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Therefore, 3B!$ 3B"$ %SSS rule&≅

This gives !B$ "B$ %C4CT&∠ ∠

Construction 11.( : To construct the perpendicular bisector of a given line

seg+ent.

Given a line seg+ent AB, we want to construct its perpendicular bisector.

Steps of Construction :

1. Taking A and B as centres and radius +ore than

1 AB, draw arcs on both sides of the line seg+ent

(

AB %to intersect each other&.

(. et these arcs intersect each other at 4 and 5.

/oin 45 %see $ig.11.(&.

. et 45 intersect AB at the point 6. Then line

465 is the re-uired perpendicular bisector of AB.

e t u s s e e h ow t h i s +e t h o d g i v e s u s t h e

perpendicular bisector of AB.

/oin A and B to both 4 and 5 to for+ A4, A5, B4

and B5.

0n triangles 4A5 and 4B5,

A4 B4 %Arcs of e-ual radii&

A5 B5 %Arcs of e-ual radii&

45 45 %Co++on&

Therefore, 3 4A5 3 4B5 %SSS rule&≅

So, A46 B46 %C4CT&∠ ∠

)ow in triangles 46A and 46B,

A4 B4 %As before&

46 46 %Co++on&



Given the base BC, a base angle, sa B and the su+ AB 7 AC of the other two sides∠

of a triangle ABC, ou are re-uired to construct it.


1. "raw the base BC and at the point B +ake an

angle, sa @BC e-ual to the given angle.

(. Cut a line seg+ent B" e-ual to AB 7 AC fro+

the ra B@.

. /oin "C and +ake an angle "C e-ual to B"C.∠

?. et C intersect B@ at A %see $ig. 11.?&.

Then, ABC is the re-uired triangle.

et us see how ou get the re-uired triangle.

Base BC and B are drawn as given. )e*t in triangle∠

AC",

AC" A"C %B construction&∠ ∠

Therefore, AC A" and then

AB B" A" B" AC

AB 7 AC B"

Alternative +ethod :

$ollow the first two steps as above. Then draw

perpendicular bisector 45 of C" to intersect B" at

a point A %see $ig 11.&. /oin AC. Then ABC is the

re-uired triangle. )ote that A lies on the perpendicular

bisector of C", therefore A" AC.

2e+ark : The construction of the triangle is not

possible if the su+ AB 7 AC D BC

Construction 11. : To construct a triangle given its base, a base angle and the

difference of the other two sides.



Given the base BC, a base angle, sa B and the difference of other two sides∠

AB AC or AC AB, ou have to construct the triangle ABC. Clearl there are

following two cases:

Case %i& : et AB EgtF AC that is AB AC is given.


1. "raw the base BC and at point B +ake an angle

sa @BC e-ual to the given angle.

(. Cut the line seg+ent B" e-ual to AB AC fro+ ra B@.

. /oin "C and draw the perpendicular bisector, sa 45 of "C.

?. e t i t i n t e r s e c t B@ a t a p o i n t A. / o i n A C

%see $ig. 11.<&. then ABC is the re-uired triangle.

et us now see how ou have obtained the re-uired triangle ABC.

Base BC and B are drawn as given. The point A lies on the perpendicular bisector of ∠

"C. Therefore, A" AC

So, B" AB A" AB AC.

Case %ii& : et AB EltF AC that is AC AB is given.

Steps of Construction : 1. Sa+e as in case %i&.

(. Cut line seg+ent B" e-ual to AC AB fro+ the

line B@ e*tended on opposite side of line seg+ent BC.

. /oin "C and draw the perpendicular bisector, sa 45 of "C.

?. et 45 intersect B@ at A. /oin AC %see $ig. 11.&.

Then, ABC is the re-uired triangle.

ou can Hustif the construction as in case %i&.

Construction 11.6 :

angles.

To construct a triangle, given its perimeter and its two base

Given the base angles, say



the triangle ABC.

B and C and BC + CA + AB, you have to construct

Steps of Construction

!. "raw a line segment, say #$ e%ual to BC + CA + AB.

&. 'a(e angles )#$ e%ual to

*. Bisect

see ig. !!.-i/0.

B and '$# e%ual to C. )#$ and '$#. )et these bisectors intersect at a point A

ig. !!.- i/

1. "raw perpendicular bisectors 23 of A# and 4S of A$.

5. )et 23 intersect #$ at B and 4S intersect #$ at C. 6oin AB and AC

see ig !!.-ii/0.

ig. !!.- ii/

Then ABC is the re%uired triangle. or the 7ustification of the construction, you

observe that, B lies on the perpendicular bisector 23 of A#.

Therefore, #B 8 AB and similarly, C$ 8 AC.

This gives BC + CA + AB 8 BC + #B + C$ 8 #$.

Again

BA# 8 A#B As in Δ A#B, AB 8 #B/ and

Σ ιµ ιλαρλψ

ABC 8 BA# + A#B 8 & A#B 8 )#$ ACB 8 '$# as re%uired.



Theore+ 19.1 : !-ual chords of a circle subtend e-ual angles at the centre.

4roof : ou are given two e-ual chords AB and C"

of a circle with centre > . ou want

to prove that A>B C>".∠ ∠

0n triangles A>B and C>",

>A >C %2adii of a circle&

>B >" %2adii of a circle&

AB C" %Given&

Therefore, 3 A>B 3 C>" %SSS rule&≅

This gives A>B C>"∠ ∠

%Corresponding parts of congruent triangles& I

Theore+ 19.( : 0f the angles subtended b the chords of a circle at the centre

are e-ual, then the chords are e-ual.

The above theore+ is the converse of the Theore+ 19.1.

if ou take A>B C>", then∠ ∠

3 A>B 3 C>" %=hJ&≅

Can ou now see that AB C"J

Theore+ 19.( : 0f the angles subtended b the chords of a circle at the centre

are e-ual, then the chords are e-ual.

The above theore+ is the converse of the Theore+ 19.1.

if ou take A>B C>", then∠ ∠

3 A>B 3 C>" %=hJ&≅

Can ou now see that AB C"J



Theore+ 19. : The perpendicular fro+ the centre of a circle to a chord bisects

the chord.

=hat is the converse of this theore+J To write this, first let us be clear what is

assu+ed in Theore+ 19. and what is proved. Given that the perpendicular fro+ the

centre of a circle to a chord is drawn and to prove that it bisects the chord. Thus in the

converse, what the hpothesis is Kif a line fro+ the centre bisects a chord of a

circleL and what is to be proved is Kthe line is perpendicular to the chordL. So the

converse is:

Theore+ 19.? : The line drawn through the centre of a circle to bisect a chord is

perpendicular to the chord.

0s this trueJ Tr it for few cases and see. ou will

see that it is true for these cases. See if it is true, in

general, b doing the following e*ercise. =e will write

the stages and ou give the reasons.

et AB be a chord of a circle with centre > and

> is Hoined to the +idMpoint 6 of AB. ou have to

p r o v e t h a t > 6 A B . / o i n > A a n d > B⊥

%see $ig. 19.1<&. 0n triangles >A6 and >B6,

>A >B %=h J&

A6 B6 %=h J&

>6 >6 %Co++on&

Therefore, 3>A6 3>B6 %Now J&≅

This gives >6A >6B ;9 %=h J&∠ ∠

Theore+ 19. : There is one and onl one circle passing through three given

nonMcollinear points

Theore+ 19.< : !-ual chords of a circle %or of congruent circles& are e-uidistant

fro+ the centre %or centres&.



)e*t, it will be seen whether the converse of this theore+ is true or not. $or this,

draw a circle with centre >. $ro+ the centre >, draw two line seg+ents > and >6

of e-ual length and ling inside the circle Then draw chords 45

and 2S of the circle perpendicular to > and >6 respectivel

6easure the lengths of 45 and 2S. Are these differentJ )o, both are e-ual. 2epeat

the activit for +ore e-ual line seg+ents and drawing the chords perpendicular to the+.

This verifies the converse of the Theore+ 19.< which is stated as follows:

Theore+ 19. : Chords e-uidistant fro+ the centre of a circle are e-ual in

length.

=e now take an e*a+ple to illustrate the use of the above results:

Theore+ 19.8 : The angle subtended b an arc at the centre is double the angle

subtended b it at an point on the re+aining part of the circle.

4roof : Given an arc 45 of a circle subtending angles 4>5 at the centre > and

4A5 at a point A on the re+aining part of the circle. =e need to prove that

4>5 ( 4A5∠ ∠

Consider the three different cases as 0n %i&, arc 45 is +inorF in %ii&,

arc 45 is a se+icircle and in %iii&, arc 45 is +aHor.

et us begin b Hoining A> and e*tending it to a point B.

0n all the cases,

B>5 >A5 7 A5>∠ ∠ ∠

because an e*terior angle of a triangle is e-ual to the su+ of the two interior opposite

angles.

Theore+ 19.; : Angles in the sa+e seg+ent of a circle are e-ual.

Again let us discuss the case %ii& of Theore+ 19.8 separatel. Nere 4A5 is an angle∠

in the seg+ent, which is a se+icircle. Also, 4A5 1O( 4>5 1O(P 189 ;9.∠ ∠

0f ou take an other point C on the se+icircle, again ou get that

4C5 ;9∠

Therefore, ou find another propert of the circle as:



Angle in a se+icircle is a right angle.

The converse of Theore+ 19.; is also true. 0t can be stated as:

Theore+ 19.19 : 0f a line seg+ent Hoining two points subtends e-ual angles at

two other points ling on the sa+e side of the line containing the line seg+ent,

the four points lie on a circle %i.e. the are concclic&.ou can see the truth of this resultas follows:

0n $ig. 19.9, AB is a line seg+ent, which subtends e-ual angles at two points C and

". That is ACB A"B∠ ∠

To show that the points A, B, C and " lie on a circle

let us draw a circle through the points A, C and B.

Suppose it does not pass through the point ". Then it

will intersect A" %or e*tended A"& at a point, sa !

%or !Q&.

0f points A, C, ! and B lie on a circle,

ACB A!B %=hJ&∠ ∠

But it is given that ACB A"B.∠ ∠

Therefore, A!B A"B.∠ ∠

This is not possible unless ! coincides with ". %=hJ&

Si+ilarl, !Q should also coincide with ".

19.8 Cclic 5uadrilaterals

A -uadrilateral ABC" is called cclic if all the four vertices

of it lie on a circle .ou will find a peculiar

propert in such -uadrilaterals. "raw several cclic

-uadrilaterals of different sides and na+e each of these

as ABC".

T h e o r e + 1 9 . 11 : T h e s u + o f e i t h e r p a i r o f o p p o s i t e a n g l e s o f a c

c l i c

-uadrilateral is 189R.



0n fact, the converse of this theore+, which is stated below is also true.

Theore+ 19.1( : 0f the su+ of a pair of opposite angles of a -uadrilateral is

189R, the -uadrilateral is cclic.

ou can see the truth of this theore+ b following a +ethod si+ilar to the +ethod

adopted for Theore+ 19.19.

$>2 2!$!2!)C! >$ "0AG2A6S 4!AS! 2!$!2 TN! )C!2T B>> >$

CASS ;TN.....=! NA! STAT!" AB>UT TN! $0GU2!.

1. Surface area of a cuboid = 2 (lb + bh + hl)

2. Surface area of a cube = 6a2

3. Curved surface area of a cylinder = 2πrh

4. Toal surface area of a cylinder = 2πr(r + h)

!. Curved surface area of a cone = πrl

6. Toal surface area of a cone = πrl + πr"r # i.e.# πr (l + r)

$. Surface area of a s%here of radius r = 4 π r 2

&. Curved surface area of a he'is%here = 2πr 2

. Toal surface area of a he'is%here = 3πr 2

1. *olu'e of a cuboid = l b h

11. *olu'e of a cube = a3

12. *olu'e of a cylinder = πr 2h



13. *olu'e of a cone = 1,3πr 2h

14. *olu'e of a s%here of radius r = 4,3π r 3

1!. *olu'e of a he'is%here = 2,3π r 3

-ere# leers l# b# h# a# r# ec. have been used in heir usual 'eanin/#de%endin/ on he cone0.

Theorem 8.1 : A diagonal of a parallelogram divides it into two congruent

triangles.



Proof : Let ABCD e a parallelogram and AC e a diagonal !see "ig. 8.8#. $serve

that the diagonal AC divides parallelogram ABCD into two triangles% namel&% ' ABC

and ' CDA. (e need to prove that these triangles are congruent.

)n ' ABC and ' CDA% note that BC ** AD and AC is a transversal.

+o% BCA , DAC !Pair of alternate angles#∠ ∠

Also% AB ** DC and AC is a transversal.

+o% BAC , DCA !Pair of alternate angles#∠ ∠

and AC , CA !Common#

+o% ' ABC ' CDA !A+A rule#≅

or% diagonal AC divides parallelogram ABCD into two congruent

triangles ABC and CDA. -

ow% measure the opposite sides of parallelogram ABCD. (hat do &ou oserve/

0ou will find that AB , DC and AD , BC

Theorem 8.2 : In a parallelogram, opposite sides are equal.

0ou have alread& proved that a diagonal divides the parallelogram into twocongruent

triangles so what can &ou sa& aout the corresponding parts sa&% thecorresponding

sides/ The& are e2ual.

+o% AB , DC and AD , BC

ow what is the converse of this result/ 0ou alread& 3now that whatever is given

in a theorem% the same is to e proved in the converse and whatever is proved inthe

theorem it is given in the converse. Thus% Theorem 8.4 can e stated as given elow :

)f a 2uadrilateral is a parallelogram% then each pair of its opposite sides is e2ual.+o

its converse is :

Theorem 8.5 : )f each pair of opposite sides of a 2uadrilateral is e2ual% then it



is a parallelogram.

Can &ou reason out wh&/

Let sides AB and CD of the 2uadrilateral ABCD

e e2ual and also AD , BC !see "ig. 8.6#. Draw

diagonal AC.

Clearl&% ' ABC ' CDA !(h&/#≅

+o% BAC , DCA ∠ ∠

and BCA , DAC !(h&/#∠ ∠

Can &ou now sa& that ABCD is a parallelogram/ (h&/

0ou have 7ust seen that in a parallelogram each pair of opposite sides is e2ual and

conversel& if each pair of opposite sides of a 2uadrilateral is e2ual% then it is a

parallelogram. Can we conclude the same result for the pairs of opposite angles/

Draw a parallelogram and measure its angles. (hat do &ou oserve/

ach pair of opposite angles is e2ual.

9epeat this with some more parallelograms. (e arrive at &et another result as

given elow.

Theorem 8. : )n a parallelogram% opposite angles are e2ual.

ow% is the converse of this result also true/ 0es. ;sing the angle sum propert& of

a 2uadrilateral and the results of parallel lines intersected & a transversal% wecan see

that the converse is also true. +o% we have the following theorem :

Theorem 8.< : )f in a 2uadrilateral% each pair of opposite angles is e2ual% then

it is a parallelogram.

There is yet another property of a parallelogram. Let us study thesame. Draw a

parallelogram ABCD and draw oth its diagonals intersecting at the point $

!see "ig. 8.1=#.

>easure the lengths of $A% $B% $C and $D.



(hat do &ou oserve/ 0ou will oserve that

$A , $C and $B , $D.

or% $ is the mid?point of oth the diagonals.

9epeat this activit& with some more parallelograms.

ach time &ou will find that $ is the mid?point of oth the diagonals.

+o% we have the following theorem :

Theorem 8.@ : The diagonals of a parallelogram

isect each other.

ow% what would happen% if in a 2uadrilateral

t h e d i a g o n a l s i s e c t e a c h o t h e r / (i l l i t e a

parallelogram/ )ndeed this is true.

This result is the converse of the result of

Theorem 8.@. )t is given elow:

Theorem 8. : )f the diagonals of a 2uadrilateral

isect each other% then it is a parallelogram.

0ou can reason out this result as follows:

ote that in "ig. 8.11% it is given that $A , $C

and $B , $D.

+o% ' A$B ' C$D !(h&/#≅

Therefore% AB$ , CD$ !(h&/#∠ ∠

"rom this% we get AB ** CD

+imilarl&% BC ** AD

Therefore ABCD is a parallelogram.

Theorem 8.8 : A 2uadrilateral is a parallelogram if a pair of opposite sides is

e2ual and parallel.

L o o 3 a t " i g 8 . 1 i n wh i c h AB , CD a n d

AB ** CD. Let us draw a diagonal AC. 0ou can show



that ' ABC ' CDA & +A+ congruence rule.≅

+o% BC ** AD !(h&/#

Let us now ta3e an eample to appl& this propert& of parallelogram.

Theorem 8.9 : The line segment joining the midpoints of two sides ofa triangle

is parallel to the third side.

!ou "an pro#e this theorem using the following

"lue:

$%ser#e &ig 8.2' in whi"h ( and & are midpoints

of )* and )+ respe"ti#ely and +D *).

- )(& - +D& )/) 0ule1≅

/o, (& D& and *( )( D+ 3hy41

Therefore, *+D( is a parallelogram. 3hy41

This gi#es (& *+.

In this "ase, also note that (& 562(D 562*+.

+an you state the "on#erse of Theorem 8.94 Is the "on#erse true4

!ou will see that "on#erse of the a%o#e theorem is also true whi"h is

stated as

%elow:

Theorem 8.57 : The line drawn through the midpoint of one side of atriangle,

parallel to another side %ise"ts the third side.In &ig 8.2, o%ser#e that( is the midpoint of

)*, line l is passsing through ( and is parallel to *+

and + *).ro#e that )& +& %y using the "ongruen"e of

- )(& and - +D&.

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