(iii) With the centre at A and radius Geometry - Construction4.8 cm
4.1 cm draw an arc to cut AX at D.
Y
(iv) At B construct ∠ABY= 75°. X
(v) With the centre at B and radius
C
4.8 cm draw an arc to cut BY at C.
(vi) Join C and D. D
Thus, ABCD is a required quadrilateral. 4.1 cm
60° 5.2 cm 75°
A B
5. When any two adjacent sides and three angles are given
Example: Construct a quadrilateral ABCD in which AB = 5 cm, BC = 3.9 cm,
∠BAD = 45o, ∠ABC = 120o and ∠BCD = 70o.
Solution:
Here, AB = 5 cm, BC = 3.9 cm, ∠ABC = 45o, ∠ABC = 120o and ∠BCD = 70o.
X
Steps of construction
(i) Draw AB = 5 cm. ZD Y
(ii) At A construct ∠BAX = 45°.
(iii) At B construct ∠ABY = 120°. C
(iv) With the centre at B and 70°
radius 3.9 cm draw an arc to 3.9 cm
cut BY at C.
(v) At C construct ∠BCZ= 70° so 45° 120°
that CZ cuts AX at D. A B
5 cm
Thus, ABCD is a required quadrilateral.
C. Construction of trapezium
In the case of regular quadrilaterals (square, rectangle, parallelogram, rhombus, etc.) and
trapezium, 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.
Facts to remember
1. Trapezium is a quadrilateral in which a pair of opposite sides are parallel.
2. To construct trapezium, we need the measurements of its four independent parts.
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Geometry - Construction
Let’s study the steps of construction of trapeziums under the following conditions.
1. When the length of three adjacent sides and one of the angles are given
Example: Construct a trapezium ABCD in which AB = 5.4 cm, BC = 4.9 cm,
AD = 4.1 cm, ∠BAD = 80o and AB//DC.
Solution:
Here, AB = 5.4 cm, BC = 4.9 cm, AD = 4.1 cm, ∠BAD = 80o and AB//DC.
Steps of construction
(i) Draw AB = 5.4 cm. X
(ii) At A, construct ∠BAD = 80o by using protractor. D CY
(iii) With the centre at A and radius 4.1 cm, draw an arc to 100° 4.9 cm
cut AX at D. 4.1 cm
(iv) At D, construct ∠ADC = 180o – 80o = 100° by using
80° 5.4 cm B
protractor. A
(v) With centre at B and radius BC = 4.9 cm, draw an arc
to cut DY at C.
(vi) Join B and C.
Thus, ABCD is the required trapezium.
2. 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 = 60°, ∠BCD = 75° and AD // BC.
Steps of construction X
D
(i) Draw a line segment AB = 5.5 cm.
60° Y
(ii) Construct ∠BAX = 60° at A. A 75° C
(iii) D lies on AX and AD // BC. B
So, at B, construct
∠ABY = 180° – 60° = 120°
(iv) With the centre at B and radius
4.5 cm draw an arc to cut BY at C.
(v) At C, construct ∠BCD = 75°. The
arm CD intersect AX at D.
Thus, ABCD is the required trapezium.
3. 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 = 45°, CD = 4.2 cm and AB // DC.
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Geometry - Construction
Steps of construction YD X
A 45° 45° C
(i) Draw a line segment AB = 5 cm.
B
(ii) At A, construct ∠BAX = 45°.
(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 = 45° 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.
4. 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.
Steps of construction
(i) Draw a line segment
AB = 4.5 cm. X D C
With centre at A and
(ii)
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. A B
(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.
5. When the lengths of all four sides are given
Example: Construct a trapezium ABCD in which AB = 6.5 cm, BC = 4.3 cm,
CD = 4 cm, AD = 5 cm and DC//AB.
Solution:
Here, AB = 6.5 cm, BC = 4.3 cm, CD = 4 cm and AD = 5 cm.
Steps of construction
(i) Draw AB = 6.5 cm.
(ii) Mark the point E on AB such that AE = 4 cm.
(iii) From E, draw an arc with radius AD = 5 cm and from B, draw another arc with
radius 4.3 cm. These two arcs intersect to each other at C. Join A, C and E, C.
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Geometry - Construction D 4 cm C
(iv) Also, from A, draw an arc with radius 5
cm and from C, draw another arc 5 cm 4.3 cm
with radius 4 cm. These two arcs 5 cm
intersect to each other at D. Join A,
D and C, D.
Thus, ABCD is a required trapezium.
A 4 cm E B
6.5 cm
EXERCISE 15.1
1. Construct the rhombus WXYZ in which-
a) WX = 4.4 cm and ∠WXY = 60° b) WX = 3.6 cm and ∠XWZ = 30°
c) Diagonals WY = XZ = 6 cm d) Diagonals WY = XZ = 5.8 cm
2. 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 = 5 cm, BC = 4.5 cm, CD = AD = 5.5 cm and, diagonal AC = 6 cm
c) AB = 5.5 cm, BC = 5.7 cm, CD = 4.7 cm, AD = 4.3 cm and ∠DAB = 60o.
d) AB= 4 cm, BC = 5 cm, CD = 5.5 cm, AD = 5.5 cm, and ∠ABC= 45°
e) AB= 4.5 cm, BC = 3.7 cm, AD = 4 cm, diagonals AC = 5.4 cm and BD = 7 cm.
f) AB= 5.2 cm, BC = 4.3 cm, AD = 3.6 cm, diagonals AC = 6.1 cm and BD = 5.8 cm.
g) AB= 5.3 cm, BC = 4.6 cm, CD = 3.9 cm, ∠A = 60o and ∠B = 45o.
h) AB= 4.5 cm, BC = 5.4 cm, AD = 4.7 cm, ∠A = 90o and ∠B = 60o.
i) AB= 5 cm, BC = 4 cm, ∠A = 60o, ∠B = 100o and ∠C = 120o.
j) AB= 4.5 cm, AD = 3.8 cm, ∠A = 90o, ∠B = 120o and ∠D = 70o.
3. Construct the trapezium ABCD in which
a) AB = 5.8 cm, BC = 4 cm, AD = 4.5 cm, ∠BAD = 60 o and AB//DC.
b) AB = 6 cm, BC = 5.1 cm, AD = 4.8 cm, ∠BAD = 75o and AB//DC.
c) AB = 5 cm, AD = 4 cm, ∠BAD = 60o, ∠ABC = 75o and AB // CD.
d) AB = 4.7 cm, BC = 5 cm, ∠BAD = 60 o, ∠ABC = 90o and AB // DC.
e) AB = 4.8 cm, diagonal AC = 5.9 cm, ∠BAC = 60o, CD = 5 cm, and AB // DC.
f) AB = 5.5 cm, BC = 7 cm, diagonal BD = 6.1 cm, ∠ABD = 45o, and AB // DC.
g) AB = 4.4 cm, diagonal AC = 6.8 cm, AD = AB= 5.2 cm and AB//DC.
h) BC = 5 cm, diagonal AD = 5.2 cm, BC = CD= 4 cm and AD//BC.
i) AB = 6 cm, BC = 5 cm, CD = 4 cm, AD = 4.5 cm, and DC//AB.
j) AB = 5 cm, BC = CD = 3 cm, AD = 4 cm, and DC//AB.
Vedanta Excel in Mathematics - Book 9 250 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
Unit 16 Geometry: Circle
16.1 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 ⊥ 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 Vedanta ICT Corner
(i) AP = PB Please! Scan this QR code or
(ii) AP = PB browse the link given below:
(iii) AP = PB
https://www.geogebra.org/m/nqdktdcc
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 ⊥ 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.
(ii) Radii of the same circle
(i) ∠OPA = ∠OPB (R) (iii) Common side
(iv) R.H.S. axiom
(ii) OA = OB (H) 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 = 90°
(ii) ∠OPA = ∠OPB = 90°
(iii) ∠OPA = ∠OPB = 90°
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 ⊥ AB AP B
Construction: O, A and O, B are joined.
Proof
Statements Reasons
1. In rt. ∠ed ∆s OAP and OBP 1.
(i) OA = OB (S) (i) Radii of the same circle
(ii) AP = PB (S) (ii) Given (P is the mid-point of AB)
(iii) OP = OP (S) (iii) Common side
(iv) ∴ ∆OAP ≅ ∆OBP (iv) S.S.S. axiom
2. ∠OPA = ∠OPB 2. Corresponding angles of congruent
triangle
3. ∠OPA = ∠OPB = 90° 3. Adjacent angles in linear pair are equal.
4. OP ⊥ 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 ⊥ AB and OQ ⊥ 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 Vedanta ICT Corner
(i) OP = OQ Please! Scan this QR code or
(ii) OP = OQ browse the link given below:
(iii) OP = OQ
https://www.geogebra.org/m/wueykd74
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 ⊥ AB, OQ ⊥ CD.
To prove: OP = OQ
Construction: O, A and O, C are joined.
Proof
Statements Reasons
1. Given
1. AB = CD 2. OP ⊥ AB and OP bisects AB; OQ ⊥ 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 ⊥ AB, OQ ⊥ 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 ⊥ AB and OP bisects AB; OQ ⊥ 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 ⊥ AB and OP bisects AB. A B
∴ AP = 1 AB = 1 × 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
OA ⊥ MN, MN = 6 cm and AX = 1 cm, find the length of
diameter of the circle. MA
X
Solution:
Here, O is the centre of the circle, OA ⊥ 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 ⊥ 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 × 12 cm = 6 cm
2 2
AQ = 1 of AB = 1 × 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 ⊥ PQ is drawn.
Proof:
Statements Reasons
1. BR = RC 1. OR ⊥ BC and OR bisects BC.
2. AR = RD 2. OR ⊥ 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 ⊥ AB and OQ ⊥ CD are drawn.
Vedanta Excel in Mathematics - Book 9 258 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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 ⊥ AB i.e. ∠OMK = 90°
1. OM joins the centre O and mid-point M
2. ON ⊥ 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
(ii) CD ⊥ AB at P AP B
To prove: (i) CM = DN (ii) CN = DM N
Proof: D
Statements Reasons
1. PM = PN 1. AP ⊥ MN and AP bisects MN
2. PC = PD 2. BP ⊥ 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
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Geometry - Circle
Example 8: 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 ⊥ AB, ON ⊥ CD are drawn and OX is joined.
Proof:
Statements Reasons
1. In DMOX and DNOX
(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. DMOX ≅ DNOX 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 ⊥ AB and ON ⊥ 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 9: 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 ⊥ MN.
Construction A and B are joined to M and N.
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Geometry - Circle
Proof:
Statements Reasons
1. In DAMB and DANB
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. DAMB ≅ DANB 2. By S.S.S. axiom
3. ∠MAB = ∠NAB i.e. ∠MAP = ∠NAP 3. Corresponding angles of congruent
triangles
congruent
4. DAMP and DANP 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. DAMP ≅ DANP
5. S.A.S. axiom
6. MP = NP
7. ∠APM = ∠APN 5. Corresponding sides of
8. ∠APM = ∠APN = 90° triangles
9. AP ⊥ MN i.e. AB ⊥ MN
6. Corresponding angles of
triangles
7. Equal angles in linear pair
9. From the statement (8)
Example 10: 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 DOCM, CM2 = OC2 – OM2 = 52 – x2 = 25 – x2 ... (i)
In rt. ∠ed DCMD, CM2 = CD2 – MD2 = 62 – (5 – x)2 = 36 – (25 – 10x + x2)
= 11 + 10x – x2... (ii)
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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 16.1
General section
1. Name the shaded portion in each figure.
a) b)
OO
B PQ
AC
2. Write down the special type of DAOB 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 ⊥ 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 ⊥ AB, OQ ⊥ 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 ⊥ PQ, OY ⊥ 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
5 cm. If OP = 3 cm, find the length of the chord. P
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 ⊥ 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. CD
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 ⊥ 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 ⊥ 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 ⊥ PQ and OY ⊥ 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 ⊥ KL, OB ⊥ 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
b) In the figure alongside, PQ and RS are two chords intersecting at S P
T in a circle with centre O. If OT is the bisector of ∠PTR, prove O
that T R
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
(ii) BE = DE DB
B
d) In the figure, L and M are the mid-points of two equal chords AB C L O Q
D
and CD of a circle with centre O. Prove that
(i) ∠OLM = ∠OML (ii) ∠ALM = ∠CML M
A
10. a) In the adjoining figure, AB is the diameter of a circle with A O B
centre O. If chord CD // AB, prove that ∠AOC = ∠BOD.
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
mid-points of PQ and RS respectively, prove that OMAN is a square. S A N R
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.
11. a) Two equal chords AB and CD of a circle with centre O A B
D
are produced to meet at E, as shown in the given figure. O E
Prove that BE = DE and AE = CE. C
b) In the given figure, O is the centre of circle ABCD. YC
If OY ⊥ PC, OX ⊥ PB and OX = OY, prove that PB = PC. D
P O
AX B
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Geometry - Circle
A
c) In the given figure, ABC is a triangle in which AB = AC. Also a D E
circle passing through B and C intersects the sides AB and AC at the
points D and E respectively. Prove that AD = AE.
O
BC
12. 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 P AM
intersecting at A and B. If MN is parallel to common chord C
AB, prove that Q
D
(i) MC = ND
BN
(ii) MD = NC
A DE
13. a) Five students Amrit, Bibika, Chandani, Dipesh, and Elina 10m
are playing in a circular meadow. Amrit is at the centre, Bibika BC
and Dipesh are inside the boundary line. Similarly, Chandani
and Elina are on the boundary of the meadow. If Amrit, Bibika,
Chandani, and Dipesh form a rectangle and the distance between
Bibika and Dipesh is 10 m, find the distance between Amrit and
Elina.
b) 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.
c) 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
and Triptee when the distance between Pooja and Shaswat and P S
the distance between Shaswat and Triptee is 30 feet each?
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Geometry - Circle
Project work and activity section
14. 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.
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. Which of the following statements is NOT true in any circle?
(A) Every chord is a diameter. (B) Every diameter is a chord.
(C) All radii are equal. (D) All diameters are equal.
2. The radius of the circle is …… the diameter.
(A) equal to B) two times (C) three times (D) half of
3. The region of circle enclosed between two radii and the corresponding arc is
(A) segment of circle (B) sector of circle (C) semi-circle (D) circle itself
4. The line joining the mid-point of a chord and the centre of circle is … to the chord.
(A) equal (B) perpendicular (C) parallel (D) inclined
5. The perpendicular drawn from the centre of a circle to a chord …. the chord.
(A) intersects (B) bisects (C) trisects (D) none
6. The equal chords of a circle are
(A) equidistance from the centre. (B) equidistance from the circumference.
(C) not equidistance from the centre. (D) parallel to each other.
7. The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance
of the chord from the centre is
(A) 23 cm (B) 24 cm (C) 12 cm (D) 3 cm
8. The length of a chord which is at a distance of 6 cm from the centre of the circle of
diameter 20 cm is
(A) 8 cm (B) 16 cm (C) 4 cm (D) 14 cm
9. In the circle with centre O, the length of each chord AB and chord CD is 16 cm, the chord
AB is at a distance of 6 cm, what is the distance of the chord CD from the centre O?
(A) 22 cm (B) 10 cm (C) 12 cm (D) 6 cm
10. The chords PQ and RS of a circle are equidistance from the centre. If the length of the
chord PQ is 12 cm, what is the length of the chord RS?
(A) 3 cm (B) 6 cm (C) 12 cm (D) 24 cm
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Unit 17 Statistics (I): Classification and
Graphical Representation of Data
17.1 Statistics –Looking backExpenditure (in Rs)
Classwork-Exercise
1. The bar graph given below shows the monthly expenditure of a family in the
last 6 months. Answer the following questions.
4500
4000
3500
3000
2500
2000
0 Jan Feb Mar Apr May Jun
Months
(i) In which month the expenditure was minimum?
(ii) In which month the expenditure was maximum?
(iii) In how many months was the expenditure more than Rs 3,000?
(iv) If the income of the family is Rs 10,000 every month, express the
expenditures of every month in percent.
2. The pie chart alongside shows the percentage of types of Walking
transportation used by 400 students to come to school.
(i) What is the total number of students in the school? Bus 40%
(ii) How many students come to school by bus? 30% Van
(iii) How many students do not come to school by van? Bicycle 10%
17.2 Statistics
The term ‘statistics’ was derived from the Latin word ‘Status’ or Italian word ‘Statista’
or German word ‘Statistiks’ all of which means the political state. In those days,
statistics was used to collect the information relating to the population, military
strength and income of the state etc. In the age of information technology, statics has
a wide range of applications. Different departments and authorities require various
facts and figures to frame policies and guidelines in order to function smoothly.
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Statistics (I): Classification and Graphical Representation of Data
Statistical information helps to understand the economic problems and formulation
of economic policies. In social science, statistics is used in the field of demography
for studying mortality, fertility, population growth rate, and so on. In science and
technology, there is a regular use of statistical tools for collecting, presenting and
analysing the observed data for various researches.
The present time is the time of information and communication. In social,
economical and technical area, we frequently require information in the form of
numerical figures. The information collected in the form of numerical figures are
called data.
Facts to remember
Statistics is a branch of mathematics dealing with collection, organisation, analysis,
interpretation, and presentation of data.
Statistics in plural form refer to data, whereas in the singular form it refers a subject.
Sir Ronald Aylmer Fisher (17 February 1890 – 29 July 1962) was
a British statistician, mathematician and biologist. Also, he was
known as the Father of Modern Statistics and Experimental Design.
Fisher did experimental agricultural research, which saved millions
from starvation. He was awarded by the Linnean Society of London’s
prestigious Darwin-Wallace Medal in 1958.
17.3 Types of data
(i) Primary data
The data collected by the investigator him/herself for definite purposes are
called primary data. These data are highly reliable and relevant.
(ii) Secondary data
The data collected by someone other than the user oneself are called secondary
data.
(iii) Raw data
The data obtained in original form are called raw data.
(iv) Array data
The data arranged in ascending or descending order are called array data.
17.4 Frequency tables
Let’s consider the following marks obtained by 10 students in a unit test in mathematics.
18, 10, 15, 15, 12, 12, 10, 15, 12, 12 → Raw data
10, 10, 12, 12, 12, 12, 15, 15, 15, 18 → Array data
Here, the marks 10 are repeated 2 times. So, 2 is the frequency of 10.
The marks 12 are repeated 4 times. So, 4 is the frequency of 12.
The marks 15 are repeated 3 times. So, 3 is the frequency of 15.
The mark 18 is repeated only one time. So, 1 is the frequency of 18.
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Statistics (I): Classification and Graphical Representation of Data
Thus, a frequency is the number of times a value occurs. Data and their frequencies
can be presented in a table called frequency table.
Marks Tally marks Frequency
10 || 2
12 |||| 4
15 ||| 3
18 | 1
Tallying is a system of recording and counting results using diagonal lines grouped in
fives. Each time five is reached, a horizontal line is drawn through the tally marks to
make a group of five. The next line starts a new group.
For example, the tally marks of frequency 5 is ||||, frequency 6 is |||| |, 7 is |||| ||,
and so on.
17.5 Grouped and continuous data
Let’s consider the following marks obtained by 20 students in a test of mathematics.
25, 38, 21, 18, 7, 28, 49, 17, 27, 36
26, 45, 32, 16, 24, 39, 20, 33, 40, 35
The above mentioned data are called individual or discrete data. Another way of
organising data is to present them in a grouped form. For grouping the given data, we
should first see the smallest value and the largest value. We have to divide the data
into an appropriate class–interval. The numbers of values falling within each class–
interval give the frequency. For example,
Marks Tally marks Frequency
0 – 10 | 1
10 – 20 ||| 3
20 – 30 |||| || 7
30 – 40 |||| | 6
40 – 50 ||| 3
Total 20
In the above series, 7 is the smallest value and 49 is the largest value. So, the data are
grouped into the interval of 0 – 10, 10 – 20, ... 40 – 50, so that the smallest and the
largest values should fall in the lowest and the highest class–interval respectively.
Let’s consider a class–interval 10 – 20.
Here, 10 is called the lower limit and 20 is the upper limit of the class–interval. The
difference between the two limit is called the length or height of each class–interval.
For example,
in 10 – 20 the length of the class–interval is 10.
Again, let’s take class–intervals, 0 – 10, 10 – 20, 20 – 30, …
Here, the upper limit of a pervious class–interval has repeated as the lower limit of the
consecutive next class–interval. Such an arrangement of data is known as grouped
and continuous data.
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Statistics (I): Classification and Graphical Representation of Data
Facts to remember
1. In individual series the items are listed singly after observation.
2. In discrete series, the discrete variables along with the corresponding frequencies
are tabulated.
3. In continuous series, the continuous variables (class intervals) along with the
corresponding frequencies are tabulated.
17.6 Cumulative frequency table
The word ‘cumulative’ is related to the word ‘accumulated’, which means to ‘pile
up’. The frequency distribution in which the frequencies are accumulated is called
cumulative frequency distribution. For example,
(i) The table given below shows the heights (in cm) of 40 students of class 9 and
the corresponding cumulative frequency table.
Height (in cm) Number of students (f) Cumulative frequency (c.f.)
110 7 7
118 4 7 + 4 = 11
120 6 11 + 6 = 17
124 10 17 + 10 = 27
130 8 27 + 8 = 35
143 5 35 + 5 = 40
Total 40
(ii) The table given below shows the marks obtained by 20 students in a mathematics
test of full marks 50 and the corresponding cumulative frequency of each class–
interval.
Marks Frequency Cumulative 1 student obtained marks 0 to less than 10.
(f) frequency (c.f) 4 students obtained marks 0 to less than 20.
0 – 10 11 students obtained marks 0 to less than 30.
10 – 20 1 1 17 students obtained marks 0 to less than 40.
20 – 30 3 1+3=4 20 students obtained marks 0 to less than 50.
30 – 40 7 4 + 7 = 11
40 – 50 6 11 + 6 = 17
Total 3 17 + 3 = 20
20
There are two types of cumulative frequency distribution.
a) Less than cumulative frequency distribution.
b) More than cumulative frequency distribution.
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Statistics (I): Classification and Graphical Representation of Data
Less than cumulative frequency distribution
Less than cumulative frequency is obtained by adding successively the frequencies of all the
previous classes including the class against which it is written. The cumulate starts from the
lowest to the highest-class intervals. For example:
Simple frequency distribution Less than cumulative frequency distribution
Ages (in years) Number of students (f) Ages (in years) Less than c.f.
0-4 25 Less than 4 25
4-8 60 Less than 8 25 + 60 = 85
8 - 12 48 Less than 12 85 + 48 = 133
12 - 16 75 Less than 16 133 + 75 = 208
16 - 20 62 Less than 20 208 + 62 = 270
Total 270
More than cumulative frequency distribution
More than cumulative frequency is obtained by adding successively the frequencies starting
from the highest to lowest class intervals. In more than, cumulative frequency distribution,
each value of c.f. refers to the total number of observations of more than or equal to the
corresponding value in discrete series or lower limit of corresponding class in continuous
series. For example.
Simple frequency distribution More than cumulative frequency distribution
Ages (in years) Number of students (f) Ages (in years) More than c.f.
0-4 25 0 or more than 0 245 + 25 = 270
4-8 60 4 or more than 4 185 + 60 = 245
8 - 12 48 8 or more than 8 137 + 48 = 185
12 - 16 75 12 or more than 12 62 + 75 = 137
16 - 20 62 16 or more than 16
62
Total 270
Worked-out Examples
Example 1: The monthly salary (in Rs) of 30 workers of a company is given below.
15000, 20000, 8000, 25000, 15000, 10000, 30000, 20000, 20000,
25000, 20000, 30000, 15000, 25000, 30000, 10000, 15000, 25000,
20000, 15000, 8000, 25000, 15000, 10000, 30000, 20000, 25000,
15000, 20000, 15000
a) Present this individual series into a discrete series with tally
bars. Also, construct a cumulative frequency table.
b) How many workers have monthly salary Rs 25,000 or less than
Rs 25,000?
Solution:
Here, minimum salary = Rs 8,000 and the maximum salary = Rs 30,000
a) Presenting the given data into discrete series with tally bars.
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Statistics (I): Classification and Graphical Representation of Data
Monthly salary Tally bars Number of Cumulative
(in Rs) workers (f) frequency (c.f.)
||
8000 ||| 2 2
10000 ||||||| 3 2+3=5
15000 |||||| 8 5 + 8 = 13
20000 ||||| 7 13 + 7 = 20
25000 |||| 6 20 + 6 = 26
30000 4 26 + 4 = 30
Total 30
b) From the above cumulative frequency table, 26 workers have monthly salary Rs 25,000
or less than Rs 25,000.
Example 3: During the medical check-up of 50 patients, their weights were recorded
as follows:
Weight in kg Below 10 Below 20 Below 30 Below 40 Below 50 Below 60
Number of 6 14 25 33 38 50
patients
Form the above table, construct the more than cumulative frequency
distribution table.
Solution:
Here,
Frequency distribution table More than cumulative frequency table
Weight in kg Number of patients (f) Weight in kg More than c.f.
0 - 10 6 0 or more than 0 (≥ 0) 44 + 6=50
10 - 20 14 - 6 = 8 10 or more than 10 (≥ 10) 36 + 8=44
20 - 30 25 - 14=11 20 or more than 20 (≥ 20) 25 + 11=36
30 - 40 33 - 25=8 30 or more than 30 (≥ 30) 17 + 8=25
40 - 50 38 - 33=5 40 or more than 40 (≥ 40) 12 + 5=17
50 - 60 50 - 38=12 50 or more than 50 (≥ 50)
12
EXERCISE 17.1
General section
1. a) Write a few paragraphs about the application of statistics.
b) Define primary, secondary, raw, and array data.
c) What do you mean by frequency of an observation in data?
d) Define cumulative frequency.
2. a) The marks obtained by 50 students of class IX in Mathematics are shown in the
table given below. Construct a cumulative frequency table to represent the data.
Marks obtained 30 40 50 60 70 80 90
Number of students 4 7 10 12 8 6 3
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Statistics (I): Classification and Graphical Representation of Data
b) The daily wages of 45 workers are given in the table below. Construct a cumulative
frequency table to represent the data.
Daily wages (in Rs) 450 500 550 600 650 700 750
Number of workers 6 4 8 10 9 3 5
3. a) The daily sales of 20 shops are recorded in the following table.
Sales (in Rs) 0-1000 1000-2000 2000-3000 3000-4000 4000-5000
Number of shops 2 5 6 4 3
(i) Construct a less than cumulative frequency distribution table.
(ii) Construct a more than cumulative frequency distribution table.
b) The values of diastolic blood pressure of 90 people of age group 20 – 40 are shown
in the table below.
Blood pressure (mmHg) 70-75 75-80 80-85 85-90 90-95
Number of people 20 25 18 17 10
(i) Construct a less than cumulative frequency distribution table.
(ii) Construct a more than cumulative frequency distribution table.
4. a) From the following frequency and less than cumulative frequency distribution table,
find the values of p, q and r.
Class interval 0-10 10-20 20-30 30-40 40-50
Frequency 5 p 10 6 r
Cumulative frequency 5 9 19 q 30
b) Find the values of a, b and c from the given frequency and less than cumulative
frequency distribution table.
Ages (in years) 4-8 8-12 12-16 16-20 20-24
Number of students
Cumulative frequency 10 15 b 25 30
10 a 45 70 c
Creative section
5. a) The marks obtained by 20 students in mathematics in a class test is given below.
15, 18, 12, 16, 18, 10, 15, 16, 15, 12, 10, 12, 15, 12, 16, 18, 12, 15, 12, 16
(i) Present this individual data in a frequency distribution table with tally marks.
(ii) Construct a cumulative frequency distribution table.
(iii) From the cumulative frequency table, find the number of students who obtained
15 or less than 15 marks.
b) The amount of money (in Rs) brought by 30 students of class IX for tiffin is given
below.
30, 20, 50, 40, 25, 50, 40, 35, 20, 50 40, 50, 35, 25, 30,
40, 50, 25, 40, 30 35, 25, 40, 50, 40, 30, 25, 50, 35, 40
(i) Draw a frequency distribution table with tally bars to represent the data.
(ii) Construct the cumulative frequency distribution table.
(iii) How many students brought less than or equal to Rs 40?
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Statistics (I): Classification and Graphical Representation of Data
6. a) The marks obtained by 40 students in mathematics in the final examination of class
9 are given below.
42, 68, 80, 45, 92, 36, 8, 17, 49, 30, 5, 26, 98, 74, 53, 65, 72, 28, 55, 46,
86, 70, 62, 27, 16, 44, 85, 59, 51, 73 66, 78, 38, 81, 97, 77, 69, 45, 33, 67
(i) Make a frequency distribution table with class interval of length 10.
(ii) Construct the less than cumulative frequency distribution table.
(iii) Construct the more than cumulative frequency distribution table.
b) The weights (in kg) of 30 teachers of a school are given below.
71, 53, 60, 50, 64, 74, 59, 67, 61, 58, 65, 70, 73, 51, 63,
66, 70, 50, 60, 70, 56, 62, 52, 67, 59, 77, 64, 78, 57, 57
(i) Make a frequency distribution table with class interval of width 5.
(ii) Construct the less than cumulative frequency distribution table.
(iii) Construct the more than cumulative frequency distribution table.
7. a) The weight (in kg) of 50 wrestlers are recorded as follows:
Weight in kg Below Below Below Below Below Below
100 110 120 130 140 150
Number of 6 14 25 33 38 50
wrestlers
(i) Construct the frequency distribution table.
(ii) Construct the more than cumulative frequency distribution table.
b) The following cumulative frequency table shows the marks obtained by 60 students
of class 9 in mathematics.
Marks ≥ 40 ≥ 50 ≥ 60 ≥ 70 ≥ 80 ≥ 90
obtained
Number of 35 30 24 16 7 4
students
(i) Construct the frequency distribution table.
(ii) Construct the less than cumulative frequency distribution table.
Project work and activity section
8. a) Measure the weights of each of your classmates. Show the weights in a frequency
distribution table of continuous class with length 5 kg. Then, construct the less
than and more than cumulative frequency table.
b) Show the marks obtained by at least 20 friends of your class in mathematics in
recently conducted exam in a frequency distribution table of continuous class
of length 10. Then, construct the less than and more than cumulative frequency
table.
c) Visit a website and write the value of p up to 50 decimal places. Make the frequency
distribution table of the digits on the value from 0 to 9 after decimal places.
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Statistics (I): Classification and Graphical Representation of Data
17.7 Graphical representation of data
We have already discussed to present data in frequency distribution tables.
Alternatively, we can also present data graphically. Different types of diagrams are
used for this purpose. Here, we shall discuss three types of diagrams: histogram,
frequency polygon and ogive.
Graphical representation
Ungrouped data Grouped data
line bar Pictograph pie histogram frequency frequency cummulative
polygon curve frequency
graph graph chart curve
1. Histogram
Statistical data can be represented by various types of diagrams, such as, bar diagram,
pie-chart, line graph, etc. A histogram is a graphical representation of a continuous
frequency distribution. For this purpose, we draw rectangular bars with class intervals
bases and corresponding frequencies as heights.
Study the following steps to construct a histogram.
(i) If the given frequency distribution has inclusive classes, change it into exclusive.
(i.e. continuous) classes.
(ii) Choose a suitable scale and mark the class intervals on the x-axis and the
corresponding frequencies on the Y-axis.
(iii) Draw adjacent rectangles on x-axis with class interval as base and the
corresponding frequencies as height.
Facts to remember
1. Bar graph is used for ungrouped data whereas histogram is used for grouped data.
2. There are gaps between the bars in bar graph but there is no gap between the bars
in histogram.
Worked-out Examples
Example 1: The table given below shows the marks obtained by 36 students in
mathematics. Represent the data by drawing a histogram.
Marks 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80
No. of students 2 3 5 8 10 5 3
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Statistics (I): Classification and Graphical Representation of Data
Solution: Y Histogram of marks obtained by 45 students
16
15
14
13
12
No. of Students 11
10
9
8
7
6
5
4
3
2
1
0
10 20 30 40 50 60 70 80 X
Marks
In the case of mid-points of class intervals are given, we should first find the class intervals
of each mid-point.
Example 2 Construct a histogram from the data given in the table below.
Mid-point 7.5 12.5 17.5 22.5 27.5 32.5 37.5
Solution: Frequency (f) 4 7 10 8 13 6 2
The class intervals of each mid-point are given in the table below.
Class interval 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40
Frequency (f) 4 7 10 8 13 6 2
Y Histogram of the given dataFrequency
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0 5 10 15 20 25 30 35 40 X
Class interval
In the case of inclusive frequency distribution, we should first change it into exclusive
(or continuous) class intervals.
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Statistics (I): Classification and Graphical Representation of Data
Example 3: The table given below shows the heights of 75 plants of a garden. Construct
a histogram to represent the data.
Heights (in inch) 1 – 9 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69
No. of plants 15 6 18 11 14 9 2
Solution:
The given frequency distribution is inclusive type. So, it should be changed into exclusive
(or continuous) class interval with the help of a correction factor.
Correction factor = Lower limit of a class interval – Upper limit of previous class interval
2
10 – 9
= 2
= 1 = 0.5
2
Then, the correction factor 0.5 is subtracted from the lower limit and added to the upper
limit of each class interval.
Height in inch (Inclusive class) Height in inch (Exclusive class) No. of plants
15
1–9 0.5 – 9.5
10 – 19 9.5 – 19.5 6
20 – 29 19.5 – 29.5 18
30 – 39 29.5 – 39.5 11
40 – 49 39.5 – 49.5 14
50 – 59 49.5 – 59.5 9
60 - 69 59.5 – 69.5 2
Histogram of heights of 75 plants
19 No. of plants
18 0 . 5 9 . 5 19.5 29.5 39.5 49.5
17 Heights (in cm) 59.5 69.5
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
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EXERCISE 17.2
General section
1. a) What is histogram?
b) Write the difference between the bar graph and histogram.
2. a) The given histogram shows the distribution No. of students 30
of marks obtained by the students of class - 25
IX in mock test of mathematics. Answer the 20
following questions. 15
10
(i) Find the number of students who obtained
more than 70 marks. 5
(ii) Find the number of students who obtained 40 50 60 70 80 90
between 50 and 80 marks. Marks obtained
(iii) Find the number of students who obtained at
most 60 marks.
b) The adjoining histogram is drawn against the No. of students 60
number of students and their weights. Study 50
it and answer the following question. 40
30
(i) How many students are there who have the 20
weight less than 30 kg? 10
(ii) In which weight group is there maximum 20 25 30 35 40 45
number of students and what is the number Weight (in kg)
of students?
(iii) Find the number of students who have the
weight at least 30 kg.
Creative section
3. a) Draw a histogram to represent the data given below in the table.
Age (in years) 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
No. of people 6 10 16 20 12 7 3
b) The marks obtained by 70 students in mathematics are given below. Construct a
histogram to represent the marks.
Marks 10–20 20–30 30–40 40–50 50–60 60–70 70–80 80–90
No. of Students 4 7 12 18 10 8 9 2
c) The daily wages of 60 workers are shown in the table given below.
Wages (in Rs) 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99
No. of people
10 15 6 17 8 4
Construct a histogram to represent the wages of the workers.
d) The values of diastolic blood pressure of 80 people of 20-50 years are shown in the
table given below.
Blood pressure (mmHg) 70 – 75 75 – 80 80 – 85 85 – 90 90 – 95 95 – 100
No. of people 10 17 23 16 9 5
Construct a histogram to show the above data.
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Statistics (I): Classification and Graphical Representation of Data
e) Draw a histogram to represent the data given below in the table.
Mid-point 12 16 20 24 28 32 36 40
Frequency 5 11 17 13 15 20 12 8
4. a) The heights (in cm) of 30 pupils are given below:
135, 140, 120, 160, 110, 107, 150, 136 102, 113,
116, 109, 129, 124, 131, 147, 142, 123, 118, 152
128, 143, 127, 134, 115, 126, 136, 147, 129, 139
Construct a frequency distribution table of continuous class with length 10 and
draw a histogram to represent the data.
b) The daily wages (in Rs) of 40 employees of a factory are given below:
415, 440, 405, 412, 402, 418, 446, 406, 435, 422,
425, 430, 428, 416, 445, 421, 449, 437, 441, 419,
432, 426, 443, 436, 424, 438, 423, 448, 414, 427,
443, 412, 448, 447, 425, 415, 445, 434, 440, 428
Construct a frequency distribution table of continuous class with length 10 and
draw a histogram to represent the data.
Project work and activity section
5. a) Measure the heights of every student in your class. Show the heights in a frequency
distribution table of continuous class of length 10. Then, present the data in a
histogram.
b) Show the marks obtained by the students of your class in mathematics in recently
conducted exam in a frequency distribution table of continuous class of length 5.
Then, show it in a histogram.
2. Frequency polygon
Frequency polygon is an alternative to a histogram that represents the similar information
about the data. It is the piecewise linear curve formed by connecting the midpoints of
the tops of the adjacent rectangles of the histogram. In other words, it is a graphical
representation of continuous data in which the frequencies of the classes are plotted by
dots against the mid-points of each class and adjacent dots are joined by straight line
segments. The frequency polygon can be constructed by the following two methods.
a) Using a histogram
Study the following steps to construct a frequency polygon using a histogram.
(i) Draw a histogram from the given data.
(ii) Mark the mid-point at the top of each rectangle of the histogram drawn.
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(iii) Also, mark the mid-point of the class interval preceding the lower class interval
and midpoint of the class interval succeeding the highest class interval with zero
frequencies.
(iv) At last, join the consecutive midpoints by start line segments to obtained the
required frequency polygon.
Worked-out Examples
Example 1: The marks obtained by the students in mathematics are given below.
Draw a frequency polygon by using histogram.
Marks obtained 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90
Number of students 6 10 18 24 20 14 12 5
Solution:
25
20
Number of students 15
10
5
0 10 20 30 40 50 60 70 80 90 100
Marks obtained
b) Without using a histogram
Study the following steps to construct a frequency polygon without using a histogram.
(i) Find the mid-value (class-mark) of each interval using the formula,
mid-value = lower limit + upper limit
2
(ii) On a graph paper, mark the mid-values along X-axis and their corresponding
frequencies along Y-axis.
(iii) At last, join the consecutive midpoints by start line segments to obtained the
required frequency polygon.
Example 2: Draw a frequency polygon for the following data without using data.
Weight (in kg) 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70
Number of people 6 12 18 20 15 9
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Solution: Mid-value (m) Number of people (f) Points
Weight (in kg) 0 (5, 0)
0 - 10 0 + 10 = 5 6 (15, 6)
10 - 20 2 12 (25, 12)
20 - 30 10 + 20 18 (35, 18)
30 - 40 2 = 15 20 (45, 20)
40 - 50 15 (55, 15)
50 - 60 20 + 30 = 25 9 (65, 9)
60 - 70 2 0 (75, 0)
70 - 80 30 + 40
2 = 35
40 + 50 = 45
2
50 + 60
2 = 55
60 + 70 = 65
2
70 + 80
2 = 75
Frequency polygon
Number of people 22
20
18 15 25 35 45 55 65 75
16 Weight in kg
14
12
10
8
6
4
2
05
EXERCISE 17.3
General section
1. a) Define frequency polygon.
b) Write down the ways of drawing the frequency polygon.
2. a) The marks obtained by 90 students of class 9 in Mathematics are shown in the
table given below. Construct a frequency polygon to represent the data by using
histogram.
Marks obtained 20-30 30-40 40-50 50-60 60-70 70-80 80-90
Number of students 3 8 10 25 20 15 9
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b) From the data given below, draw a histogram and frequency polygon imposed on it.
Data 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 6 10 16 20 15 8 5
3. a) Construct a frequency polygon to represent the data without using histogram.
Weight (in kg) 10-20 20-30 30-40 40-50 50-60 60-70
Number of people 10 15 18 20 25 15
b) In a study of dental problem, the following data was recorded. Present the data in a
frequency polygon without using histogram.
Age (in years) 0-10 10-20 20-30 30-40 40-50 50-60 60-70
No. of people 5 14 25 30 35 40 20
Creative section
4. a) The marks obtained by 40 students in mathematics in the final examination of class
9 are given below.
65, 40, 24, 18, 36, 45, 75, 89, 60, 70, 15, 99, 52, 85, 64, 44, 28, 23, 37, 58,
46, 55, 27, 16, 49, 73, 90, 67, 95, 80, 34, 94, 65, 26, 41, 63, 58, 25, 75, 81
(i) Make a frequency distribution table with class interval of length 10.
(ii) Construct the frequency polygon for the data.
b) The following are the ages (in years) of patients admitted in a month in a hospital.
25, 18, 61, 23, 35, 60, 46, 57, 70, 15, 12, 72, 39, 51, 30, 20,
32, 71, 42, 59, 29, 19, 38, 24, 48, 40, 32, 54, 49, 28, 68, 45
(i) Construct a frequency distribution table of continuous class with length 10.
(ii) Draw a frequency polygon to represent the data.
Project work and activity section
5. a) Collect the ages of family members of your 10 classmates. Show the ages in a
frequency distribution table of class of length 10. Then, construct the frequency
polygon.
b) Show the marks obtained by the students of your class in mathematics in recently
conducted exam in a frequency distribution table of continuous class of length 5.
Then, represent the data in the frequency polygon.
3. Ogive (Cumulative frequency curve)
The sum of the frequencies of all the values up to a given value is known as
cumulative frequency. It is denoted by c.f. Ogive is a graphic presentation of the
cumulative frequency distribution of continuous series.
In the case of a continuous series, if the upper limit (or the lower limit) of each
class interval is taken as x-coordinate and its corresponding c.f. as y-coordinate and
the points are plotted in the graph, we obtain a curve by joining the points with
freehand. Such curve is known as ogive or cumulative frequency curve.
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Statistics (I): Classification and Graphical Representation of Data
Since ‘less than’ c.f. and ‘more than’ c.f. are the two types of cumulative frequency
distributions, there are two types of ogive. They are less than ogive and more than
ogive.
(i) Less than ogive (or less than cumulative frequency curve)
When the upper limit of each class interval is taken as x-coordinate and its
corresponding frequency as y-coordinate, the ogive so obtained is known as
less than ogive (or less than cumulative frequency curve). Obviously, less than
ogive is an increasing curve, sloping upwards from left to right and has the
shape of an elongated S.
(ii) More than ogive (or more than cumulative frequency curve)
When the lower limit of each class interval is taken as x-coordinate and its
corresponding frequency as y-coordinate, the ogive so obtained is known as
more than ogive (or more than cumulative frequency curve). More than ogive
is a decreasing curve sloping downwards from left to right and has the shape of
an elongated S, upside down.
17.8 Construction of less than ogive and more than ogive
Study the following steps to construct a less than ogive.
(i) Make a less than cumulative frequency table.
(ii) Choose the suitable scale and mark the upper class limits of each class interval
along X-axis and cumulative frequencies along Y-axis.
(iii) Plot the coordinates of (the upper limit, less than c.f.) on the graph.
(iv) Join the point by freehand and obtain a less than ogive.
In the case of more than ogive, we should prepare the more than cumulative
frequency table. The lower class limits of each class interval are marked on x-axis.
Then, the process is similar to the construction of less than ogive.
Worked-out Examples
Example 1: Look at the given ‘less than’ ogive and answer Cumulative 50
the following questions. No. of students 40
(i) How many students are there in all? 30
(ii) How many students have obtained less 20
10
than 20 marks?
(iii) In which class interval of obtained 10 20 30 40 50
Marks obtained
marks, there is the maximum number
of students?
Solution:
(i) Total number of students is 40
(ii) Number of students who obtained less than 20 marks is 10.
(iii) The class interval of marks obtained by maximum number of students is (20 - 30).
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Example 2: The table given below shows the marks obtained by 50 students in
Mathematics. Construct (i) less than ogive and (ii) more than ogive.
Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
No. of students 3 8 10 17 6 2 4
Solution: Less than cumulative frequency table. Coordinates (x, y)
No. of students (f) Upper limit less than c.f. (10, 3)
Marks (20, 11)
0 – 10 3 10 3 (less than 10) (30, 21)
10 – 20 8 20 11 (less than 20) (40, 38)
20 – 30 10 30 21 (less than 30) (50, 44)
30 – 40 17 40 38 (less than 40) (60, 46)
40 – 50 6 50 44 (less than 50) (70, 50)
50 – 60 2 60 46 (less than 60)
60 - 70 4 70 50 (less than 70)
The less than ogive
Less than cumulative frequencies 60
50 (70, 50)
(60, 46)
40 (50, 44)
(40, 38)
30
20 (30, 21)
10 (20, 11)
0 (10, 3)
10 20 30 40 50 60 70
Upper limits
More than cumulative frequency table
Marks No. of students (f) Lower limit more than c.f. Coordinates
(0, 50)
0 – 10 3 0 47 + 3 = 50 (more than 0) (10, 47)
(20, 39)
10 – 20 8 10 39 + 8 = 47 (more than 10) (30, 29)
(40, 12)
20 – 30 10 20 29 + 10 = 39 (more than 20) (50, 6)
(60, 4)
30 – 40 17 30 12 + 17 = 29 (more than 30)
40 – 50 6 40 6 + 6 = 12 (more than 40)
50 – 60 2 50 4 + 2 = 6 (more than 50)
60 - 70 4 60 4 (more than 60)
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Statistics (I): Classification and Graphical Representation of DataMore than cumulative frequencies
The more than ogive
60
(0, 50)
50 (10, 47)
40 (20, 39)
30 (30, 29)
20
(40, 12)
10 (50, 6) (60, 4)
0 10 20 30 40 50 60 70
Lower limits
If we draw both the less than ogive and more than ogive of a distribution on the same
graph paper, they intersect at a point. The foot of the perpendicular drawn from the point of
intersection of two ogives to the x-axis gives the value of median of the distribution.
For example,
Marks No. of students (f) Upper limit Less than c.f. Lower limit More than c.f.
0 – 10 3 10 30 50
10 – 20 8 20 11 10 47
20 – 30 10 30 21 20 39
30 – 40 17 40 38 30 29
40 – 50 6 50 44 40 12
50 – 60 2 60 46 50 6
60 – 70 4 70 50 60 4
60
Less than cumulative frequencies 50 (0, 50) (70, 50)
(10, 47) (60, 46)
(50, 44)
40 (20, 39) (40, 38)
30 (30, 29)
20 (30, 21)
10 (20, 11) (40, 12)
(50, 6) (60, 4)
(10, 3)
0 10 20 30 40 50 60 70
Upper limits
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From the graph, the perpendicular drawn from the point of intersection of two ogives meets
x-axis at 32.65 (approx.) units from the origin. So, the required median of the given
distribution is 32.65.
EXERCISE 17.4
General section
1. a) Looking at the given less than cumulative frequency curve, answer the following
questions. 60
(i) How many people are participated in the 50
survey? 40
(ii) How many people are there, who are less Cumulative 30
than 30 years? numbers of people 20
10
(iii) How many people are there in the age group
(30 - 40) years? 10 20 30 40 50 60
Ages (in years)
b) Looking at the given less than ogive curve, Numbers of students 35 Y
answer the following questions. 30
25
(i) How many students are there who obtained 20
less than 50 marks? 15
10
(ii) In which class interval do the marks obtained
by the maximum member of students fall? 5
(iii) Find the number of students who obtained 10 20 30 40 50 60 70
marks in the interval (10 - 20)? Marks obtained less than
2. a) Looking at the more than ogive curve given Numbers of students 24
alongside, answer the following questions. 20
16
(i) How many students are there? 12
8
(ii) How many students are there who obtained
more than 20 marks? 4
10 20 30 40 50 60
Marks obtained (more than)
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90
b) Look at the given more than ogive and answer Numbers of workers 75
the following questions. 60
(i) How many workers are there altogether? 45
30
(ii) How many workers have monthly salary more 15
than Rs 15,000?
5 10 15 20 25 30 35
Creative section Monthly salary (in Rs 1,000)
3. a) Draw a ‘less than’ ogive from the data given below.
Data 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
Frequency 5
8 15 10 6 3
b) Construct a ‘more than’ ogive from the data given below.
Marks 20–30 30–40 40–50 50–60 60–70 70–80 80–90 90–100
No. of students 6 10 16 15 10 8 4 3
c) The table given below shows the marks obtained by 60 students in mathematics.
Construct (i) less than ogive and (ii) more than ogive.
Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60
No. of students
4 10 20 15 6 5
4. a) Draw ‘less than’ ogive and ‘more than’ ogive on the same coordinate axes and find
the median of the distribution.
Data 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
Frequency 6 12 18 20 15 9
b) The table given below shows the daily wages 40 workers of a factory.
Wages (in Rs) 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75 75 – 80
No. of workers 4 6 10 12 5 3
Draw ‘less than’ ogive and ‘more than’ ogive on the same coordinate axes and find
the median of the distribution.
5. a) The following are the marks obtained by the students in mathematics in an
examination.
49, 23, 64, 35, 44, 60, 79, 20, 41, 50
32, 34, 25, 57, 40, 54, 23, 60, 63, 68
(i) Make a frequency distribution table of class interval 10.
(ii) Construct a less than ogive from the above data.
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b) The following are the ages (in years) of patients admitted in a month in a hospital.
15, 12, 23, 35, 46, 57, 18, 12, 39, 51
32, 42, 59, 18, 38, 48, 40, 32, 54, 49
(i) Make a frequency distribution table of class size 10.
(ii) Construct a more than ogive from the above data.
OBJECTIVE QUESTIONS
Let’s tick (√) the correct alternative.
1. The correction factor of a data with classes 10 – 19, 20 – 29, 30 – 39, … is
(A) 0.5 (B) 1 (C) 1.5 (D) 2
2. The number of times an observation occurs in the given data is called
(A) tally marks (B) data (C) frequency (D) cumulative frequency
3. The data that can take values between the class intervals is
(A) individual (B) discrete (C) continuous (D) frequency
4. In a frequency distribution, if the lower limit of a class is 20 and width of the class is
4, what is its upper limit?
(A) 12 (B) 18 (C) 22 (D) 24
5. The mid-value of a class interval 40-60 is
(A) 45 (B) 50 (C) 55 (D) 57
6. Which of the following is the graphical representation of ungrouped data?
(A) histogram (B) frequency polygon (C) pie chart (D) ogive
7. Which of the following is the graphical representation of grouped data?
(A) histogram (B) pie chart (C) bar graph (D) pictograph
8. Histogram is a graph of a …………… frequency distribution.
(A) continuous (B) discontinuous (C) discrete (D) individual
9. A line graph for the continuous frequency distribution is
(A) frequency polygon (B) histogram (C) less than ogive (D) more than ogive
10. The frequency polygon can be drawn by
(A) using histogram (B) without using histogram (C) only (A) (D) both (A) and (B)
Vedanta ICT Corner
Please! Scan this QR code or
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Statistics (II): Measures of Central Tendency
Unit 18 Statistics (II): Measures of
Central Tendency
18.1 Central tendency –Looking back
Classwork-Exercise
1. Let’s say and write the answers as quickly as possible.
a) Average of 2 and 4 is ................. b) Average of 1, 7 and 13 is .................
c) Average of 4, 6, 8, 10 is ................. d) Average of 3 and x is 5, x = .................
e) Average of m and 8 is 6, m = .................
2. a) Σfx = 50, n = 5, x = .............. b) Σfx = 85, n = 17, x = ..............
d) Σfx = 200, x =20, n = ..............
c) Σfx = 120, x =15, n = ..............
e) x =12, n = 15, Σfx =.............. f) x =18, n = 20, Σfx =..............
3. a) Median of 4, 6, 9 is ..................... b) Median of 5, 8, 10, is .....................
c) Median of 5, 1, 4, 3, 7 is ............. d) Median of 10, 8, 15, 17, 12 is .............
18.2 Measures of central tendency
The measure of central tendency gives a single central value that represents the
characteristics of entire data. A single central value is the best representative of the
given data towards which the values of all other data are approaching.
Average of the given data is the measure of central tendency. There are three types
of averages which are commonly used as the measure of central tendency. They are:
mean, median and mode.
18.3 Arithmetic mean
Arithmetic mean is the most common type of average. It is the number obtained by
dividing the sum of all the items by the number of items.
sum of all the items
i.e., mean = the number of items
(i) Mean of non–repeated data
If x represents all the items and n be the number of items, then mean ( x ) = Σx
n
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Statistics (II): Measures of Central Tendency
Worked-out Examples
Example 1: Calculate the average of the following marks obtained by 7 students of a
class in mathematics.
87, 63, 45, 72, 95, 35, 79
Solution:
Here, Σx = 87 + 63 + 45 + 72 + 95 + 35 + 79 = 476 and n = 7
Now, mean ( x) = Σx = 476 = 68
n 7
Example 2: If the average of the following wages received by 5 workers is Rs 400, find
the value of p.
300, 380, p, 420, 460.
Solution:
Here, Σx = 300 + 380 + p + 420 + 460 = 1560 + p and n = 5
Now, a verage = Σx
n
or, 400 = 1560 + p or, 1560 + p = 2000 or, p = 440
5
Hence, the required value of p is Rs 440.
Example 3: In a discrete data, if Σfx = 400 + 20a and Σf = 20 + a, find the mean ( x ) .
Solution:
Here, Σfx = 400 + 20a and Σf = 20 + a, x = ?
Now, x = Σfx = 400 + 20a = 20(20 + a) = 20
Σf 20 + a 20 + a
Example 4: In a discrete series, the mean is 50, Σfx = 100p + 150 and N = 4p – 15, find
the values of p and N.
Solution:
Here, mean ( x ) = 50, Σfx = 100 p + 150 and N = 4p – 15
Σfx 100p + 150
Now, x = N or, 50 = 4p – 15
or, 50 (4p – 15) = 100p + 150
or, 200p – 750 = 100p + 150
or, 100p = 900
∴ p = 9
Also, N = 4p – 15 = 4 × 9 – 15 = 21
Hence, the required values of p is 9 and N is 21.
(ii) Mean of individual repeated data (Mean of a frequency distribution)
In the case of repeated data, follow the steps given below to calculate the mean.
– Draw a table with 3 columns.
– Write down the items (x) in ascending or descending order in the first column
and the corresponding frequencies in the second column.
– Find the product of each item and its frequency (fx) and write in the third
column.
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Statistics (II): Measures of Central Tendency
– Find the total of f column and fx column.
– Divide the sum of fx by the sum of f (total number of items), the quotient is the
required mean.
Example 5: From the table given below, calculate the mean mark.
Marks 14 17 28 35 40
5 10 15 8 12
No. of students
Solution:
Calculation of mean marks:
Marks (x) No. of students (f) fx
14 5 70
17 10 170
28 15 420
35 8 280
40 12 480
Total N = 50 Σfx =1420
Now, mean mark ( x ) = Σfx = 1420 = 28.4
N 50
Hence, the required mean mark is 28.4.
Example 6: If the mean of the data given below be 32, find the value of p.
Solution: x 10 20 30 40 50 60
f 5 8 10 9 p 1
x f fx
10 5 50
20 8 160
30 10 300
40 9 360
50 p 50p
60 1 60
Total N = 33 + p Σfx = 930 + 50p
Now, mean ( x ) = Σfx
N
930 + 50p
or, 32 = 33 + p
or, 1056 + 32p = 930 + 50p
or, 18p = 126
or, p = 7
Hence, the required value of p is 7.
EXERCISE 18.1
General section
1. a) The marks obtained by 6 students out of 50 full marks are given below. Find the
average marks.
48, 26, 36, 14, 42, 38
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Statistics (II): Measures of Central Tendency
b) The daily wages (in Rs) of 5 workers are given below. Calculate the average wage.
360, 500, 450, 540, 400
c) The ages of Anupa, Bimlesh, Chandra, Dipika and Elina are 12, 18, 13, 16 and 6
years respectively. Find their average age.
d) Find the mean from the data given below.
72, 46, 54, 80, 99, 62, 56
2. a) The average age of 5 students is 9 years. Out of them the ages of 4 students are 5, 7,
8 and 15 years. What is the age of the remaining student?
b) If 7 is the mean of 3, 6, a, 9 and 10, find the value of a.
c) Find x if the mean of 2, 3, 4, 6, x, and 8 is 5.
3. a) In a discrete series, Σfx = 15p + 800, Σf = p + 10 and mean ( x ) = 20, find the value
of p.
b) In a discrete data, if Σfx = 12 + 34n, N = 12 + 3n and x = 10, find the n and N.
c) In a discrete series, if Σfx = 45 + 60k and N = 9 + 12k, find the mean ( x ).
d) In a discrete series, if the sum of frequencies (Σf) = a2 + 4a and Σfx = 25a2 + 100a,
find the mean ( x ).
Creative section
4. a) Find the mean of the data given in the table.
Marks obtained 40 50 60 70 80 90
No. of students 2 7 4 8 6 3
b) The ages of the students of a school are given below. Find the average age.
Age (in years) 5 8 10 12 14 16
No. of students 20 16 24 18 25 15
c) The weight of the teachers of a school are given below. Find the average weight.
Weight (in kg) 50 55 60 65 70 75 80
No. of teachers 2 5 7 10 4 3 1
d) Compute the arithmetic mean from the following frequency distribution table.
Height (in cm) 58 60 62 64 66 68
No. of teachers 12 14 20 13 8 5
5. a) Find the missing frequency p for the following distribution whose mean is 50.
x 10 30 50 70 90
f 17 p 32 24 19
b) The mean of the data given below is 17. Determine the value of m.
x 5 10 15 20 25 30
f 2 5 10 m 4 2
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Statistics (II): Measures of Central Tendency
c) The heights of plants planted by the students of class IX in the afforestation program
are given below. The average height of the plant is 36 cm.
Height (in cm) 5 15 25 35 45 55
Number of plants 8 5 12 p 24 16
(i) Find the value of p. (ii) Find the total number of plants.
d) The speeds of vehicles recorded in a highway during 15 minutes on a day is given
in the following table. The average speed of the vehicles was 54 km per hour.
Speed (km/hr) 10 30 50 70 90
No. of vehicles 7 k 10 9 13
(i) Find the value of k.
(ii) Find the total number of vehicles counted during the time.
6. a) Find the value of p, for the following distribution whose mean is 10.
x 3 5 8 p 16 21
f 8 4 9 10 3 6
b) Find the missing value of k if the mean of the following distribution is 21.
x 9 14 17 k 28 34
f 12 15 16 25 13 19
c) If the mean of the following data is 14, find the value of m.
x m 10 15 20 25
f7 6 8 4m
d) If the mean of the following data is 55, determine the value of ‘a’.
x 10 30 50 70 89
f 7 8 a 15 a
18. 4 Median
Let’s take a series: 25, 13, 7, 10, 16, 22, 19
Now, arranging the series in ascending order.
7 10 13 16 19 22 25
Thus, when we arrange the given series in ascending order, the fourth item 16 falls
exactly at the middle of the series. Therefore, 16 is called the median of the series.
In this way, median is the value that falls exactly at the middle of an array data.
i.e., Median = the middle value of a set of arry data
(i) Median of ungrouped data
To find the median of an ungrouped data, arrange them in ascending or descending
order. Let the total number of observation be n.
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Statistics (II): Measures of Central Tendency
⇒ If n is odd, the median is the value of the n + 1 th
2
observation.
⇒ If n is even, the median is the average of n th n th
2
2 and + 1 the observations.
Worked-out Examples
Example 1: The heights of 7 students (in cm) are given below. Find the median
height.
115, 119, 125, 110, 140, 135, 128
Solution:
Arranging the heights in ascending order, we have,
110, 115, 119, 125, 128, 135, 140
Here, n= 7 th th
Now, the position of median = n+ 1 term = 7+1 term = 4th term
2 2
Thus, median lies at the 4th position.
∴ Median = 125.
Example 2: The daily wages (in Rs) of 12 workers are given below. Calculate the median
wage.
400, 350, 450, 520, 250, 375, 480, 550, 380, 555, 600, 580
Solution:
Arranging the wages in ascending order, we have,
250, 350, 375, 380, 400, 450, 480, 520, 550, 555, 580, 600
Here, n = 12 th th
Now, the position of median = n+1 term = 12 + 1 term = 6.5th term
2 2
6.5th term is the average of 6th and 7th terms.
450 + 480 930
∴ Median = = 2 = 465
2
Hence, median wage is Rs 465.
Example 3: If the given data is in ascending order and median is 34, find the value of x.
10, 15, 26, 3x + 1, 39, 44, 53
Solution:
Here,
The given data in ascending order is 10, 15, 26, (3x + 1), 39, 44, 53
Number of observations (n) = 7 th th
Now, the position of median = n+1 term = 7 + 1 term = 4th term
2 2
∴ Median =3x + 1
But, by question median = 34
or, 3x + 1 = 34
or, 3x = 33
∴ x = 11
Hence, the required value of x is 11.
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Statistics (II): Measures of Central Tendency
(ii) Median of Discrete series
To compute the median of a discrete series of frequency distribution, we should
display the data in ascending or descending order in a cumulative frequency table.
Then, the median is obtained by using the formula,
N + 1 th
2
Median = value of observation.
Example 4: Compute the median from the table given below,
Weight (in kg) 25 30 35 40 45 50
No. of students 4 7 10 4 3 2
Solution:
Cumulative frequency table:
Weight in kg (x) No. of students (f) c.f.
25 4 4
30 7 11
35 10 21
40 4 25
45 3 28
50 2 30
Total N = 30
N + 1 th 30 + 1 th
2 2
Now, position of median = term = term = 15.5th term
In c.f. column, the c.f. just greater than 15.5 is 21 and its corresponding values is 35.
∴ Median = 35 kg.
18.5 Quartiles
Quartiles are the values that divide the data arranged in ascending or descending
order into four equal parts. Hence, there are three quartiles that divide a distribution
into four equal part.
x1 x2 x3 x4 x5 x6 x7
1st quartile (Q1) 2nd quartile (Q2) 3rd quartile (Q3)
⇒ The first or lower quartile (Q1) is the point below which 25 % of the observations
lie and above which 75 % of the observations lie.
⇒ The second quartile (Q2) is the point below which 50 % of the observations lie
and above which 50 % of the observations lie. Of course, the second quartile is
the median.
⇒ lTieheanthdiradboorveupwpheircqhu2a5rt%ileo(Qf t3h) eisotbhseeprvoaintitobnesloliwe.which 75 % of the observations
If N be the number of observations in ascending (or in descending) order of a
distribution, then in the case of discrete data th
The position of the first quartile (Q1) = N + 1 term
4
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