX = ‘R = 130° [? XY // QR, corresponding angles]

ii) ‘Q = 180° (‘P ‘R) = 180° 150° = 30° A
? ‘PXY = ‘Q = 30° [ ? XY // PQ, corresponding angles]

2. In the given 'ABC, P, Q and R are the mid - point of sides AB, BC

and CA respectively. If the perimeter of 'PQR is 15 cm. find the P R
perimeter of 'ABC.

Solution: 21AC, QR = 12AB and PR = 21BC BQ C
i) PQ =

[line segment that joins the mid points of two sides is half of the third side in triangle.]

ii) Perimeter of 'PQR = 15 cm

or, P21(QAC QR PR = 15 cm
or, AB BC) = 15 cm

? AB BC AC = 30 cm AD

Hence, the perimeter of 'ABC is 30 cm.

3. In the given figures, AD // BC and P is the mid - point of AB. If P QR
BC = 8cm and QR = 3cm, find the length of PQ and AD.

Solution: [? AP = PB and PQ // BC] B C
i) In 'ABC, AQ = QC
[PQ joints the mid - points of the sides AB and AC]
PQ = 21BC

= 1 u 8 cm
2
= 4 cm

Vedanta Excel in Mathematics Teachers' Manual - 9 98

ii) In 'ACD; CR = RD [?AQ = QC and QR // AD]

? QR = 21AD [? Q and R joint the mid - point of AC and CD]

or, 3 cm = 21AD

? AD = 6 cm

4. In the adjoining figures, P and Q are the mid - point of the

sides AB and AC of 'ABC respectively. X is a point on PQ.

Prove that AX = XD.

Solution:

Given: In 'ABC, P and Q are the mid - point of the sides AB and
AC respectively. X is a point on PQ.

To prove: AX = XD

Proof

Statements Reasons
1. In 'ABC, PQ // BC 1. PQ joins the mid - points of the sides AB and AC
2. In 'ABD; AX = XD 2. PX // BD and P is the mid point of AB

Proved

5. In the adjoining triangle XYZ, A and B are the mid - point
of the sides XY and YZ respectively. P is any point on XZ.
Prove that AB bisects PY at Q.

Solution:

Given: In 'XYZ; A and B are the mid - points of sides XY and
YZ respectively. P is any point on XY.

To prove: AB bisects PY at Q

Proof

Statements Reasons
1. In 'XYZ, AB // XZ 1. AB joins the mid - points of XY and YZ
2. In 'PXY, YQ = QP 2. AP // XP, and A is the mid - point of side XY
3. AB bisects PY at Q 3. From (2)

Proved

6. In the adjoining equilateral triangle PQR. X, Y and Z are
the middle points of the sides PQ, QR and RP respectively.
Prove that XYZ is also an equilateral triangle.

Solution:

Given: 'PQR is an equilateral triangle. X, Y and Z are the mid -
points of the sides PQ, QR and RP respectively.

To prove: 'XYZ is also an equilateral triangle

Proof

Statements Reasons
1. PQ = QR = RP 1. 'PQR is an equilateral triangle

99 Vedanta Excel in Mathematics Teachers' Manual - 9

2. YZ = 12PQ, ZX = 21QR and XY = 21RP 2. ? Line joins the mid - points of any
two sides of the triangle is half of the

third side

3. YZ = ZX = XY 3. From (1) and (2)

4. 'XYZ is also the equilateral triangle 4. From (3)

Proved

7. In the given figure, AB // DC. If E is the mid - point of BC and
F is the mid - point of AC, prove that G is the mid - point of
AD.

Solution:
Given: AB // DC, E is the mid - point of BC and F is the mid -

point of AC.
To prove: G is the mid - point of AD.
Proof

Statements Reasons

1. In 'ABC, FE // AB 1. EF join the mid - points of sides BC and

AC

2. AB // DC 2. Given

3. AB // FE // DC i.e, AB // GE // DC 3. From (1) and (2)

4. In 'ADC, G is the mid - point of AD 4. GE // DC and F is the mid - point of AC

Proved

8. In the adjoining figure, AB // CD // PQ and AP = PC. Prove

that: AB CD.
Solution:

Given: AB // CD // PQ and AP = PC

To prove: AB = CD

Proof

Statements Reasons

1. Q is mid - point of BC 1. ? In 'ABC, AP = PC and PQ // AB

2. PQ = 12AB 2. PQ joins the mid - point of AC and BC in
3. Q is mid - point of AD 'ABC

4. PQ = 12CD 3. ? In 'ADC, AP = PC and PQ // CD
5. AB = CD
4. PQ joins the mid - point of AC and AD in
'ABC

5. From (ii) and (iv)

Proved

9. In the figure alongside, AD // PQ // BC and DQ = QC. Prove that
AD BC = 2PQ.

Solution:

Given: AD // PQ // BC and DQ = QC

To prove: AD BC = 2PQ

Proof

Statements Reasons
1. In 'ADC, AR = RC 1. DQ = QC and RQ // AD

Vedanta Excel in Mathematics Teachers' Manual - 9 100

2. RInQ'=AB21CA,DAP ?AD = 2RQ 2. RQ joins the mid - points of side AC and DC
3. PB
= 3. AR = RC and PR // BC
4. ? PR joins the mid - point of sides AB and
4. PR = 21BC ? BC = 2PR
AC
5. AD BC = 2RQ 2PR 5. Adding (ii) and (iv)
6. Whole part axiom, PR RQ = PQ
vi. AD BC = 2PQ

10. In the figure alongside, P is the mid - point of BC. A Proved
PQ // CA and QR // BC. Prove that BC = 4QR.
QR C
Solution:
P
Given: P is mid - point of BC, PQ // CA and QR // BC B

To prove: BC = 4QR

Proof

Statements Reasons
1. PQ // CA and BP = PC
1. In 'ABC, AQ = BQ 2. QR // BC and AQ = BQ, from (1)
3. From (1) and (2)
2. In 'ABD, AR = RP
4. P is the mid - point of BC
3. QR = 21BP
5. From (3) and (4)
4. BP = 21BC

5. QR = 1 u 1 BC
2 2
? BC = 4QR

11. In the given 'ABC, AX and BY are medians, Z is a point A Proved
on BC such that YZ // AX. Prove that BC = 4CZ Z
B
Solution:

Given: In 'ABC, AX and BY are median. Z is a point on BC Y
and YZ // AX. C

To prove: BC = 4CZ X

Proof

Statements Reasons

1. X is mid - point of BC and Y is the 1. Being AX and BY the median of 'ABC
mid - point of AC

2. Z is mid - point of CX 1 2. In 'ACX, AY = YC and YZ // AX
i.e, CZ = ZX i.e, CZ 2
? CX = 2CZ = CX

3. CX = 1 BC 3. BX = CX
2

4. 2CZ = 12BC 4. From (2) and (3)
?BC = 4CZ

Proved

101 Vedanta Excel in Mathematics Teachers' Manual - 9

12. In the given figure, A is the mid - point of QR and B is the C Q
mid - point of PA. Prove that PC = 13PQ. B A

Solution: C R
B Q
P A

Given: A is the mid - point of QR and B is the mid - point of PA R
To prove: PC = 13PQ
Construction: AD // BC is drawn where D is on PQ.

Proof: P

Statements Reasons
1. In 'PAD; PC = CD 1. PB = AB and BC // AD
2. In 'QCR, CD = QD 2. QA = AR and AD // CR
3. PC = CD = QD 3. From (1) and (2)
4. PC = 21PQ 4. From (3)

Proved

13. In the given right angled triangle ABC, right angled at
12AC.
B, P is the mid - point of AC. Prove that BP =
Solution:

Given: ABC is a right angled triangle, right angled at B, P

is the mid - point of AC.
1
To prove: BP = 2 AC.

Proof:

Statements Reasons
1. In 'ABC, PQ // AB and AP = CP
1. BQ = QC 2. PQ // AB, corresponding angles
3.
2. ‘PQC = ‘ABC = 90° i. From (1)
ii. Both are right angles, From (2)
3. In 'PQC and 'PQB iii. Common side
4. By S.A.S axiom
i. CQ = BQ (S) 5. Corresponding sides of congruent triangles
6. ?AP = PC
ii. ‘PQC = ‘PQB (A)
7. From (5) and 6
iii. PQ = PQ (S)

4. 'PQC # 'PQB

5. PC = BP

6. PC = 1 AC
2

7. BP= 1 AC
2

Proved

Vedanta Excel in Mathematics Teachers' Manual - 9 102

14. In the given 'ABC, AP is the bisector of ‘BAC. If BO A AP A
and OQ // AC, prove that BQ = QC
O
Solution: PQ

Given: In 'ABC, AP is the bisector of ‘BAC. BO A AP and B A C
OQ // AC. D

To prove: BQ = QC.

Construction: BO is produced to meet AC at D. O C
Proof: B PQ

Statements Reasons
1. In 'AOB and 'AOD 1.
i. ‘BAO = ‘OAD (A) i. Given
ii. AO = AO (S) ii. Common side
iii. ‘AOB = ‘AOD iii. Both are right angles
2. 'AOB # 'AOD 2. By A.S.A. axiom
3. BO = OD 3. Corresponding side of congruent triangle are equal
4. BQ = QC 4. In 'BCD, BO = OD and OQ // DC.

Proved

15. In the given trapezium PQRS, A and B are the mid - point of

the diagonals QS and PR respectively. Prove that (i) AB // SR
12(SR
(ii) AB = PQ)
Solution:

Given: In trapezium PQRS, A and B are the mid - point of the

diagonals QS and PR(iri)esApBec=tiv21el(yS.R
(i) AB // SR
To prove: PQ)

Construction: QB is produced to meet SR at C.

Proof:

Statements Reasons

1. In 'PQB and 'CBR 1.
i. ‘QPB = ‘BRC (A) i. PQ // CR and alternate angles
ii. PB = BR (S) ii. B is the mid - point of PR
iii. ‘PBQ = ‘CBK (A) iii. Vertically opposite angles are equal
2. 'PQB # 'CBR
3. BQ = BC and PQ = CR 2. By A.S.A. axiom

3. Corresponding sides of congruent
triangles are equal

4. AB // SC i.e, AB // SR and AB = 1 SC 4. In 'QSC, AB joins the mid - point of
2 sides QS and QC

5. SC = SR CR 5. Subtraction axiom
6. From (3), (4) and (5)
6. AB = 1 (SR - PQ)
2

Proved

103 Vedanta Excel in Mathematics Teachers' Manual - 9

16. In the adjoining trapezium PQRS, X and Y are the mid PQ
- point of PS and QR respectively. Prove that: XY =
1
(PQ SR) 2

Solution: SR
PQ
Given: X and Y are the mid - point of PS and QR
X
rXeYsp=ec21tiv(PeQly in trapezium PQRS.
SR)
To prove:

Construction: QX is joined and produced to meet RS T S R

Proof: produced at T.

Statements Reasons

1. In 'PXQ and 'TXS 1.
i. PQ // TR and alternate angles
i ‘XPQ = ‘TSX (A) ii. Given
iii. Vertically opposite angles are equal
ii. PX = SX (S) 2. By A.S.A. axiom
3. Corresponding sides of congruent triangle
iii. ‘PXQ = ‘TXS (A) 4. XY join the mid - points of sides QT and QR

2. 'PXQ # 'TXS respectively in 'QTR
5. Whole part axiom
3. PQ = TS and QX = TX 6. From (3), (4) and (5)

4. XY // TR and XY = 1 TR
2

5. TR = TS SR

6. XY = 1 (PQ SR)
2

1S7o.lut(Ii1no)nth:MeQad=jo12inSiRng figure, PQRS and M(iiQ) NMONa=re12rQecSt.angles. Prove that: Proved

P S

Given: PQRS and MQNO are rectangles M O R
1 1 Q N
To prove: (i) MQ = 2 SR (ii) MN = 2 QS

Proof:

Statements Reasons
1. ? ‘ONQ = ‘SRQ = 90°, corresponding angles
1. ON // SR 2. Diagonals of rectangle bisect each other
3. From (1) and (2)
2. O is mid - point of QS 4. From (2) and (3)

3. N is mid - point of QR 5. ON = MQ
6. MO // PS and QO = OS
4. ON = 1 SR 7. From (3) and (6)
2
8. ? PR = QS, diagonals of rectangles
5. MQ = 1 SR
2

6. M is the mid - point of PQ

7. MN // PR and MN = 1 PR
2

8. MN = 1 QS
2

Proved

Vedanta Excel in Mathematics Teachers' Manual - 9 104

18. In the figure alongside, M is the mid - point of BC, Q is the P A
mid - point of MR and AB // NM // CQ. B
Prove that: (i) PR = 3PM (ii) AB = 4CQ N

Solution: C
M
Given: M is the mid - point of BC, Q is the mid - point of MR R
and AB // NM // CQ. Q

To prove: (i) PR = 3PM (ii) AB = 4CQ

Proof:

Statements Reasons
1. In 'ABC, AB // NM and BM = MC
1. AN = NC 2. MN joins the mid - point of BC and AC respectively

2. MN = 1 AB 3. In 'NMR, NM // CQ and MQ = QR
2 4. CQ joins the mid - point NR and MR respectively

3. NC = CR 5. From (2) and (4)

4. CQ = 1 NM 6.
2 i. PB // CQ and alternate angles
ii. Given
5. CQ = 1 u 1 AB iii. Vertically opposite angles
2 2 7. By A.S.A. axiom
? AB = 4CQ 8. Corresponding sides of congruent triangle are equal
9. Given and statement (8)
6. In 'PBM and 'CQM 10. From (9)

i. ‘PBM = ‘MCQ (A)

ii. BM = MC (S)

iii. ‘BMP = ‘CMQ (A)

7. 'PBM # 'CQM

8. PM = MQ

9. PM = MQ = QR

10 PR = 3PM

Proved

105 Vedanta Excel in Mathematics Teachers' Manual - 9

Unit Geometry - Similarity

12

Allocated teaching periods 3

Competency

- To identify the geometric figures of similar shapes and solve the related problems

Learning Outcomes

- To indicate the condition of similarity on polygons (triangles and polygons) and
solve the problems related problems

Level-wise learning objectives

S.N. LEVELS OBJECTIVES
1. Knowledge (K)
- To define similarity on polygons
2. Understanding (U) - To tell the conditions of similarity of triangles
- To recall the Pythagoras theorem
3. Application (A)
4. High Ability (HA) - To find the sides of similar triangles or polygons
- To show the similarity of given triangles

- To prove the theorems based on similarity of triangles

- To prove the Pythagoras theorem by the virtue of the
similarity of triangles

- To explore the required relations based similar triangles

Required Teaching Materials/ Resources

Various photos of same shapes, card boards of different sizes but same shapes, leafs of
shame tree, models of enlargement and reduction etc

Pre-knowledge: Congruent triangles, similar triangles

Teaching Activities
1. For warm-up, show the pair of square/circular sheet of paper, Nepali flags, leafs, photos

etc and discuss upon the properties so observed
2. Divide the students into 4 groups and do the following activities
- Provide a pair of similar triangles to group-A
- Provide a pair of similar quadrilaterals to group-B
- Provide a pair of similar pentagons to group-C
- Provide a pair of similar hexagons to group-D

Then tell them to discuss in own group, measure the corresponding sides and angles,
their relations and present the own result in the class.
3. Also, ask the following questions
(i) Are the corresponding sides of similar triangles equal?
(ii) Are the corresponding sides of similar triangles proportional?
(iii) Are the corresponding angles of similar triangles equal?
(iv) Are the corresponding angles of similar polygons equal?
(v) Are the corresponding sides and perimeter of similar polygons equal?
(vi) Are the corresponding sides and diagonals of similar polygons equal?

Vedanta Excel in Mathematics Teachers' Manual - 9 106

4. Discuss on the conditions of similarity of triangles
5. Solve the problems from the exercise with discussion.
6. Focus on project work

Solution of selected problems from Vedanta Excel in Mathematics

1. In the given figure, 'ABC a 'ABD, ‘BAD = A

‘ACB, find the length of BD. B D8 cm4 cm
AA
Solution:

Here, 'ABC a ABD C

? AB = AB
BD BD
[Corresponding sides of similar triangle

are proportional] 4 cm
4 cm
4 cm 8 cm
or, BD = 4 cm B DB C

8 cm

? BD = 2 cm A
Hence, the length of BD is 2 cm.

2. In the adjoining figure, AB // QC, PR = 2RQ and QC = P RQ
3 cm. Find the length of AP with the suitable reasons. 3 cm

Solution: B C

Statements Reasons

1. In 'APR and 'QCR 1.

i. ‘APR = ‘RQC i. AB // QC and alternate angles

ii. ‘PAR = ‘RCQ ii. Same as (i)

iii. ‘ARP = QRC iii. Vertically opposite angles are equal

2. 'APR a 'QCR 2. By A.A.A. axiom

AP = PR Corresponding sides of similar triangle are
QC QR proportional
3. 2 RQ 3.
AP RQ
or, 3 cm = ? AP = cm

3. In a given right angled triangle ABC right angled at B, BD A

AC and ‘CBD = ‘BAC. Prove that:

(i) 'ABC a 'BCD (ii) 'BCD a 'ABD
BC2 AC
(iii) BC2 = AC.CD (iv) BD2 = AD.CD (v) BC2 = AD
Solution:

Given: 'ABC is a right angled triangle in which ‘ABC = 90q,

BD A AC and ‘CBD = ‘BAC

To prove: (i) 'ABC a 'BCD (ii) 'BCD a 'ABD (iii) BC2 = AC.CD

(iv) BD2 = AD.CD (v) BC2 = AC
BC2 AD

107 Vedanta Excel in Mathematics Teachers' Manual - 9

Proof

Statements Reasons
1.
1. In 'ABC and 'BCD i. Both are right angles
i. ‘ABC = ‘BCD ii. Common angles
ii. ‘ACB = ‘BCD 2. By A.A fact
2. 'ABC a 'BCD 3. Corresponding sides of similar triangle are proportion

3. AB = BC = AC From 2nd and 3rd ration
BD CD BC 4.
i. Both are right angles
? BC2 = AC.CD ii. Given
5. By A.A. fact
4. In 'BCD and 'ABD 6. Corresponding sides of similar triangle are proportion
i. ‘BDC = ‘ADB
ii. ‘DBC = ‘BAD From 2nd and 3rd ration
7. From (3) and (6)
5. 'BCD a 'ABD

6. BC = BD = CD
AB AD BD

? BD2 = AD.CD

7 BC2 AC.CD AC
BD2 = AD.CD = AD

Proved

4. In the given figure, 'ADE = ‘ACB and ‘DAC = ‘BAE. A B
Prove that, AD.BC = AC.DE. E
D
Solution: C

Given: ‘ADE = ‘ACB and ‘DAC = ‘BAE.

To prove: AD.BE = AC.DE

Proof

Statements Reasons
1. Given
1. ‘DAC = ‘BAE 2. Adding ‘EAC on both sides of (1)
3.
2. ‘DAE = ‘BAC i. Given
3. In 'ADE and 'ABC ii. From (2)
i. ‘ADE = ‘ACB 4. By A.A axiom
ii. ‘DAE = ‘BAC 5. Being corresponding sides of similar

4. 'ADE a 'ABC triangles

5. AE = DE = AD 6. From (5), taking last two ration
AB BC AC

6. AD.BC = AC.DE

Proved

5. In the given figure, AB = DC, AB // DC and M is the mid -

point of AB. Prove that:

(i) 'AOM a 'COD (ii) CO = 2AO

Solution:

Given: AB = DC, AB // DC and M is the mid - point of AB.

To prove: (i) 'AOM a 'COD (ii) CO = 2AO

Vedanta Excel in Mathematics Teachers' Manual - 9 108

Proof

Statements Reasons
1.
1. In 'AOM and 'COD i. AB // DC and alternate angles
i. ‘OAM = ‘OCD ii. Vertically opposite angles
ii. ‘AOM = ‘COD 2. By A.A. axiom
3. Corresponding sides of similar
2. 'AOM a 'COD
triangle
3. AO = OM = AM
CO OD CD 4. Given

4. AB = 2AM and AB = CD

5. AO = AM ? CO = 2AO
CO 2AM

6. In the given figure, AB // MN // DC. If AB = x, DC = y and MN = z. Proved
1 1 1z.
Prove that x y = A D
Solution:
M
Given: AB // MN // DC, AB = x, DC = y and MN = z. x y

To prove: 1 1 = 1 B z
x y z NC
Proof

Statements Reasons
1.
1. In 'ABC and 'MNC i. AB // MN, corresponding angles
ii. Same as (i)
i. ‘ABC = ‘MNC iii. Common angles
ii. ‘BAC = ‘NMC 2. By A.A.A. axiom
iii. ‘ACB = ‘MCN
Corresponding sides of similar triangles
2. 'ABC a 'MNC
4.
3. AB = BC i. MN // DC, corresponding angles
MN NC ii. Same as 4(i)
BC iii. Common angles
or, x = NC ? NC = z BC 5. By A.A.A. axiom
z x 6. Corresponding sides of similar triangles

4. In 'BCD and 'BNM
i. ‘BCD = ‘BNM
ii. ‘BDC = ‘BMN
iii. ‘DBC = ‘MBN

5. 'BCD a BNM

6. DC = BC
MN BN
BC
or, y = BN ? BN = z BC
z y

7. NC BN = zBC 11 7. Adding (3) and (6)
x y

8. BC = zBC 11 ? 1 1 = 1 8. NC BN = BC
x y x y z

Proved

109 Vedanta Excel in Mathematics Teachers' Manual - 9

Extra Questions M A
x 4 cm
1. In the given figure, 'ABC aMNC, find the value of x.
[Ans: 1.6 cm] C 2 cm N 3 cm B

X 15
P
2. In the adjoining figure, XZ = 15 cm, PZ = 10 cm QZ = 6 cm
cm and XY // PQ. Find the length of YQ
[Ans: 3 cm] 10 cm

Y Q 2 cm Z

A

3. In the figure given alongside 'ABC a 'ADC. Find the x cm (x 3)cm

DE

value of x. (x 2)cm (x 4)cm
[Ans: 6]
BC

A

4. In the given figure, ‘BAC = 90q, AD A BC, CD = 9 cm 4 cm
and BC = 12 cm, find the length of AB.
[Ans: 6 cm] B D12 9cmcm C

B. Pythagorean Theorem

1. IntheadjoiningrightangledtriangleABC,DisanypointonAB.
Prove that AB2 AD2 = BC2 CD2

Solution: Reasons
1. Using Pythagoras theorem in 'ABC
Statements 2. Using Pythagoras theorem in 'ADC
1. AB2 AC2 = BC2 3. From (1) and (2)

? AC2 = BC2 AB2 Proved
2. AD2 AC2 = CD2

? AC2 = CD2 AD2
3. BC2 AB2 = CD2 AD2

? BC2 CD2 = AB2 AD2

2. In the given right angled triangle ABC, M is the mid - point of
BC. Prove that AC2 = AM2 3CM2

Solution: Reasons
1. By Pythagoras theorem
Statements
1. AB2 BM2 = AM2

Vedanta Excel in Mathematics Teachers' Manual - 9 110

2. AB2 CM2 = AM2 2. BM = CM and from (1)
? AB2 = AM2 CM2
3. By Pythagoras theorem
3. AB 2 BC2 = AC2 4. From (3), BC = 2 cm
4. AB2 (2cm)2 = AC2 5. From (2) and (4)
5. AM2 CM2 4 cm2 = AC2

? AC2 = AM2 3 cm2

3. In the given figure, X is the mid - point of PQ. Prove that R Proved
PQ2 =(RX2 PR2)
Q

X

Solution: P

Statements Reasons
1. PQ = 2PX 1. X being the mid - point of PQ

? PQ2 = (2PX)2 = 4PX2 2. In 'PRX, by Pythagoras theorem
2. PX2 = RX2 PR2 3. From (1) and (2)
3. PQ2 = 4(RX2 PR2)

Proved

4. ABC is an isosceles triangle in which AB = AC = 2BC. If AD is an A

altitude of the triangle. Prove that 4 AD2 = 15 BC2.

Solution:

B DC

Statements Reasons

1. BD = CD = 1 BC 1. Altitude of isosceles triangle bisects its
2 base

2. AD2 BD2 = AB2 2. In 'ABD, by Pythagoras theorem
or, AD2 (12BC)2 = (2BC)2 1
or, 4AD2 BC2 = 16BC2 BD = 2 BC and AB = 2BC

? 4AD2 = 15BC2

Proved

5. In the given figure, diagonals of a quadrilateral PQRS are P S

intersected at T at right angle.

Prove that: PQ2 SR2 = QR2 PS2 T
Solution:

Given: The diagonals of the quadrilateral PQRS intersect at T at Q R
a right angle.

To prove: PQ2 SR2 = QR2 PS2

111 Vedanta Excel in Mathematics Teachers' Manual - 9

Proof

Statements 1. Reasons
1. In 'PQT, QT2 = PQ2 PT2 By using Pythagoras therorm

In 'QTR, QT2 = QR2 TR2

Thus, PQ2 PT2 = QR2 TR2

? PQ2 = QR2 PT2 TR2 2. Same as (1)
2. In 'PST, ST2 = PS2 PT2

In 'STR, ST2 = SR2 TR2

Thus, PS2 PT2 = SR2 TR2

? SR2 = PS2 PT2 TR2 From (1) and (2)
3. PQ2 SR2 = QR2 PT2 TR2 PS2 PT2 TR2 3.

? PQ2 SR2 = QR2 PS2

Proved

6. In the figure alongside, O is any point W ZW Z
interior to the rectangle. OP OQ
Prove that OX2 OY2 = OW2 OZ2.

Solution:

Given: O is any point interior to the X YX Y
rectangle WXYZ.

To prove: OX2 OY2 = OW2 OZ2

Construction: PQ // XY is drawn through O where P is on WX and Q is on YZ

Proof

Statements Reasons
1. OP A XW and OQ A YZ
1. PQ // XY and ‘X = ‘Y
2. OP2 = OW2 PW2 and OP2 = OX2 PX2 = 90q
or, OW2 PW2 = OX2 PX2
? OX2 = OW2 PX2 PW2 2. In 'POW and 'POX

3. OQ2 = OZ2 QZ2 and OQ2 = OY2 QY2 using Pythagoras
or, OZ2 QZ2 = OY2 QY2 therorm
? OY2 = OZ2 QZ2 QY2
3. In 'QOZ and 'QOY
4. OX2 OY2 = OW2 PX2 PW2 (OZ2 QZ2 QY2)
= OW2 OZ2 PX2 PW2 QZ2 QY2 using Pythagoras
therorm
5. OX2 OY2 = OW2 OZ2 QY2 PW2 PW2 QY2
? OX2 OY2 = OW2 OZ2 4. Subtracting (3) from
(2)

5. PX = QY, QZ = PW

Proved

Vedanta Excel in Mathematics Teachers' Manual - 9 112

Unit Parallelogram

13 Allocated teaching periods 10

Competency
- To prove the theorems and properties of parallelograms and verify the other
properties by inductive method.

Learning Outcomes
- To prove the theorems and properties of parallelograms
- To verify the other properties of parallelogram by experimental or inductive
method.

Level-wise learning objectives

S.N. LEVELS OBJECTIVES
1. Knowledge (K) - To define parallelogram
- To recall the types of parallelogram
2. Understanding (U) - To tell the properties of parallelogram
- To recall the relations between corresponding parts of

parallelogram (rectangle, square and rhombus).
- To calculate the size of unknown angles based on

properties of parallelograms
- To prove the theorems on properties of parallelogram

3. Application (A) - To verify the properties of the parallelogram by inductive
method

- To prove more theorems by drawing realted diagrams/

4. High Ability (HA) figure

Required Teaching Materials/ Resources
Colourful chart-paper, scale, scissors, pencil, geo-board, rubber bands, tangram, marker,
models of measuring tape, ICT tools etc

Pre-knowledge: Congruency of triangles, types of parallelograms, mid-point theorem etc.

Teaching Activities
1. Present the parallelograms on the geo-board by using rubber bands or make

parallelograms using a rectangular piece of paper along with students and recall about
parallelograms.
2. With the figure on the chart-paper, prove the theorems with students.
3. Rename the same theorem and give to students to prove them.
4. Select the students randomly and call to prove the theorem on the board.

113 Vedanta Excel in Mathematics Teachers' Manual - 9

Solution of selected problems from Vedanta Excel in Mathematics

1. In the given figure, PQRS is a rhombus and P S
SRM is an equilateral triangle. If SN A RM and

‘PRS = 55°, find the size of ‘QSN.

Solution: Q O 55° N M
(i) In rhombus PQRS, ³SOR = 900 [Diagonals of R
R
rhombus bisect each other perpendicularly] Q

? ³OSR = 1800 – (900 + 550) = 450

(ii) In equilateral 'SRN; ³SRN= 600 [Each angle of equilateral triangle]

? ³RSN = 1800 – (900 + 600) = 300

Hence, ³QSN = ³OSR +³RSN = 450 + 300 = 750

2. In the adjoining figure, PQRS is a rectangle. If S
OP = (2x – 3) cm and OR = (x + 1) cm, find the length of (x+1)cm
diagonal QS.
(2x–3)cm O
Solution: P

(i) OP = OR [Diagonals of rectangle bisect each other]

or, 2x – 3 = x + 1 ?x = 4 and PR = OP + OR = (2x – 3)+(x+1) = 10 cm

(ii) PR= QS [Diagonals of rectangle are equal] ?QS = 10 cm

3. RACE is a rectangle in which diagonal RC = 18 cm, R E
OA = (p + q) cm and OE = 3p cm, find the values of p and q. 3p cm C

Solution: (p+q)cmO C
A

(i) AE = RC = 18 cm [Diagonals of rectangle are equal]

(ii) OA= OE = 9 cm [Diagonals of rectangle bisect each other]

Now, OE = 9 cm or, 3p = 9 cm ?p = 3 cm

Again, OA = 9 cm or, p + q = 9 cm or, 3 cm + q = 9 cm ?q = 6 cm

D

4. In the adjoining figure ABCD is a square and ABE is an E
equilateral triangle. Find the measure of ‘ADE and ‘DCE.

Solution:

(i) ³DAB = 900 [Being an angle of the square]

(ii) ³EAB = 600 [Being an angle of the equilateral triangle] A B

? ³DAE = 900 – 600 = 300

(iii) ³ADE = ³AED [ AD = AE]

Now, ³ADE + ³AED + ³DAE=1800

or, 2³ADE +300 = 1800 ?³ADE = 750

(iv) ³BCE = 750 [Same as the above process] ?³DCE= 900 – 750 = 150

5. In the given figure, ABCD is a parallelogram. AP bisects
‘A. Prove that DP = BC.

Solution:
Given: In the parallelogram ABCD; AP bisects ³A.

Vedanta Excel in Mathematics Teachers' Manual - 9 114

To prove: DP = BC
Proof:

³DAP = ³BAP [AP bisects ³A]

³APD = ³BAP [DC//AB, alternate angles]

³DAP = ³APD [From (1) and (2)]

(4) DP = AD [From (3), the base angles of 'ADP are equal]

(5) AD = BC [Opposite sides of parallelogram are equal]

(6) DP = BC [From (4) and (5)]

6. In the given figure, EXAM is a parallelogram, the bisector of E P M

‘A meets the mid-point of EM at P. Prove that AX = 2AM.

Solution:

Given: In the parallelogram EXAM; the bisector of ³A meets the mid- X A
point of EM at P.

To prove: AX = 2AM

Proof:

(1) ³XAP = ³MAP [AP is the bisector of ³A]

(2) ³XAP = MPA [EM//XA, alternate angles]

(3) ³MAP = MPA [From (1) and (2)]

(4) PM = AM [From (3), the base angles of 'ADP are equal]

(5) ME = 2PM [P is the mid-point of ME]

(6) ME = AX [Opposite sides of parallelogram are equal]

(7) AX = 2AM [From (4),(5) and (6)] AB

7. In the adjoining figure, ABCD is a rhombus in which CD is

produced to E such that CD = DE. Prove that ‘EAC = 90°.

Solution: E DC

Given: In the rhombus ABCD; CD is produced to E such that CD = DE.

To prove: ³EAC = 900
Proof:

(1) AD = CD [Adjacent sides of the rhombus are equal]

(2) CD = DE [Given]

(3) AD = CD = DE [From (1) and (2)]

(4) ³AEC = ³EAD [From (3), AD = DE]

(5) ³DAC = ³ACE [From (3), AD = CD]

(6) ³AEC + ³EAC +³ACE = 1800 [Sum of angles of triangle]

(7) ³EAD + ³EAC +³DAE = 1800 [From (4), (5) and (6)]

(8) ³EAC + ³EAC = 1800 [From (7), ³EAD + ³DAE = ³EAC]

(9) ³EAC = 900 [From (8)]

115 Vedanta Excel in Mathematics Teachers' Manual - 9

8. In the given figure, P and R are the mid–points of the sides AB
and BC of ∆ABC respectively. If PQ // BC,
prove that BP = RQ.

Solution:

Given: In 'ABC; P and R are the mid-points of the sides AB and BC
respectively PQ // BC.

To prove: BP = RQ

Proof:

(1) AQ = QC [In 'ABC;PQ//BC and AP = BP]

(2) QR //AB [AQ = QC and BR = RC]

(3) PQRB is a parallelogram [From (2) and PQ // BC]

(4) BP = RQ [Opposite sides of the //gm PQRB]

9. In the given quadrilateral PQRS, the mid–points of the
sides PQ, QR, RS and SP are A, B, C and D respectively.
Prove that ABCD is a parallelogram.

Solution:

Given: In quadrilateral PQRS; the mid-points of sides PQ, QR, RS

and SP are A, B, C and respectively.

To prove: ABCD is a parallelogram

Construction: Diagonal PR of the quadrilateral PQRS is drawn
Proof:

(1) In 'PQR, AB//PR and AB = 1 PR [AB joins the mid points of PQ and QR]
2 [CD joins the mid points of RS and SP]

(2) In 'PSR, DC//PR and AB = 1 PR
2

(3) AB//DC and AB = DC [From (1) and (2)]

(4) BC//AD and BC =AD [From (3)]

(5) ABCD is a parallelogram [From (3) and (4)]

10. In the adjoining figure, P, Q, R and S are the mid-points of AB, BC,

CD and AD respectively. Prove that PQRS is a parallelogram.

Solution:

Given: P, Q, R and S are the mid-points of sides AB, BC, CD and AD

respectively

To prove: PQRS is a parallelogram

Construction: AC is joined.

Proof:

(1) In 'ABC, PQ//AC and PQ = 1 AC [PQ joins the mid points of AB and BC]
2

(2) In 'ADC, SR//AC and SR = 1 AC [SR joins the mid points of AD and CD]
2

(3) PQ//SR and PQ = SR [From (1) and (2)]

(4) QR//PS and QR = PS [From (3)]

(5) PQRS is a parallelogram [From (3) and (4)]

Vedanta Excel in Mathematics Teachers' Manual - 9 116

11. In the given figure, P, Q, R and S are the mid-points of AB, BC, CD A
and AD respectively. Prove that PQRS is a parallelogram.

Solution: P S
Given: P, Q, R and S are the mid-points of sides AB, BC, CD and AD C

respectively QR
BD
To prove: PQRS is a parallelogram

Construction: BD is joined.

Proof:

(1) In 'ABD, PS//BD and PS = 1 [PS joins the mid points of AB and AD]
BD 2

(2) In 'BCD, QR//AC and PS = 1 [QR joins the mid points of BC and CD]
BC 2

(3) PS//QR and PS = QR [From (1) and (2)]
(4) PQ//SR and PQ = SR [From (3)]

(5) PQRS is a parallelogram [From (3) and (4)]

12. In the given figure, PQRS is a square. A, B, C and D are
the points on the sides PQ, QR, RS and SP respectively.
If AQ = BR = CS = DP, prove that ABCD is also a square.

Solution:

Given: PQRS is a square. AQ = BR = CS = DP

To prove: ABCD is the square.

Proof:

(1) AQ = BR = CS = DP [Given]

(2) PA = BQ = CR = SD [Remaining parts of equal sides]

(3) In 'AQB and 'BRC [From (1)]
(i) AQ = BR [Both are right angles]
(ii) ³AQB = ³BRC [From (2)]
(iii) BQ = CR

(4) 'AQB # 'BRC [By S.A.S axiom]

(5) AB = BC and ³ABQ = ³BCR [Corresponding parts of congruent triangles]

(6) ³CBR + ³BCR = 900 [Sum of acute angles in rt. ³ed 'BCR]

(7) ³CBR + ³ABQ = 900 [From (5) and (6)

(8) ³CBR + ³ABQ + ³ABC = 1800 [Being the parts of a straight angle]

?³ABC = 900 [By S.A.S axiom]
(9) 'BCR # 'CDS, 'CDS # 'AQB
(10) BC = CD and CD = AD [Corresponding sides of congruent triangles]
(11) ABCD is a square [ From (5), (9) and (10)]

117 Vedanta Excel in Mathematics Teachers' Manual - 9

13. In the given parallelogram PQRS, PA bisects ‘P and RB

bisects ‘R. Prove that PA // BR.

Solution:

Given: In the parallelogram PQRS, PA bisects ³A and RB bisects ³R

To prove: PA//BR

Proof:

(1) ³SPQ = 2³APQ and ³QRS = 2³BRS [Given]

(2) ³SPQ = ³QRS [Opposite angles of parallelogram]

(3) ³APQ = ³BRS [From (1) and (2)]

(4) ³APQ = ³SAP [SR//PQ, alternate angles]

(5) ³BRS = ³SAP [From (3) and (4)]

(6) PA//BR [Corresponding angles are equal]

14. ABCD is a parallelogram. P and Q are two points on the

diagonal BD such that DP = QB. Prove that APCQ is a

parallelogram.

Solution:

Given: In the parallelogram ABCD, P and Q are two points on the

diagonals BD such that DP = QB

To prove: APCQ is a parallelogram

Proof:

(1) OA = OC and OD = OB [Diagonals of parallelogram bisect each other]

(2) DP = QB [Given]

(3) OA = OB and OP = OQ [OD-DP = OP, OB – QB =OQ ]

(4) APCQ is a parallelogram [From (3), diagonals are bisected each other]

15. ABCD is a parallelogram. DE A AC and BF A AC. Prove

that BEDF is a parallelogram.

Solution:

Given: ABCD is a parallelogram. DEAAC and BFAAC

To prove: BEDF is a parallelogram

Proof:

(1) In 'DAE and 'BCF

(i) ³AED = ³BFC [Both are right angles]

(ii) ³DAE= ³BCF [DA//BC, alternate angles]

(iii) AD = BC [Opposite sides of parallelogram]

(2) 'DAE # 'BCF [By A.A.S. axiom]

(3) AP = QC [Corresponding sides of congruent triangles]

(4) DE//FB [³AED = ³BFC, alternate exterior angles are equal]

(5) DF = EB, DF//EB [From (3) and (4)]

(6) BEDF is a parallelogram [From (3), (4) and (5)]

Vedanta Excel in Mathematics Teachers' Manual - 9 118

16. In parallelogram PQRS, the bisectors of ‘PQR and ‘PSR P M S
meet the diagonal at M and N respectively. Prove that R
MQNS is a parallelogram.
D
Solution: N
Q
Given: In //gm PQRS, the bisectors of ³PQR and ³PSR meet the

diagonal PR at M and N respectively i.e., ³PQM = ³RSN and Q

³PSN = ³RSN

To prove: MQNS is a parallelogram

Proof:

(1) In 'PQM and 'RSN

(i) ³PQM = ³RSN [Given and ³PQR= ³PSR]

(ii) PQ = RS [Opposite sides of parallelogram]

(iii) ³QPM= ³SRN [PQ//SR, alternate angles]

(2) 'PQM # 'RSN [By A.S.A. axiom]

(3) MQ = SN, ³PMQ = ³SNR [Corresponding parts of congruent triangles]

(4) MQ//SN [From(3), alternate exterior angles are equal]

(5) MS=QN, MS//QN [From (3) and (4)]

(6) MQNS is a parallelogram [From (3), (4) and (5)] A

17. In the given figure, ABCD is a parallelogram. If P and Q are

the points of trisection of diagonal BD, prove that PAQC is P

a parallelogram. BC
Solution:

Given: In parallelogram ABCD, P and Q are the points of trisections of the diagonal BD. i.e.,

BP = PQ = QD

To prove: PAQC is a parallelogram

Proof:

(1) In 'ABP and 'CDQ

(i) AB = CD [Opposite sides of parallelogram]

(ii) ³ABP = ³CDQ [AB//CD, alternate angles]

(ii) BP =QD [Given]

(2) 'ABP # 'CDQ [By S.A.S. axiom]

(3) AP = CQ, ³APB = ³CQD [Corresponding parts of congruent triangles]

(4) AP//QC [From(3), alternate exterior angles are equal]

(5) AQ = PC, AQ//PC [From (3) and (4)]

(6) PAQC is a parallelogram [From (3), (4) and (5)]

18. In the figure, ABCD is a parallelogram. M and N are the
mid-points of the sides AB and DC respectively. Prove
that
(i) MBCN is a parallelogram
(ii) DMBN is a parallelogram
(iii) DB and MN bisect each other at O.

119 Vedanta Excel in Mathematics Teachers' Manual - 9

Solution:

Given: In parallelogram ABCD, M and N are the mid-points of the sides AB and DC respectively

To prove: (i) MBCN is a parallelogram (ii) DMBN is a parallelogram

(iii) DB and MN bisect each other

Proof:

(1) AB = DC and AB // DC [Opposite sides of parallelogram]

(2) MB = CN and MB//CN [M and N are the mid-points of AB and DC]

(3) MN = BC and MN//BC [From (2)]

(4) MBCN is a parallelogram [From (2) and (3)]

(5) DM = BN and DM//BN [MB=DN and MB//DN]

(6) DMBN is a parallelogram [From (5)]

(7) DB and MN bisect each other [Being the diagonals of parallelogram DMBN]

19. In the given parallelogram PQRS, M and N are the mid-
points of the sides PQ and SR respectively. Prove that

(i) PNRM is a parallelogram (ii) QA = AB = BS

Solution:

Given: In parallelogram ABCD, M and N are the mid-points of the

sides PQ and SR respectively

To prove: (i) PNRM is a parallelogram (ii) QA = AB = BS

Proof:

(1) PM = NR and PM // NR [PQ = SR, PQ//SR and given]

(2) PN = MR and PM//MR [From (1)]

(3) PNRM is a parallelogram [From (1) and (2)]

(4) QA = AB [In 'PQB, PM = QM and MA // PB]

(5) AM = BS [In 'SAR, SN = RN and BN// AR]

(6) QA = AB = BS [From (4) and (5)]

20. ABCD is a square. P and Q are any points on the sides AB
and BC respectively. If AQ = DP, prove that AQ and DP are
perpendicular to each other.

Solution:

Given: In square ABCD, P and Q are the points on the sides AB and BC respectively

To prove: AQ and DP are perpendicular to each other. AQ = DP

Proof:

(1) In 'DAP and 'ABQ

(i) ³DAP = ³ABQ (R) [Both are right angles]

(ii) DP = AQ (H) [Given]

(iii) DA = AB (S) [Adjacent sides of the square ABCD]

(2) 'DAP #'ABQ [By R.H.S. axiom]

(3) ³ADP = ³BAQ [Corresponding angles of congruent triangles]

(4) ³DAQ + ³ADP = 900 [From (3) and ³DAP = ³DAQ + ³QAP = 900]

(5) ³AOD = 900 [From (4), remaining angle of 'AOD]

Hence, AQ and DP are perpendicular each other.

Vedanta Excel in Mathematics Teachers' Manual - 9 120

21. In the figure along side, ABCD and DEFC are parallelograms.
Prove that
(i) AE = BF (ii) 'ADE # 'BCF

Solution:

Given: ABCD and DEFC are parallelograms.

To prove: (i) AE = BF (ii) 'ADE # 'BCF

Proof:

(1) AB = DC, AB//DC [Opposite sides of parallelogram ABCD]

(2) DC = EF, DC//EF [Opposite sides of parallelogram DEFC]

(3) AB = EF and AB//EF [From (1) and (2)]

(4) AE = BF and AE//BF [From (3)]

(6) In 'ADE and 'BCF

(i) AE = BF (S) [From (4)]

(ii) AD = BC (S) [Opposite sides of parallelogram ABCD]

(iii) DE= CF (S) [Opposite sides of parallelogram DEFC]

(7) 'ADE #'BCF [By S.S.S. axiom]

22. In the given quadrilateral ABCD, AD = BC, P and Q are the DN C

mid-points of the diagonals AC and BD respectively. If M and N PQ
are the mid-points of the sides DC and AB respectively, prove

that PMQN is a parallelogram.

Solution: A MB

Given: In quadrilateral ABCD, AD = BC, P and Q are the mid-points of the diagonals AC

and BD respectively, M and N are the mid-points of the sides DC and AB respectively.

To prove: PMQN is a rhombus

Proof: 1
2
NP//DA and NP = DA [Applying mid-point theorem in 'CDA]

QM//DA and QM = 1 DA [Applying mid-point theorem in 'BDA]
2

(1) NP//QM and NP = QM [From (1) and (2)]

(2) PM//NQ and PM = NQ [From (3)]

(3) NP = PM = QM = NQ [AD = BC]

(4) PMQN is a rhombus [From (5)] DC

23. In the adjoining figure, ABCD is a parallelogram. AS, P S
BS, CQ and DQ are the bisectors of ‘A, ‘B, ‘C and ‘D R
respectively. Prove that PQRS is a rectangle. Q

Solution: A B
Given: In parallelogram ABCD; AS, BS, CQ and DQ are the

bisectors of ³A, ³B, ³C and ³D respectively.
To prove: PQRS is a rectangle

121 Vedanta Excel in Mathematics Teachers' Manual - 9

Proof:

(1) ³DAB = 2³DAS = 2³BAS, ³ABC = 2³ABS = 2³CBS, ³BCD = 2³BCQ = 2³DCQ

and ³ADC = 2³ADQ = 2³CDQ [Given]

(2) ³DAB +³ABC = 1800 [AD//BC, co-interior angles]

(3) 2³BAS + 2³ABS = 1800 [From (1) and (2)]

?BAS + ³ABS = 900

(4) ³BAS + ³ABS + ³ASB = 1800 [Sum of angles of triangle ABS]

(5) ³ASB = 900 [From (3) and (4)]

(6) ³CQD = ³SPQ = ³SRQ = 900 [Same as above process]

(7) PQRS is a rectangle [From (5) and (6)]

24. Prove that the diagonals of a rectangle are equal.

Solution:

Given: ABCD is a rectangle in which AC and BD are the diagonals

To prove: AC = BD

Proof:

(1) In 'ABC and 'BCD

(i) AB = CD (S) [Opposite sides of rectangle are equal]

(ii) ³ABC = ³BCD (A) [Both are right angles]

(iii) BC = BC (S) [Common side]

(2) 'ABC # 'BCD [By S.A.S. axiom]

(3) AC = BD [Corresponding sides of congruent triangles]

25. If the diagonals of a rhombus are equal, prove that it is a square.

Solution:

Given: In rhombus ABCD, AC = BD

To prove: ABCD is a square

Proof:

(1) In 'ABC and 'BCD

(i) AB = CD (S) [Opposite sides of rectangle are equal]

(ii) BC = BC (S) [Common side]

(iii) AC = BD (S) [Given]

(2) 'ABC # 'BCD [By S.S.S. axiom]

(3) ³ABC = ³BCD [Corresponding angles of congruent triangles]

(4) ³ABC + ³BCD = 1800 [AB//DC and co-interior angles]

(5) ³ABC = 900 [From (3) and (4)]

(6) ABCD is a square [From (5) and AB = BC = CD = A]

A D
C
26. Prove that the diagonals of a rhombus bisect each other B
perpendicularly.

Solution:
Given: In rhombus ABCD, the diagonals AC and BD intersect at O.

Vedanta Excel in Mathematics Teachers' Manual - 9 122

To prove: Diagonals AC and BD bisect each other perpendicularly i.e., OA = OC, OB = OD

and ³AOB = 900

Proof:

(1) In 'AOB and 'AOD

(i) AB = AD (S) [Sides of rhombus are equal]

(ii) ³BAO = ³DAO (A) [Diagonal AC bisects ³A]

(iii) OA = OA (S) [Common side]

(2) 'AOB # 'AOD [By S.A.S. axiom]

(3) OB = OD [Corresponding sides of congruent triangles]

(4) ³AOB =³AOD [Corresponding angles of congruent triangles]

(5) ³AOB = ³AOD = 900 [Being linear pair]

(6) 'AOB # 'BOC [Same as above process]

(7) OA = OC [Corresponding sides of congruent triangles]

(8) AC and BD bisect each other perpendicularly at O [From (3), (5) and (7)]

S

27. In the adjoining quadrilateral PQRS, ‘P = ‘R and
‘Q = ‘S. Show that PQRS is a parallelogram.

Solution: P R
C
Given: In quadrilateral PQRS, ³P = ³R and ³Q = ³S

To prove: PQRS is a parallelogram

Proof: Q
(1) ³P = ³R and ³Q = ³S [Given]

(2) ³P+³R+³Q+³S=3600 [Sum of angles of quadrilateral]

(3) ³R+³R+³S+³S=3600 [From (i) and (ii)]

?³R +³S=1800

(4) PS//QR [From (3) co-interior angles are supplementary]

(5) PQ//SR [Same as above process]

(6) PQRS is a parallelogram [From (4) and (5)] D

28. In the given quadrilateral ABCD, AO = OC and

BO = OD. Prove that ABCD is a parallelogram. O
Solution:

Given: In quadrilateral PQRS, OA = OC and OB = OD A B
To prove: ABCD is a parallelogram

Proof:

(1) In 'AOD and 'BOC

(i) OA= OC (S) [Given]

(ii) ³AOD = ³BOC (A) [Vertically opposite angles are equal]

(iii) OD = OB (S) [Given]

(2) 'AOD # 'BOC [By S.A.S. axiom]

(3) AD = BC and ADO = ³OBC [Corresponding parts of congruent triangles]

(4) AD//BC [From (3), alternate angles are equal]

(5) AB//DC and AB=DC [From (3) and (4)]

(6) ABCD is a parallelogram [From (3),(4) and (5)]

123 Vedanta Excel in Mathematics Teachers' Manual - 9

29. In the adjoining parallelogram ABCD, P and Q are the
mid-points of the sides AD and BC respectively. Prove that
BP and QD trisect the diagonal AC at X and Y respectively.

Solution:

Given: In parallelogram ABCD, P and Q are the mid-points of the

sides AD and BC respectively

To prove: BP and QD trisect AC at X and Y i.e., AX = XY = YC

Proof:

(1) PD = BQ and PD // BQ [AD = BC, AD//BC and given]

(2) PB = DQ and B//DQ [From (1)]

(3) AX = XY [In 'ADY, AP = PD and PX // DY]

(4) XY = YC [In 'BCX, BQ = QC and YQ// XB]

(5) AX = XY = YC [From (3) and (4)]

30. In the given parallelogram PQRS, M and N are the mid- P Q
points of the sides SR and QR respectively. If the diagonal PR N
and QS intersect at H, prove that MRNH is a parallelogram. H R

Solution: B

Given: In parallelogram PQRS, M and N are the mid-points of the S M

sides SR and QR respectively

To prove: MRNH is a parallelogram

Proof:

(1) HP = HR and HQ = HS [Diagonals of parallelogram bisect each other]

(2) PS//QR and PQ//SR [Opposite sides of parallelogram PQRS]

(3) In 'QSR, HN// SR [HN joins mid-points of the sides QS and QR]

(4) In 'PSR, HM// PS [HM joins mid-points of the sides PR and SR]

(5) HN//SR and HM//QR [From (2), (3) and (4)] A

(6) MRNH is a parallelogram [From (5)]

31. In the figure alongside, ABCD is a quadrilateral in which OA D O
= OC, OB = OD and ‘ AOB = ‘ AOD = ‘ DOC = ‘ COB = C
90°. Prove that ABCD is a rhombus.

Solution:

Given: In quadrilateral ABCD, OA = OC, OB = OD and ³AOB = ³AOD = ³DOC = ³COB = 900

To prove: ABCD is a rhombus

Proof:

(1) In 'AOB and 'BOC

(i) OA= OC (S) [Given]

(ii) ³AOB = ³BOC (A) [Both are right angles]

(iii) OB = OB (S) [Common side]

(2) 'AOB # 'BOC [By S.A.S. axiom]

(3) AB = BC [Corresponding sides of congruent triangles]

(4) 'BOC # 'COD and 'COD # 'AOB [Same as above process]

(5) BC = CD = AD [Corresponding sides of congruent triangles]

(6) ABCD is a rhombus [From (3) and (5)]

Vedanta Excel in Mathematics Teachers' Manual - 9 124

32. In the adjoining figure, EFGH is a H G Q
parallelogram and P is the mid-point of P
FG. EP and HG are produced to meet at Q.
Prove that AQ = 2HG.

Solution:

Given: In parallelogram EFGH, P is the mid-point E F

of FG. EP and HG produced meet at Q.
To prove: HQ = 2HG
Proof:
(1) In 'EFP and 'GPQ
(i) ³PEF = ³PQG (A) [HQ//EF, alternate angles]
(ii) ³EPF = ³GPQ (A) [Vertically opposite angles are equal]
(iii) FP = GP (S) [Given]
(2) 'EFP # 'GPQ [By A.A.S. axiom]
(3) EF= GQ [Corresponding sides of congruent triangles]

(4) EF = HG [Opposite sides of parallelogram EFGH]

(5) HQ = 2HG [From (3) and (4)]

33. In the adjoining figure, ABCD is a rectangle and P, A P B
Q, R and S are the mid-points of AB, BC, CD and DA Q
respectively. Prove that PQRS is a rhombus. C
Solution: S

Given: In rectangle ABCD; P, Q, R and S are the mid-points of

AB, BC, CD and DA respectively DR
To prove: PQRS is a rhombus
Proof:
Proof:
(1) SA = SD = BQ = QC [S and Q are mid-points of AD and BC]
(2) DR = CR = AP = BP [R and P are mid-points of DC and AB]
(3) In 'SDR and 'RCQ
(i) SD = QC (S) [From (1)]
LL ³SDR = ³RCQ (A) [Both are right angles]
(iii) DR = CR (S) [From (2)]
(4) 'SDR # 'RCQ [By S.A.S. axiom]
(5) SR = QR [Corresponding sides of congruent triangles]
(6) 'RCQ # 'PQB # 'ASP [Same as above process]
(7) QR = PQ = PS [Corresponding sides of congruent triangles]

(8) PQRS is a rhombus [From (5) and (7)] D GC

34. In the given square ABCD, E, F, G and H are the mid-points F
of AB, BC, CD and DA respectively. Prove that EFGH is also a H B

square.
Solution:
Given: In square ABCD; E, F, G and H are the mid-points of AB, BC, A E

CD and DA respectively
To prove: EFGH is a square
Proof:
(1) In 'HAE and 'EBF
(i) HA = BF (S) [H and F are mid-points of AD and BC, AD =BC]
(ii) ³HAE = ³EBF (A) [Both are right angles]

125 Vedanta Excel in Mathematics Teachers' Manual - 9

(iii) AE = BE (S) [E is the mid-point of AB
(2) 'HAE # 'EBF [By S.A.S. axiom]
(3) HE = EF and ³AEH = ³BEF [Corresponding parts of congruent triangles]
(4) 'EBF# 'FCG # 'HDG [Same as above process]
(5) EF = FG = HG [Corresponding sides of congruent triangles]
(6) ³AEH = ³BEF = 450 [Acute angles of isosceles right angled triangles]
(7) ³AEH +³BEF+ ³HEF= 1800 [Being parts of straight angle]
(8) ³HEF= 900 [From (6) and (7)]
DQ C

(9) PQRS is a square [From (3), (5) and (8)]

35. In the given figure, ABCD is a parallelogram. If P and Q A PB
are the mid-points of the sides AB and DC respectively,
A
prove that RC = 2 AQ. P

Solution:

Given: In parallelogram ABCD; P and Q are the mid-points of R

AB and DC respectively

To prove: RC = 2AQ

Proof:

(1) QC = AP and QC//AP [DC = AB, DC//AB and given]

(2) AQ = PC and AQ//PC [From (1)]

(3) DA = AR 1 RC [DQ = QC and AQ//PC]
2 [AQ joins the mid-points of DR and DC]
In 'DRC, AQ =

?RC = 2 AQ

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

Solution:

Given: AB = AC, PB = CM and PQ//AM B QO C

To prove: PM and QC bisect each other. i.e., OP = OM and OQ = OC

Proof:

(1) ³ABC = ³ACB [AB = AC] M

(2) ³PQB = ³ACB [PQ//AC, corresponding angles]

(3) ³ABC= ³PQB [From (1) and (2)]

(4) PB = PQ [From (3)]

(5) PB = CM [Given]

(6) PQ = CM [From (4) and (5)]

(7) In 'POQ and 'COM

(i) ³POQ = ³COM (A) [Vertically opposite angles are equal]

(ii) ³OPQ = ³OMC (A) [PQ//CM, alternate angles]

(iii) PQ = CM (S) [From (6)]

(8) 'POQ # 'COM [By A.A.S. axiom]

(9) OP = OM and OQ = OC [Corresponding sides of congruent triangles]

Hence, PM and QC bisect each other at O.

Vedanta Excel in Mathematics Teachers' Manual - 9 126

Unit Circle

14

Allocated teaching periods 10
Competency

- To prove the properties of circle theoretically and experimentally and solve the
related problems

Learning Outcomes
- To introduce the circle, prove the theorems on circle theoretically and verify them
by induction method.

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

- To define circle

- To identify the parts of the circle

- To tell the relation between the line segment joining

1. Knowledge (K) the centre of the circle and the mid-point of the chord.

- To write the relationship between the chords which

are equidistance form the centre of the circle

- To find the length of the radius / diameter of the circle

2. Understanding (U) - To calculate the length of the chord
- To verify the properties of the circle

- To prove the theorems on the circle

3. Application (A) - To find the distance between the parallel lines

- To show the various relations of parts of circle based

on the properties and theorems

- To represent the properties of triangles, quadrilaterals

and circle diagrammatically and logically.

4. High Ability (HA) - To establish the required relations using the properties

of triangles, quadrilaterals and circle in the logical

way.

Required Teaching Materials/ Resources
Geometric instruments, geo-board, rubber bands, card-boards, scissors, pencils, marker,
ICT tool etc.

Pre-knowledge: Circle adn its parts
Teaching Activities
1. Give/ask real life examples of circular objects
2. Display circle and its parts in chart paper or card board or geo-board or ICT tools like

geo-gebra and discuss
3. Experimentally verify and then prove (if possible) the following relations under

discussion

127 Vedanta Excel in Mathematics Teachers' Manual - 9

(i) The radius of the circle are equal
(ii) The perpendicular drawn from the centre of a circle to a chord bisects the chord.
(iii) The line perpendicular to the chord passes through the centre of the circle
(iv) The line segment joining the mid-point of a chord and the centre of circle is perpendicular

to the chord.
(v) The line Equal chords are equidistance from the centre.
(vi) The chords which are equidistance form the centre of a circle are equal

Solution of selected problems from Vedanta Excel in Mathematics

1. In the adjoining figure, O is the centre of both circle. If

PX = 3 cm and AQ = 8 cm, find the length of XY. O
P X A YQ
Solution:

Here, O is the centre of circle, PX = 3 cm and AQ = 8 cm

Now, (i) AP = AQ = 8 cm [Q OAAPQ]

(ii) AX = AP – PX = 8 cm – 3 cm = 5 cm

(iii) XY = 2 AX = 2× 5 cm = 10 cm [Q OAAXY]

2. In the figure alongside, O is the centre of a circle. AB = 20 cm, A O B
CD = 16 cm and AB // CD. Find the distance between AB and CD. C D
O
Solution: A P B
C D
Here, O is the centre of circle, AB = 20 cm, CD = 16 cm and AB//CD C
D
Construction: OP A CD is drawn and OC is joined
B
Then, OC = radius = 10 cm
1 D
Now, CP = 2 CD = 8 cm [Q OPACD]
B
In right angled triangle COP; OP = OC2 – CP2 = 102 – 82 = 6 cm R

Thus, the distance between the AB and CD is 6 cm

3. In the given figure, O is the centre of a circle. AB and CD are two O
parallel chords of lengths 16 cm and 12 cm respectively. If the radius

of the circle is 10 cm, find the distance between the chords. A

Solution:

Here, O is the centre of circle, AB = 16 cm, CD = 12 cm, radius = 10 cm and AB//CD

Construction: OMA AB and ONA CD are drawn and, A and C are joined to O.
1 N
Now, AM = 2 AB = 8 cm [QOM A AB] C O

In right angled triangle AOM; OM = OA2 – AM2 = 52 – 42 = 3 cm M
Again, CN = 12CD = 6 cm
[Q ONACD] A

In right angled triangle CON; ON = OC2 – CN2 = 52 – 32 = 4 cm

Now, MN = OM + ON = 6 cm + 8 cm = 14 cm P
Thus, the distance between the AB and CD is 14 cm.

4. In the adjoining figure, O is the centre of a circle, PQ and RS are two O
equal and parallel chords. If the radius of the circle is 5 cm and the QB
distance between the chords is 8 cm, find the length of the chords.

Vedanta Excel in Mathematics Teachers' Manual - 9 128

Solution: PR

Here, O is the centre of circle, PQ = RS and PQ//RS

Construction: OMA PQ and ONA RS are drawn and, P and O joined. M O N
1
Now, OM = 2 MN = 4 cm [Equal chords are equidistance from the centre]

Again, in right angled triangle POM; OP2 – OM2 = 52 – 42 = 3 cm QS

?PQ = 2PM = 2× 3 cm = 6 cm

Hence, the length of each of chords PQ and RS is 6 cm.

5. In a circle of radius 5 cm, AB and CD are two parallel chords of length 8 cm and 6
cm respectively. Calculate the distance between the chords, if they are

(i) on the same side of the centre (ii) on the opposite side of the centre.

Solution:

Here, O is the centre of circle, radius = 5 cm, AB = 8 cm, CD = 6 cm O

and AB//CD A M
B
(i) If AB and CD lie on the same side of the centre
CN D
Construction: OMA AB and ONA CD are drawn and, A and C are

joined to O.
1
Now, AM = 2 AB = 4 cm [Q OMAAB]

In right angled triangle AOM; OM = OA2 – AM2 = 52 – 42 = 3 cm

Again, CN = CD = 3 cm [Q ONACD]

In right angled triangle CON; ON = OC2 – CN2 = 52 – 32 = 4 cm

Now, MN = ON – OM = 4 cm – 3 cm = 1 cm

Thus, the distance between the AB and CD is 1 cm.

(ii) If AB and CD lie on the opposite side of the centre CND

Construction: OMA AB is drawn and MO is produced to meet CD O

at N. So, ONA CD [Q AB//CD]
1 [Q OMAAB] AM
Now, AM = 2 AB = 4 cm B

In right angled triangle AOM; OM = OA2 – AM2 = 52 – 42 = 3 cm
1
Again, CN = 2 CD = 3 cm [ONACD]

In right angled triangle CON; ON = OC2 – CN2 = 52 – 32 = 4 cm

Now, MN = OM + ON = 3 cm + 4 cm = 7 cm

Thus, the distance between the AB and CD is 7 cm.

6. AB and CD are two parallel chords of a circle such that AB = 10 cm and
CD = 24 cm. If the chords are on the opposite sides of the centre and the distance

between them is 17 cm, find the radius of the circle. CND
Solution:

Here, O is the centre of circle, AB = 10 cm, CD = 24 cm and AB//CD O
AB and CD lie on the opposite side of the centre

Construction: OMA AB is drawn and MO is produced to meet CD at N. So, A M B

129 Vedanta Excel in Mathematics Teachers' Manual - 9

ONA CD [Q AB//CD]

Let the radius of the circle OA = OC = r and OM = x then

ON = MN – OM = (17 – x) cm
INCnoNrwig=, hAt21MaCnDg=le=12d1Atr2Biac=nmg5lecAmOM;
[Q OMAAB] AM2 = x2 + 52 = x2 + 25
[Q ONACD]

OA2 = OM2 +

In right angled triangle CON; OC2 = ON2 + CN2 = (17 – x)2 + 122 = 433 – 34x + x2

As OA = OC, OA2 = OC2 or, x2 + 25 = 433 – 34x + x2 ? x = 12

Again, OA2 = r2 = x2 + 25 = 122 + 25 = 169 ? r = 13

Hence, the radius of the circle is 13 cm.

7. In the given figure, O is the centre of the circle, chords MN and M
RS are intersecting at P. If OP is the bisector of ‘MPR, prove that S

MN = RS. OP

Solution: N
Given: O is the centre of the circle, chords MN and RS intersect at P and R

OP is the bisector of ³MPR

To prove: MN = RS MS

Construction: OAA MN and OBA RS are drawn. A
Proof: P
(1) In 'AOP and 'BOP
O BN

L ³OAP = ³OBP (A) [Both are right angles] R
LL ³OPA = ³OPB (A) [OP is the bisector of ³MPR]

(iii) OP = OP (S) [Common side]

(2) 'AOP # 'BOP [By A.A.S axiom]

(3) OA = OB [Corresponding sides of congruent triangles]

(4) MN = RS [From (3), chords equidistance from centre]

P

8. In the figure alongside, PQ and RS are two chords intersecting at T S O
in a circle with centre O. If OT is the bisector of ‘PTR, prove that T
(i) PT = RT (ii) ST = TQ

Solution: Q

Given: O is the centre of the circle, chords PQ and RS intersect at T and R
OT is the bisector of ³PTR P

To prove: (i) PT = RT (ii) ST = TQ

Construction: OAA PQ and OBA RS are drawn. SA

Proof: TO

(1) In 'AOT and 'BOT QB

(i) ³OAT = ³OBT (A) [Both are right angles] R
(ii) ³OTA = ³OTB (A) [OP is the bisector of ³MPR]

(iii) OT = OT (S) [Common side]

(2) 'AOT # 'BOT [By A.A.S axiom]

(3) OA = OB and AT = BT [Corresponding sides of congruent triangles]

Vedanta Excel in Mathematics Teachers' Manual - 9 130

(4) PQ = RS [From (3), chords equidistance from centre]
(5) AP = BR [From (4) and OAAPQ, OBARS]
(6) PT = RT [Adding (3) and (5), AT + AP = BT + BR]
(7) ST = TQ
[Subtracting (6) from (4), RS – RT = PQ – PT]

9. In the given figure, O is the centre of the circle. Two equal
chords AB and CD intersect each other at E. Prove that (i) AE
= CE (ii) BE = DE A C
O

Solution: E
Given: O is the centre of the circle, equal chords AB and CD intersect DB

at E

To prove: (i) AE = CE (ii) BE = DE

Construction: OMA AB and ONA CD are drawn, O and E are joined AO C

Proof: MN

(1) In 'MOE and 'NOE E B
D
(i) ³OME = ³ONE (R) [Both are right angles]

(ii) OE = OE (H) [Common side]

(iii) OM = ON (S) [Equal chords are equidistance form the centre]

(2) 'MOE # 'NOE [By R.H.S axiom]
(3) ME = NE [Corresponding sides of congruent triangles]

(4) AM = CN [AB = CD and OMAAB, ONACD]
(5) AE = CE [Adding (3) and (4), ME + AM = NE + CN]

(6) AB = CD [Given]

(7) BE = DE [Subtracting (5) from (6), AB – AE = CD – CE]

B

10. In the figure, L and M are the mid-points of two equal chords AB CL
and CD of a circle with centre O. Prove that Q
O
(i) ‘OLM = ‘OML (ii) ‘ALM = ‘CML M D
Solution: A
Given: O is the centre of the circle, L and M are the mid-points of equal

chords AB and CD respectively

To prove: (i) ³OLM = ³OML (ii) ³ALM = ³CML
Proof:

(1) OL A AB, OM A CD [AL = BL and AM = DM]
(2) ³OLA = ³OMC [From (1), both are right angles]
(3) OL = OM [Equal chords are equidistance from the centre]

(4) ³OLM = ³OML [From (3)]
(5) ³ALM = ³CML [Subtracting (4) from (2)]

11. In the adjoining figure, AB is the diameter of a circle with centre A O B
D
O. If chord CD // AB, prove that ‘AOC = ‘BOD.

Solution: C

Given: O is the centre of the circle, AB is the diameter and chord CD//AB

131 Vedanta Excel in Mathematics Teachers' Manual - 9

To prove: ³AOC = ³BOD [OC = OD] P
Proof: [AB//CD and alternate angles]
[AB//CD and alternate angles]
(1) ³OCD= ³ODC [From (1), (2) and (3)]
(2) ³OCD = ³AOC
(3) ³ODC = ³BOD
(4) ³AOC = ³BOD

12. In the given figure, equal chords PQ and RS of a circle with centre
O intersect each other at right angle at A. If M and N are the mid- M O

points of PQ and RS respectively, prove that OMAN is a square. S AN R

Solution: Q

Given: O is the centre of the circle, equal chords PQ and RS intersect at right angle at A. M

and N are the midpoints of PQ and RS

To prove: OMAN is a square

Proof:

(1) OMAPQ [OM joins the centre O and mid-point M of the chord PQ]

(2) ONARS [ON joins the centre O and mid-point N of the chord RS]

(3) ³MAN = 900 [Given]

(4) ³MAN = 900 [Remaining angle of the quadrilateral OMAN]

(5) OM = ON [Equal chords of a circle are equidistance from the centre]

(6) OMAN is a square [From (1), (2), (3), (4) and (5)]

X

13. In the adjoining figure, two chords WX and WY are equally W OZ
inclined to the diameter at their point of intersection. Prove that Y
the chords are equal.

Solution:
Given: O is the centre of the circle, chords WX and WY are equally

inclined to the diameter WZ at W. X
To prove: Chords WX and WY are equal

Construction: OMA WX and ONA WY are drawn W OZ
Proof:

(1) In 'MOW and 'NOW Y
(i) ³ OMW = ³NOW (A) [Both are right angles]

(ii) ³ OWM = ³OWN (A) [Given]

(iii) OW = OW (S) [Common side]

'MOW # 'NOW [By A.A.S axiom]

(3) OM = ON [Corresponding sides of congruent triangles]

(4) WX = WY [From (3) and OMAXY, ONAYZ]

14. In the figure alongside, AB is a diameter. Two chords AD and BC A D B
are equal. Prove that AD // BC. C O

Solution:
Given: O is the centre of the circle, AB is the diameter, and chords AD

and BC are equal

Vedanta Excel in Mathematics Teachers' Manual - 9 132

To prove: AD//BC D
Construction: OMA AD and ONA BC are drawn M
Proof: A OB

(1) In 'MOA and 'NOB N
(i) ³ OMA = ³ONB (R) [Both are right angles] C

(ii) OA = OB (H) [Radii]

(iii) OM = ON (S) [Equal chords are equidistance from the centre]

(2) 'MOA # 'NOB [By R.H.S axiom]

(3) ³ OAM = ³OBN [Corresponding angles of congruent triangles]

(4) AD//BC [From (3), alternate angles are equal] P

A M

15. In the given figure, AB and BC are equal chords of the circle with
centre O. If OM A AB, ON A BC,OM and ON are produced to meet the O B
circumference at P and Q respectively. Prove that:
N

(i) AP = CQ (ii) ‘PAM = ‘QCN. CQ

Solution:

Given: O is the centre of the circle, chords AB and BC are equal, OMAAB, ONABC, OM

and ON are produced to meet the circumference at P and Q respectively.

To prove: (i) AP = CQ (ii) ³PAM = ³QCN

Proof:

(1) OP = OQ [Radii]

(2) OM = ON [Equal chords are equidistance from the centre]

(3) PM = QN [Subtracting (2) from (1)]

(4) In 'AMP and 'CNQ

(i) PM = QN (S) [From (3)]

(ii) ³ AMP = ³CNQ (A) [Given]

(iii) AM = CN (S) [AB = BC, OMAAB and ONABC]

(5) 'AMP # 'CNQ [By S.A.S axiom]

(6) AP = CQ and ³PAM = ³QCN [Corresponding parts of congruent triangles]

16. Two equal chords AB and CD of a circle with centre O are A B
produced to meet at E, as shown in the given figure. Prove D
that BE = DE and AE = CE. O E
E
Solution: C

Given: O is the centre of the circle, equal chords AB and CD are

produced to meet at E AM
O
To prove: (i) BE = DE (ii) AE = CE B

Construction: OMAAB and ONACD are drawn

Proof: C ND
(1) In 'MOE and 'NOE

(i) ³ OME = ³ONE (R) [Both are right angles]

(ii) OE = OE (H) [Common side]

(iii) OM= ON (S) [Equal chords are equidistance from the centre]

133 Vedanta Excel in Mathematics Teachers' Manual - 9

(2) 'MOE # 'NOE [By R.H.S axiom]
(3) ME = NE [Corresponding sides of congruent triangles]
(4) MB =ND [AB=CD, OMAAB and ONACD]
(5) BE = DE [Subtracting (4) from (3), ME – MB = NE – ND]
(6) AM = CN [AB=CD, OMAAB and ONACD]
(7) AE = CE [Adding (3) and (6), ME+AM = NE+CN]

17. In the given figure, O is the centre of circle ABCD. D YC
If OY A PC, OX A PB and OX = OY, prove that PB = PC.
P O
A XB
YC
Solution:

Given: O is the centre of the circle, OYAPC, OXAPB and OX = OY P D O
To prove: PB = PC A X
Construction: O and P are joined B
Proof:

(1) In 'POX and 'POY

(i) ³ OXP = ³OYP (R) [Both are right angles]

(ii) OP = OP (H) [Common side]

(iii) OX=OY (S) [Given]

(2) 'POX # 'POY [By R.H.S axiom]

(3) PX = PY [Corresponding sides of congruent triangles]

(4) AB = CD [OXAAB , OYACD and OX = OY]

(5) BX = CX [From (4), AB = CD and OXAAB , OYACD]

(6) PB = PC [Adding (3) and (5), PX+PB = PY+PC]

18. In the given figure, ABC is a triangle in which AB = AC. Also a circle A
passing through B and C intersects the sides AB and AC at the points
DE
D and E respectively. Prove that AD = AE.
O
Solution: BC

Given: In triangle ABC, AB = AC. The circle passing through B and C A

intersects the side AB and AC at D and E respectively. DE

To prove: AD = AE O
BC
Construction: OPAAD and OQACE are drawn

Proof:

(1) AB = AC [Given]

³ ABC = ³ACB [From (1)]

³ OBC = ³OCB [OB = OC, the radii]

³ ABO = ³ACO [Subtracting (3) from (2)]

(5) In 'POB and 'QOC

(i) ³ OPB = ³OQC (A) [Both are right angles]

(ii) ³ PBO = ³QCO (A) [From (4), ³ ABO = ³ACO]

(iii) OB = OC (S) [Radii]

'POB # 'QOC [By A.A.S axiom]

Vedanta Excel in Mathematics Teachers' Manual - 9 134

(7) OP = OQ [Corresponding sides of congruent triangles]
(8) BD = CE [OPABD , OQACE and OP = OQ]
(9) AD = AE [Subtracting (8) from (1), AB – BD = AC - CE]

19. In the given figure, O is the centre of a circle, AB the diameter, AC A
the chord and OM A AC. Prove that: (i) OM // BC (ii) BC = 2 OM OM

Solution:

Given: O is the centre of circle, AB the diameter, AC the chord and OMAAC B C
To prove: (i) OM//BC (ii) BC = 2 OM

Proof:

(1) AM = MC [OMAAC and OM bisects the chord AC]

(2) OA =OB [Radii]

(3) OM//BC and BC = 2 OM [OM joins the mid-points of AB and AC] C
Q
20. In the adjoining figure, P is the centre of a circle. If PQ // BC,
prove that: (i) AC = 2AQ (ii) PQ A AC. A PB

Solution:

Given: P is the centre of circle, AB the diameter, C is a point on the

circumference and PQ//BC

To prove: (i) AC = 2AQ (ii) PQAAC

Proof:

(1) AP = BP [Radii]

(2) AQ = QC i.e., AC = 2AQ[In 'ABC; AP = BP and PQ//BC]
(3) PQAAC [PQ joins the mid-point of chord AC and the centre of the circle]

A

21. In the given figure, two circles with centres P and Q intersect PQ of
at A and B. Prove that the line joining the two centres B
the circles is the perpendicular bisector of the common chord. Q

Solution:

Given: Circles with centres P and Q intersect at A and B A
To prove: PQ is the perpendicular bisector of AB i.e., POAAB and OA = OB

Construction: A and B are joined to P and Q P
Proof:

(1) In 'PAQ and 'PBQ B
(i) AP = BP (S)
[Radii of circle with centre P]

(ii) AQ = BQ (S) [Radii of circle with centre Q]

(iii) PQ = PQ (S) [Common side]

(2) 'PAQ # 'PBQ [By S.S.S axiom]

(3) ³APQ = ³BPQ [Corresponding angles of congruent triangles]

(4) POAAB and OA = OB [In 'PAB, PA = PB and ³APQ = ³BPQ]

135 Vedanta Excel in Mathematics Teachers' Manual - 9

22. In the given figure, P and Q be the centres of two intersecting A X B
circles and AB // PQ. Prove that AB = 2 PQ. B

Solution: P Q
Given: Circles with centres P and Q intersect at X and Y and AB//PQ N
Q
To prove: AB = 2PQ A MX
Construction: PM AAX and QNABX are drawn

Proof: [AB//PQ and by construction] P
(1) PQNM is a rectangle

(2) AX = 2MX and BX=2NX [PM AAX, QNABX and bisect the AX and BX]

(3) AB = AX+BX [Whole part axiom]

(4) AB = 2(MX+NX)=2MN [From (2) and (3)]

(5) AB=2PQ [From (4),MN = PQ; the opposite sides of rectangle]

23. In the given figure, two circles with centres P and Q are AM
intersecting at A and B. If MN is parallel to common chord C
AB, prove that

(i) MC = ND (ii) MD = NC PQ

Solution: D
BN
Given: Circles with centres P and Q intersect at A and B and MN//

AB AM

To prove: (i) MC = ND (ii) MD = NC C
Q
Construction: A and B are joined to P and Q P
Proof: D
BN
(1) In 'PAQ and 'PBQ
(i) AP = BP (S) [Radii of circle with centre P]

(ii) AQ = BQ (S) [Radii of circle with centre Q]

(iii) PQ = PQ (S) [Common side]

(2) 'PAQ # 'PBQ [By S.S.S axiom]

(3) ³APQ = ³BPQ [Corresponding angles of congruent triangles]

(4) POAAB and OA = OB [In 'PAB, PA = PB and ³APQ = ³BPQ]
(5) PQACD [From (4), AB//MN]

(6) CX = DX [PXACD and PX bisects CD]

(7) MX = NX [QXAMN and QX bisects MN]

(8) MC = ND [Subtracting (6) from(7), MX – CX = NX – DX]

(9) MD = NC [Adding CD in (8), MC + CD = ND + CD]

24. Prove that the diameter of a circle perpendicular to one of the two parallel chords

of a circle is perpendicular to the other and bisects it. A

Solution:

Given: O is the centre of circle, AB the diameter, chord PQ is M O
perpendicular to the diameter AB and PQ//RS PN
Q
To prove: AB is perpendicular bisector of RS R S
Proof:
B

(1) ONARS [OMAPQ and PQ//RS]

(2) RN = NS [ONARS and ON bisects RS]

(3) AB is perpendicular bisector of RS [From(1) and (2)]

Vedanta Excel in Mathematics Teachers' Manual - 9 136

25. Prove that the line joining the mid-points of two parallel chords of P
a circle passes through the centre of the circle.

Solution: O
M
Given: O is the centre of circle, AB//CD, M and N are the mid-points of A N B
the chords AB and CD respectively C
D

To prove: MN passes through the centre O

Construction: P is a point such that PMAAB and O does not lie on PM

Proof:

(1) OMAAB [OM joins the mid-point of AB]

(2) PMAAB [By assumption]

(3) ³OMB = ³PMB [From (1) and (2)]

(4) PM passes through O [From (3)]

(5) ONACD and PNACD [ON joins the mid-point of CD and AB//CD]

(6) MN passes through the centre O [From (4) and (5)] A

26. Prove that a diameter of a circle which bisects a chord of the circle O
also bisects the angle subtended by the chord at the centre of the
M
circle. C B D
Solution:

Given: O is the centre of circle, AB is the diameter, CM = DM

To prove: ³COM = ³DOM

(1) In 'COM and 'DOM

(i) OC = OD (S) [Radii of the circle]

(ii) CM = DM (S) [Given]

(iii) OM = OM (S) [Common side]

(2) 'COM # 'DOM [By S.S.S axiom]

(3) ³COM = ³DOM [Corresponding angles of congruent triangles]

27. Of two chords, the chord nearer to the centre of circle is longer. NC
Solution:
D
Given: O is the centre of circle; AB and CD are two chords such that AB A O B
> CD. OMAAB and ONACD M

To prove: OM < ON

Construction: A and C are joined to O [OMAAB and OM bisects AB, ONACD and
ON bisects CD]
(1) AM = 1 AB and DN = 12CD
2

(2) OA2 = OM2 + AM2 and OD2 = ON2 + DN2 [Using Pythagoras theorem in rt.
(3)
OM2 + AM2 = ON2 + DN2 [From (2)] ³ed 'AOM and 'DON]

i.e., OM2 – ON2 = DN2 – AM2

(4) DN < AM or, DN2 < AM2 [AB > CD i.e., CD < AB and from (1)]

(5) OM2 < ON2 or, OM < ON [From (3) and (4)]

137 Vedanta Excel in Mathematics Teachers' Manual - 9

28. Five students Amrit, Bibika, Chandani, Dipesh and Elina are 10mA DE
playing in a circular meadow. Amrit is at the centre, Bibika and BC
Dipesh are inside the boundary line. Similarly, Chandani and Elina
are on the boundary of the meadow, If Amrit, Bibika, Chandani
and Dipesh form a rectangle and the distance between Bibika and
Dipesh is 10 m, find the distance between Amrit and Elina.

Solution:

(i) AC = BD = 10 m [Diagonals of rectangle are equal]

(ii) AC = AE = 10 m [Radii of the circle]

Hence, the distance between Amrit and Elina is 10 m.

29. Three students Pooja, Shaswat and Triptee are playing a game by O
standing on the circumference of a circle of radius 25 feet drawn in T

a park. Pooja throws a ball to Shaswat and Shaswat to Triptee and
Triptee to Pooja. What is the distance between Pooja and Triptee P S

when the distance between Pooja and Shaswat and the distance between Shaswat
and Triptee is 30 feet each?

Solution: O
Here, OP = OT = OS = 25 feet, PS = ST = 30 feet T
Since, OP = OT and PS = ST. So, OPST is a kite in which diagonal OS
M
bisects the diagonal PT at M at a right angle. PS
Let OM = x feet then MS = OS – OM = (25 – x) feet

Now,

From rt. ³ed 'OPM, PM2 = OP2 – OM2 = 252 – x2 = 625 – x2 ... (i)

From rt. ³ed 'PMS, PM2 = PS2 – MS2 = 302 – (25 – x)2 = 900 – (625 – 50x + x2)

= 275 + 50x – x2 ... (ii)

From (i) and (ii), we get

625 – x2 = 275 + 50x – x2 ?x = 7

Also, from (i); PM2 = = 625 – 72 = 576 ?PM = 24 feet

Again, PT = 2×PM = 2×24 feet = 48 feet Q
Hence, the distance between Pooja and Triptee is 48 feet. 100m B
O 60m
30. The diameter of a circular ground with centre at O is 200 m. Two
vertical poles P and Q are fixed at the two points in the circumference PC
of the ground. Find the length of a rope required to tie the poles
tightly at a distance of 60 m from the centre of the ground. A

Solution:

Let the poles P and Q are fixed at the points A and B on the circumference and OCAAB.
Then the required length of rope to tie the poles is AB. and diameter = 200 m
?radius = OA = OB = 100 m and OC = 60 m

Now,

From rt. ³ed 'OBC, BC2 = OB2 – OC2 = 1002 – 602 = 6400 ?BC = 80 m

Also, AB = 2BC = 2×80m = 160 m [OCAAB and OC bisects AB]

31. In the adjoining figure, OAB is an isosceles triangle and a circle O

with O as the centre cuts AB at C and D. Prove that AC = DB. AC DB
Solution:
Given: O is the centre of circle; OAB is an isosceles triangle

Vedanta Excel in Mathematics Teachers' Manual - 9 138

To prove: AC = BD

Construction: OMAAB is drawn O
M
Proof:

(1) In 'AOM and 'BOM [Both are right angles] AC DB
(i) ³OMA = ³OMB (A) [Base angles of isosceles triangle]
(ii) ³OAM = ³OBM (A) [Given]
(iii) OA = OB (S)

'AOM # 'BOM [By A.A.S axiom]
(3) AM = BM [Corresponding sides of congruent triangles]

(4) CM = BD [OMACD and OM bisects CD]

(5) AC = BD [Subtracting (4) from (3), AM – CM = BM – BD]

32. In the figure alongside, MN is the diameter of a circle with centre A C

O. If BD = CD, prove that ‘ OAD = ‘ OCD. M OD N
Solution: B
Given: O is the centre of circle; MN is the diameter, BD = CD

To prove: ³OAD = ³OCD AC
Construction: O and B are joined

Proof: M ODN
B
(1) In 'COD and 'BOD

(i) OC = OB (S) [Radii of the circle]

(ii) OD = OD (S) [Common side]

(iii) CD= BD (S) [Given]

(2) 'COD # 'BOD [By S.S.S axiom]
(3) ³OCD = ³OBD [Corresponding angles of congruent triangles]
(4) ³OAD = ³OBD [OA = OB]
(5) ³OAD = ³OCD [From (3) and (4)]

33. In the figure alongside, A and B are the centres of two C
intersecting circles. If CD intersects AB perpendicularly at P,
prove that M
(i) CM = DN
A PB
N

D

(ii) CN = DM

Solution:

Given: A and B are the centres of the intersecting circles. CD intersects AB perpendicularly

To prove: (i) CM = DN

(ii) CN = DM

Proof:

(1) PM = PN [APAMN and AP bisects MN]

(2) PC = PD [BPACD and BP bisects CD]

(3) CM = DN [Subtracting (1) from (2); PC – PM = PD – PN]

(4) CN = DM [Adding MN in (3), CM + MN = DN + MN]

139 Vedanta Excel in Mathematics Teachers' Manual - 9

Unit Geometry - Construction

15

Allocated teaching periods 5

Competency

- To construct square, rectangle, rhombus, parallelogram, quadrilateral and trapezium and

analyze them.
Learning Outcomes

- To construct square, rectangle, rhombus, parallelogram, quadrilateral and trapezium.

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

- To sketch the rough figure and construct square,

1. Application (A) rectangle, rhombus, parallelogram, quadrilateral and

trapezium under the given information

Required Teaching Materials/ Resources
Geometric instruments, geo-board, pencils, marker, ICT tool etc.

Pre-knowledge: properties of square, rectangle, rhombus, parallelogram, quadrilateral and

trapezium.
Teaching Activities
1. Discuss about the procedures of constructions of square and give the different side /

diagonal lengths of squares to construct in groups.
2. Discuss about the procedures of constructions of rectangle and rhombus and give the

different information to construct in groups.
3. Explain the constructions procedures of quadrilateral and give the different information

to construct in groups.

4. Explain the constructions procedures of trapezium and give the different information to

construct in groups.

Solution of selected problems from Vedanta Excel in Mathematics

1. Construct a parallelogram ABCD in which diagonals AC = 4.6 cm, BD = 5.8 cm and
they bisect each other making an angle of 30q.

Steps of construction XD O C
(i) Draw a diagonal AC = 4.6 cm, A BY
(ii) Draw the perpendicular bisector of AC and

mark its mid-point O.
(iii) At O, construct ‘AOX = 30q and produce

XO to Y.
(iv) Here, O is also the mid-point of BD. With

centre at O and radius OB = OD = 2.9 cm
(12 of BD) cut OY at B and OX at D.
v) Join A, D; B, C; A, B and C, D.
Thus, ABCD is the required parallelogram.

Vedanta Excel in Mathematics Teachers' Manual - 9 140

2. Construct a trapezium ABCD in which AB = 5.5 cm, BC = 4.5 cm,
‘DAB = 45q, ‘BCD = 60q and AD // BC.

Steps of construction

(i) Draw a line segment AB = 5.5 cm. X
D
(ii) Construct ‘BAX = 45q at A.
45° Y
(iii) D lies on AX and AD // BC. A 60° C

So, at B, construct B

‘ABY = 180q – 45q = 135q
(iv) With the centre at B and radius

4.5 cm draw an arc to cut BY at C.
(v) At C, construct ‘BCD = 60q. The

arm CD intersect AX at D.

Thus, ABCD is the required trapezium.

3. Construct a trapezium ABCD in which AB = 4.8 cm, diagonal AC = 5.9 cm,
‘BAC = 60q, CD = 5 cm and AB // DC.

Steps of construction Y D X
60° C
(i) Draw a line segment AB = 4.8 cm.

(ii) At A, construct ‘BAX = 60q.

(iii) With centre at A and radius
5.9 cm, draw an arc to cut AX at C.

(iv) Join B and C.

(v) As AB // DC, alternate angles BAC
and ACD are equal. So, construct
‘ACY = 60q at C.

(vi) With centre at C and radius 5 cm,
draw an arc to cut CY at D.

(vii) Join D and A.

Thus, ABCD is the required trapezium.

A 60° B

4. Construct a trapezium ABCD in which AB = 4.4 cm, diagonal AC = 6.8 cm,
AD = BC = 5.2 cm and AB // DC.

Steps of construction
(i) Draw a line segment AB = 4.4 cm.
(ii) With centre at A and radius 6.8 cm, draw an arc.
(iii) With centre at B and radius 5.2 cm, draw another arc to intersect the previous arc at C.

141 Vedanta Excel in Mathematics Teachers' Manual - 9

(iv) Join A, C and B, C.
(v) At C, construct ‘ACX = ‘BAC.
(vi) With centre at A and radius 5.2 cm, draw an arc to cut CX at D.
(vii) Join A and D.

Thus, ABCD is the required trapezium.

XD C

AB

5. Construct a quadrilateral ABCD in which AB = 5 cm, BC = 5.6 cm,

CD = 4.5 cm, AD = 5.4 cm and the diagonal BD = 6.5 cm.

Steps of construction D

(i) Draw AB = 5 cm. 4.5 cm

(ii) From A, draw an arc with C

radius AD = 5.4 cm and

from B draw another arc 5.4 cm
6.5 cm
with radius BD = 6.5 cm. 5.6 cm

These two arcs intersect

each other at D.

(iii) From B, draw an arc with

radius BC = 5.6 cm and A 5cm B
from D, draw another arc

with radius DC = 4.5 cm. These two arcs intersect to each other at C.

Join A, D; B, C and D, A.

Thus, ABCD is the required quadrilateral.

Vedanta Excel in Mathematics Teachers' Manual - 9 142

Unit Trigonometry

16

Allocated teaching periods 7

Competency

- To solve the related problem based on trigonometric ratios.
Learning Outcomes

- To introduce trigonometric ratios

- To verify the fundamental trigonometric ratios from the right angled triangle for

the given reference angle

- To find the trigonometric ratios of some standard angles

- To solve the right angled triangle by using trigonometric ratios

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

- To define trigonometry

1. Knowledge (K) - To tell the six trigonometric ratios

- To identify the trigonometric ratio based on the

reference angle and given sides

- To tell the trigonometric ratios of some standard angles

- To find the trigonometric ratios of given reference

2. Understanding (U) angle in terms of sides of the right angled triangle

- To convert a trigonometric ratio in to another

- To find the value of expression involving trigonometric

ratios with standard angles

- To find the unknown length of side of triangle when

reference angle is a standard angle

- To verify the fundamental trigonometric ratios from

the right angled triangle for the given reference angle

3. Application (A) - To find the trigonometric ratios of some standard

angles from right angled triangle

- To solve the right angled triangle by using trigonometric

ratios

- To prepare a report on investigating the application of

4. High Ability (HA) trigonometry, ratios and identities

Required Teaching Materials/ Resources
Colourful chart-paper, models of right angled triangles, geometric instruments, colourful
markers, chart paper with trigonometric ratios of standard angles up to 900, ICT tool etc

143 Vedanta Excel in Mathematics Teachers' Manual - 9

Pre-knowledge: right angled triangles, Pythagorean triplets and name of parts its sides

A

Teaching Activities

1. Recall Pythagorean triplets with examples

2. Show a right angled triangle and ask to identify BC
its hypotenuse, base and perpendicular

3. Discuss about perpendicular, hypotenuse and base in right angled triangle with respect

to one of the acute angle as the reference angle

4. Ask the students to make the possible ratios of perpendicular (p), base (b) and

hypotenuse (h) then explain about following six trigonometric ratios

(i) sinT = p (ii) cosT = b (iii) tanT = p
h h b

(iv) cosecT = h (v) secT = h (vi) cotT = b
p b p

5. Divide the students in to 5 groups. Give them the right angled triangles with different

names and ask to identify the perpendicular, base and hypotenuse.

Also, ask them to express all six trigonometric ratios in terms of sides of the triangles

6. With discussion, give guidelines to solve the problems given in the textbooks.
7. Discuss on establishing the relationships of trigonometric ratios
8. With discussion, explain the trigonometric ratios of some standard angles up to 900
9. Tell the students to draw the table of trigonometric ratios of standard angles on a chart

paper and past on the mathematics corner of the classroom or mathematics lab as project

work.

10. Encourage the students to evaluate the trigonometric expressions involving the standard
angles

11. Discuss on the solution of right angles triangles which is supportive for the next class to
solve the problems related to height and distance

Solution of selected problems from Vedanta Excel in Mathematics
BC C
1) If sin (90° − D) = CA , write down the ratio of sinD.
Solution:

Here, (90° − D) = BC = p 90°D
Sin CA b

? For reference angle (90° − D) B A

Perpendicular (p) = BC and hypotenus (h) = CA

Also, ‘C =90° − (90° − D) = D

For reference angle D, perpendicular (p) = AB, base (b) = BC and hypotenuse (h) = CA

? SinD = p' = AB X
h' CA E 26 ft
W
4 6 ft
3
2) From the adjoining figure, show that tanE = .
Solution:
Z
i) In rt. angled ∆XYZ. Y 24 ft

XY = XZ2 – YZ2 = 262 – 242 = 10 ft

Vedanta Excel in Mathematics Teachers' Manual - 9 144

ii) In rt. angled ∆XYZ,

WY = XY2 – XW2 = 102 – 62 = 8 ft

? tanE = WY(p) =86 ft = 4
XW(b) ft 3

3) In the adjoining figure ABCD is a rhombus in which AC = 6 A T D
cm, BD = 8 cm and ‘ADC = T. Find the values of sinT and O
tanT.

Solution: 12AC = 1 u 6 cm = 3 cm BC
i) OA = 2 = 4 cm
( Diagonal of rhombus do bisect each
OD = 21BD = 1 u 8 cm other at a right angle )
2

and ‘AOD = 90°

ii) In rt.angled 'AOD; AD = OA2 OD2 = 32 42 = 5 cm

? sinT = OA(p) = 3 and tanT = OA(p) = 3
AD(h) 5 OD(b) 4

4) In the circle given alongside, O is the centre, M is the mid - point of O
chord AB. If AB = 12 cm and CM = 8 cm and ‘OAM = T, find the
T
values of sinT and cosT. AM B

Solution: 21AB = 1 u 12 cm = 6 cm
i) AM = 2

ii) OM A AB [OM joins the centre O and the mid - point M of the chords AB.

iii) In rt. angled 'AOM; AO = OM2 AM2 = 82 62 = 10 cm

? sinT = OM(p) = 8 cm = 4 , cosT = AM = 6 cm = 3
OA(h) 10 cm 5 OA 10 cm 5

5) In the adjoining figure, AB = 6, BC = 8 cm, ‘ABC = 90°,
BD A AC and ‘ABD = T. Find the value of sinT.

Solution:

i) In rt ‘ed ∆ ABC, AC = AB2 + BC2 = 62 + 82 = 10

ii) In ∆ ABC and ∆ ABD,

(a) ‘ABC = ‘BDA [Both are right angles]

(b) ‘BAC = ‘BAD [Common angle]

(c) ‘ACB = ‘ABD [Remaining angles]

? 'ABC ~ 'ABD

iii) AC = BC = AB [corresponding angles of similar angle]
AB BD AD

or, 10 = 8 = 6
6 BD AD

145 Vedanta Excel in Mathematics Teachers' Manual - 9

From 1st and 2nd ratios, we get
10 8 24
6 = BD ? BD = 5

From 1st and 3rd ratios, we get
10 6 18
6 = AD ? AD = 5

iv) In 1st angled 'ABD,

sinT = AD(p) = 18/5 = 3
AB(h) 4 5

6) In the given 'ABC, ‘B = 90°, and ‘CAB = 45°. If AB = x cm. Prove that sin 45° = 1 2

Solution: C

Given: In 'ABC, ‘B = 90°, and ‘ACB = 45°, AB = x cm
To prove: sin 45° = 1
2
Proof:

i) ‘ACB = 180q (‘A ‘B) = 180° (45q 90q) = 45q B 45q A

ii) AB = BC [‘A = ‘C = 45q]

iii) AC = AB2 BC2 = x2 x2 = 2x cm [By using pythagoras theorem]

iv) In rt.angled 'ABC; sin 45q = BC = x =1
AC 2x 2
A

7) In the adjoining equilateral 'ABC, AD A BC and AC = 2a units. 2a

Prove that sin 60q = 3
Solution: 2

Given: ABC is an equilateral triangle. AD A BC and AC = 2a units. BD C

To prove: sin 60° = 3
2
Proof:

i) ‘C = 60q [Being an angle of equilateral triangle]
1 1
ii) CD = 2 BC = 3 u 2a = a [Median of equilateral ' bisects its base]

iii) In rt. ‘ed 'ABC; AD = AC2 CD2 = (2a)2 a2 = 3a cm [By Pythagorom theorem]

iv) sin60q = AD(p) = 3a = 3
AC(h) 2 2

? sin60q = 3 proved DC
2

8) In the given figure, ABCD is a square and ‘CPM = 90q, P 30q
AM
if CM = 4 cm, PD = 3 cm. Find the side of square.

Solution: cos30q = PC(b) = 3 = CP B
i) In rt. angled 'CMP; CM(h) 2 4

? CP = 2 3 cm

ii) In rt. angle 'PCD; CD = CP2 DP2 = (2 3)2 ( 3)2 = 3 cm
Hence, the side of square ABCD is 3 cm.

Vedanta Excel in Mathematics Teachers' Manual - 9 146