Oasis Math 9

Hence, perpendicular drawn from any vertex to its opposite side is called the
altitude of a triangle.

Note: • A triangle has three altitudes.
• All three altitudes intersect at the same point.

Median: In the given triangle ABC, D, E and F are the mid-points of A

BC, AC and AB respectively. AD, BE and CF are its medians. FE
Hence, the straight line joining a vertex to the mid–point of

opposite side is the median of a triangle. BDC

Note: • A triangle has three medians.
• The three medians always intersect each other at the same point.

Some theorems related on properties of triangles

Theorem 11.3

Base angles of an isosceles triangle are equal.

Or,

Angles opposite to equal sides of an isosceles triangle are equal.

Experimental Verification:

Draw two isosceles triangles of different shapes and sizes making AB = AC in both
triangles.

To verify : ∠B = ∠C AC
Measure ∠B and ∠C in both figures

and tabulate in the given table.

Figure ∠B ∠C Result B C A B
(i) ...... º ...... º ∠B = ∠C Fig. (i) Fig. (ii)
(ii) ...... º ...... º ∠B = ∠C

Conclusion: Hence, base angles of an isosceles triangles are equal.

Theoretical Proof. A

Given : In ∆ ABC, AB = AC

To prove : ∠ABC = ∠ACB

Construction : From A draw AD⊥ BC. BC
Proof: D

Statements Reasons
1. In ∆ABD and ∆ADC 1.
(i) ∠ADB = ∠ADC (R) (i) Being both right angles

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(ii) AB = AC(H) (ii) Given

(iii) AD = AD (S) (iii) Common side
2. ∆ ABD ≅ ∆ADC 2. By R.H.S axiom
3. ∠ABD = ∠ACD 3. Corresponding angles of congruent
i.e. ∠ABC = ∠ACB
triangles

Hence, base angles of an isosceles triangle are equal.
Converse: Triangle having any two angles equal is an isosceles triangle.

Experimental Verification

Draw two triangles of different shapes and size having ∠B = ∠C.

AA

B CB Figure. (ii) C
Figure. (i)

To verify : AB = AC

Observations : Measure the length of AB and AC of both triangles and tabulate in
the given table.

Figure AB AC Remarks
(i) .........cm .........cm AB = AC

(ii) .........cm .........cm AB = AC

Conclusion: Triangle having two angles equal is an isosceles triangle.

Theoretical Proof : A

Given : In ∆ABC, ∠ABC = ∠ACB
To prove : AB = AC

Construction : Draw AD⊥BC BD C
Proof.

Statements Reasons
1. In ∆ABD and ∆ADC
1.

(i) ∠ABD = ∠ACD (A) (i) Given

(ii) ∠ADB = ∠ADC (A) (ii) Being both right angles

(iii) AD = AD (S) (iii) Common side

2. ∆ABD ≅ ∆ADC 2. By A.A.S. axiom
3. AB = AC 3. Corresponding sides of congruent triangles

Hence, triangle having any two angles equal is an isosceles triangle.

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Worked Out Examples

Example: 1

In the given figure, AB||CD. Find the value of x and y. A B
Solution: y D

700

In the given figure, Cx
∆ACE is an isosceles triangle. E
∴ ∠ACE = ∠AEC = x

In ∆ACE, ∠ACE + ∠AEC+∠CAE = 180º [Sum of three angles of triangle]

or, x + x + 70º = 180º

or, 2x + 70º = 180º

or, 2x = 180º – 70º

or, 2x = 110º
or,
∴ x = 110º
2

x = 55º

Again, x = y [Alternate angles]
∴ y = 55º

Example: 2 A
In ∆ABC, if AB = AC and AD ⊥ BC, prove that BD = DC.

Given : In ∆ABC, AB = AC and AD ⊥ BC

To Prove : BD = DC B DC
Proof :

Statements Reasons
1. In ∆ ABD and ∆ADC 1.
(i) ∠ADB = ∠ ADC (R) (i) Being right angles
(ii) AB = AC (H) (ii) Given
(iii) AD = AD (S) (iii) Common sides
2. ∆ABD ≅ ∆ ADC 2. By R.H.S. axiom
3. BD = DC 3. Corresponding sides of congruent triangles

Hence, proved

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Example: 3 A

Prove that the perpendiculars drawn from the vertices of equal

angles of an isosceles triangle to the opposite sides are equal. M N
Given : ∆ABC is an isosceles triangle where AB = AC, CM⊥AB

and BN⊥AC. BC
To prove : CM = BN.

Proof:

Statements Reasons
1. In ∆BMC and ∆BNC 1.
(i) ∠BMC = ∠BNC (A) (i) Being both right angles
(ii) ∠MBC = ∠NCB (A) (ii) Base angles of an isosceles triangle
(iii) BC = BC (S) (iii) Common side
2. ∆BMC ≅ ∆BNC 2. By A.A.S. axiom
3. CM = BN 3. Corresponding sides of congruent triangles

R Hence, proved

Example: 4 A

In the given figure, RQ⊥AQ and PQ⊥BQ, RQ = QA P

and QB = PQ, prove that PR = AB. Q
B
Given : In the given figure, RQ⊥AQ, PQ⊥BQ, RQ = QA

and QB = PQ.

To prove: PR = AB.
Proof:

Statements Reasons

1. ∠AQR = ∠PQB 1. Being both right angles

2. ∠AQR + ∠AQP = ∠PQB+∠AQP 2. Adding common angle AQP on statement (1)

3. ∠PQR = ∠AQB 3. From statement (2), whole part axiom

4. In ∆PQR and ∆AQB 4.
(i) RQ = AQ (S) (i) Given

(ii) ∠PQR = ∠AQB (A) (ii) From statement (3)

(iii) PQ = QB (S) (iii) Given

5. ∆PQR ≅ ∆AQB 5. By S.A.S. axiom

6. PR = AB 6. Corresponding sides of congruent triangles

Hence, proved

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Exercise 11.4



1. (a) What are the three types of triangle on the basis of sides?
(b) Which two angles of an isosceles triangle are equal?
(c) What is the magnitude of each angle of an equilateral triangle?
(d) How many altitudes can be drawn in a triangle?

(e) How many median can be drawn in a triangle? A

(f) In the given figure, E is the mid point of BC and AD⊥BC, DE C
write the names of median and altitude of ∆ABC.

B

(g) In the given figure, PA⊥QR, QB⊥PR and RC⊥PQ, what P
are the three altitudes of ∆PQR? CB

(h) Are all three altitudes of a triangle are equal? Q AR

(i) On which type of triangle, all three altitudes are equal in length?

2. Find the value of unknown angles in each of the given figures.

(a) A (b) A (c) A D

z Dz y

D x 70º yC y x 55º z E
B 50º B x
C
C

PQ B P QA B
z 50º y
(d) Ry (e) (f) 55º

70º Rx y x
xS TS CD

T E D A E
xy
(g) A (h) y A (i) z

50º B 65º C
D

Bx y E B x 35º D
C C

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A

3. (a) In the given figure, AB = AC and AD⊥BC.
Prove that (i) BD = DC (ii) ∠BAD = ∠CAD.

(iii) What is the conclusion of this proof? B D C
P
A

(b) In the given figure, AD ⊥ BC and

BD = DC. Prove that AB = AC.
Also, write conclusion of this B
proof. D C

(c) In the given figure, PS bisects ∠QPR and PS ⊥ QR. Prove
that PQ = PR. What is the conclusion of this proof?

(d) In the given figure, EF=EG, and FH = HG. E Q R
Prove that EH ⊥ FG: What is the conclusion S
of this proof? H Y
G C
F A A
D C
4. (a) M is the middle point of side BC of an X D
isosceles ∆ABC. MX ⊥ AB and MY ⊥ AC, C B
MX = MY.
M
Prove that : AX = AY and BX = CY.
E


(b) In the given figure, AC⊥BE, AB=DE and B
∠ABC = ∠CDE. Prove that BC = CD.

A



(c) In the given figure, AB = AC and BD = EC.

Prove that (i) ∆ABD ≅ ∆AEC (ii) AD = AE. B EC
(iii) ∆ADE is an isosceles triangle. D

A

(d) In the given figure, AO = OD and BO = OC, O
prove that (i) ∆AOB ≅ ∆COD (ii) AB||CD.
BA
B M

5. (a) In the adjoining figure ∠BAC = ∠BDC, D
∠MBC = ∠MCB. Prove that AB = DC. C



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A D
B C
(b) In the given figure, ∠CAD =
∠ADB and ∠BAC = ∠BDC, prove D
that AC=BD. CE


A

(c) In the given figure, ∆ABC and ∆DCE

are two equilateral triangles. Prove that

AE = BD. B

Answer (b) x = 50º, y = 80º, z = 100º (c) x = 55º, y = 70º, z = 70º
1. Consult your teacher. (e) x = 65º, y = 115º (f) x = 70º, y = 125º
2. (a) x = 110º, y = 70º, z = 40º (h) x = 70º, y = 105º (i) x = 65º, y = 65º, z = 50º
(d) x = 40º, y = 40º, z = 70º 4. Consult your teacher. 5. Consult your teacher.
(g) x = 65º, y = 115º
3. Consult your teacher.

12.6 Similarity



Look at the photographs shown above. They have the same shapes but different
sizes. Such photographs are called similar figures. Thus, the figures having the same
shapes are similar.
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Now let's discuss it with geometrical shape.

(a) (b)

Two line segments

are similar Two circles are similar.

(c) A P (d) A P S D (e) A P B
Q

S R
D C
B CQ Q R
RB C

∆ABC and ∆PQR are ABCD and PQRS are Rectangles ABCD and PQRS
similar triangles similar figures are similar figures

Let's try to find the answer to the following questions:

• Are all circles similar?

• Are all squares similar?

• Are all rectangles similar?

• Are all equilateral triangles similar?

• Are all line segments similar?

Similar Triangles: P

Lets compare given triangles.

A

B C R
Q

Here, both triangles are different in size but similar in shape.

Hence, ∆ABC and ∆PQR are similar triangles.

Conditions of similarity of two triangles

I. Two triangles are similar if all the angles of one triangle are respectively equal to
corresponding angles of another triangle.

AP

550 550 Here, ∠A = ∠P (A)

∠B = ∠Q (A)

650 C ∠C = ∠R (A)
B 600
600 650 R
Q

Then, ∆ABC is similar to ∆PQR.

Symbolically, ∆ABC ~ ∆PQR.

What happens when two triangles are similar?

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Take the ratio of opposite sides of two equal angles:
BC
Ratio of opposite sides of ∠A and ∠P is QR

Ratio of opposite sides of ∠B and ∠Q is AC
PR

Ratio of opposite sides of ∠C and ∠R is AB
PQ

If two triangles are similar, these ratios

BC = AC = AB
QR PR PQ

Hence, if two triangles are similar their corresponding sides are proportional.

II. Two triangles are similar if any two corresponding sides of triangles are

proportional and the angle included by them are equal.

P

500 X

500 PQ 6cm 3
3cm XY 4cm 2
4.5cm
6cm Here, = =
4 cm

Q RY Z PR = 4.5cm = 3
XZ 3cm 2
PQ PR
XY = XZ and ∠P = ∠X

then, ∆PQR ~ ∆XYZ.

III. Two triangles are similar if their corresponding sides are proportional.
A

D

8cm 4cm
F 2cm E
6cm Here, AB = 6cm = 2
3cm DE 3cm

C 4cm B AC 8cm
DF 4cm
i.e., AB = BC = AC = =2
DE EF DF
BC = 4cm =2
then, ∆ABC ~ ∆DEF. FE 2cm

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Remember !
From above three activities, we conclude that two triangles are said to be
similar in the following cases:

I. AAA t hat is all angles of one tri a n g l e III. All three sides of a triangle are
are respectively equal to all angles of proportional to all three sides of
another triangle.
other triangle.

PA PA

Q RB CQ RB C

PQ = PR = QR
AB AC BC

II. If two sides of a triangle are proportional to two sides of the other triangle
and angles included by them are equal.

XA

Y ZB C AXZC = YZ and ∠Z = ∠C
BC

Worked Out Examples

Example: 1 A P
Make the given pair of triangles similar. 60º 72º

Solution: B 72º 48º C Q 60º 48º R
Here,

Given : ∠A = 60º, ∠B = 72º, ∠C = 48º, ∠P = 72º, ∠Q = 60º, ∠R = 48º

To prove : ∆ABC ~ ∆PQR

Proof:

Statements Reasons
1. In ∆ABC and ∆PQR 1.
i) ∠A = ∠Q (A) i) Being both 60º°
ii) ∠B = ∠P (A) ii) Being both 72º°
iii) ∠C = ∠R (A) iii) Being both 48º
2. ∆ABC ~ ∆PQR 2. By A.A.A.

Hence, proved.

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Example: 2

Make the following pair of triangles similar.

P
A

4cm 10cm
6cm5cm

B 6cm 8cmCQ R
3.6cm
Given: In ∆ ABC and ∆PQR 12cm

AB = 4 cm, AC = 5 cm, BC = 6 cm, PQ = 8 cm, QR = 12 cm, PR = 10 cm,

To prove : ∆ABC ~ ∆PQR

Proof :

Statements Reasons
1. Taking the ratio of AB and PQ
1. AB = 4 cm = 1
PQ 8 cm 2

2. BC = 6 cm = 1 2. Taking the ratio of BC and QR
QR 8 cm 2 3. Taking the ratio of AC and PR

3. AC = 5 cm = 1
PR 10 cm 2

4. AB = QBCR = AC 4. From statements 1, 2, and 3
PQ PR 5. Being corresponding sides proportional

5. ∆ABC ~ ∆PQR

Hence, proved.

Example: 3

Make the following pair of triangles similar. S
60º 4.8cm
P
60º 8cm

Solution: Q RT U

Given : In ∆PQR and ∆STU, ∠P = ∠S = 60º, PQ = 6 cm, ST = 3.6 cm, PR = 8 cm, SU = 4.8 cm
To prove : ∆PQR ~ ∆STU

Proof :

Statements Reasons
1 . Taking the ratio of PQ and ST
1. PSQT = 6 cm = 5
3.6 cm 3

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2. PR = 6 cm = 5 2. Taking the ratio of PR and SU
SU 8 cm 3
3. From statement 1 and 2.
3. PSQT = PR
SU 4. Both being 60º
5. Being two sides proportional and angle
4. ∠ QPR = ∠TSU
between them equal
5. ∆PQR ~ ∆TSU Hence, proved.

Example: 4

In the adjoining figure AB||PQ, AP = 1.8 cm, PC = 3.6 cm, A 1.8cm
BC = 6.3 cm, PQ = 2.2 cm. Prove that ∆ABC ~ ∆PQC and find the P
length of AB and QC.
2.2 cm 3.6 cm

Solution: B Q C
Given
: In ∆ABC and ∆PQC, AB||PQ, AP = 1.8 cm, 6.3 cm

PC = 3.6 cm, PQ = 2.2 cm, BC = 6.3 cm.

To Prove : ∆ABC ~ ∆PQC,

To Calculate : The value of QC and AB

Proof.

Statements Reasons

1. In ∆ABC and ∆PQC 1.

(i) ∠ACB = ∠PCQ (A) (i) Common angles

(ii) ∠BAC = ∠QPC (A) (ii) Corresponding angles on AB||PQ

(iii) ∠ABC = ∠PQC (A) (iii) Same as (ii)

2. ∆ABC ~ ∆PQC 2. By A.A.A.

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3. AC = BC = AB 3. Corresponding sides of the similar tri-
PC QC PQ angles are proportional.
Hence, proved.
or, 5.4 = 6.3 = AB
3.6 QC 2.2

Again, taking first two ratios.

5.4 = 6.3
3.6 QC
or,
QC = 6.3 × 3.6
5.4

= 4.2cm.

Taking last two ratios.

or, AB = 5.4
2.2 3.6

or, AB = 5.4 × 2.2 = 3.3cm
3.6
∴ AB = 3.3 cm, QC = 4.2 cm.



Example: 5 4 cm A 10 cm
B
In the adjoining figure,∠BAC = ∠ADC. Prove that C D
∆ABC ~ ∆ABD and find the length of BC.
Given : In the given figure, ∠BAC = ∠ADB, AB = 4 cm, 8 cm
BD = 8 cm, AD = 10 cm
To Prove : ∆ABC ~ ∆ABD.
To Find : The length of BC.

Proof:

Statements Reasons

1. In ∆ABC and ∆ABD. 1.

(i) ∠BAC = ∠ADB (A) (i) Given

(ii) ∠ABC = ∠ABD. (A) (ii) Common angles

iii) ∠ACB = ∠BAD (A) iii) Remaining angles of the triangles

2. ∆ABC ~ ∆ABD 2. By A.A.A.

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AB AC BC 3. Corresponding sides of similar triangles
BD AD AB
3. = =

or, 4 = AC = BC
8 10 4
Taking first and last ratios

or, 4 = BC = 16
8 4

or, BC = 16 = 2 cm.
8

Hence, proved

Example: 6 F
E
In the given figure, AB||CD||EF, AB = 6 cm, CD = x cm, EF = 10 cm,

BD = 4 cm and DE = y cm. Find the value of x and y. B 4 cm
Given : AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and 6 cm D
10 cmxy
DE = y cm and AB||CD|EF

To find : Value of x and y AC

Statements Reasons
1. Alternate
1. In ∆ABD and ∆DEF
(i) Alternative angles on AB||FE
(i) ∠ABD = ∠DEF (A) (ii) Alternative angles on AB||FE
(iii) Vertically opposite angles
(ii) ∠BAD = ∠DFE (A) 2. By A.A.A. facts
3. Ratios of corresponding
(iii) ∠ADB = ∠EDF (A)
sides of similar triangles are
2. ∴ ∆ABD ~ ∆DEF proportional.
4.
3. ByD=AEFB or, 4 y =160ccmm ∴ y= 20 cm (i) Corresponding angles in AB||CD
cm 3
(ii) Corresponding angles in AB ||CD
4. In ∆ABE and ∆CDE,
(iii) Common angles
(i) ∠ABE = ∠CDE (A) 5. By A.A.A facts
6. Corresponding sides of similar
(ii) ∠BAE = ∠DCE (A)
triangles are proportional
(iii) ∠BEA = ∠CED (A)

5. ∴∆ABE ~ ∆CDE

6. CD = DBEE or, x 20 DE) = 3 20 230)
AB = 6 = 3(BD + (4 +
15
∴x 4 cm

Hence, proved.

208 Oasis School Mathematics-9

Example: 7 DC

In parallelogram ABCD, M is the mid-point of AB the O
diagonal AC intersects DM at O. MB

Prove that: DO = 2 MO. A
Given:

(i) M is the mid-point of AB i.e. AM = MB.

(ii) AC and DM intersect at point O.

To prove: DO = 2 MO

Proof:

Statements Reasons
1. In ∆AOM and ∆DOC 1.
(i) Vertically opposite angles
(i) ∠AOM = ∠DOC (A) (ii) Being alternative angles
(iii) Being alternative angles
(ii) ∠OAM = ∠OCD (A) 2. By A.A.A. facts
3. Corresponding sides of similar triangles are
(iii) ∠OMA = ∠ODC (A)
2. ∆AOM~ ∆DOC proportional
4. Being AB = CD i.e. opposite sides of
3. OM = AM
OD CD parallelogram
5. Whole parts axiom.
4. OM = AM
OD AB 6. Being AM = MB

5. OM = AM
AM + BM
OD

6o.r,ODODMOMO = = A 2AMAMAM+MAM

∴ DO = 2 MO
Hence, proved.

Note: Areas of similar triangles are proportion to squares on their corresponding sides.

Exercise 11.5

1. (a) Are all line segments similar?
(b) Are all circles similar?
(c) Are all squares similar?
(d) Are all triangles similar?

Oasis School Mathematics-9 209

(e) Which type of triangles are always similar?
(f) Is it necessary that the size of two similar figures are always same?

(g) In the given figure, ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R are ∆ABC and ∆PQR
similar triangles?

P
A

A P 75º 80º R

B CQ R 25º

(h) Are the given triangle ABC and PQR similar?

B 75º 80º C 25º
Q

(i) Are all congruent figures, also similar? Justify your answer.

(j) What is the relation among the sides of similar figures?

2. Make the following pairs of triangles similar.

(a) A P (b) A P
35º
62º 35º C Q 85º

B 81º C Q 62º 81º R B 85º P
7.2 cm
3. Make the following pairs of triangles similar. R
CQ R
(a) A P (b) A
58º 12 cm
58º
7.8 cm
6 cm
4 cm
5.2 cm
5 cm

3 cm

B CQ RB

4. Make the following pairs of triangles similar.

(a) A P A P

10.5 c m (b) 7.6 cm 2.6 cm
4.5 cm 5.2 cm
3 cm
8.4 cm
4.2 cm
7 cm Q 7.5 cm R CQ R
C B A
B 5 cm 3.8 cm

P

5. (a) In the given figure, ∆ABC ~ ∆PQR. What is the 5 cm
length of PR?
CQ 10 cm R
B 4 cm

210 Oasis School Mathematics-9

4 cm A
(b) In the adjoining figure, ∆ABC ~ ∆MNC. Find the N

length of MN. B 2 cm C
M 5 cm
(c) In the given figure, ∆ABC ~ ∆ AXY,
AY = 10cm, AC = 15cm and AX = 6cm. Find the C

length of the side AB. Y

A XB A
(x–3)cm (x–4)cm
(d) In the given figure, AD = xcm, BD = (x –2)cm, AE = (x –3)cm, 2 cm (x–2)cm
x cmDE
CE = (x – 4)cm. Find the value of x.

AB C

6. (a) In the figure, prove that ∆ABC ~ ∆PQC P x cm
and hence find the length of AB.

C 1.5 cm Q B A
4 cm

(b) In the given figure, ∆ ABC ~ ∆ ADE, AD = BD and BE = EC. D

Find the ratio of DE and AC. B C
E

A B
(c) In the given figure, AB//MN, AO = 6cm, O
4 cm 3 cm
ON = 3cm, OB = 8cm. Find the length of PM 8 cm6 cm N
OM.

(d) In the given figure prove that ∆ PQR ~ ∆PQS and find 10 cm
the length of PR. R8 cm
A
Q S
D C
7. (a) In the given diagram ∠A = 90º, DE⊥BC. Prove that B E
∆ABC ~ ∆DE. Also write the proportionality of their
corresponding sides. C



(b) In the given ∆ABC, ∠A = 90º, AD ⊥ BC. Then A
prove that ∆ABC ~ ∆ABD ~ ∆ADC. D

B

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8. In the given figure AB//CD. Prove that AE.EC = BE.ED. A B
E

9. In ABC, ∠A = 90º, AD ⊥ BC. Prove that: A C D
D C
(i) AB2 = BC.BD

(ii) AC2 = BC.DC B
(iii) AD2 = BD.DC

10. In the given figure AB//EF//DC, prove that A E D
C
(i) ∆BEF ~ ∆BDC BF
E
(ii) ∆ECF ~ ∆CAB

(iii) ∆ABE ~ ∆EDC

(iv) Find the length of EF

11. In the given figure PQRS is a parallelogram. Prove that P M NS
R
(i) ∆EQR ~ ∆EMN Q
(ii) EM = EN = MN B
C
EQ ER QR

A M
O
12. In a parallelogram ABCD, M is the mid-point AB, AC

and DM are intersected at O. Prove that CO = 2AO. D

Answer
1. Consult your teacher 2. Consult your teacher. 3. Consult your teacher. 4. Consult your teacher.
5. (a) 12.5 cm (b) 1.6 cm (c) 9 cm (d) 6 6. (a) 5.33 cm (b) 1:2 (c) 4 cm (d) 5 cm

11.7 Pythagoras Theorem

The Right Angled Triangle
In the given figure, ABC is a right-angled triangle

in which ∠C = 90° and ∠B be the angle of reference.
The side opposite to 90° is the hypotenuse (h).
The side opposite to the angle of reference is the perpendicular (p).
The side adjacent to the angle of reference is the base (b).

212 Oasis School Mathematics-9

Remember ! A

• Side opposite to right–angle is 'h'. ph
• Side opposite to angle of reference is 'p'. CB
• Side adjacent to the angle of reference is 'b'.
b
The Pythagoras Theorem

∆ABC is a right angled triangle, right-angled at C.

Here, AC = 4 units and BC = 3 units and AB = 5 units

and squares on side AB, BC and AC are ARSB, BXYC Q

and APQC respectively. Y

P C
A
We know that, Area of a square = (side)2 X
∴ Area of square on side AB = AB² = 52 = 25
Area of square on side AC = AC2 = 42 = 16 45 3 B

Area of square on side BC = BC2 = 32 = 9

Now, 25 = 16 + 9 RS

i.e. area of square ARSB = area of square APQC + area of square BXYC

or AB2 = AC2 + BC2

i.e. square of its hypotenuse = sum of the squares of two sides of a right angled triangle.

Hence, Pythagoras theorem states that in a right-angled triangle, the area of the square
described on the hypotenuse is equal to the sum of the areas of the squares described
on the other two sides.

Note: • The longest side of the right–angled triangle is hypotenuse.
• The angle opposite to the longest side of a right-angled triangle is a

right–angle.
• In every right angled triangle, h2 = p2 + b2


Theorem-11.4

Pythagoras Theorem B
Theoretical proof:

Given : In a right angled triangle ∠ABC = 90º.

To prove : AC2 = AB2 + BC2 A NC
Construction : Draw BN ⊥ AC

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Proof :

Statements Reasons
1. In ∆ANB and ∆ABC
1.

(i) ∠ANB = ∠ABC (A) (i) Being both 90º

(ii) ∠BAN = ∠BAC (A) (ii) Being common angle

(iii) ∠ABN = ∠BCA (A) (iii) Remaining angles in triangles BAN and
BAC

2. ∆ANB ~∆ABC 2. By A.A.A condition

3. AB = AN 3. Corresponding sides of the similar triangles
AC AB

or, AB2 = AC.AN

4. Similarly, BC2 = AC.NC 4. By making ∆ABC and ∆BNC similar

5. AB2 + BC2 = AC.AN + AC.NC 5. Adding 3 and 4

or, AB2 + BC2 = AC(AN + NC) Hence, proved.

= AC . AC

= AC2


Pythagorean Triplets

If BC2 = AB2 + AC2, where BC, AB and AC are three sides of ∆ABC, then the triangle
must be right angled triangle.

For example: C

If BC = 5 units, AB = 3 units and AC = 4 units, then

BC2 = AB2 + AC2 AB
or, 52 = 32 + 42

or, 25 = 25

∴ The relation BC2 = AB2 + AC2 is satisfied.

But if we take AB = 2 and AC = 3, BC = 5 then 52 ≠ 22 + 32

i.e the relation is not satisfied.

Hence, the Pythagorean triplets are those values which satisfy the Pythogorean
relation p2 + b2 = h2.

Pythagorean triplets can be obtained by substituting (n = 1, 2, 3, ….. in

2n + 1, 2n (n + 1) and 2n(n + 1) + 1

214 Oasis School Mathematics-9

n 2n + 1 2n (n + 1) 2n (n + 1) + 1 Remarks

1 345 32 + 42 = 52

2 5 12 13 52 + 122 = 132

3 7 24 25 72 + 242 = 252

4 9 40 41 92 + 402 = 412

Do You Know!
Pythagoras lived in 6th Century BC and was a teacher in Samos, Babylon and Egypt.
The Pythagorean triplets were already known in Babylonian times.

Worked Out Examples

Example: 1

In a right angled triangle, the hypotenuse is 25cm and one of the other sides is 24 cm,

find the third side. A

Solution: 24 cm 25 cm
Let, ∆ABC be a right angled triangle in which ∠B = 90º

Here, AC = 25 cm BC
AB = 24 cm

BC = ?

Using Pythagoras theorem, AC2 = AB2 + BC2

or, BC2 = AC2 – AB2

or, BC = AC2 – AB2

= 252 – 242

= 625 – 576

= 49

= 7 cm.

∴ The third side is 7 cm. A

Example: 2 x
D
In a right angled triangle ABC, AB = 6cm, BC = 8cm. Find the values 6 cm y
of x and y. 8 cm

Solution: B C
In ∆ABC, AB = 6cm, BC = 8cm

We have,

AC2 = AB2 + BC2

Oasis School Mathematics-9 215

x2 = (6)2 + (8)2

x2 = 36 + 64

x2 = 100

x = 10cm
12 AC
Now, area of ∆ABC = × BD [Area of trianlge = 1 base × height]
= 12 x × y 2

Again =, a r =e12a 211o0f×∆×6Ay×B 8 C= =5 y 12c m A2 B × BC

= 24cm2

Now, 5y = 24

y = 24
5

= 4.8cm.

Example: 3 A

A 5m long ladder reaches 3m above the ground level. How far is its foot 3m
B
from the foot of the wall? 5m
Solution:

Let, AC be the ladder, AB be the wall. BC be the distance C
between the foot of the ladder from the foot of the wall.

Here, AC = 5 m, AB = 3m ?

Using the pythagoras theorem,

BC = AC2–AB2

BC = 52 – 32

= 25 – 9

= 16

= 4 m
∴ The foot of the ladder is 4 m away from the foot of the wall.

216 Oasis School Mathematics-9

Example: 4 A

In ∆ABC, ∠B = 90°, BD⊥AC, prove that: c
(i) bh = ca (ii) 1 = 1 + 1
D b
h2 c2 a2 h
Solution:
Ba
In ∆ ABC, C
AC = b, AB = c, BC = a, BD = h

We know that, area of ∆ = 1 base × height
2
1
Are of ∆ABC = 2 AB × BC

∆ABC = 1 c.a. …………… (i)
2
1
Again, area of ∆ABC = 2 BD×AC

or, ∆ABC = 1 h.b. …………(ii)
2

From (i) and (ii) 1 bh = 1 ca
22
∴ bh = ca ……………….(iii)

Squaring both sides of (iii) we get,

b2h2 = c2a2

or, (c2 + a2)h2 = c2a2 (∵ b2 = c2 + a2.)

or, c2h2 + a2h2 = c2a2

Dividing both sides by c2a2h2 then,

c2h2 + a2h2 = c2a2
∴ a2c2h2 a2c2h2 c2a2h2

1 + 1 = 1 Proved.
a2 c2 h2

Exercise 11.6

1. (a) Which is the longest side of a right angled triangle?
(b) Write the relation among 'p', 'b' and 'b' of a right angled triangle.
(c) What is the meaning of Pythagorean Triplets?
(d) Give two examples of Pythagorean Triplets?

Oasis School Mathematics-9 217

2. Determine whether the followings are right angled triangle or not?

(a) (b) A (c) A

A

5 cm 6 cm 6 cm 41 cm
4 cm
40 cm

BC BC BC
4 cm 3 cm 9 cm

3. Calculate the unknown length of the sides of given right angled triangles.
A
P (c) X
(a) (b)

x cm 10 cm x cm
3.1 cm x cm
15 cm
Y
B x cm CQ R 25 cm Z
7.4 cm

4. Find the values of 'x' and 'y' in the given figures:

(a) A A (c) P

(b) 10

6 cm 5 cm Dx x y S 2 cm
QR
BD x cmC y
y cm

10 cm B 12 cm C

5. (a) A field is 40 m long and 30 m wide. If a path runs diagonally on the rectangular
field, how long is the foot path?

(b) A ladder 6.4 m long is to be placed so that it just reaches a window 5.7 m from
the ground. How far from the wall is the foot of the ladder?

6. (a) Find the value of x and y in the given figures. A 26 cm B

8 cm 24 cm
(b) In the given figure, ∆ADC is a right angled
triangle, show that ∆ABC is a right angled D 6 cm C
triangle.

7. (a) Three sides of a right angled triangle are 3cm, (x+1)cm and (x+2) cm. If (x+2)cm
is the greatest value, find the sides of the triangle.

(b) If xcm, (2x+2)cm and (2x+3) cm are three sides a right angled triangle, find the

length of all three sides. A

8. (a) In the given triangle ABC, D is any D
point on AB.

Prove that: AB2 – BD2 = AC2 – CD2. BC

218 Oasis School Mathematics-9

A

(b) In the given triangle ABC, M is the mid–point of BC.
Prove that AC2 = AM2 + 3CM2

B MC

9. (a) The sides of a triangle are m2 + n2, m2 – n2 and 2mn, where m and n are positive
integers. Prove that the triangle is a right angled triangle.

(b) What is the length of diagonal in the given cuboid H G
if AB = 16 cm, BC = 25 cm and AE = 9 cm? D C

E F
A B



Answers

1. Consult your teacher

2. (a) No. (b) No (c) Yes 3. (a) 5 2 cm (b) 8.02 cm c) 20

4. (a) x = 8cm, y = 4.8cm (b) x = 13cm, y = 4.61cm (c) x = 10cm, y = 5 2 cm

5. (a) 50cm (b) 2.91m, 6. (a) x = 13cm, y = 11.5cm

7. (a) 3cm, 4cm, 5cm (b) 5cm, 12cm, 13cm, 9. (b) 31.02 cm

Oasis School Mathematics-9 219

Unit Quadrilateral and
12 Parallelogram

12.1 Warm-up Activities

Discuss the following questions in your class.
• What is the sum of four angles of a quadrilateral?
• What are the types of quadrilateral?
• Draw one figure of each type of quadrilateral and find their differences.

12.2 Quadrilateral A B
D C
In the given figure, ABCD is a plane figure bounded by 4
sides AB, BC, CD and DA. It has 4 angles A, B, C and D,
Hence, a quadrilateral is a plane figure bounded by 4 sides.

Types of Quadrilateral

Parallelogram Rectangle Square Rhombus

Trapezium

Kite

Different types of quadrilateral are shown in the above figures.

Quadrilateral having opposite sides parallel is a D C
parallelogram. In the figure, AD||BC and AB||DC. So, it B
is parallelogram.

Features: • Opposite sides are parallel. A
• Opposite sides are equal.

• Opposite angles are equal.
• Diagonals bisect each other.

220 Oasis School Mathematics-9

Rectangle A D
C
Parallelogram having each angle 90° is called a rectangle. In
the given parallelogram ABCD, ∠B = 90º. So, it is a rectangle. D
C
Features: • Opposite sides are parallel. B
• Opposite sides are equal.

• All angles are equal.

• Each angle is 90º.

• Diagonals bisect each other.
• Diagonals are equal in length.



Rhombus

Rhombus is a parallelogram having adjacent sides equal. A
In the figure ABCD, AB||DC, AD||BC and AB = BC.

Here, AB and BC are adjacent sides adjusted at B. They are
equal. So it is rhombus. Moreover BC and CD, CD and DA, B
DA and AB are also adjacent sides and they are also equal.

Features: • Opposite sides are parallel.
• All sides are equal.
• Opposite angles are equal.
• Diagonals bisect each other at right angle.



Square

Rhombus having one angle 900 is a square. A D
B C
Or, Rectangle having adjacent sides equal is a square. In the
figure, ABCD is a rectangle and AB = BC. So, it is a square. A D
B
Features: • All sides are equal.
C
• Each angle is 90º.

• Diagonals are equal in length.

• Diagonals bisect each other at right angle.

Kite

Kite is a quadrilateral having two adjacent pairs of sides equal.
In the figure, AB = AD
BC = CD, so ABCD is a kite.

Oasis School Mathematics-9 221

Features: • Two pairs of adjacent sides are equal.

• Diagonals intersect each other at right angle.
• Longest diagonal bisect the shortest diagonal


• Longest diagonal bisect the vertex angle
through which it passes.

Trapezium D

Quadrilateral having a pair of opposite sides parallel is called A
trapezium.

In the given figure, BC

ABCD is a trapezium because AD||BC. If the non–parallel sides of trapezium are
equal then the trapezium is called isosceles trapezium.



Quadrilateral

Trapezium Kite Parallelogram

Rectangle Square Rhombus

The comparative study of their property is listed below.

Remember !

• Every rectangle is a parallelogram.
• Every square is a rectangle.
• Every square is a rhombus.
• Every rhombus is a parallelogram.
• Every square is a parallelogram.

Theorem 12.1

Two line segments which join the ends of two equal and parallel lines towards the
same side are themselves equal and parallel.

222 Oasis School Mathematics-9

Theoretical Proof:
: AB = CD and AB||DC. AC and BD are A B
Given joined. D

To prove : AC = BD, AC||BD. C
Construction : Join A and D.

Proof:

Statements Reasons

1. In ∆ABD and ∆ADC 1.
(i) AB =DC (S) (i) Given
(ii) ∠BAD = ∠ADC (A) (ii) Alternate angles
(iii) AD = AD (S) (iii) Common side.
2. ∆ABD ≅ ∆ ADC 2. By S.A.S. axiom
3. AC = BD 3. Corresponding sides of congruent triangles
4. ∠ADB = ∠CAD 4. Corresponding angles of congruent triangles

5. AC||BD 5. From (4) being alternate angles equal

Hence, two line segments which join the ends of two equal and parallel lines towards
the same side are themselves equal and parallel.

Experimental Verification:

Draw two sets of parallel lines AB and CD such that AB = CD. Join AC and BD.

AB AB

CD CD
Figure : (i) Figure : (ii)

To verify : AC = BD and AC||BD

Take the measurement of AC, BD, ∠ACD and ∠BDC. Tabulate the measurement
below.

Fig. AC BD ∠ACD ∠BDC ∠ACD + ∠BDC Remarks

i. ..... cm .... cm ...... º ..... º 180º AC = BD, AC||BD

ii. ..... cm .... cm ..... º ..... º 180º AC = BD, AC||BD

Conclusion: Hence, the line joining the same sides of two equal and parallel lines
are equal and parallel.

Oasis School Mathematics-9 223

Theorem 12.2

Two straight lines which join the ends of two equal and parallel straight lines
towards the opposite sides bisect each other.

Theoretical Proof. AB
O
Given : AB = CD, AB||CD, AD and BC are joined. D

To prove : AO = OD and BO = OC. C
Proof :

Statements Reasons
1. In ∆AOB and ∆COD 1.
(i) ∠ABO = ∠OCD (A) (i) Being alternate angles
(ii) AB = CD (S) (ii) Given
(iii) ∠BAO = ∠CDO (A) (iii) Being alternate angle.
2. ∆AOB ≅∆ ∆COD 2. By A.S.A. axiom.
3. AO = OD and OB = OC 3. Corresponding sides of congruent

triangles

Hence, two straight lines which join the ends of two equal and parallel lines towards
the opposite sides bisect each other.

Experimental Verification:

The lines joining the opposite sides of two equal and parallel lines bisect

each other.

Draw the two sets of equal and parallel lines AB and CD. Join AD and BC.

A C AB

O
O

B D CD
Figure (i) Figure (ii)

To verify: AO = OD, BO = OC

Measure the length of AO, OD, BO, OC and tabulate below.

Observation:

Figure AO OD BO OC Remarks
i) AO = OD, BO = OC
ii) "

Conclusion: Hence, two lines which join the ends of the opposite sides of two equal
and parallel lines bisect each other.

224 Oasis School Mathematics-9

Properties of parallelogram

Theorem 12.3 AD

Opposite sides and angles of a parallelogram are equal.

Given : ABCD is a parallelogram. BC

To Prove : AB = DC and AD = BC, ∠ABC = ∠ADC and ∠BAD = ∠BCD.

Construction : Join A and C.

Proof:

Statements Reasons
1. In ∆ABC and ∆ADC
1.

i) ∠BAC =∠ACD (A) i) Alternate angles
ii) AC = AC (S) ii) Common side

iii) ∠ACB = ∠CAD (A) iii) Alternate angles

2. ∆ABC ≅ ∆ADC 2. By A.S.A. axiom
3. AB = DC, BC = AD 3. Corresponding sides of congruent triangles

4. ∠ABC = ∠ADC 4. Corresponding angles of congruent triangles
5. Similarly, ∠DAB = ∠BCD 5.As statement 4, joining BD

Hence opposite sides and angles of a parallelogram are equal.

Converse I
If the opposite sides of a quadrilateral are equal, it is a parallelogram.

Given : ABCD is a quadrilateral where AB = DC and AD = BC.

To Prove : ABCD is a parallelogram. A D

Construction : Join AC. BC
Proof :

Statements Reasons

1. In ∆∆ABC and ∆ADC 1.

(i) AB = DC (S) (i) Given

(ii) BC = AD (S) (ii) Given.

(iii) AC = AC (S) (iii) Common side

2. ∆ABC ≅ ∆ADC 2. By S.S.S. axiom

3. ∠BAC = ∠ACD and 3. Corresponding angles of congruent triangles
∠ACB = ∠DAC

4. AB||DC and AD||BC 4. From statement 3, being alternate angles equal

5. ABCD is a parallelogram 5. From statement 4, being opposite sides parallel

Hence, proved.

Oasis School Mathematics-9 225

Converse II
If the opposite angles of a quadrilateral are equal, it is a parallelogram.

Given : ABCD is a quadrilateral, where A D
C
∠BAD = ∠BCD and ∠ABC = ∠ADC.

To Prove : ABCD is a parallelogram. B

Proof :

Statements Reasons

1. ∠BAD + ∠ABC + ∠BCD + ∠ADC = 360º. 1. Sum of four angles of a quadrilateral

2. ∠BAD = ∠BCD 2. Given

3. ∠ABC = ∠ADC 3. Given

4. ∠BCD + ∠ADC + ∠BCD + ∠ADC = 360º 4. From statement 1, 2 and 3

5. 2∠BCD + 2∠ADC = 360º 5. From statement 4

6. ∠BCD + ∠ADC = 180º 6. From statement 5
7. AD||BC 7. From statement 6, being sum of

8. Similarly, AB||DC co-interior angles 180º
9. ABCD is a parallelogram 8. As statement 7
9. From statements 7 and 8

Hence, proved.

Theorem 12.4

Diagonals of a parallelogram bisect each other. AD

Given : (i) ABCD is a parallelogram where O
AB||DC and AD||BC BC

(ii) AC and BD are diagonals.

To prove : AO = OC , BO = OD

Proof :

Statements Reasons
1. In ∆AOD and ∆BOC 1.
(i) ∠OAD = ∠OCB (A) (i) Alternate angles on AD||BC
(ii) AD = BC (S) (ii) Opposite sides of a parallelogram

(iii) ∠ADO = ∠OBC (A) (iii) Being alternate angles

2. ∆AOD ≅ ∆BOC 2. By A.S.A. axiom.
3. AO = OC, BO = OD. 3. Corresponding sides of congruent triangles

Hence, proved.



226 Oasis School Mathematics-9

Converse

If the diagonals of a quadrilateral bisect each other, it is a parallelogram.

Given : ABCD is a quadrilateral, where AO = OC, BO = OD. A D

To Prove : ABCD is a parallelogram. O
Proof : BC

Statements Reasons
1. In ∆AOB and ∆DOC 1.
(i) Given
(i) AO = OC (S) (ii) Vertically opposite angles
(ii) ∠AOB = ∠DOC (A) (iii) Given
(iii) BO = OD (S) 2. By S.A.S. axiom.
2. ∆AOB ≅ ∆DOC 3. Corresponding angles of congruent triangles
3. ∠BAO = ∠OCD 4. From statement 4, being alternate angles equal.
4. AB||DC 5. As statement 4
5. Similarly, AD||BC 6. From statements (4) and (5)
6. ABCD is a parallelogram
Hence, proved.


Remember !

• Opposite sides and angles of a parallelogram are equal.
• Diagonals of a parallelogram bisect each other.
• If the opposite sides of a quadrilateral are equal, then it is a parallelogram.
• If the opposite angles of a quadrilateral are equal, then it is a parallelogram.
• If the diagonals of a quadrilateral bisect each other, it is a parallelogram.

Worked Out Examples A D E
Example: 1 x ya

Find the value of unknown angles in the given figure. 70º z C
Solution: Here,
B

∠ADC = ∠ABC [Opposite angles of a parallelogram]

∴ y = 70º

Again, ∠ABC + ∠BAD = 180º [Sum of co-interior angles]

or, 70º + x = 180º

or, x = 180º – 70º = 110º

Oasis School Mathematics-9 227

Again, ∠BAD = ∠BCD [Opposite angles of a parallelogram]
or, 110º = z [Corresponding angles]
∴ z = 110º
Again,
∠BAD = ∠CDE
x = a

∴ 110º = a

a = 110º

Example: 2

Find the values of unknown angles in the given figure. A E B
105º x
B
Solution: C

Here, ∠DAE = 105º and CB = CE D y
C

then, ∠CEB = ∠CBE = x

Now, ∠DAE +∠ CBE = 180º [Co-interior angles]

105º + x = 180º

or, x = 180º – 105º = 75º

Again, ∠BEC = ∠ECD [Alternate angles]

or, x = y

or, 75º = y

∴ y = 75º

Example: 3

In the given figure, ABCD is a rectangle. Prove that AC = BD. A

Solution:

Given : ABCD is a rectangle. AC and BD are its diagonals.

To prove : AC = BD D

Proof :

Statements Reasons
1. In ∆ADC and ∆BDC 1.
(i) AD = BC (S) (i) Being opposite sides of a rectangle
(ii) Being both right angles
(ii) ∠ADC = ∠BCD (A) (iii) Common side
(iii) DC = DC (S)
2. ∆ADC ≅ ∆BDC 2. By S.A.S. axiom
3. AC = BD 3. Corresponding sides of congruent

triangles

Hence, proved.

228 Oasis School Mathematics-9

Example: 4 E

In the given figure, ABCD is a parallelogram. AE = CF, prove A B
that EBFD is a parallelogram.
C
Solution: D F
Given : ABCD is a parallelogram AE = CF.

To prove : EBFD is a parallelogram.

Proof :

Statements Reasons
1. ∠DAC = ∠ACB 1. Alternate angles
2. ∠EAD = ∠BCF 2. Supplements of equal angles of statement (i)
(Exterior Alternate Angles)
3. In ∆AED and ∆BCF 3.
(i) AE = CF (S) (i) Given
(ii) ∠EAD = ∠BCF (A) (ii) From statement 2
(iii) AD = BC (S) (iii) Opposite sides of a parallelogram
4. ∆AED ≅ ∆BCF 4. By SAS axiom
5. ED = BF
6. ∠AED = ∠BFC 5. Corresponding sides of congruent triangles.
6. Corresponding angles of congruent triangles
7. ED // BF
8. EBFD is a parallelogram 7. From statement 6, being alternate angles equal
8. From statement 5 and 7

Hence, proved.

Exercise 12.1 A 6 cm B

1. (a) In the given figure, ABCD is a parallelogram, 5 cm
if AB = 6cm and BC = 5cm, what is the
length of AD and DC? D C
P
(b) In the given figure, PQRS is a parallelogram, 3 cm 5 cm Q
if OP = 3cm and OQ = 5cm, find the length S
of OR and OS. O

R

(c) What is a quadrilateral called whose opposite sides are equal, opposite angles
are equal and non of its angle is 900?

(d) What is a parallelogram called if its all sides are equal and none of its angle is
900?

Oasis School Mathematics-9 229

(e) Are all rectangles a parallelogram? Justify your answers. B
D
(f) If the diagonals of a rhombus are equal, then what is it called?

A

(g) In the given figure, AB = CD and AB//CD, what is

the relation between AC and BD? C

(h) What happens when the opposite sides of two equal and parallel lines are joined?

(i) ABCD is a parallelogram, write the relation between A B
(i) AD and BC (ii) AB and DC

(iii) ∠A and ∠C (iv) ∠A and ∠B (v) ∠B and ∠D. D C

2. From the following figures, find the values 'x', 'y' and 'z'.

(a) A 110º yB (b) P y 3xº Q (c) M y

D x S 75º (2y+3)cm N3 cm
x O
C R P (x+2)cm

(d) A x y B (e) P (3x+20)º Q (f) (4x+5)cm

EF

6 cm O 4 cm S (4x–10)ºR H (y+5)cm
DC
G
(3x +8)cm
3. From the following figures, find the values 'x', 'y' , 'z'.

(a) P Q (b) E
45º F (c)
Px S
zG
OA D zy

x a T

x 92º 50º Q 40º 70º
25º
S R R

B 40º yº C

B (e) My 25º x N (f) W M
(d) A xX
30º 110º N
x

Dy Q y

E C Pz O ZY

A B
C
4. (a) In the given figure, ABDC is a rhombus. 40º
If ∠ADB = 40º, find the size of ∠CDE. D

E

230 Oasis School Mathematics-9

(b) One angle of a parallelogram is 70º. Find the other angles of the parallelogram.
(c) The side DC of a parallelogram ABCD is produced to E. If the exterior angle

BCE is 65º, find the angles of the parallelogram.

A B
C
(d) ABCD is a parallelogram and the bisectors of ∠A and P
∠B meet at P. Show that ∠APB = 90º.

D

(e) In the given figure, ABCD is a Ay z x B
square, ∆EDC is an equilateral
triangle. Find the value of x, y E
and z.
DC

5. (a) In the given figure, ABCD is a rhombus. Prove that A B

(i) AO = OC and BO = OD (ii) AC ⊥ BD O
DC

A B
C
(b) In the given figure, ABCD is a parallelogram where

AC = BD, prove that ABCD is a rectangle.

D

6. (a) In the figure, ABCD is a parallelogram, M and N are the middle points of the

sides DC and AB respectively. Prove that AN B
(i) MCBN is a parallelogram.

(ii) DMBN is a parallelogram. D MC
(iii) DB and MN bisect each other.

(b) In the given figure ABCD is a parallelogram D C
F
with E and F any two points on the diagonal
AC such that AE = FC. Prove that DEBF is a B
parallelogram.
E
A

(c) In the given figure, ABCD is a parallelogram, AF B

DF⊥AC and BE⊥AC, BF and DE are joined.

Prove that: DEBF is a parallelogram. D E
C

Oasis School Mathematics-9 231

Answer

1. Consult your teacher 2. (a) x = 110º, y = 70º (b) x = 25º, y = 105º (c) x = 3 cm, y = 5cm

(d) x = 4cm, y = 6cm (e) x = 30º (f) x = 3m, y = 2cm

3. (a) x = 42º (b) x = 95º, y = 95º, z = 40º (c) x = 25º, y = 40º, z = 110º, a = 70º

(d) x = 105º, y = 75º e) x = 25º, y = 65º, z = 25º (f) x = 115º, y = 25º

4. (a) 140º (b) 70º, 110º, 70º, 110º (c) 65º, 115º, 65º, 115º (e) x = 15º, y = 75º, z = 150º

12.3 Mid–Point Theorem

A line joining the middle points of the two sides of a triangle is parallel to the third
A
side and one half of it.

Given : In a triangle ABC, P and Q are the mid P Q
points of side AB and AC respectively.
R

To Prove : PQ || BC and PQ = 1 BC. B C
2

Construction : Produce PQ up to R making PQ = QR and join R to C.

Proof:

Statements Reasons
1. In ∆APQ and ∆QRC 1.
(i) AQ = QC (S) (i) Given
(ii) Vertically opposite angles
(ii) ∠AQP = ∠RQC (A) (iii) By construction
(iii) PQ = QR (S) 2. By S.A.S. axiom
2. ∆APQ ≅ ∆QRC 3. Corresponding sides of congruent triangles
3. AP = RC 4. Given
4. AP = PB 5. From statements 3 and 4
5. PB = RC 6. Corresponding angles of congruent triangles
6. ∠APQ = ∠QRC 7. From statements 6, being alternate angles equal
7. AP||RC 8. From statements (5) and (7) being one pair of opposite
8. PBCR is a parallelogram
sides equal and parallel.
9. PR||BC .ie. PQ||BC 9. Being opposite sides of a parallelogram
10. Opposite sides of a parallelogram
10. PR = BC 11. By construction

11. PQ = 1 PR 12. From statements (10) and (11)
2

12. PQ = 1 BC
2

Hence, the line joining the mid-points of two sides of a triangle is parallel to the
third side and one half of it.

232 Oasis School Mathematics-9

Converse of Mid-point Theorem

A line passing through mid-point of one side of a triangle and parallel to other side

bisects the third side. A

Given : In ∆ABC, M is the mid-point of AB and
MN||BC
M NP

To prove : N is the mid-point of AC (AN = NC). B C
Construction: Draw CP||BA and produce MN to P.
Proof:

Statements Reasons

1. MBCP is a parallelogram 1. Being MN||BC and BA||CP

2. BM = CP 2. Opposite sides of a parallelogram

3. BM = AM 3. Given

4. AM = CP 4. Form statements 2 and 3

5. In ∆AMN and ∆PCN (i) Vertically opposite angles
(i) ∠ANM = ∠PNC (A) (ii) Alternate angles
(ii) ∠MAN = ∠PCN (A) (iii) From statement 4
(iii) AM = CP (S) 6. By A.A.S. axiom
6. ∆AMN ≅ ∆PCN

7. AN = NC 7. Corresponding sides of congruent triangles

Hence, a line passing through the mid-point of one side of a triangle and parallel to
other side bisects the third side.

Worked Out Examples

Example: 1

Find the values of x and y in the given figure. A

Solution: D 2.5cm y E
B 65º C
In the given figure, D and E are the mid-points of AB and
AC respectively. Then using mid-point theorem, x

DE = 1 BC and DE//BC Oasis School Mathematics-9 233
2

or, 2.5 = 1 x
or, 2
5 = x

∴ x = 5cm

Again, DE||BC [Using mid-point theorem]

or, ∠AED = ∠ACB [Alternate angles]
∴ x = 5cm, y = 65º.

Example: 2 A

Find the values of x and y in the given figure.

Solution: EF G
4 cm y
In the given figure, E is the mid point of AB, EF||BD.

Then, F is the mid-point of AD. Using mid point theorem, BC
x D 10 cm
1
EF = 2 BD

or, 4 = 1 x
or, 2
x = 8cm

Again, in ∆ADC, F is the mid-point of AD, FG||DC [∵ EF||BD]
Then G is the mid-point of AC.

Now, using mid-point theorem,

FG = 1 DC
2
or,
or, y = 1 × 10cm
2
y = 5cm

Example: 3 A PB

In the given figure, AD||PQ||BC, DQ = QC. Prove that,

(i) QR = 1 AD. (ii) AD + BC = 2PQ. R
2

Given : In the given figure, AD||PQ||BC and DQ = QC. DC
Q
To Prove : (i) QR = 1 AD (ii) AD + BC = 2PQ
2
Proof:

Statements Reasons
1. ∆ADC, Q is the mid point of DC 1. Given
2. RQ||AD 2. Given
3. R is the mid-point of AC. 3. The line joining the mid point of a side of a triangle

4. QR = 21AD parallel to the base bisects the third side
5. In ∆ABC, PR||BC 4. From statements (2) and (3)
6. P is the mid point of AB
5. Given
6. From statements (3) and (5)

234 Oasis School Mathematics-9

7. PR = 12BC i.e. BC = 2PR 7. From (5) and (6)
8. AD + BC = 2(QR + PR) = 2 PQ 8. Adding statements (4) and (7).

Hence, proved.

Exercise 12.2 A

1. (a) In the given figure, what is the relation P Q X
of PQ with BC? B C B
Z
(b) In the given figure, A is the mid point of XY and AB//YZ. A
Y
Then, what is the next result? R

2. Find the value of unknown size of the angles. A 146º Y
A y
(a) (b) P (c) N 130º
z C
P x 70º Q 110º M M

B 60º yC Qz x 20º R x
y B
N

3. Find the length of unknown sides in each of the following figures.

(a) A (b) E (c) P

G S 6.1cm T

P 6.2 cm
z

8.4cm
y cm

x
z
5.6 cmD
5.8 cm

x cm
2.6 cm
BC
Q H Q Uy R
F

A A 3.8 cm B

(d) P 3.2 cm 5R.2 cm (e) E 3.2 cm xF

yC G

B y cm DC
y

4. (a) In the given figure, ABC is an equilateral triangle of A
side length 8 cm and P, Q and R are the mid points of PQ
AB, AC and BC respectively. Find the length of PQ, PR BC
and RQ.
R
A

(b) In ∆ABC PQ//BC, AP = PB. If the
length of AC and BC are 8 cm and P
Q

7.2cm respectively, find the length of BC
AQ and PQ.

Oasis School Mathematics-9 235

A

5. In ∆ ABC, P and Q are the mid–points of side AB and AC

respectively. AR meets PQ at S and BC at R. P Q
C
Prove that AS = SR. A S
B

R

6. In the given figure, E is the mid point E F
G
of AB, EG//BD, prove that F is the B
mid point of AC. C
D

A D
Q
7. In the figure, PQ//AD//BC and AP = PB. P
S
Prove that AS = SC and DQ = QC. B C
A
8. In the given figure, ABCD is a trapezium, prove P D
Q
that: PQ = 1 (AD + BC). PB C
2

9. In the given figure, T is the mid point of PQ, T U R
TU//QR and US//PQ, prove that QS = SR. S
S
Q A D
R
10. In the given figure, A, B, C and D are the P C
mid–points PS, PQ, QR and RS respectively. Prove B
that ABCD is a Parallelogram.
Q

11. Prove that lines joining the middle-point of opposite sides of a quadrilateral bisect
each other.
A PB

12. In the figure, P,Q, R, S are the mid–points of side AB, SQ
AC, DC and BD respectively prove that PQRS is a
parallelogram.

13. In the figure ABCD is a parallelogram, M and N DR C
are the mid points at DC and AB respectively. DM C

Prove that; Q

(i) DMBN is a parallelogram. P B
(ii) AP = PQ = QC. A

N

Answer

1. Consult your teacher 2. (a) x = 60° y = 70° z = 50° (b) x = 110°, y = 50°, z = 50°

(c) x = 96°, y = 50°, z = 96° 3. (a) x = 5.2 cm (b) y = 4.2 cm (c) x = 11.2 cm, y = 12.2 cm, z = 11.6 cm

(d) x = 6.4 cm, y = 2.6 cm, z = 3.1 cm e) x = 1.9 cm, y = 6.4 cm
4. (a) PQ = 4 cm, QR = 4 cm, PR = 4 cm (b) AQ = 4 cm, BC = 3.6 cm

236 Oasis School Mathematics-9

Unit

13 Circle

13.1 Warm-up Activities

Discuss the following in your class.

• Draw a circle and show its different parts like circumference, diameter, radius
• What is the relation of diameter with radius?
• What is the relation of diameter with circumference?
• Draw a circle and measure its circumference.

13.2 Circle P S
OR
Basic definitions
Q
If a point moves in a plane in such a way that its distance
from a fixed point is always the same, the locus of the point is
called a circle. It is denoted by .

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

Circumference

The total length of the rim of the circle is circumference. In oth- O
er words, the perimeter of the circle is the circumference of the
circle. Q

Radius Radius O

A line segment which joins the centre of circle to any point of Q
the circumference is a radius. It is denoted by 'r'.
Chord B
Chord O
A
A line segment which joins the two points of the circumfer-
ence is a chord. AB and CD are the chords. CD

Diameter diameter C
AO
A chord which passes through the centre of the circle is
called a diameter. i.e. a diameter is the longest chord. AC is a Chord D
diameter.

Note: Diameter = 2 × Radius = 2r.

Oasis School Mathematics-9 237

Arc C

Any part of the circumference of a circle is called the arc. It AB
is denoted by ' '. If the arc is less than half of the circumfer-
ence, it is called a minor arc and if it is more than half of the Minor arc
circumference, it is a major arc.

Note: In the above figure, AB is a minor arc and ACB is the major arc.

Semi–circle A

A diameter of a circle divides a circle into two DB
equal parts, each part is called semi–circle. O

In the figure DAB and DCB are two semi–circles of the circle. C
Major
Sector
O sector
A region enclosed by any two radii and an arc of a circle is called Q
a sector of the circle. If the sector is less than semi circle, it is a
minor sector and if it is more than semi circle, it is a major arc. P

Segment Minor sector

A region enclosed by an arc and a chord of a circle is Major segment
called a segment of the circle. If the segment is less than O
half of circle, it is called a minor segment and if it is
more than half of the circle, it is called a major segment. AB
Minor segment
Concentric circles

Two or more than two circles having the same centre are O
called the concentric circles and each circle belongs to the
same family of the circles. The portion between two circles I
is called annulus region. II
III
Line of centres
Annulus region

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

Intersecting circles A O'
O
When two circles intersect each other at two different
points, they are said to be intersecting circles. The line B
which joins their points of intersection is their common A
chord.
In the given figure, two circles intersect at A and B and B
AB is a common chord. O

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

238 Oasis School Mathematics-9

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

Central angle Central angle
B
An angle at the centre of the circle is a central angle. O

Centre is the vertex for the angle. Here, AOB is the cen- A
tral angle.
P

Some points to remember about circles:

1. One and only one circle can be drawn with a given radius.

2. A line cannot intersect a circle in more than two points.

3. Radii of the same circle or equal circles are equal.

4. Two circles are equal if their radii are equal.

5. The longest chord of a circle is a diameter.

Some important theorems related to the circle:
1. A perpendicular drawn from the centre of a circle bisects the chord.
2. A line passing through the centre of a circle and bisecting a chord which is not

a diameter, is perpendicular to the chord.
3. Equal chords of a circle are equidistant from the centre.
4. Chords which are equidistant from the centre of a circle are equal.

Tangent to a circle T
P
A line which intersects the circle one and only point is called R
a tangent to the circle. The point at which the tangent touches
the circle is called the point of contact.
In the given figure RT is a tangent and P is a point of contact.

Secant of a circle YB

A line that intersects a circle in two distinct points is called a
secant of the circle. In the given figure, AB is a secant which A X
intersects a circle at X and Y.

Theorem 13.1

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

Experimental Verification

Draw two circles of different radii with centre O and also draw a chord AB of
different lengths and draw OP ⊥ AB in each circle. B

O
OP

A PB

Fig. I Fig. IIA



Oasis School Mathematics-9 239

To verify : AP = PB
Now, measure the length of AP and PB with the help of ruler and tabulate them as

follows:
Observation:

Figure AP PB Remarks
........... cm AP = PB
(i) ........... cm

(ii) ........... cm ........... cm AP = PB

Conclusion : Hence, perpendicular drawn from the centre of a circle to a chord
bisects it.

Theoretical Proof: O
Given : O is the centre of circle and AB is a chord and OP⊥AB. A PB
To prove : AP = PB
Construction : Join O and A, O and B.

Proof :

Statements Reasons

1. In ∆APO and ∆BPO 1.

i) ∠OPA = ∠OPB (R) i) Being both 90º

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

iii) OP = OP (S) iii) Common sides.

2. ∴∆APO ≅ ∆ BPO 2. By R.H.S. axiom.

3. AP = PB 3. Corresponding sides of congruent triangles.

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

Hence, proved.

Theorem 13.2

The line joining the middle point of the chord and the centre of a circle is perpendic-
ular to the chord.

Experimental Verification

Draw two circles with different radii with centre O. Also draw a chord AB of
different length in each circle. Determine the middle point M of the chord AB and
join OM.
B

O
OM

A MB

Fig. I Fig. II A



240 Oasis School Mathematics-9

To Verify : OM ⊥ AM
Now, measure ∠OMA and ∠OMB with the help of protractor and tabulate them as

follows:

Figure ∠OMA ∠OMB Remarks
(i) 90º 90º OM ⊥ AB

(ii) 90º 90º OM ⊥ AB

Conclusion : The line joining the middle point of the chord and the centre of a circle
is perpendicular to the chord.

Theoretical Proof :

Given : O is the centre of a circle and M is the middle O
point of the chord AB and OM is joined. A MB

To prove : OM⊥AB

Construction : Join O and A, O and B.

Proof :

Statements Reasons
1. In ∆OMA and ∆OMB 1.
(i) Radii of the same circle
(i) OA = OB (S) (ii) M is middle point of AB
(iii) common sides
(ii) AM = MB (S) By S.S.S. axiom
Corresponding angles of congruent triangles
(iii) OM = OM (S) Being linear pairs
From (3) and (4)
2. ∴∆ OMA ≅ ∆OMB 2. From (5)

3. ∠OMA = ∠OMB 3.

4. ∠OMA + ∠OMB = 180º 4.

5. ∠OMA = ∠OMB = 90º 5.

6. OM ⊥ AB 6.

Hence, the line joining the middle point of the chord and the centre of a circle is

perpendicular to the chord.


Theorem 13.3

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

Given : 'O' is the centre of the circle. CM is the perpendicular bisector of AB.

To prove : CM passes through centre 'O' . C

Construction: Let's suppose CM does not pass through O
'O'. Join OM.

AM B

Oasis School Mathematics-9 241

Proof

Statements Reasons

1. ∠CMA = 90º 1. Being CM, perpendicular bisector of AB

2. ∠OMA = 90º 2. The line joining the centre of the circle
to the mid point of the chord perpendic-
ular to it

3. ∠CMA = ∠OMA 3. From (1) and (2)

4. Statement 3 is not possible 4. Since OM and CM do not lie on the same
straight line

5. Statement 3 is possible if 5. From statement (4)
CM and OM lie on the same
straight line.

Hence, perpendicular bisector of a chord passes through the centre of the circle.

Theorem 13.4

Equal chords of a circle are equidistant from the centre.

Experimental Verification

Draw two circles with different radii with centre O. Also draw equal chords AB and
CD in each circle. From the centre O, draw OM and ON perpendicular to AB and
CD in each circle respectively.

AC AC

M ON O N
M

BD

To verify : OM = ON Fig. I BD
Fig. II

Now measure the length of OM and ON with the help of ruler and tabulate them as
follows.

Observation:

Figure OM ON Remarks

(i) ........... cm ........... cm OM = ON

(ii) ........... cm ........... cm OM = ON

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

Theoretical Proof: AC

Given : O is the centre of a circle, AB = CD, OM⊥AB and ON⊥CD. M N
To prove : OM = ON.
O

Construction : Join A and O, C and O. BD

242 Oasis School Mathematics-9

Proof :

Statements Reasons

1. AM = 1 AB 1. The perpendicular drawn from centre of circle
2 bisects the chord

2. CN = 1 CD 2. Same as statement 1
2

3. AM = CN 3. Being AB = CD and from St. no. (1) and (2)

4. In ∆AMO and ∆CNO 4.

(i) ∠AMO = ∠CNO (R) (i) Being both right angles.

(ii) AO = CO (H) (ii) Radii of the same circle.

(iii) AM = CN (S) (iii) From statement (3).

5. ∆AMO ≅ ∆CNO 5. By R.H.S. axiom.

6. ∴ OM = ON 6. Corresponding sides of congruent triangles.

Hence, equal chords of a circle are equidistant from the centre of a circle. Hence,
proved.

Theorem 13.5

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

Experimental verification:

Draw two circles with centre O with different radii. Also draw two perpendiculars
OM and ON from centre O to AB and CD in each circle.

M B
A
AC

M ON O

BD C D
Fig. I N

Fig. II

To verify : AB = CD

Now measure the length of AB and CD with the help of ruler and tabulate them as
follows.

Table.

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

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

Conclusion: Hence, chords which are equidistant from the centre of a circle
are equal.

Oasis School Mathematics-9 243

Theoretical Proof: AC
Given : O is the centre of circle. AB and CD are two
MN
chords OM⊥AB, ON ⊥ CD and OM = ON. O
To prove : AB = CD.
Construction : Join A and O, C and O. BD

Proof

Statements Reasons
1.
1. In ∆ AMO and ∆CNO (i) Being both 90º
(ii) Radii of the same circles.
(i) ∠AMO =∠CNO (R) (iii) Given
2. By R.H.S. axiom.
(ii) AO = OC (H) 3. Corresponding sides of congruent triangles.
4. The perpendicular drawn from centre of circle
(iii) OM = ON (S)
bisects the chord.
2. ∆AMO ≅ ∆CNO 5. Same as (4)

3. AM = CN 6. From (3) , (4) and (5)

4. AM = 1 AB
2

5. CN = 1 CD
2

6. AB = CD

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

Worked Out Examples

Example: 1

The radius of a circle is 5cm and the length of a chord is 8 cm. Calculate the distance of
the chord from the centre.

Solution:

Let O be the centre of the circle, AB be the chord and OM be the O
distance between the centre and the chord. Here, AB = 8cm, radius
(OA) = 5cm
Since, OM ⊥ AB A MB

So, AM = 1 AB [Perpendicular drawn from the centre of the
2 circle bisects the chord]

= 21 × 8 = 4 cm
In right angled ∆OMA

OA2 = OM2 + AM2

244 Oasis School Mathematics-9