Vedanta Excel in Mathematics Book 9 Final (2078)

Geometry - Similarity

P

c) In the adjoining figure, ∆PQR ~ ∆ABR, QR = 20 cm, A
B
QB = 8 cm, AB = 6 cm and PR = 15 cm. Find the

length of PQ and AR. Q R

10 cm P
R
d) In the adjoining figure, ∆PEM ~ ∆REM, ‘EPM = ‘RME. 8 cm
Find the length of EM.

EM

A

e) In the given figure, ∆ABC ~ ∆ABD, ‘BAD = ‘ACB, 4 cm
find the length of BD.
D8 cm C
B

2. In the givn similar polygons, AB = 6 cm, D C S
PQ = 3 cm, AC = 8 cm, BF = 9 cm and E T
PS = 4.5 cm. If ‘A = ‘P and ‘B = ‘Q, find the
lengths of PR and AD and QU. 8 cm R
4.5 cm
F 9 cm B
UQ
6 cm P 3 cm
A

Creative section - A DA
x cm
3. a) In the figure alongside AB // DE. Show that
∆ABC ~ ∆DCE and find the value of x. C
9 cm B
E
10 cm
5 cm

A6 cm 10 cm
8Dcm
b) In the adjoining figure, ‘BAD = ‘ACB. Prove that C
∆ABC ~ ∆ABD and find the lengths of AD and BD.

B

c) In the figure alongside, ‘ABC = ‘ACP. Prove that P
∆ABC ~ ∆APC and find the lengths of AB and PB.

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Geometry - Similarity

d) In the given figure, AX A BC. Prove that ∆ABC ~ ∆AXC 3 cm A 4 cm
and find the lengths of CX and AX.
X C
B A

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

reasons. B C

4. a) A rectangular ground is 40 ft. long and 30 ft. wide. If the lengths of the map of the
ground is 8 cm, find the width of the ground in the map.

b) In a photo frame, the outer breadth is treble than that of inner breadth. If the outer
length of the frame is 18 cm, find the inner length of the frame.

c) A boy of height 1.2 m is standing in front of the tree of height 19.5 m. What will
be the length of the shadow of the boy when the length of the shadow of the tree is
32.5 m?

Creative section - B

5. a) In the given figure, AB // CD. Prove that
(i) ∆AOB ~ ∆COD
(ii) AB.DO = CD.AO

b) In the adjoining figure, ‘P = ‘S. P S
Prove that R
(i) ∆POR ~ ∆QOS Q
(ii) PR.OQ = QS.OR A
(iii) PO.OQ = OR.OS
M
c) In the given right angled triangle ABC, ‘ABC = 90° and N
NM A AC. Prove that B

(i) ∆ABC ~ ∆AMN C
(ii) BC.AM = AB.MN

d) In the adjoining figure, ‘PQS = ‘PRQ. Prove that
(i) ∆PQS ~ ∆PQR
(ii) PQ2 = PS.PR

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Geometry - Similarity

e) In the given right-angled triangle ABC right-angled at B,
BD A AC and ‘CBD = ‘BAC. Prove that

(i) ∆ABC ~ ∆BCD (ii) ∆BCD ~ ∆ABD

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

f) In the trapezium ABCD, AB // DC and the diagonals AC
and BD intersect each other at O.
Prove that AB.DO = CD.BO

A B
DE
g) In the adjoining figure, ‘ADE = ‘ ACB, and
‘DAC = ‘BAE. Prove that AD.BC = AC.DE C

6. a) In the given figure, AB = DC, AB // DC, M is the
mid-point of AB. Prove that
(i) 'AOM ~ 'COD (ii) CO = 2AO

b) In the adjoining diagram, AD // BC and AD = BC. AM E
ND
Prove that
C
(i) 'EBC ~ 'EMN (ii) BE.MN = AD.ME D

B F
CE
A

c) In the given figure, ABCD is a parallelogram. Prove
that
(i) 'ADF ~ 'CEF
(ii) BC.EF = AF.CE

B

12.3 Pythagoras Theorem

The given right angled triangle, ABC is right angled at
B. The side AC opposite to the right angle is called its
hypotenuse. Similarly, the sides AB and BC are called
the perpendicular and the base respectively.

Pythagoras was a Greek Mathematician who lived
around 500 B.C. and discovered a significant fact about
right–angled triangles known as ‘Pythagoras Theorem’.

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Geometry - Similarity

Theorem 12

The square of the hypotenuse of a right–angled triangle is equal to the sum of the squares
of the remaining two sides.

Experimental verification
Step1: Three right–angled triangle ABC of different shapes and sizes and right angled at B

are drawn.

Step 2: The length of the sides of each triangle are measured and tabulated in the table.

Fig. AB BC CA AB2 BC2 CA2 AB2 + BC2 Result

(i) CA2 = AB2 + BC2

(ii) CA2 = AB2 + BC2

(iii) CA2 = AB2 + BC2

Conclusion: The square of the hypotenuse of a right–angled triangle is equal to the sum of
the squares of the remaining two sides.

Theoretical proof

Given: ∆ ABC is a right–angled triangle in which
‘ABC = 90°.

To prove: CA2 = AB2 + BC2

Construction: BD A AC is drawn.

Proof

Statements Reasons
1.
1. In ∆s ABC and BCD (i) Both of them are right angles.
(ii) Common angles
(i) ‘ABC = ‘BDC (A) (iii) Remaining angles of the triangles
(ii) ‘BCA = ‘BCD (A) (iv) A.A.A. axiom
(iii) ‘BAC = ‘DBC (A) 2. Corresponding sides of similar triangles
(iv) ? ∆ABC a ∆BCD
3. Same as above
2. oCBrAC, B=C2 BC 4. Corresponding sides of similar triangles
C=DCA.CD
5. Adding the statements (2) and (4)
3. ∆ABC a ∆ABD
Proved
4. oCArAB, A=B2 AB
A=DCA.AD

5. AB2 + BC2 = CA.AD + CA.CD

or, AB2 + BC2 = CA (AD + CD)

or, AB2 + BC2 = CA.CA

or, AB2 + BC2 = CA2

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Geometry - Similarity

12.4 Pythagorean Triple

Let h, p and b are the hypotenuse, perpendicular, and base of a right–angled triangle
respectively.
From Pythagoras theorem,

h2 = p2 + b2

Suppose, h = 5, p = 4 and b = 3,
52 = 42 + 32

or, 25 = 16 + 9
or, 25 = 25

? The relation h2 = p2 + b2 is satisfied by the particular values of h, p and b, i.e., h = 5,
p = 4, and b = 3.

On the other hand, let’s take any other values of h, p, and b, say h = 8, p = 4,
b= 5,

82 = 42 + 52
or, 64 = 16 + 25
or, 64 z 41

Thus, the relation h2 = p2 + b2 is not satisfied by h = 8, p = 4 and b = 5.

In this way, to satisfy the Pythagoras theorem, the hypotenuse (h), perpendicular
(p) and the base (b) of a right–angled triangle should have the particular values in
order. These values of h, p, and b are called Pythagorean Triples.

We can use the following expressions to find the Pythagorean Triple.

2n + 1 Ÿ gives the value of perpendicular (or base)

2n (n + 1) Ÿ gives the value of base (or perpendicular)

2n (n + 1) + 1 Ÿ gives the value of hypotenuse

Where 'n' is the natural number.

Now, let's use the above expressions to find a few sets of Pythagorean Triples.

n p = (2n + 1) b = 2n (n + 1) h = 2n (n + 1) + 1 Result
13 4 5 p2 + b2 = h2
25 12 13 p2 + b2 = h2
37 24 25 p2 + b2 = h2

and so on.

When a set of Pythagorean triple is known, we can find another set of triple by an
alternative process. In this process, each number of a triple is multiplied by any
other same natural number to get another set of triple.

For example,

3, 4 and 5 is a triple, then 3 × 2 = 6, 4 × 2 = 8 and 5 × 2 = 10 is another set of
triple.

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Geometry - Similarity

Worked-out examples

Example 1: In the adjoining figure, find the unknown y 8 cm
length of the sides of the triangle. 3.6 cm

Solution:
Here, CD = BC – BD = 10 cm – 3.6 cm = 6.4 cm

In right angled ∆ACD, using Pythagoras Theorem,

x = AD = AC2 – CD2 = 82 – 6.42 = 23.04 = 4.8 cm

Again in right angled ∆ ABC, using Pythagoras Theorem,
y = AB = BC2 – AC2 = 102 – 82 = 100 – 64 = 36 = 6 cm

Hence, x = 4.8 cm and y = 6 cm.

Example 2: In the figure alongside, show that ‘ BAC = 1
right angle.

Solution:

In right angled ∆ADC, using Pythagoras Theorem,

AC = AD2 + DC2

= 32 + 42 = 9 + 16 = 25 = 5 cm

Now, in ∆ ABC,
AB2 + AC2 = 122 + 52 = 144 + 25 = 169 sq. cm
BC2 = (13)2 = 169 sq. cm.
? BC2 = AB2 + AC2

? Pythagoras theorem is satisfied by the sides of ∆ ABC being the side BC as hypotenuse.

Here, the opposite angle of the hypotenuse BC is ‘BAC.
? ‘BAC = 1 right angle. Proved.

EXERCISE 12.2
General section

1. a) In a right-angled triangle ABC, right angled at B, name the hypotenuse, perpendicular,
and the base.

b) In a right-angled triangle PQR, right angled at R, what are the sides PQ, PR, and QR
called?

c) In a right-angled triangle XYZ, right angled at X, write the relation between its sides
using Pythagorus Theorem.

2. a) Test for Pythagoras theorem (h2 = p2 + b2) and list the Pythagorean triple.

(i) 3, 4, 5 (ii) 5, 8, 10 (iii) 6, 8, 10 (iv) 9, 12, 15

(v) 8, 10, 14 (vi) 9, 16, 20 (vii) 8, 15, 17 (viii) 4.5, 6, 7.5

b) (2n + 1), 2n(n + 1) and [2n(n + 1) + 1] are the sides of a triangle ABC. Substitute
n = 4 and prove that ∆ ABC is a right–angled triangle.

c) From a Pythagorean triple 3, 4, and 5, obtain three more consecutive sets of triples.

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Geometry - Similarity

3. Calculate the unknown lengths of the sides of these triangles.

a) X b) P 3 cm S c) A

12 cm x cm y cm x cm 17 cm
10 cm

YP Z Q 13 cm R B 6 cm D y cm C

Creative section E
R
M

4. a) In the adjoining figure, show that ‘MPR = 1 right angle.

P 4 cm C

D3 cm
13 cm
b) In the given figure, show that ∆ABC is a right angled triangle.

A
12 cm

X B
Y
c) In the right angled triangle given alongside,
XZ : XY = 3 : 4 and YZ = 16 cm. Find XZ and XY.

Z

5. a) In the given figure, ABCD is a square.
Show that BD = AB 2

b) In the adjoining right angled triangle ABC, D is any point
on AB. Prove that AB2 – AD2 = BC2 – CD2

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

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Geometry - Similarity

d) In the adjoining figure, ‘PQR = 90° and QS A PR. Prove
that PQ2 + RS2 = QR2 + PS2

R Q
X
e) In the given figure, X is the mid-point of PQ. Prove
that PQ2 = 4 (RX2 – PR2)

P

6. a) If x be the length of a side of an equilateral triangle, show that the altitude of the
triangle is x 2 3 .

b) ABC is an isosceles triangle in which AB = AC = 2 BC. If AD is an altitude of the
triangle, prove that 4 AD2 = 15 BC2.

7. a) A ladder 10 m long is inclined on a wall and touches the wall at the height of
8 m from the ground. Calculate the distance between the wall and the foot of the
ladder.

b) Two poles of heights 9 m and 14 m respectively stand vertically on the ground. If the
distance between their feet is 12 m, find the distance between their tops.

Project work

8. a) Select the smallest right-angled triangle in the middle of the following diagram and
draw its outline using a ruler. Draw the outlines of squares standing on each side of
the right-angled triangle. Count the number of triangle formed inside each square.
Then verify that h2 = p2 + b2.

b) Make a group of your friends. Take a Pythagorean triple of your own choice. Draw the
right angled triangle using the measurement of your triple and complete the squares
along each side. Draw the square grids of same size in each square and verify the
Pythagoras theorem by counting the numbers of square boxes.

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Unit Geometry - Parallelogram

13

13.1 Special types of Quadrilaterals

The quadrilateral ABCD given alongside is a closed plane figure Q
bounded by four line segments called the sides of the quadrilateral. P
AC is a diagonal of the quadrilateral.

DP A AC and BQ A AC are drawn.

Therefore, DP = h1, and BQ = h2 are the heights of 'ACD and 'ABC respectively.
Here, area of quadrilateral = Area of ('ACD + 'ABC)

= 1 AC × h1 + 1 AC × h2 = 1 AC (h1 + h2)
2 2 2

Thus, area of quadrilateral = 1 diagonal × sum of perpendiculars drawn from
2

opposite vertices to the diagonal

Trapezium, parallelogram, rectangle, square, rhombus, and kite are the special types

of quadrilaterals.

Trapezium

The quadrilateral in which any two opposite sides are
parallel is called a trapezium. In the adjoining figure, PQRS
is a trapezium in which PQ // SR.

Properties of trapezium

(i) The parallel sides of a trapezium are called its bases. In the figure, PQ and SR

are the bases.

(ii) The non-parallel sides of a trapezium are called its legs. In the figure, PS and

QR are the legs.

(iii) The straight line segment that joins the mid-points of legs is called the median

of the trapezium. XY is the median.

(iv) The perpendicular distance between the parallel sides (bases) of a trapezium is

called its altitude (height). h is the height of the trapezium.

(v) Area of trapezium = 1 Height × Sum of the bases = 1 h (PQ + SR)
2 2

Parallelogram

The quadrilateral in which opposite sides are parallel
is called a parallelogram. In the figure, ABCD is a
parallelogram.

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Geometry - Parallelogram

(i) The opposite sides of a parallelogram are equal. h
(ii) The opposite angles of a parallelogram are equal. b
(iii) The diagonals of a parallelogram bisect each other.
(iv) The area of a parallelogram = base × height = b × h.

Rectangle, square, and rhombus are some special types of parallelograms. Let's
discuss the important properties of these special types of parallelograms.

Rectangle

A rectangle is a parallelogram in which each angle is 90°.
(i) The opposite sides of a rectangle are equal.
(ii) The diagonals of a rectangle are equal.
(iii) The diagonals of a rectangle bisect each other.
(iv) The area of a rectangle = length × breadth = l × b.

Square

A parallelogram in which all sides are equal and each angle is 90° is called a square.

(i) The diagonals of a square are equal.

(ii) Each diagonal bisects the vertical angles.

(iii) Diagonal bisect each other perpendicularly.

(iv) The triangles formed by the diagonals are congruent. l

(v) The area of a square = (side)2 = l2 or = 1 (diagonal)2
2

Rhombus

A rhombus is a parallelogram in which all sides are equal.

(i) The opposite angles of a rhombus are equal. d1
d2
(ii) Each diagonal bisects the vertical angles.

(iii) The diagonals are not equal and bisect each other

perpendicularly.

(iv) The triangles formed by the diagonals are congruent.

(v) The area of a rhombus = 1 × product of diagonals
2

= 1 × d1 × d2
2

Theorem 13

The straight line segments that join the ends of two equal and parallel line segments
towards the same sides are also equal and parallel.

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Geometry - Parallelogram

Given: AB = CD and AB // CD. The ends A, C and B, D
are joined.

To prove: AC = BD
AC // BD

Construction: B and C are joined.

Proof

Statements Reasons
1. In ∆s ABC and BCD
1.

(i) AB = CD (S) (i) Given
(ii) AB // CD and alternate angles
(ii) ‘ABC = ‘BCD (A) (iii) Common side
(iv) S. A. S. axiom
(iii) BC = BC (S)

(iv) ? ∆ABC # ∆BCD

2. AC = BD 2. Corresponding sides of congruent triangles

3. ‘ACB = ‘CBD 3. Corresponding angles of congruent triangles

4. AC // BD 4. Alternate angles being equal

Proved

Theorem 14

The straight line segments that join the ends of two equal and parallel line segments
towards the opposite sides bisect each other.

Given: AB = CD and AB // CD. The opposite ends A, D and
B, C are joined. Let, AD and BC intersect at O.

To prove: AD and BC bisect each other at O.
i.e. AO = OD and BO = OC

Proof

Statements Reasons

1. In ∆s AOB and COD 1.
(i) AB // CD and alternate angles
(i) ‘ABO = ‘OCD (A) (ii) Given
(iii) AB // CD and alternate angles
(ii) AB = CD (S) (iv) A. S. A. axiom

(iii) ‘BAO = ‘ODC (A)

(iv) ? ∆AOB # ∆COD

2. AO = OD and BO = OC 2. Corresponding sides of congruent triangles

3. AD and BC bisect each other at O. 3. From statement (2)

Proved

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Geometry - Parallelogram

Theorem 15

The opposite angles and sides of a parallelogram are equal.

Given: ABCD is a parallelogram in which AB // DC and
AD // BC.

To prove: (i) ‘ABC = ‘ADC, ‘BAD = ‘BCD
(ii) AB = DC, AD = BC

Construction: Diagonal AC is drawn.

Proof

Statements Reasons
1.
1. In ∆s ABC and ACD (i) AB // DC and alternate angles

(i) ‘BAC = ‘ACD (A) (ii) Common side

(ii) AC = AC (S) (iii) AD // BC and alternate angles

(iii) ‘ACB = ‘CAD (A) (iv) A. S. A. axiom
2. Corresponding angles of congruent triangles
(iv) ? ∆ABC # ∆ACD 3. Drawing the diagonal BD and same as above

2. ‘ABC = ‘ADC in ∆s ABD and BCD
4. Corresponding sides of congruent triangles
3. ‘BAD = ‘BCD
Proved.
4. AB = DC and AD = BC

Converse (I) Theorem 15

If the opposite sides of a quadrilateral are equal, the quadrilateral is a parallelogram.
Given: ABCD is a quadrilateral in which AB = DC and

AD =BC.
To prove: ABCD is a parallelogram, i.e. AB // DC and AD // BC.
Construction: Diagonal AC is drawn.
Proof

Statements Reasons
1.
1. In ∆s ABC and ACD (i) Given
(ii) Given
(i) AB = DC (S) (iii) Common side
(iv) S. S. S. axiom
(ii) BC = AD (S) 2. Corresponding angles of congruent triangles

(iii) AC = AC (S) 3. From statement (2), alternate angles being
equal
(iv) ? ∆ABC # ∆ACD
4. Opposite sides are parallel
2. ‘BAC = ‘ACD and Proved

‘ACB = ‘CAD

3. AB // DC and AD // BC

4. ABCD is a parallelogram

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Geometry - Parallelogram

Converse (II) of Theorem 15

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

Given: ABCD is a quadrilateral in which ‘ABC = ‘ADC and
‘BCD = ‘BAD.

To prove: ABCD is a parallelogram, i.e., AB // DC and AD // BC.

Proof

Statements Reasons

1. ‘A + ‘B + ‘C + ‘D = 360° 1. Sum of the angles of a quadrilateral
2. ‘A + ‘B + ‘A + ‘B = 360q 2. Given, ‘A = ‘C and ‘B = ‘D.

or, 2 (‘A + ‘B) = 360°

or, ‘A + ‘B = 180° 3. Being the sum of co-interior angles 180°
3. ? AD // BC 4. Same as above
4. Similarly, ‘A + ‘D = 180° 5. Same as reason (3)
5. ? AB // DC 6. Opposite sides are parallel
6. ABCD is a parallelogram

Theorem 16 Proved

The diagonals of a parallelogram bisect each other.

Given: ABCD is a parallelogram. Diagonals AC and BD intersect
at O.

To prove: AC and BD bisect each other at O, i.e., AO = OC and
BO = OD.

Proof

Statements Reasons
1.
1. In ∆s AOB and COD (i) AB // DC and alternate angles
(ii) Opposite sides of a parallelogram
(i) ‘OAB = ‘OCD (A) (iii) AB // DC and alternate angles
(iv) A. S. A. axiom
(ii) AB = DC (S) 2. Corresponding sides of congruent triangle.
3. From statement (2)
(iii) ‘OBA = ‘ODC (A)
Proved
(iv) ? ∆AOB # ∆COD
2. AO = OC and BO = OD
3. AC and BD bisect each other at O.

Converse of Theorem 16

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

Given: ABCD is a quadrilateral in which diagonals AC and
BD bisect each other at O.
? AO = OC and BO = OD

To prove: ABCD is a parallelogram, i.e., AB // DC, AD // BC,
AB = DC, AD = BC.

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Geometry - Parallelogram

Proof

Statements Reasons
1. In ∆s AOB and COD
1.
(i) AO = OC (S) (i) Given
(ii) Vertically opposite angles
(ii) ‘AOB = ‘COD (A) (iii) Given
(iv) S. A. S. axiom
(iii) BO = OD (S) 2. Corresponding sides of congruent triangle
3. Corresponding angles of congruent triangles
(iv) ? ∆ AOB # ∆ COD 4. From statements (3), alternate angles are equal
2. AB = DC 5. AD and BC join the ends of two equal and parallel
3. ‘OAB = ‘OCD
4. AB // DC lines towards the same side.
5. AD = BC and AD // BC 6. Opposite sides are equal and parallel.

6. ABCD is a parallelogram Proved

Worked-out examples

Example 1: In the given figure, PQRS is a parallelogram. P S
Solution: If PT = SR, ‘ PSR = 4x° and ‘ QPT = x°, find the x° 4x°
value of ‘ QRS.
Q R
T

(i) ‘PQR = ‘PSR = 4x° [Opposite angles of parallelograms are equal]

(ii) PQ = PT [ PQ = SR and PT = SR]

(iii) ‘PQT + ‘PTQ = 4x° [ PQ = PT]
(iv) ‘PQT + ‘PTQ + ‘QPT = 180° [Sum of angles of 'PQT]

or, 4x° + 4x° + x° = 180°

or, 9x° = 180°

? x = 20°

(v) ‘QRS = 180° – ‘PQR [ PQ // SR and co-interior angles]

= 180° – 4 × 20° = 100° A D
35° C
Example 2: In the figure alongside, ABCD is a rhombus.
If ‘DAC = 35°, find the measure of ‘ABC. B

Solution:

(i) ‘DAC = ‘ACD = 35° [AD = CD, sides of rhombus]
(ii) ‘ADC + ‘DAC + ‘ACD = 180° [Sum of angles of 'ADC]
or, ‘ADC + 35° + 35° = 180°

? ‘ADC = 110°
(iii) ‘ABC = ‘ADC [Opposite angles of rhombus]
? ‘ABC = 110°

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Geometry - Parallelogram

Example 3: In the figure alongside, WXYZ is a square. ZY
If ‘ZNX = 115°, find the sizes of ‘WZN and ‘ZMY.
M
Solution: 115°

(i) ‘WZN + ‘ZWN = ‘ZNX [‘ZNX is the exterior angle in 'ZWN] W NX
or, ‘WZN + 90° = 115°

? ‘WZN = 25°

(ii) ‘ZWY = ‘XWY=45° [Diagonal of square bisects its vertical angle]

(iii) ‘ZMY = ‘WZM+‘ZWM [In 'WZM, ‘ZMY is the exterior angle]

? ‘ZMY = 25° + 45° = 70°

Example 4: P, Q, R and S are the mid–points of the sides AB, BC,
CD and DA of the quadrilateral ABCD respectively.
Prove that PQRS is a parallelogram.

Solution:

Given: P, Q, R and S are the mid–points of the sides AB, BC,
CD and DA of the quadrilateral ABCD respectively.

To prove: PQRS is a parallelogram.

Construction: Diagonal AC of the quadrilateral ABCD is drawn.

Proof

Statements Reasons

1. In ∆ABC, PQ // AC and PQ = AC 1. PQ joins the mid–points of the sides AB and
BC.

2. In ∆ACD, SR // AC and SR = AC 2. SR joins the mid–points of the sides AD and
DC.

3. ? PQ // SR and PQ = SR 3. From the statements (1) and (2)
4. SP // RQ and SP = RQ 4. From the statement 3.
5. PQRS is a parallelogram 5. Being opposite sides equal and parallel

Proved

Example 5: In the given figure, PQRS is a parallelogram in which P TS
QT is the bisector of ‘PQR. Prove that PT = SR. Q R
Solution:
Given: In parallelogram PQRS,QT is the bisector of ‘PQR.
To prove: PT = SR.
Proof

Statements Reasons
1. ‘PQT = ‘TQR 1. QT being bisector of ‘PQR.

2. ‘PTQ = ‘TQR 2. PS // QR and alternate angles.
3. ‘PQT = ‘PTQ 3. From the statements (1) and (2)
4. PQ = PT 4. From the statement 3.
5. PQ = SR 5. Opposite sides of parallelogram
6. PT = SR 6. From statements (4) and (5).

Proved

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Geometry - Parallelogram

Example 6: In the figure, ABCD is a parallelogram. The A Q
D
Solution: diagonal BD is produced in either sides to
Given: C
To prove: the points P and Q such that BP = DQ. Prove
Proof
that: B
(i) ‘BCP = ‘DAQ

(ii) AQ // PC. P

In parallelogram ABCD, the diagonal BD is produced to the points P and Q
such that BP = DQ.

(i) ‘BCP = ‘DAQ (ii) AQ // PC.

Statements Reasons
1. AD // BC, alternate angles
1. ‘DBC = ‘ADB 2. Supplements of equal angles, ‘DBC = ‘ADB
3.
2. ‘PBC = ‘ADQ (i) Given
3. In 'BPC and 'ADQ (ii) From statement(2)
(i) BP = DQ (S) (iii) Opposite sides of parallelogram ABCD
(ii) ‘PBC = ‘ADQ (A) (iv) By S.A.S, axiom
(iii) BC = AD (S) 4. Corresponding angles of congruent triangles
(iv) 'BPC # 'ADQ
4. ‘BCP = ‘DAQ and 6. From statements (4), alternate angles ‘BPC
and ‘AQD are equal.
‘BPC = ‘AQD Proved
5. AQ // PC

Example 7: In the given parallelogram ABCD, M and N are the A M B
mid-points of sides AB and DC respectively. Prove X C
Solution: that (i) ANCM is a parallelogram and Y
Given: D N
To prove: (ii) BX = XY = DY.
Proof
In parallelogram ABCD, M and N are the mid-points of sides AB and DC
respectively.

(i) ANCM is a parallelogram (ii) BX = XY = DY.

Statements Reasons
1. AM = NC and AM // NC 1. AB = DC, AB // DC and given

2. MC = AN and MC // AN 2. From statement (1)
3. ANCM is a parallelogram 3. From statements (1) and (2)
4. BX = XY 4. In 'BAY, BM = MA and MX // AY
5. DY = XY 5. In 'DCX, DN = NC and YN // XC
6. BX = XY = DY 6. From statements (4) and (5)

Proved

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Geometry - Parallelogram

Example 8: In a trapezium ABCD, if AB // DC and AD = BC, A B
prove that (i) ‘ADC = ‘BCD and (ii) AC = BD. D EC

Solution: In trapezium ABCD, AB // DC and AD = BC.
Given:

To prove: (i) ‘ADC = ‘BCD (ii) AC = BD

Construction: BE // AD is drawn to meet DC at E.

Proof

Statements Reasons
1. ABED is a parallelogram 1. By construction and given
2. AD = BC and AD = BE
2. AD = BE = BC 3. Base angles of the isosceles 'BEC
3. ‘BEC = ‘BCE
4. AD // BE, corresponding angles
i.e. ‘BEC = ‘BCD 5. From statements (3) and (4)
4. ‘ADC = ‘BEC 6.
5. ‘ADC = ‘BCD (i) Given
6. In 'ADC and 'BCD (ii) From statement (5)
(i) AD = BC (S) (iii) Common side
(iv) By S.A.S. axiom
(ii) ‘ADC = ‘BCD (A) 7. Corresponding sides of congruent triangles
(iii) DC = DC (S)
(iv) 'ADC # 'BCD Proved
7. AC = BD

Example 9: In the figure alongside, ABCD is a parallelogram.
If 2MO = OD, prove that M is the mid-point of BC.

Solution:

Given: ABCD is a parallelogram in which BC // AD and
To prove: AB // DC, 2 MO = OD
M is the mid-point of BC, i.e. BC = 2 MC

Proof:

Statements Reasons

1. In ∆ AOD and ∆ MOC 1.
(i) (i) AD // BC and alternate angles
(ii) ‘ADO = ‘CMO (ii) Vertically opposite angles
(iii) (iii) AD // BC and alternate angles
(iv) ‘AOD = ‘COM (iv) A.A.A. axiom

2. (i) ‘DAO = ‘MCO 2. (i) Corresponding sides of similar triangles

∆AOD ~ ∆MOC

AD = OD = AO
MC MO CO

(ii) BC = 2MO (ii) AD = BC (opposite sides of parallelogram)
MC MO and OD = 2MO (given)

or, BC = 2 MC (iii) From statement (ii)
(iii) M is the mid-point of BC

Proved

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 217 Vedanta Excel in Mathematics - Book 9

Geometry - Parallelogram

EXERCISE 13.1 A B
General section C

1. a) In the figure alongside, AB // DC and AB = DC. Write down the Q
relation between AD and BC.

D
P

b) In the given figure, PQ // RS and PQ = RS. What is the relation

between PS and QR? R S
T
B
S
c) In the figure alongside, BEST is a parallelogram. Write down Y
the relation between BT and ES, BE, and TS.

E
Z

d) In the parallelogram WXYZ, diagonals, ZX and WY intersect O X
at O. Write the relation between OW, OY and OZ, OX. W D

A O C
P S
2. a) In the given figure, ABCD is a rectangle. AC and BD are

diagonals. If the length of diagonal AC is 10 cm, what is the

length of diagonal BD? B

b) In the square PQRS given alongside, what is the value of ‘QPR?

Q R
R D

c) In the given figure, READ is a rhombus. If ‘RED = 40°, what is the A
measure of ‘RDE? Z

E Y

d) In the given rectangle WXYZ, O is the point of intersection W

of its diagonal WY and XZ. If ‘XOY = 120°, find the size of O

‘OWZ. X

3. Calculate the size of unknown angles:

a) A yD b) P Q c) L p+q 2p–50° E d) A D
5y x
3x a 60°

2x zw b 4y p+10° r O
B C ES R I K 2x

e) L 115° B CE

x D f) D E F g) P T h) A F
x 110° b x y
y A 100° S
ET
55° 50° E
D

aa 60°
CQ BC
A R
B

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4. a) In the given figure, ABCD is a parallelogram. Geometry - Parallelogram
If ‘BCD = 125° and AB = AE, find the size of
‘BAE. 110°

b) In the figure alongside, DBCE is a parallelogram.
If ‘AFE =110° and ‘FCE = 40°, find the value
of ‘DBC.

c) In the adjoining figure, PQRS is a parallelogram. H ND25°
Find the value of ‘SMR. A
M
d) In the given figure, HEAD is a parallelogram. If E
AM A HE, AN A HD and ‘NAD = 25°, find the
measure of ‘MAE and ‘MAN.

5. a) In the given figure, ABCD is a rhombus. If ‘ADB = 40°,
find the size of ‘CDE.

b) In the given figure, ABCD is a rhombus. If
‘BCD = 114°, find the measure of ‘ABD.

c) In the adjoining figure, ABCD is a rhombus. If CD
‘OAB = 30°, find the measures of ‘ODC and
‘ABC. O
30°
d) In the given figure, PQRS is a rectangle.
If ‘POS = 124°, find the sizes of ‘PQO and BA
‘OQR. PS

124°
O

QR

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Geometry - Parallelogram

e) In the figure alongside, PQRS is a square. If ‘SMR = 70°, SR
find the measures of ‘PST and ‘STQ.
70° Q
f) In the given figure, FIVE is a square. If PQ // FV and M E
‘QPR = 75°, find the measure of ‘FIR and ‘FPI. Q
PT
P FP V

75°
R

M

I
S

g) In the given figure, PQRS is a rhombus and 55° M
SRM is an equilateral triangle. If SN A RM and RN
‘PRS = 55°, find the size of ‘QSN.

Q

6. a) In the given figure, ABCD is a parallelogram. If AB = 3x cm, 3x cm A 19 cm D

BC = (5y – 1) cm, CD = 12 cm, and AD = 19 cm, find the 12 cm

values of x and y. B (5y – 1) C

b) In the figure alongside PQRS is a parallelogram. If (x – y) cmP (x + y) cm S 10 cm
PQ = (x – y) cm, QR = 20 cm, RS = 10 cm, and Q 20 cm R
PS = (x + y) cm, find the value of x and y.

L P
5cm
c) In the adjoining figure, LMNP is a rhombus. If
MN = (2a – 3) cm, OM = (3b + 2) cm, OP = 5 cm, and (3b+2)Ocm
perimeter = 28 cm, find the values of a and b.
M (2a – 3)cm N
d) In the adjoining figure, PQRS is a rectangle. If
OP = (2x – 3) cm and OR = (x + 1) cm, find the length S R
of diagonal QS. (x+1)cm Q

(2x–3)cm O
P

e) 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 3p cm C
and q.
(p+q)cmO
A

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Geometry - Parallelogram

D C

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

A A B
B E
b) In the given figure, ABCD is a square and BEC is
an equilateral triangle. Find ‘AEB and ‘DAE.

D C
P
Creative section -A Q
A
8. a) In the adjoining figure, PQ = AB and PQ // AB. B
D
Prove that (i) AP = BQ (ii) AP // BQ.
C
b) In the given quadrilateral, AB = DC and BC = AD. A C
Prove that the quadrilateral ABCD is a parallelogram. B
B
c) In the figure alongside, diagonals AC and BD of the D S
quadrilateral bisect each other at O. Prove that ABCD is a O
parallelogram. R

A

P

d) In the adjoining quadrilateral PQRS, ‘P = ‘R and ‘Q = ‘S.
Prove that PQRS is a parallelogram.

Q

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

E P M

b) In the given figure, EXAM is a parallelogram, the
bisector of ‘A meets the mid-point of EM at P. Prove that
AX = 2AM.

XA

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Geometry - Parallelogram AB
DC
c) In the adjoining figure, ABCD is a rhombus in which CD is
produced to E such that CD = DE. Prove that ‘EAC = 90°.

E

d) 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.

Creative section -B

10. a) 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.

b) 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.

A

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

QR
BD

d) 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.

11. a) In the given parallelogram PQRS, PA bisects ‘P and RB
bisects ‘R. Prove that PA // BR.

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Geometry - Parallelogram

b) ABCD is a parallelogram. P and Q are two points on the
diagonal BD such that DP = QB. Prove that APCQ is a
parallelogram.

c) ABCD is a parallelogram. DE A AC and BF A AC. Prove
that BEDF is a parallelogram.

P S
M
d) In parallelogram PQRS, the bisectors of ‘PQR and
‘PSR meet the diagonal at M and N respectively. Prove N
that MQNS is a parallelogram.

QR

e) In the given figure, ABCD is a parallelogram. If P and A D
Q are the points of trisection of diagonal BD, prove that Q
PAQC is a parallelogram.
P

BC

12. a) 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.

b) 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

c) 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.

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Geometry - Parallelogram

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

f) In the given quadrilateral ABCD, AD = BC, P and Q are the DN C

13. a) mid-points of the diagonals AC and BD respectively. If M and PQ
b) N are the mid-points of the sides DC and AB respectively,
c)
d) prove that PMQN is a rhombus. A M B
e) D C
f) (Hint: Apply mid-point theorem in 'BCD, 'ACD, 'ABD and ' ABC)

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

Prove that the diagonals of a rectangle are equal. A B

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

Prove that the diagonals of a rhombus bisect each other perpendicularly.

Prove that the diagonal of a parallelogram divides it into two congruent triangles.

If P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively of
a quadrilateral ABCD, prove that PQRS is also a parallelogram.

Prove that the line segments joining the mid-points of the diagonals of a trapezium
is parallel to parallel sides and equal to half their difference.

Project work

14. a) Take two rectangular sheets of paper of the same size. Fold one sheet through one
diagonal and another sheet through other diagonal. Then, cut out each sheet of
papers through diagonals.

(i) Are two diagonals equal?

(ii) Do these diagonals bisect each other?

b) Take a square sheet of paper and fold it through both diagonals.

(i) Are two diagonals equal?

(ii) Do these diagonals bisect each other perpendicularly?

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Unit Geometry - Circle

14

14.1 Circle and its various parts P6 P5 P4
P7 P3
A circle is a plane curved consisting of all points that have the O
same distance from a fixed point, called the centre. In the figure, P8 P1 P2
O is the centre of a circle.
Of course, when the number of sides of a polygon is increased B A
without a limit, the sides merge into one line and the polygon C
becomes a circle.
Let’s study about the various parts of a circle. O
D
(i) Circumference

The perimeter of a circle is called its circumference. It is
the total length of the curved line of the circle.

(ii) Radius

It is the line segment that joins the centre of a circle and
any point on its circumference. In the figure, OA is the
radius. The plural form of radius is radii. All the radii of a
circle are equal,
i.e., OA = OB = OC = OD = ...

(iii) Diameter R

A line segment that passes through the centre of a circle

and joins any two points on its circumference is called the

diameter of the circle. In the figure PQ, RS, etc., are the P OQ
diameters. The length of a diameter is two times radius.

So, PQ = 2OQ or 2OP and RS = 2OR or 2OS. A diameter S
divides a circle into two halves.

X

(iv) Semi-circle
A diameter divides a circle into two halves and each half is P
called the semi-circle. In the figure, PXQ and PYQ are the OQ

semi-circles.

(v) Arc Y
X
A part of the circumference of a circle is called an arc. It is

denoted by a symbol ‘ ‘. In the figure, PQR is the minor OQ
and PSR is major arcs of the circle. A minor arc is less than P

half of the circumference and a major arc is greater than

half of the circumference. Y

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Geometry - Circle Q
P
(vi) Chord
O
The line segment that joins any two points on the R
circumference of a circle is called the chord of the circle.
In the figure, PQ and RS are any two chords of the circle. S
Diameter is the longest chord of a circle.
PQ
(vii) Segment O

The region enclosed by a chord and corresponding arc is
called the segment of a circle. In the figure, the shaded
region is the minor segment and non-shaded region is the
major segment of the circle. A minor segment is less than
half of a circle whereas the major segment is greater than
half of the circle. The minor and the major segments of a
circle are known as the alternate segments to each other.

(viii) Sector O

The region enclosed between any two radii and the Q
corresponding arc of a circle is called a sector of the circle. P
In the figure, OPQ is a sector of a circle.
Q CZ R
(ix) Concentric circles Y
B
Two or more circles are said to be concentric circles if
they have the same centre but different radii. In the figure, P XAO
circles PQR, XYZ, and ABC are three concentric circles.

(x) Intersecting circles P

If two circles intersect each other at two points, they are O O'
said to be intersecting circles. In the figure, two circles Q
are intersecting each other at P and Q. PQ is the common
chord of these two intersecting circles.

(xi) Tangent to a circle P O Q
T
A line that intersects the circle exactly at one point is
called a tangent to the circle. The point at which the
tangent touches the circle is called the point of contact. In
the figure, PQ is the tangent and T is the point of contact.

(xii) Secant of a circle A BQ
P O
A line that intersects a circle in two distinct points is
called a secant of the circle. In the figure, PQ is the secant
of the circle.

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Geometry - Circle

14.2 Theorems related to chords of a circle

Theorem 17

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

Experimental verification
Step 1: Three circles with centre O and different radii are drawn.
Step 2: In each circle, a chord AB of different lengths is drawn.
Step 3: OP A AB is drawn in each circle.

A
A

O P P
O O
AP B
B B

(i) (ii) (iii)

Step 4: The lengths of AP and PB are measured and the results are tabulated.

Figure AP PB Result
(i) AP = PB
(ii) AP = PB
(iii) AP = PB

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

Theoretical proof

Given: O is the centre of a circle. AB is the chord of the circle O
and OP A AB.

To prove: AP = PB AP B

Construction: O, A and O, B are joined.
Proof

Statements Reasons
1.
1. In rt. ‘ed 's OAP and OBP (i) Both of them are right angles.
(i) ‘OPA = ‘OPB (R) (ii) Radii of the same circle
(iii) Common side
(ii) OA = OB (H) (iv) R.H.S. axiom
2. Corresponding sides of congruent triangle
(iii) OP = OP (S)
Proved
(iv) ?'OAP # 'OBP

2. AP = PB

i.e. OP bisects AB

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Geometry - Circle

Converse of Theorem 17

The line joining the mid-point of a chord and the centre of a circle is perpendicular to the
chord.

Experimental verification

Step 1: Three circles with centre O and different radii are drawn.

Step 2: In each circle chord AB of different lengths is drawn and the mid-point of the chord
is marked as P.

Step 3: O and P are joined. A

A

O P P
O O
AP B
B B

(i) (ii) (iii)

Step 4: ‘OPA and OPB are measured and the results are tabulated.

Figure ‘OPA ‘OPB Result
(i) ‘OPA = ‘OPB = 90q
(ii) ‘OPA = ‘OPB = 90q
(iii) ‘OPA = ‘OPB = 90q

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

Theoretical proof

Given: O is the centre of a circle and AB is the chord. P is the mid- O
point of AB. O and P are joined.

To prove: OP A AB AP B
Construction: O, A and O, B are joined.

Proof

Statements Reasons
1. In rt. ‘ed 's OAP and OBP
(i) OA = OB (S) 1.
(ii) AP = PB (S) (i) Radii of the same circle
(iii) OP = OP (S) (ii) Given (P is the mid-point of AB)
(iv) ? 'OAP # 'OBP (iii) Common side
(iv) S.S.S. axiom
2. ‘OPA = ‘OPB
2. Corresponding angles of congruent
triangle

3. ‘OPA = ‘OPB = 90q 3. Adjacent angles in linear pair are equal.

4. OP A AB 4. From the statement 3.

Proved

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Geometry - Circle

Theorem 18

Equal chords of a circle are equidistant from the centre.

Experimental verification
Step 1: Three circles with centre O and different radii are drawn.
Step 2: Two equal chords AB and CD of different lengths are drawn in each circle.
Step 3: OP A AB and OQ A CD are drawn in each circle.

AC A C C Q
Q A
PO Q P D
BD BO
O
(i) P

D B

(ii) (iii)

Step 4: The lengths of OP and OQ are measured and the result is tabulated.

Figure OP OQ Result
(i) OP = OQ
(ii) OP = OQ
(iii) OP = OQ

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

Theoretical proof AC

Given: O is the centre of a circle. Chords AB = CD and PO Q
B D
OP A AB, OQ A CD.

To prove: OP = OQ

Construction: O, A and O, C are joined.

Proof

Statements Reasons
1. Given
1. AB = CD 2. OP A AB and OP bisects AB; OQ A CD

2. 2 AP = 2 CQ and OQ bisects CD.

i.e. AP = CQ 3.
(i) Both of them are right angles.
3. In rt. ‘ed 's OAP = OCQ (ii) Radii of the same circle
(iii) From statements 2
(i) ‘OPA = ‘OQC (R) (iv) R.H.S. axiom
4. Corresponding sides of congruent
(ii) OA = OC (H)
triangles
(iii) AP = CQ (S)
Proved
(iv) ? 'OAP # 'OCQ

4. OP = OQ

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Geometry - Circle

Converse of Theorem 18

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

Experimental verification

Step 1: Three circles with centre O and different radii are drawn.

Step 2: Two lines OP and OQ are drawn from the centre O such that OP = OQ in each
circle.

Step 3: Two chords AB and CD perpendicular to OP and OQ respectively are drawn in each

circle.

A

AC CD

O AO O C
BD
B B
(i) (ii)
D
(iii)

Step 4: The lengths of chords AB and CD are measured and the results are tabulated.

Figure AB CD Result
(i) AB = CD
(ii) AB = CD
(iii) AB = CD

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

Theoretical proof A C

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

circle. OP A AB, OQ A CD and OP = OQ. POQ

To prove: AB = CD

Construction: O, A and O, C are joined. BD

Proof

Statements Reasons

1. In rt. ‘ed 's OAP and OCQ 1.
(i) Both of them are right angle
(i) ‘OPA = ‘OQC (R) (ii) Radii of the same circle
(iii) Given
(ii) OA = OC (H) (iv) R.H.S. axiom

(iii) OP = OQ (S) 2. Corresponding sides of congruent
(iv) ? 'OAP # 'OCQ triangles

2. AP = CQ 3. OP A AB and OP bisects AB; OQ A CD
and OQ bisects CD.
3. 2 AB = 2 CD
i.e. AB = CD Proved

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Geometry - Circle

Worked-out examples

Example 1: The radius of a circle is 10 cm and the length of a chord of the circle is
16 cm. Find the distance of the chord from the centre of the circle.

Solution:

Let O be the centre of the circle. AB be the chord of the circle such that AB O
= 16 cm. OP be the distance between the centre and the chord. P

Here, OP A AB and OP bisects AB. A B

? AP = 1 AB = 1 u 16 cm = 8 cm
2 2
Also, radius OA = 10 cm.

Now, in rt. ‘ed 'OAP, OP = OA2 – AP2 = 102 – 82 = 100 – 64 = 36 = 6 cm
So, the required distance of the chord from the centre is 6 cm.

Example 2: In the given figure O is the centre of the circle. If O N
Solution: OA A MN, MN = 6 cm and AX = 1 cm, find the length of
diameter of the circle. MA
X
Here, O is the centre of the circle, OA A MN.

MN = 6 cm and AX = 1 cm

? AM = 1 MN = 1 × 6 cm = 3 cm
2 2
Let, OA = x, then OX = OM = (x + 1) cm.

Now,

In rt. ‘ed 'OAM, OM2 = OA2 + AM2

or, (x + 1)2 = x2 + 32

or, x2 + 2x + 1 = x2 + 9

or, 2x = 8 ?x=4

? The length radius OM (r) = (x + 1) cm = (4 + 1) cm = 5 cm

and diameter (d) = 2r = 2 × 5 cm = 10 cm.

Example 3: In the adjoining figure AB and CD are two chords of a

circle such that AB = 6 cm, CD = 12 cm and AB // CD. If O
the distance between AB and CD is 3 cm, find the

radius of the circle. CP D
AQ B
Solution:

OP A CD is drawn and it is produced to meet AB at Q.
? OQ is also perpendicular to AB ( AB // CD)
Let the radius of the circle OA = OC = r and OP = x cm. So, OQ = (x + 3) cm

Also, CP = 1 of CD = 1 u 12 cm = 6 cm
2 2

AQ = 1 of AB = 1 u 6 cm = 3 cm
2 2

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Geometry - Circle

In rt. ‘ed 'OQA,
OA2 = OQ2 + AQ2 = (x + 3)2 + 32 = x2 + 6x + 9 + 9 = x2 + 6x + 18
In rt. ‘ed 'OPC,
OC2 = OP2 + CP2 = x2 + 62 = x2 + 36
As, OA = OC, OA2 = OC2
? x2 + 6x + 18 = x2 + 36
or, 6x = 18
or, x = 3
? OC2 = r2 = x2 + 36

r2 = 32 + 36

r = 45
= 6.71 cm

So, the radius of the circle is 6.71 cm.

Example 4: If a line PQ intersects two concentric circles at

the points A, B, C, and D as shown in the figure,

prove that AB = CD.
O
Solution:
Given: O is the centre of two concentric circles. Line PQ P A B R C D Q

intersects two concentric circles at A, B, C, and D.

To prove: AB = CD

Construction: OR A PQ is drawn.

Proof:

Statements Reasons
1. BR = RC 1. OR A BC and OR bisects BC.
2. AR = RD 2. OR A AD and OR bisects AD.
3. AR – BR = RD – RC 3. Subtracting (1) from (2)
4. AB = CD 4. AR = AB + BR and RD = RC + CD

Proved

Example 5: In the given figure, O is the centre of the circle. Two equal chords AB and CD
intersect at E. Prove that OE is the bisector of ‘BED.

Solution: D

Given: O is the centre of the circle. AQ

Chord AB = chord CD and they intersect at E. E O
B
To prove: OE is the bisector of ‘BED. P
C

Construction: OP A AB and OQ A CD are drawn.

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Geometry - Circle

Proof

Statements Reasons

1. OP = OQ 1. Equal chords are equidistant from the

centre of circle.

2. O lies in the locus of the bisector of ‘BED. 2. O is equidistant from the arms of
‘BED.

3. OE is the bisector of ‘BED 3. From statement (2)

Example 6: In the given figure, equal chords AB and CD of a circle Proved

with centre O intersect each others at right angle at K. If M A C
M KB
and N are the mid-points of AB and CD respectively, prove ON

that MONK is a square. D

Solution:

Given: (i) O is the centre of circle

(ii) Equal chords AB and CD intersect at right angle at K

(iii) M and N are the mid-points of AB and CD respectively.

To prove: MONK is a square.

Proof

Statements Reasons
1. OM A AB i.e. ‘OMK = 90°
1. OM joins the centre O and mid-point M
2. ON A CD i.e. ‘ONK = 90° of chord AB.
3. ‘MKN = 90°
4. ‘MON = 90° 2. ON joins the centre O and mid-point N
5. OM = ON of chord CD.

6. MONK is a square. 3. Given

4. Remaining angle of quadrilateral MONK

5. fErqoumalthcehocrednstroef. a circle are equidistant

6. From statements (1), (2), (3), (4) and (5).

Proved

Example 7: In the given figure, A and B are the centres of two intersecting circles. If CD

intersects and AB perpendicularly at P. Prove that (i) CM = DN (ii) CN = DM.

Solution: C

Given: (i) A and B are the centres of two intersecting circles. M

To prove: (ii) CD A AB at P AP B
(i) CM = DN (ii) CN = DM N

Proof: D

Statements Reasons
1. PM = PN 1. AP A MN and AP bisects MN
2. PC = PD 2. BP A CD and BP bisects CD
3. PC – PM = PD – PN 3. Subtracting (1) from (2)
4. CM = DN 4. Remaining parts of PC = PD
5. PC + PN = PD + PM 5. Adding PC = PD and PN = PM
6. CN = DM 6. From (5) by whole part asiom

Proved

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 233 Vedanta Excel in Mathematics - Book 9

Geometry - Circle

Example 8: In the figure alongside, AB is the diameter of a circle PR
with centre O. If RX = QX prove that ‘OPX = ‘ORX. OX B

Solution:

Given: (i) O is the centre of circle, AB is a diameter and RX = QX. A

To prove: ‘OPX= ‘ORX DQ
Construction O and Q are joined.

Proof: Reasons

Statements 1.
1. In 'QOX and 'ROX

(i) OQ = OR (S) (i) Radii of the same circle

(ii) QX = RX (S) (ii) Given

(iii) OX =OX (S) (iii) Common side
2. 'QOX # 'ROX
3. ‘OQX = ‘ORX 2. By S.S.S. axiom

4. ‘OPQ = ‘OQP 3. Corresponding angles of congruent
i.e. ‘OPX = ‘OQX triangles

5. ‘OPX = ‘ORX 4. Base angles of isosceles 'POQ

5. From statements (3) and (4)

Proved

Example 9: Prove that the longer chord of a circle is nearer to the C ND
centre. A
O
Solution: MB

Given: (i) AB and CD are two unequal chords such that AB > CD

(ii) OM A AB and ON A CD

To prove: OM < ON

Construction A and C are joined to O.

Proof:

Statements Reasons

1. AM = 1 AB and CN = 1 CD 1. OM A AB and OM bisects AB, ON A CD
2 2 and ON bisects CD.

2. OA2 = AM2 + OM2 and OC2 = CN2 + ON2 2. Using pythagoras theorem in rt.
‘ed 'AOM and 'CON

3. AM2 + OM2 = CN2 + ON2 3. From (2), OA2 = OC2 = r2
? OM2 – ON2 = CN2 – AM2

CN < AM

4. ? OM2 – ON2 < O 4. From (1) and AB > CD, i.e. CD < AB
or, OM2 < ON2

? OM < ON

Proved

Vedanta Excel in Mathematics - Book 9 234 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Circle

Example 10: In the adjoining figure, two equal chords AB and A M
CD of a circle with centre O meet at an external
B
point X. Prove that: (i) BX = DX and (ii) AX =CX. O D X
Solution:

Given: (i) O is the centre of the circle. C N

(ii) Equal chords AB and CD meet at an external
point X.

To prove: (i) BX = DX (ii) AX = CX

Construction OM A AB, ON A CD are drawn and OX is joined.

Proof:

Statements Reasons
1. In 'MOX and 'NOX
(i) ‘OMX = ‘ONX (R) 1.
(ii) OX = OX (H)
(iii) OM = ON (S) (i) Both are right angles.

(ii) Common side.

(iii) Equal chords are equidistant from the
centre.

2. 'MOX # 'NOX 2. By R.H.S. axiom

3. MX = NX 3. Corresponding sides of congruent
triangles

4. MB = 1 AB and ND = 1 CD 4. OM A AB and ON A CD
2 2

5. MB = ND 5. AB = CD (Given) and from statement (4)

6. MX – MB = NX – ND 6. Subtracting (5) from (3)

7. BX = DX 7. Remaining parts of MX and NX.

8. BX + AB = DX + CD 8. Adding equal chords AB and CD on both
9. AX = CX sides of (7)

9. By whole part axiom

Proved

Example 11: In the figure alongside, two circles with centres A M
and B intersect at M and N. The common chord MN A PB
intersects AB at P. Prove that AB is perpendicular
bisector of MN.

Solution: N
Given:
(i) A and B are centres of two intersecting circles.

(ii) The common chord MN intersecting AB at P.

To prove: AB is perpendicular bisector of MN

i.e. MP = NP and AB A MN.

Construction A and B are joined to M and N.

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Geometry - Circle

Proof:

Statements Reasons
1. In 'AMB and 'ANB
1.

(i) AM = AN (S) (i) Both are right angles.

(ii) BM = BN (S) (ii) Common side.

(iii) AB = AB (S) (iii) Common side.
2. 'AMB # 'ANB 2. By S.S.S. axiom

3. ‘MAB = ‘NAB i.e. ‘MAP = ‘NAP 3. Corresponding angles of congruent
triangles
congruent
4. 'AMP and 'ANP 4. congruent

(i) AM = AN (S) (i) Radii of the same circle Proved

(ii) ‘MAP = ‘NAP (A) (ii) From statement (3)

(iii) AP = AP (S) (iii) Common side
5. 'AMP # 'ANP
5. S.A.S. axiom
6. MP = NP
7. ‘APM = ‘APN 5. Corresponding sides of
8. ‘APM = ‘APN = 90° triangles

6. Corresponding angles of
triangles

7. Equal angles in linear pair

9. AP A MN i.e. AB A MN 9. From the statement (8)

Example 12: Anjali tied her three pets cat, dog and rabbit to a stake by

three ropes of 5 feet each. They were moving in a circular

path always keeping the ropes tight. Once, the dog was O R
equidistant from cat and rabbit and the distance between M
5 ft.
the dog and cat was 6 feet. Find the distance between the 6 ft.
cat and rabbit at the same time. 6 ft.
C D

Solution:

Let C, D and R be the position of cat, dog and rabbit respectively and O be the centre of
circular path.

? OC = OD = OR = 5 ft, CD = RD = 6 ft.

Since OC = OR and CD = RD. So, OCDR is a kite in which the diagonal OD bisects the
diagonal CR at M at a right angle.

Let OM = x ft. then MD = OD – OM = (5 – x) ft.

Now,

In rt. ‘ed 'OCM, CM2 = OC2 – OM2 = 52 – x2 = 25 – x2 ... (i)

In rt. ‘ed 'CMD, CM2 = CD2 – MD2 = 62 – (5 – x)2 = 36 – (25 – 10x + x2)

= 11 + 10x – x2... (ii)

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Geometry - Circle

From (i) and (ii), we get,
25 – x2 = 11 + 10x – x2

or, 14 = 10x ? x = 1.4
Also from (i); CM2 = 25 – (1.4)2 = 23.04 ? CM = 4.8 ft.
? CR = 2 × CM = 2 × 4.8 ft = 9.6 ft
Hence, the distance between the cat and the rat was 9.6 ft.

EXERCISE 14.1

General section
1. Name the shaded portion in each figure.

a) b)

OO

B PQ
AC

2. Write down the special type of 'AOB in each figure with appropriate reasons.

a) b) B c)

O O O
AB A
AB C

B

3. a) In the given figure, O is the centre of the circle and AB is the P
chord. If OP A AB, write the relation between AP and PB. O

A

b) In the adjoining figure, O is the centre of the circle and PQ is its Q
chord. If PX = XB, write the relation between OX and PQ. OX

P

A O C

c) O is the centre of the given circle. If OP A AB, OQ A CD and

OP = OQ, write the relation between AB and CD. PQ

BD

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Geometry - Circle

S
Y

R O
XQ
d) O is the centre of the given circle. If OX A PQ, OY A RS and

PQ = RS, write the relation between OX and OY. P

4. a) In the adjoining figure, the radius of the circle OX = 13 cm and X O Y
the length of a chord XY = 10 cm. Find the distance of the chord P
from the centre of the circle.

b) In the adjoining figure, O is the centre of a circle with radius O
P
5 cm. If OP = 3 cm, find the length of the chord. M N

c) Find the length of a chord which is at a distance of 9 cm from the centre of the circle

of diameter 30 cm. C

d) In the given figure, O is the centre of a circle and CD is a diameter. If O
OE A AB, CD = 20 cm and AB = 16 cm, find the length of ED.

AE B
D

5. a) In the figure alongside, O is the centre of two concentric circles. O

If AP = 5 cm and CD = 6 cm, find the length of AC. AC P DB

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

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

Creative section - A

6. a) 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

b) In the given figure, O is the centre of a circle. AB and CD are C D

two parallel chords of lengths 16 cm and 12 cm respectively. If

the radius of the circle is 10 cm, find the distance between the O

chords. AB

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Geometry - Circle

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

7. a) 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.

b) Two chords AB and CD are 3.9 cm apart and lie on the opposite sides of the centre
of the circle. If AB = 1.4 cm and CD = 4 cm , find the radius of the circle.

c) 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.

8. a) In the adjoining figure, O is the centre of the circle and AB is a O
P
chord. If OP A AB, prove that AP = BP. A B

b) In the figure given alongside, O is the centre of the circle and M is O Y
the mid-point of the chord XY, prove that OM A XY.
M
X

PR

c) In the given figure, O is the centre of the circle and PQ and RS X Y
are two chords. If PQ = RS, OX A PQ and OY A RS, show that
O
OX = OY.
QS

d) In the given figure, O is the centre of the circle, KL and MN are two L N
chords. If OA A KL, OB A MN and OA = OB, prove that KL = MN. O

A B

Creative section - B KM
M
9. a) In the given figure, O is the centre of the circle, chords MN and
RS are intersecting at P. If OP is the bisector of ‘MPR, prove that S
MN = RS.
OP

N

R

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Geometry - Circle

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

(ii) ST = TQ

c) In the given figure, O is the centre of the circle. Two equal A C
chords AB and CD intersect each other at E. Prove that O

(i) AE = CE E
DB
(ii) BE = DE
B

d) In the figure, L and M are the mid-points of two equal chords AB C L O Q

and CD of a circle with centre O. Prove that

(i) ‘OLM = ‘OML (ii) ‘ALM = ‘CML M

A
D

10. a) In the adjoining figure, AB is the diameter of a circle with O B

centre O. If chord CD // AB, prove that ‘AOC = ‘BOD. A

C D
P

b) In the given figure, equal chords PQ and RS of a circle with centre M O
O intersect each other at right angle at A. If M and N are the
R
mid-points of PQ and RS respectively, prove that OMAN is a square. S A N

Q

X

c) 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. D

d) In the figure alongside, AB is a diameter. Two chords AD and BC OB

are equal. Prove that AD // BC. A

C

e) In the given figure, AB and BC are equal chords of the circle with A P
centre O. If OM A AB, ON A BC,OM and ON are produced to meet M

the circumference at P and Q respectively. Prove that OB

(i) AP = CQ and (ii) ‘PAM = ‘QCN. N
CQ

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Geometry - Circle

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

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

AX B

A

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

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

b) In the adjoining figure, P is the centre of a circle. If PQ // BC, BC
C
prove that (i) AC = 2AQ and (ii) PQ A AC. A
Q

PB

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

b) In the given figure, P and Q be the centres of two intersecting A X B
circles and AB // PQ. PQ

Prove that AB = 2 PQ.

c) 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
AB, prove that C
PQ
(i) MC = ND
D
(ii) MD = NC BN

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Geometry - Circle

14. a) 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.

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

c) Prove that a diameter of a circle which bisects a chord of the circle also bisects the
angle subtended by the chord at the centre of the circle.

d) Of two chords, the chord nearer to the centre of circle is longer.

15. a) Five students Amrit, Bibika, Chandani, Dipesh, and Elina are
playing in a circular meadow. Amrit is at the centre, Bibika
and Dipesh are inside the boundary line. Similarly, Chandani A10m DE
and Elina are on the boundary of the meadow. If Amrit, Bibika, B C
Chandani, and Dipesh form a rectangle and the distance between
Bibika and Dipesh is 10 m, find the distance between Amrit and Elina.

b) Three students i.e., Pooja, Shaswat, and Triptee are playing a

game by standing on the circumference of a circle of radius 25 feet O
drawn in a park. Pooja throws a ball to Shaswat and Shaswat to T

Triptee and Triptee to Pooja. What is the distance between Pooja S
and Triptee when the distance between Pooja and Shaswat and P

the distance between Shaswat and Triptee is 30 feet each?

c) 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 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.

Project work

16. a) Draw a large circle in a chart paper. Label its parts and paste in ‘Math Corner’ of your
classroom.

b) Draw three circles and draw the chords AB and PQ of your own choice. By drawing
the perpendicular bisectors of the chords, find the centre of circle in each circle.

c) A ground is in the shape of a circle. You need to fix a pole at it’s centre. How do
you find the centre of this ground? Write a report to show your process by steps and
discuss in the class. Is your process accepted by your friends and your teacher?

d) Without taking any measurement, draw two equal intersecting circles with a pencil
compass. Join the point of intersection of these two circles and also join their centres.
Prove that the common chord is the perpendicular bisector of the line joining the
centres of the circle. Now, verify it by the actual measurements.

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Unit Geometry - Construction

15

15.1 Construction of quadrilaterals C

A quadrilateral has ten parts in all: four sides, four angles, and two D B
diagonals. To construct a quadrilateral, we shall usually need data about A
five specified parts of it. However, in the case of regular quadrilaterals
(square, rectangle, parallelogram, rhombus, etc.), the measurements of a
few required number of parts may be sufficient.

Of course, to construct any type of quadrilateral, we should know its properties and we
apply these properties in its construction.

Let’s study the steps of construction of different quadrilaterals under the following conditions.

A. Construction of square

Let’s construct squares under the following conditions.

1. When a side of a square is given

Example: Construct a square ABCD in which AB = 4 cm.
X Y
Steps of construction C

(i) Draw a line segment AB = 4 cm. D

(ii) At A and B construct ‘BAX = ‘ABY = 90q

(iii) With centres at A and B and radius 4 cm,

cut AX at D and BY at C.

(iv) Join C, D.

Thus, ABCD is the required square. 90° 4 cm 90°
A B
2. When the length of a diagonal is given

Example: Construct a square ABCD in which YX
diagonal BD = 4.6 cm.

Steps of construction A

(i) Draw the diagonal BD = 4.6 cm. 45° 4.6 cm 45°
(ii) Diagonals bisect the angles of a square. So, B D

construct ‘DBX = ‘BDY = 45q.
(iii) Let, BX and DY intersect at A.
(iv) With centres at B and D and radius equal to AB

(or AD), draw two arcs intersecting at C.
(v) Join B, C and D, C.

Thus, ABCD is the required square.

C

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Geometry - Construction

Alternative process: X
A
Steps of construction:
OD
(i) Draw the diagonal BD = 4.6 cm
C
(ii) Draw the perpendicular bisector XY of BD and Y

name the intersecting point by O.

(iii) With centre at O and radius B

DB = 1 BD = 1 × 4.6 cm = 2.3 cm, draw the arcs
2 2

to cut OX at A and OY at C.

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

B. Construction of rectangle

Let’s construct the rectangle under the following conditions.

1. When two adjacent sides of a rectangle are given

Example: Construct a rectangle ABCD in Y X 3.5 cm
which AB = 4.5 cm and D C
BC = 3.5 cm. 4.5 cm
90° 90°
Steps of construction A B

(i) Draw a line segment AB = 4.5 cm.
(ii) At A and B, construct ‘ABX = ‘BAY = 90q.
(iii) With the centres A and B and radius

3.5 cm, draw two arcs to cut AY at D and
BX at C.
(iv) Join C and D.
Thus, ABCD is the required rectangle.

2. When the diagonals and an angle made by them are given.

Example: Construct a rectangle ABCD in which diagonals AC = BD = 5 cm and
they bisect each other making 60q angle.

Steps of construction X
D
(i) Draw AC = 5 cm and draw its perpendicular bisector
to find its mid-point O.

(ii) At O, construct ‘AOX = 60q and produce XO to Y.

(iii) With centre at O and radius 2.5 cm (12 of BD) draw two A 60° C
arcs to cut OX at D and OY at B. O

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

BY

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Geometry - Construction

3. When length of a side and a diagonal of a rectangle are given
Example : Construct a rectangle ABCD in whch AB = 5.2 cm and diagonal

AC = 6.5 cm. D X
Steps of construction C

(i) Draw AB = 5.2 cm

(ii) At B, construct ‘ABX = 90q 6.5 cm

(iii) With centre at A and radius 6.5 cm, draw
arc to cut BX at C and join A, C.

(iv) With centre at A and radius equal to BC, 90°
B
draw an arc. A 5.2 cm

(v) With centre at A and radius equal to AB,

draw another arc to cut previous arc drawn from A at D.

(vi) Join D, A and C, D. Thus, ABCD is the required rectangle.

C. Construction of rhombus

Let’s construct the rhombus under the following conditions.
1. When a side and angle made by two adjacent sides are given
Example: Construct a rhombus ABCD in which AB = 4.5 cm and ‘ ABC = 45q.

Steps of construction

(i) Draw AB = 4.5 cm D 4.5 cm X
(ii) At B, construct ‘ABX = 45q.
C
(iii) With centre at B and radius 4.5 cm
4.5 cm, cut BX at C
4.5 cm
(iv) With centres at A and C and radius
4.5 cm, draw two arcs intersecting A 4.5 cm 45°
each other at D. B

(v) Join A, D and C, D. Thus, ABCD X
is the required rhombus.

2. When two diagonals are given D

Example: Construct a rhombus ABCD in which diagonals
AC = 5.4 cm and BD = 4.2 cm.

Steps of construction A OC
B
(i) Draw AC = 5.4 cm and draw its
perpendicular bisector XY. Mark
the mid-point of AC as O.

(ii) With centre at O and radius 2.1 cm

(21 of BD), draw two arcs to cut OX at D and OY at B.
(iii) Join A, B; B, C; C, D and D, A.

Thus, ABCD is the required rhombus.

Y

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Geometry - Construction

D. Construction of parallelograms

Let’s construct the parallelogram under the following conditions.

1. When two adjacent sides and angle made by them are given

Example: Construct a parallelogram ABCD in which AB = 4.5 cm, BC = 3.5 cm and

‘ABC = 60q. X
Steps of construction

(i) Draw a line segment AB = 4.5 cm. D 4.5 cm
(ii) At B, construct ‘ABX = 60q. C

(iii) With the centre at B and radius BC = 3.5 cm 3.5 cm
3.5 cm
draw an arc to intersect BX at C.

(iv) With centre at C and radius CD = AB = 4.5 cm,

draw an arc. A 4.5 cm 60° B
(v) With centre at A and radius AD = BC = 3.5 cm,

draw another arc to cut the previous arc at D.

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

Thus, ABCD is the required parallelogram.

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

Example: Construct a parallelogram ABCD in which base AB = 5 cm, diagonal
X
AC = 6.3 cm, and ‘BAC = 30q. D 5 cm C
Steps of construction

(i) Draw base AB = 5 cm 6.3 cm
(ii) At A, construct ‘BAX = 30q

(iii) With centre at A and radius AC = 6.3 cm,

draw an arc to cut AX at C.

(iv) With centre at C and radius CD = AB = 5 cm, 30° B
draw an arc. A 5 cm

(v) With centre at A and radius AD = BC, draw

another arc to cut the previous arc at D.

(vi) Join A, D, and B, C.

Thus, ABCD is the required parallelogram.

3. When the length of a diagonal and an angle made by them are given

Example: Construct a parallelogram ABCD in which diagonals AC = 5 cm,
and 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 = 5 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
(21 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 - Book 9 246 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Construction

E. Construction of trapezium

Let’s construct the trapezium under the following conditions.

1. When two adjacent sides and two angles are given
Example: Construct a trapezium ABCD in which AB = 5.5 cm, BC = 4.5 cm,
‘DAB = 60q, ‘BCD = 75q and AD // BC.

Steps of construction

(i) Draw a line segment AB = 5.5 cm. X
D
(ii) Construct ‘BAX = 60q at A.
60° Y
(iii) D lies on AX and AD // BC. A 75° C

So, at B, construct B

‘ABY = 180q – 60q = 120q
(iv) With the centre at B and radius

4.5 cm draw an arc to cut BY at C.
(v) At C, construct ‘BCD = 75q. The

arm CD intersect AX at D.

Thus, ABCD is the required trapezium.

2. When two sides, a diagonal and angle made by the diagonal with one given

side are given
Example: Construct a trapezium ABCD in which AB = 5 cm, diagonal AC = 6.5 cm,

‘BAC = 45q, CD = 4.2 cm and AB // DC.

Steps of construction Y D X
45° C
(i) Draw a line segment AB = 5 cm.

(ii) At A, construct ‘BAX = 45q.

(iii) With centre at A and radius
6.5 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 = 45q at C.

(vi) With centre at C and radius 4.2 cm,
draw an arc to cut CY at D.

(vii) Join D and A.

Thus, ABCD is the required trapezium.

A 45° B

3. When three sides and diagonal are given

Example: Construct a trapezium ABCD in which AB = 4.5 cm, diagonal AC = 6 cm,
AD = BC = 5 cm and AB // DC.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 247 Vedanta Excel in Mathematics - Book 9

Geometry - Construction

Steps of construction
(i) Draw a line segment AB = 4.5 cm.
(ii) With centre at A and radius 6 cm, draw an arc.
(iii) With centre at B and radius 5 cm, draw another arc to intersect the previous arc at C.
(iv) Join A, C and B, C.
(v) At C, construct ‘ACX = ‘BAC.
(vi) With centre at A and radius 5 cm, draw an arc to cut CX at D.
(vii) Join A and D.

Thus, ABCD is the required trapezium.

XD C

AB

F. Construction of quadrilateral

Let’s construct the quadrilateral under the following conditions.

1. When all four adjacent sides and one of the diagonal are given

Example: Construct a quadrilateral ABCD in which AB = 5.4 cm, BC = 5.1 cm,

CD = 4.9 cm, AD = 6.2 cm and the diagonal BD = 5.8 cm.

Steps of construction D

(i) Draw AB = 5.4 cm. 4.9 cm

(ii) From A, draw an arc with C
radius AD = 6.2 cm and
from B draw another arc 6.2 cm
with radius BD = 5.8 cm. 5.8 cm
These two arcs intersect 5.1cm
each other at D.

(iii) From B, draw an arc with

radius BC = 5.1 cm and A 5.4 cm B
from D, draw another arc

with radius DC = 4.9 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 - Book 9 248 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur

Geometry - Construction

2. When all four sides and the angle between any two adjacent sides are given

Example: Construct a quadrilateral ABCD in which AB = 5.2 cm, BC = 5 cm,

CD = 4.2 cm, AD = 4 cm and ‘ABC = 60°. X
C
Steps of construction

(i) Draw AB = 5.2 cm. D 4.2 cm

(ii) At B construct ‘ABX = 60°.

(iii) With the centre at B and radius

5 cm draw an arc to cut BX at C. 4 cm
5 cm
(iv) From A, draw an arc with radius

4 cm and from C, draw another arc

with radius 4.2 cm. These two arcs

intersect to each other at D. 60° B
Join C, D and A, D.
A 5.2 cm

Thus, ABCD is a required quadrilateral.

EXERCISE 15.1

1. Construct the squares ABCD in which-

a) AB = 4.5 cm b) BC = 5.1 cm

c) Diagonal BD = 5.8 cm d) Diagonal AC = 6.2 cm

2. Construct the rectangles PQRS in which-

a) PQ = 4.8 cm and QR = 4.2 cm
b) QR = 5.6 and RS = 4.5 cm
c) Diagonals PR = QS = 7 cm and they bisect each other making an angle of 45q.
d) Diagonals PR = QS = 5.6 cm and they bisect each other making an angle of 60q.

e) Side PQ = 5 cm and diagonal PR = 6.3 cm

f) Side QR = 4.7 cm and diagonal RP = 6 cm

3. Construct the rhombus WXYZ in which-

a) WX = 4.4 cm and ‘WXY = 60q b) WX = 3.6 cm and ‘XWZ = 30q

c) Diagonals WY = XZ = 6 cm d) Diagonals WY = XZ = 5.8 cm

4. Construct the parallelogram ABCD in which-

a) AB = 4.5 cm, BC = 3.5 cm, and ‘ABC = 60q

b) AB = 5.4 cm, AD = 4.6 cm, and ‘BAD = 30q

c) Base AB = 5.2 cm, diagonal AC = 6.7 cm, and ‘BAC = 30q

d) Base AB = 4.8 cm, diagonal BD = 6.4 cm, and ‘ABD = 45q

e) Diagonals AC = 4.6 cm, BD = 5.8 cm, and they bisect each other making an angle
of 30q.

f) Diagonals AC = 5.2 cm, BD = 6.2 cm, and they bisect each other making an angle
of 60q.

Approved by Curriculum Development Centre, Sanothimi, Bhaktapur 249 Vedanta Excel in Mathematics - Book 9

Geometry - Construction

5. Construct the trapezium ABCD in which-
a) AB = 4 cm, BC = 5.3 cm, ‘BAD = 60q, ‘BCD = 90q and AD // BC.
b) AB = 5.5 cm, BC = 4.5 cm, ‘BAD = 45q, ‘BCD = 60q and AD // BC.
c) AB = 4.8 cm, diagonal AC = 5.9 cm, ‘BAC = 60q, CD = 5 cm, and AB // DC.
d) AB = 6.4 cm, diagonal AC = 7.5 cm, ‘BAC = 30q, CD = 6 cm, and AB // DC.
e) AB = 4 cm, diagonal BD = 5.6 cm, ‘ABD = 60q, BC = 5.5 cm, and AB // DC.
f) AB = 4.4 cm, diagonal AC = 6.8 cm, AD = BC = 5.2 cm.
g) AB = 6 cm, diagonal AC = 7.6 cm, CD = 6.5 cm, BC = 4.5 cm.

6. Construct the quadrilateral ABCD in which-
a) AB = 5 cm, BC = 5.6 cm, CD = 4.5 cm, AD = 5.4 cm and, diagonal BD = 6.5 cm
b) AB = 4.2 cm, BC = 5.1 cm, CD = 5.4 cm, AD = 3.8 cm and, diagonal BD = 4.6 cm
c) AB = BC = CD = 4.5 cm, AD = 5.6 cm, and diagonal AC = 7.1 cm
d) AB = 5.5 cm, BC = 5.7 cm, CD = 4.7 cm, AD = 4.3, and ‘BAD = 60°.
e) AB = 4 cm, BC = 5 cm, CD = 5.5 cm, DA = 5.5 cm, and ‘ABC = 45°.
f) AB = BC = 5.5 cm, CD = DA = 4.5 cm, and ‘DAB = 75°.

Project work

7. a) Make the groups of your friends. Then, using matchsticks or wire or thread, prepare
the model of each of square, rectangle, rhombus, parallelogram, and trapezium on
the paper sheet. Then, present in classroom. Also, make a short report of preparation
of models.

b) Draw a quadrilateral in your choice. Measure the length of its sides. Can you draw
the same quadrilateral by using a ruler and compass? Are these measurements
sufficient to draw a quadrilateral? Discuss about the sufficient measurements for
construction of quadrilateral then write a note and present in classroom.

Objective Questions
1. The sum of interior angles of any triangle is-

(A) 1 right angle. (B) 2 right angles (C) 3 right angles (D) 4 right angles

2. In a triangle if the measurement of any two angles are 400 and 700 then triangle is

(A) Equilateral triangle (B) Isosceles triangle only

(C) Scalene triangle (D) Isosceles and acute angled triangle

3. The relation among ‘x, ‘y and ‘z in the figure given alongside is- z

(A) ‘x = ‘y + ‘z (B) ‘z = ‘x + ‘y

(C) ‘x = ‘y – ‘z (D) ‘x + ‘y + ‘z = 1800 xy

4. In ∆ABC, AB < BC < CA, then the greatest angle is-

(A) ‘ABC (B) ‘BAC (C) ‘ACB (D) None

5. Under which condition, ∆ABC does not exist?
(A) AB + BC > AC (B) BC + AC > AB
(C) AB + BC < AC (D) AC – BC < AB

Vedanta Excel in Mathematics - Book 9 250 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur