X Mat Ch 2 Full Notes (E)

Circles 3

T .B page 60
Q6) i)

Ans)
Join PQ.
Now ABQP and PQDC becomes
cyclic quadrilaterals.

Let ∠A = x° ,
Then ,

∠PQD = x° (any outer angle of a cyclic quadrilateral is
equal to the inner angle at the opposite vertex)

Since PQDC is cyclic ,∠C = 180°− x°
Now ,
∠A + ∠C = x° + 180°− x°

So, ∠A + ∠C = 180°
Since ∠A + ∠C = 180° we can say AB parallel to CD

( co-interior angles are supplementary)

Since AB parallel to CD , and given AC = BD,
ABDC is an isosceles trapezium.
So ABDC is a cyclic quadrilateral.

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 4

Q6) ii)

Ans)

Draw PQ & RS

Let ∠BAP = x°, then ∠PQS = x° (any outer angle of a cyclic
quadrilateral is equal to the inner angle at the opposite vertex)
Since ∠PQS = x° , then ∠PRS =180°−x° (opposite angles of a

cyclic quadrilateral are supplementary)
Since ∠PRS =180° − x°, then ∠SDC = 180° − x° (any outer
angle of a cyclic quadrilateral is equal to the inner angle at
the opposite vertex)

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 5

So ,
∠BAP + ∠SDC = x° + 180° − x°
= 180°

So considering quadrilateral ABDC,

∠A + ∠D = 180°
∠B + ∠C = 360°− 180° = 180°
Since opposite angles of quadrilateral ABDC are
supplementary,
ABDC is a cyclic quadrilateral.

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 1

Online Class – X − 23 12 / 08 / 2021
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Two chords
* Any two diameters of a circle intersect at the centre, and the

length of the four pieces are equal.

PA = PB = PC = PD
(Radii of a circle are equal)

* When two chords which are not diameters intersect within
the circle we get four pieces which are not equal.

* Finding the relation between the four parts PA, PB ,PC & PD,
when the chords AB & CD intersect within the circle at P.
Consider two chords AB & CD which
are not diameters of the circle.
The chords AB & CD intersect at P.

Draw AC & BD

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 2

We can see, All angles made by an arc on
∠A = ∠D its alternate arc are equal

∠C = ∠B

Consider △ APC & △ DPB

Since two angles of both triangles are equal, third angles are

also equal.
So △ APC & △ DPB are similar triangles.

Since sides opposite to equal angles of similar triangles are in

proportion.
PC PA
PB = PD

Cross multiplying we get,

PA × PB = PC × PD

Here PA, PB are parts of the chord AB and
PC, PD are parts of the chord CD.

So we can say,

If two chords of a circle intersect within the circle, then
the products of the parts of the two chords are equal.

PA × PB = PC × PD

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 3

Q ) In the figure two chords AB and CD intersect at a point P .

PB = 2 cm, PC = 3 cm , PD = 4 cm . Find the length of AB .

Ans)

PA × PB = PC × PD

PA × 2 = 3 × 4
12
PA = 2

PA = 6 cm

AB = PA + PB

= 6 + 2 = 8cm

Geometrical interpretation
We can interpret the product of two lengths as an area.
So,

PA × PB = Area of the rectangle with sides PA and PB

PC × PD = Area of the rectangle with sides PC and PD
So the relation PA × PB = PC × PD can be put in geometric
language as below:

If two chords of a circle intersect within a circle, then
the rectangles formed by the parts of the same chord

have equal area.

PA × PB = PC × PD

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 4

Assignment
The chords AB and CD of a circle intersect at a point P .
If PA = 9 cm , PD = 12 cm , AB = 13 cm ,
find the lengths of PB , PC and CD ?

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

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Assignment Answer 9 cm 12 cm
Given PA = 9 cm , 13 cm
So PB = AB − PA
= 13 − 9
= 4 cm

PA × PB = PC × PD

9 × 4 = PC × 12 36
9X4 12
PC = 12 = = 3cm

CD = PC + PD
= 3 + 12 = 15cm

Q) In the figure two chords AB and CD 8 cm
intersect at a point P . 5 cm
PA = 5cm, PB = 12 cm , PC = 8 cm .
Find the length of PD .

12 cm
?

Ans) PA × PB = PC × PD

5 × 12 = 8 × PD
5 X 12
PD = 8

= 60
8

= 7.5 cm

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 2

Construction 3

Q1) Draw a rectangle of width 5 centimetres and height 3
centimetres.
Draw a rectangle of the same area with width 6 centimetres.

Ans)
Steps :
1. Draw a rectangle of width 5 cm and height 3 cm .
2. Let the name of the rectangle be ABCD .
3. Extend AB to E such that BE = 3cm .
4. Since given length of new rectangle is 6 cm ,extend CB to P
such that BP = 6cm .
5. Join AP & EP to get △ AEP .
6. Draw perpendicular bisectors of AP & EP, they intersect
at a point say M. With M as centre draw a circle which
passes through A, E & P.
7. Let this circle intersect BC at N .
8. Now we get two chords AE & PN .
On the compass measure BN , mark this measurement on
BE as BR.
9. With PB & BR as length and breadth complete the rectangle
BRQP .

Now area of rectangle ABCD & area of rectangle BRQP
are same.

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 3

Assignment
Q2) Draw a rectangle of length 4 centimetres and width 3

centimetres .
Draw another rectangle of the same area with one side 5
centimetres .

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

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Assignment Answer

Q) Draw a rectangle of length 4 centimetres and width 3
centimetres.
Draw another rectangle of the same area with one side 5
centimetres.

Ans)

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 2

Finding the relation between parts of two intersecting chords −
when one chord is a diameter and other chord is perpendicular
to the diameter

In the picture given,
AB is a diameter and CD is a chord perpendicular to AB.

Since the chords AB & CD
intersect at P.

PA × PB = PC × PD . . . . . . (1)
We know, the perpendicular
drawn from the centre of a circle
bisects the chord.
So AP bisects CD
∴ PC = PD

Substituting in (1) we have,

PA × PB = PC × PC
PA × PB = PC2

The product of the parts into
which a diameter of a circle is cut
by a perpendicular chord, is equal
to the square of half the chord.

PA × PB = PC2

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 3

The relation PA × PB = PC 2 can be put in geometric
language as below:

The area of the rectangle formed of parts into which a
diameter of a circle is cut by a perpendicular chord is
equal to the area of the square formed by half the chord.

PA × PB = PC2

Since AB is the diameter , by considering the semi circle
we can observe above relation as ,

PA × PB = PC2

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 4

Q1) In the picture PA = 4cm, PB = 2cm, find PC ?

Ans) Given PA = 4cm, PB = 2cm
PA × PB = PC2
4 × 2 = PC2
PC2 = 8
∴ PC = √ 8 cm

Q2) In the picture PA = 6cm, PB = 4cm, find PC ?

Ans) Given PA = 6 cm, PB = 4 cm
PA × PB = PC2
6 × 4 = PC2
PC2 = 24
∴ PC = √ 24 cm

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 5

Q3) In the picture PA = 9cm, PC = 6cm, find PB ?

Ans) Given PA = 9 cm, PC = 6 cm

PA × PB = PC2

9 × PB = 62

9 × PB = 36
36
PB = 9

∴ PB = 4 cm

Let PA = a , PB = b
PA × PB = PC2
a × b = PC2
PC2 = ab
PC = √ ab

If PA = a , PB = b

then PC = √ ab

We can use this relation to draw lines of
irrational lengths like √ 8 , √ 15 , √ 24 etc .

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 6

Construction 4

Q1) Draw a line of length √ 12 cm.

Ans)

Find two numbers whose product is 12.

4,3 6,2 12 , 1 are the numbers.

Choose any pair, let us choose 6 , 2 .

Steps
1 : Draw line AB of length 6 + 2 = 8 cm .
2 : Mark the mid point .
3 : Draw a semicircle with AB as diameter .
4 : Mark the point P on AB such that AP = 6cm and

PB = 2cm.
5 : Through P draw line CP perpendicular to AB.

6 × 2 = PC2
PC2 = 12

PC = √ 12 cm

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 7

Q2) Draw a square of area 15cm2 .
Ans) Given , area of square = 15cm2

∴ side of square = √ 15 cm

Find two numbers whose product is 15.

5,3 15 , 1 are the numbers.

Choose any pair, let us choose 5 , 3 .

Steps

1 : Draw line AB of length 5 + 3 = 8 cm.

2 : Mark the mid point.

3 : Draw a semicircle with AB as diameter .

4 : Mark the point P on AB such that AP = 5cm and

PB = 3cm.

5 : Through P draw line CP perpendicular to AB.

5 × 3 = PC2

PC2 = 15

PC = √ 15 cm

6 : Extend line PB, measure PC on the compass,

using this measurement with P as centre draw an

arc on this line and mark the point Q.

7 : With the same measurement draw arcs by keeping

the compass at Q & C to obtain the point R.

Complete the square PQRC.

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 8

Assignments
Q1) Draw a line of length √ 7 cm .
Q2) Draw a square of area 8 cm2 .
Q3) Draw a square of area 24 cm2 .

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

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Online Class – X − 26 24 / 08 / 2021

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Answers of last class assignment

Q1) Draw a line of length √ 7 cm.

Ans) Find two numbers whose product is 7.

7 , 1 are the numbers. 7×1=7

Draw line AB of length 7 + 1 = 8 cm and do the construction

as per the steps given under construction 4 (Q1) .

Q2) Draw a square of area 8 cm2 .

Ans) Given , area of square = 8cm2
∴ side of square = √ 8 cm

Find two numbers whose product is 8.

8 , 1 are the numbers. 8×1=8

Draw line AB of length 8 + 1 = 9 cm and do the construction

as per the steps given under construction 4 (Q2) .

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 2

Q3) Draw a square of area 24 cm2 .

Ans) Given , area of square = 24cm2

∴ side of square = √ 24 cm

Find two numbers whose product is 24.

Let us take 6 & 4 6 × 4 = 24

Draw line AB of length 6 + 4 = 10 cm and do the construction

as per the steps given under construction 4 (Q2) .

Q ) Draw an equilateral triangle of side √ 12 cm .

Ans) Find two numbers whose product is 12.

Let us take 6 & 2 6 × 2 =12

1 . Draw line AB of length 6 + 2 = 8 cm.

2 . Mark the mid point.

3 . Draw a circle with AB as diameter.

4 . Mark the point P on AB such that AP = 6 cm and

PB = 2 cm.

5 . Through P draw line PQ perpendicular to AB.

now PQ = √ 12 cm

6 . Measure PQ on the compass, using this

measurement draw arcs with P & Q as centres

to intersect at R.

7 . Complete triangle PQR

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 3

Construction 5
Q) Draw a rectangle of width 5 centimetres and height

3 centimetres and draw a square of the same area.

Ans) Steps:
1. Draw a rectangle of width 5 cm and height 3 cm .
2. Let the name of the rectangle be ABCD .
3. Extend AB to Y such that BY = 3cm.
4. Draw the perpendicular bisector of AY.
5. Mark the midpoint of AY as M .
6. Now draw a semicircle below ,with AY as diameter.
Extend the side CB of the rectangle downwards to meet
the semi circle at P.
This line BP is the side of the required square.
7. Extend line BY, measure BP on the compass,
using this measurement with B as centre draw an arc on
this extended line and mark the point R.
8. With the same measurement draw arcs by keeping
the compass at P & R . Let these arcs meet at Q.
Complete the square BPQR .

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 4

5 cm 3cm
3cm

Q) Draw a square of area 5 square centimetres in three
different ways.
(Recall Pythagoras theorem)

Ans)
Method 1 ( Using PA × PB = PC2 )

Area of square = 5 cm2 ∴ Side of square = √ 5 cm
Find two numbers whose product is 5.

5 , 1 are the numbers.
Draw line AB of length 5 + 1 = 6 cm and do the construction
as per the steps given under construction 4 (Q2)

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 5

Method 2 ( Using Pythagoras Theorem)
In a right triangle ,
(Hypotenuse) 2 = (Base) 2 + (Altitude) 2
If base = 2cm & Altitude = 1cm, then
(Hypotenuse) 2 = (2) 2 + (1) 2
(Hypotenuse) 2 = 4 + 1 = 5

Hypotenuse = √ 5

1. Draw line AB of length 2 cm.
2. Draw a line perpendicular to AB at B, mark BC = 1cm
3. Join AC, now AC = √ 5 cm .

By Pythagoras Theorem

AC = √ 22+ 12 = √ 4 +1 = √ 5

4. Draw perpendiculars at A & C, measure AC on the compass,
with this measurement draw arcs on these perpendiculars
and mark the points D & E .

5. Complete the square ACDE .
Area of square ACDE = 5 cm2 .

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 6

Method 3 ( Using Pythagoras Theorem)
In a right triangle ,
(Hypotenuse) 2 = (Base) 2 + (Altitude) 2
so, (Altitude) 2 = (Hypotenuse) 2 − (Base) 2
If base = 2cm & hypotenuse = 3cm , then
(Altitude) 2 = (3) 2 − (2) 2
=9−4
=5
Altitude = √ 5

1. Draw line AB of length 2 cm.
2. Draw a perpendicular at B.
3. Measure 3 cm on the compass, with A as centre draw

an arc on this perpendicular and mark point C.
Now BC = √ 5 cm .

4. Draw perpendiculars at B & C, measure BC on the
compass, draw arcs on these perpendiculars with this

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

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measurement and mark the points E &D .
5. Complete the square BCDE .

Area of square BCDE = 5 cm2 .

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

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Text Book Page 67

Q1) In the picture, chords AB and CD of the circle are extended
to meet at P.

i ) Prove that the angles of ΔAPC
and ΔPBD, formed by joining
AC and BD, are the same.

ii ) Prove that PA × PB = PC × PD.
iii ) Prove that if PB = PD, then ABDC is an isosceles trapezium.

Ans)
i) Consider △ PBD and △ APC

∠P is common

ABDC is a cyclic quadrilateral

So,

∠PBD = ∠C

Outer angle at any vertex of a cyclic

quadrilateral is equal to the inner

∠PDB = ∠A angle at the opposite vertex.

∴ The angles of ΔAPC and ΔPBD are same.

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

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ii) ΔAPC and ΔPBD are similar triangles. So sides opposite to

equal angles are proportional.
PB PD
∴ PC = PA

Cross multiplying we have

PA x PB = PC x PD

iii) PA x PB = PC x PD . . . . . . (1)

Given PB = PD , substituting in (1) we have
PA x PB = PC x PB

PA = PC
∴ ΔPAC is an isosceles triangle.
∴ ∠A =∠C Angles opposite to the equal sides of isosceles

triangle are equal.

Since ABDC is a cyclic quadrilateral

∠C + ∠ABD = 180° ∠A + ∠ABD = 180°

Since the co-interior angles are supplementary,

AC is parallel to BD

AB = PA – PB = PC – PD = CD AB = CD

AC parallel BD & AB = CD
∴ ABDC is an isosceles trapezium.

Q5) In the picture, a line through the centre of a circle cuts a
chord into two parts:

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 3

What is the radius of the circle? 5 cm r cm
Ans) Extend both ends of OP to meet the

circle at C & D.
Since the chords AB and CD intersect
within the circle at P,

PA x PB = PC x PD

Let the radius of circle be ‘ r ’ cm. 4 cm 6 cm
So, PC = (r + 5)cm

PD = (r − 5)cm

∴ 4 x 6 = (r + 5) ( r − 5)
24 = r2 − 52
24 = r2 − 25
r2 = 24 + 25
r2 = 49

r = √ 49
r = 7cm

Q6) In the picture, a line through the centre of a circle meets a
chord of the circle:

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 4

What are the lengths of the two pieces of the chord?
Ans) Extend both ends of OP to meet the circle at C & D.

Given radius is 7 cm
So , OC = OD = 7
PD = OD − OP
=7−3 = 4
PC = OC + OP
= 7 + 3 = 10

Since the chords AB and CD intersect
within the circle at P,
PA x PB = PC x PD

Let PA = x & PB = y

So, x × y = 10 × 4

xy = 40 . . . . . . . . . . (1)

From the figure

x + y = 13 . . . . . . . . . .(2)

We know,

(x − y)2 = (x + y)2 − 4xy

From equations (1) & (2) we have

(x − y)2 = (13)2 − 4 × 40

x − y = √ 169−160

= √9 = 3

x + y = 13 . . . . . . (2)

x − y = 3 . . . . . . (3)

(2) + (3) 2x = 16
16
x= 2 =8

Substituting value of x in equation (2)

8 + y = 13

y = 13 − 8 = 5

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi

Circles 5

∴ PA = 8 cm , PB = 5 cm

Cecilia Joseph, St. John De Britto’s A. I. H. S, Fortkochi