Maths Without Use Of Pen_pencil _2-reduced-flattened.pdfmaths Without Use O.pdf

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(a+b+c)2 = a2 +b2 + c2 +2ab + 2bc +2ca

2.

a3+b3+c3 - 3abc = [a+b+c][a2+b2+c2-ab-bc-ca]  abc  [(a-b)2+(b-c)2+(c-a)2] 2  

=

59 15

p 4- e t 35 it 91 io 5, n 86 k 85 in g 9

a2+b2+c2 = 0  a = 0, b = 0 & c = 0

3.

e.g. If (a-3)3 + (b-2)2 + (c-4)2 = 0

-3

then a-3 = 0  a = 3, b = 2, c = 4

If a2+b2+c2+3 = 2(a+b-c), find a+b+c

4.

.co

1.

m

ALGEBRA PDF BY S.KUMAR (MATHS GURU) BHAKLANIA

Solution :- a2 + b2 +c2 +3 = 2a +2b -2c  a2+b2+c2 - 2a -2b+2c +3 = 0

 a2 -2a+1+b2-2b+1+c2+2c+1 = 0  (a-1)2 + (b-1)2 + (c+1)2 = 0

ALGEBRA

1 x

28

w Ph .c o .

m

a = 1, b = 1, c = -1 a+b+c = 1+1-1 = 1

97

If x   2 , then find the value of x99 

1 x100

1

w



w

Solution :- x   2  x2 - 2x+1 x  (x-1)2 = 0  x = 1

1 x

So If x   2 Put x = 1 x 99 

1 1 99 + (1)100 = 2 100 = (1) x 1

1

99 99 & x  (1)100 = (1)  (1)100 1 - 1 = 0

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

Special use in trigonometry if sin  + cosec  =2, then sinn  + cosecen  = 2 if cos  + sec  = 2, then conn  + secnn  = 2 if tan  +cot  = 2, then tann  + cotn = 2 1 x

If x   2 , then put (x = -1)



1  1 x  2 x  - 2 x x 

m



2

2

1  1 x  2 x  2 x x 

91 5

2

2

1  1  x  4   x2  2   2 x x   4

3

1  1 1  x  3   x    3 x   x  x x  3



1

i 91 tio 5, n 86 k 85 in 9- g. 35 c o

2

1

1

pe t

If x  = a  x3 + 3 = a3 - 3a x x 1

3

1

1

4-

m



1 1 1   x    3 x   3 =  x x x  

w Ph .c o .

x3 

35

e.g. x   5  x 3  3 = 125 - 3  5 = 110 10 x x

1

97

1

28

e.g. If x  = a  x 3  3 = a3 + 3a x x

w



w

If x  = 5  x 3  3 = 125 + 3  5 = 140 x x x

1 3 x

x

1 3 x

x2 

1 92  7 x2

x2 

1  9  2  11 x2

x4 

1  49  2  47 x4

x4 

1  112  2  119 4 x

x8 

1  2209  2  2207 x8

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Q.

1

 2  1 If x   3 find  x  x 2  x





1

1

Solution :- x   3  x 2  2  7 x x

1

1

1

Q.

If x 4 

1 1  119 Find x  4 x x

Sol. x2+  x

1 1 x4  4  2 = 2 = x x

1 1  x2  2  2 x x

w Ph .c o .

m

Solution x2 - 3x + 1= 0 Divide by x

2  x 

1 =9-2=7 x2

w Q.

97

28

1 1 =0  x+ =3 x x

w

x-3+

4  x 

1 x4

35

If x2 - 3x + 1 = 0, Find x 4 

4-

Q.

1 119  2 = 11

pe t

= 11  2  13

i 91 tio 5, n 86 k 85 in 9- g. 35 c o

 2     and  x  x 2    x  x   x  x   5  3 = 3 5     

1 = 49 - 2 = 47 x4

1 x

If x+ =3, Find x 5  Solution -

x

1 =? x5

1 1 1 = 3  x 2  2 = 7  x 4  4 = 49 - 2 = 47 x x x

1  4 1   x  x 4   x  x  = 47  3 = 141

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m

1 2  72  5 x2

Ph. 97284-35915, 86859-35915

91 5

1 x

and x   x 2 

1 1  = 141 x 3 x5

x5 + x3 + x 5+

1  3 1  x  3  = 141 - 18 = 123 5 = 141 -  x  x 

If x2 - 3x + 1 = 0 then find the value of

2.

If 2a 

1 x4

(2) x 3 

1 x3

(3) x 6 

1  5 , then find 3a

(a) 8a 3 

1 27 a 3

(b) 27 a 3 

1 1 x3  3 = ? 4 = 119, find x x

4.

If x +

1 = 5 Find x 1 x5

(5) x3/2 

1 x 3/2

35

(b) x3/2 

m

1 x

If x  = 1

1 x7

pe t

If x 4 



(4) x 7 

1 =? 8a 3

3.

(a) x 5 

1 x6

4-

w Ph .c o .

 x2 + 1 = x  x2 - x + 1 = 0

w

e.g. If x +

28

97

w

& (x3 +1) = (x + 1) (x2 - x + 1) = 0 [ x2 - x + 1 = 0] x3 = -1

1 = 1, then find x54 + x 48 + x12 +x 6 x 1

Solution :- x  = 1  x3 = -1 x x54 + x48 + x12 + x6 = (x3)18+ (x3)16 + (x3)4+ (x3)2 = (-1)18+ (-1)16+ (-1)4+ (-1)2 =1+1+1+1=4

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1 1 x5  5 3/2 (6) x x

91 5

(1) x 4 

i 91 tio 5, n 86 k 85 in 9- g. 35 c o

1.

m

MOCK QUESTIONS

1 = -1 x

x+

Q.

If x +

 x3 = + 1

1 = x

 x3 +

3

1 = 3 3 - 3 3 =0 x3

m



1 = x

Solution -

18 12 6 3 , find x + x + x +1

x+

1 = x

6 3  x +1=0

91 5

e.g. If x +

i 91 tio 5, n 86 k 85 in 9- g. 35 c o

 x6 + 1 = 0

x18 + x12 + x6 +1 = 0 = x12 (x6+1) + (x6 + 1) = x12  0 + 0 = 0  [x6 + 1 = 0] Q.

1 1 = 1, Find x17  17 x x

If x +

1 1 x18 x Solution - x  = 1  x17  17 =  18 x x

x 

1

=

x

6



x

 1

1

6

1 x

= x = 1

35

x

3 6

w Ph .c o .

If x  = 3  x 3  3 = 3 3 - 3 3 = 0  x6 + 1 = 0 x x 1

28

Q.



 1

x

4-

=

x

m

3 6

x 

pe t

x

1

97

w

e.g. If x x  = 3 , Find x18 + x12 + x6 +1 x

w

Solution- x  = 3  x6 + 1 = 0 x x18 + x12 + x6 +1 = x12 (x6+1) + (x6 + 1) = x12  0 + 0 = 0

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[ x6+1 = 0]

Ph. 97284-35915, 86859-35915

Square Root In Algebra 7  4 3 =?



Always Try to make 2ab from Root 2ab = 4 3  ab = 2 3 a = 2, b = 3  a2 + b2 = 4 + 3 = 7

38  5 3 

Q.

x

1 76  10 3 = 2 2

3 , then find 2

42 3 1 = 2 4



1 (5 3 +1) 2

1 x + 1 x

1

3 2 3  2 2

3 1 2



3 1 =

pe t

=

76  10 3 =

3  1 x = 2

Solution - x 

=2+ 3

m

13  4 3 = ?



2

91 5

Q.

2 3

i 91 tio 5, n 86 k 85 in 9- g. 35 c o



Solution: - 7  4 3 =

3

4-

3 1 3 1 2 3 + = = 2 2 2

If a = 998, b = 998, & c = 997 find a3 + b3 + c3 - 3abc = ?

28

Q.

=

w Ph .c o .

 1 x + 1 x

3 1 2

35

m

in same way 1  x =

97

Solution :- a3 + b3 +c3 - 3abc =

abc [(a-b)2+(b-c)2+(c-a)2] 2

w

 998  998  997    0  1  1 2

w

= 

= 2993

Q.

If a + b+ c = 8, then find (a-4)3 + (b-3)3 +(c-1)3 - 3(a-4)(b-3)(c-1) Solutions :- Put a = 4, b = 3, c = 1 4+3+1=8 we get 0

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Q.

If a2 + b2 +c2 = ab + bc + ca then

ac =? b

Solution ;- a2 +b2+c2= ab + bc + ca  a2 + b2 +c2 - ab - bc - ca = 0 1 [(a-b)2+(b-c)2+(c-a)2] = 0 2

ac aa = =2 b a

or by hit and trial method put a = b = c =1

Q.

11 = 2 1

91 5

Ans. is

m

 a=b=c 

nk i 59 n g - 3 .c 5 o



If a2 +b2 +2b+4a + 5 = 0, find

a b ab

,8 68

If a+b+c = 9 & ab+bc+ca = 26, then find a3+b3+c3-3abc =? Solution :- [a+b+c]2 = a2 +b2 +c2 + 2 (ab + bc +ca)  [9]2 = a2 + b2 + c2 +2 (26)  a2 +b2 + c2 = 29 3 3 3 2 2 2  a + b +c - 3abc = (a+b+c) [a +b +c - (ab +bc +ca)] = 9 [29-26] = 27 or by hit and trial method put a = 4, b = 3, and c = 2

97

w w

w Ph .c o .

Q.

a b 1 = ab 3

59 15

Put

m 28 p 4- e t 3

iti o

Solution :- a2 + b2 + 2b + 4a + 5 = 0  a2 + 4a +4 +b2 +2b +1 = 0  [a+2]2 +[b+1]2 = 0 a = -2, b= -1

a3 + b3 + c3 - 3abc =

1 (a+b+c) [(a-b)2 + (b-c)2 + (c-a)2] 2

1 2

=  9 [1 + 1 + 4] = 27

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1

If x = 3 + 2 , then Find x 3  3 x Solution :- x = 3 + 2 

1 = x

3 x 

3-

1 2  x x = 2 3

1 = = x 3 24 3 6 3 18 3

2p

m

Q.

1

2p 1 p2  2 p  1  or =4 p2  2 p  1 4 2p

p2  2 p  1 =8 p

1

91 5

Solution :-

m 28 p 4- e t 35 it 91 io 5, n 86 k 85 in 9- g. 35 c o

1 2 If p 2  2 p  1 = , then find p  p 2 4

Q.

1

p - 2 + p = 8  p + p = 10 1  p2 + p 2 = 100 - 2 = 98

Q.

If

a b b + + = 1, then find the value of 1 a 1 b 1 c

1 1 1 + + 1 a 1 b 1 c

(b) 2

w Ph .c o .

(a) 1

Solution:- Put a = b = c =

(d) 4

1 4

97

Ans. is 4 Find the value at x = 16 of the following expression x4 - 17x3 + 17x2 - 17x + 14 Solution:- x4 - 17x3 + 17x2 - 17x +14 = x4 - 16x3 - x3 + 16x2 + x2 - 16x -x +14 Now put x = 16 = 164-164-163-163-162-162-16+14 = - 2

w

w

Q.

(c) 3

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Ph. 97284-35915, 86859-35915

Some Selected Questions Q.

1

If x  = 3 , find the value of x (i) x72 +x66 +x54+x36+x24+x6+1 (ii) x206 +x200+x90+x84+x18+x12+x6+1 1

Q.

If x = a  a  b (a) 2a3

3

1 3

 + a 

2

a b

(b) -2a3



Solution:-

3



1 3

59 15

-3



2

.co

x72 + x66 +x54 +x36 +x24 +x6 +1 [x6]12 +[x6]11 +[x6]9+[x6]6+[x6]4+x6+1 = [-1]12+[-1]11+[-1]9 +[-1]6+[-1]4+0 =1-1-1+1+1=1 x206 + x200 +x90+ x84 +x18+x12+x6+1 x200[x6+1]+x84[x6+1]+x12[x6+1] + x6+1 0 + 0 + 0 + 0 = 0 [ x6 + 1 = 0]

p 4- e t 35 it 91 io 5, n 86 k 85 in g 9

(ii)

m

Solution :- (i) x+ = 3 = x6+1= 0 x

then find the value of x3 +3bx -2a

(c) 0

2

x = a a b

3

1 3

 + a 

(d) 1

2

a b

3



1 3

cubing both sides, we get

 



m

x3 = a  a 2  b 3 + a  a 2  b 3 1 3

w Ph .c o .

+ 3 a a b

3

 a 

2

a b

3

28



2



1 3



(x)

97

 x3 = 2a +3[a2-(a2+b3)]1/3 [x]

w w

= 2a +3 [-b3]1/3 x = 2a -3bx  x3 +3bx - 2a = 0

Q.

1

2

If x = 2 + 2 3 + 2 3 , then value of x3 - 6x2 + 6x +1 is (a) 3 (b) 2 (c) 1 (d) 0 Solution:-

1

2

1

2

x = 2 + 2 3 + 2 3  x -2 = 2 3 + 2 3 cubing both sides, be get x3 - 8 - 3.2x [x-2] = 2 + 22 +3  21/3  22/3 (x-2)  x3 - 8 - 6x2 +12x = 2 + 4 + 6 (x -2) www.competitionking.com

Ph. 97284-35915, 86859-35915

 x3 - 6x2 + 6x = 2  x3 - 6x2 + 6x +1 = 2+1 = 3

If a = 3+ 2 2 , then find the value of Solution:-

1

a = 3 + 2 2  = 3 -2 2 a

1 1 a6  a4  a2  1 Now = a3+a + + 3 = 6 + 198 = 204 3 a a a

1 5x =5, then find the value of 2 3x 6 x  20 x  1



Q.

1 =5  6x2 +1 = 15x 3x

5x 5x 5x 1 = = = 6 x  20 x  1 15 x  20 x 35 x 7 2

1 x

If x + = 99, then find of value of Solution:-

1

m

100 x 100 x 100 x 1 = = = 2 x  102 x  2 198 x  10 x 300 x 3 2

w Ph .c o . 1 x

If x   3 , then find the value of x 3 

w w

Solution: x

Q.

1 = x

97

Q.

100 x 2 x  102 x  2 2

x+ = 99  x2 + 1 = 99x x

2x2+2 = 198x 

-3

Solution :- 2x+

p 4- e t 35 it 91 io 5, n 86 k 85 in g 9

If 2x+

28

Q.

m

1 =6 a

59 15

a+

a6  a4  a2  1 a3

.co

Q.

2

1 x3

2

1   1  x  x  =  x  x  + 4 = 13

1 3 13  x  x3 = 13 13 - 3 13 = 10 13

1 

 

1 x

2 If x  = 5, then find the value of 2  x  x 2 

Solution:-

1

1

1

 2     2  x  x2  = 2  x  x   x  x      

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2

2

1  1   x  x  =  x  x  +4 = 2 5+4 = 29

 x

1 = x

29

1

x+

1

x = 3+ 2 2  = 3- 2 2 x

1 =6 x 3

= 216 - 3 [6] = 198 If

1   x  x 

p 4- e t 35 it 91 io 5, n

1 1  x + 3 = x   - 3 x x  3

Q.

g.

Solution:-

a b  =1, then find the value of a3 +b3 b a

Solution:-

15

1

If x = 3  2 2 ,then find the value of x 3  3 x

a b  = 1  a2 +b2 = ab b a

86

Q.

co m

 2  2  x  x 2  = 2  5  29 = 10 29  

p3  3 p 2  3 p  1

3 Solution:p3  3 p 2  3 p  1 = = 101 -1 = 100

3

p 3  13  3 p ( p  1) =

3

( p  1)3 = p - 1

97

If

28

3

a b c 1 1 1     = 1, then find the value of 1 a 1 b 1 c 1 a 1 b 1 c

w w

Q.

If p = 101, then find the value of

w Ph .c o .

Q.

m

& a3 + b3 = (a+b) (a2 -ab+b2) = (a+b) (ab -ab) = 0

Solution:-

a b c   =1 1 a 1 b 1 c

a b 1 1 1  1 = 1+3 = 4 1 a 1 b 1 c

a 1 a b 1 b c 1 c   =4 1 a 1 b 1 c

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Ph. 97284-35915, 86859-35915

1 1 1   =4 1 a 1 b 1 c

Q.

If x + y = 2z, then find the value of

z x + yz xz

If a+b+c = 10, a2 +b2+c2 = 38 & abc = 30, then find the value of

(ii) a3 +b3+c3 Solution:-

(i)

1 1 1 bc  ac  ab   = a b c abc

59 15

1 1 1   a b c

-3

(i)

p 4- e t 35 it 91 io 5, n 86 k 85 in g 9

Q.

.co

z x x z xz + yz = = =1 xz xz xz xz

m

Solution:- x+y = 2z  x -z = z - y

(a+b+c)2 = a2+b2+c2+ 2(ab+bc+ca)  (1012 = 38 +2(ab+bc+ca)  ab +bc +ca = 31 

1 1 1 31   = a b c 30



(a-b)2 = a2 + b2 - 2ab

28

97

13  4 3

w w

Q.

w Ph .c o .

m



(ii) a3 +b3+c3 - 3abc = (a+b+c) [a2+b2+c2-ab-bc-ca] a3+b3+c3 -3  30 = 10 [38-31] = 70 a3+b3+c3 = 160 (a+b)2 = a2+b2+2ab

2ab = 4 3 ab = 2 3 1

a= 2 3 b=1 a + b = 12 + 1 = 13 2

2

13  2 3 =

2



2

3 1 = 2 3 1

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Q.

38  5 3 =

1 76  10 3 = 2 2

76  10 3

Now 2ab = 10 3 ab = 5 3 a2 + b2 = 75 +1 = 76

m



4 6 x  8 x  12 , then find + 2 3 x  8 x  12

x = 8

x

4 6 4 2 3 = = 2 3 2 3

12 2 3

8  12 2 3

applying C & D , we get

x 8 12  2  3 3 3 2 = = x 8 12  2  3 3 2

-i

x 8 = 12 2 3

x  12 = x  12

97

w w

Applying C & D, we get

28

w Ph .c o .

m

Simillary

59 15

Solution:-





.co

If x 

1 5 3  1 Ans. 2

76  10 3 =

-3

Q.

1 2

p 4- e t 35 it 91 io 5, n 86 k 85 in g 9



 8 

8

 3 2 = 3 2

2 3

2

3 3

- ii

Adding (i) of (ii), we get x 8 3 3 2 3 2 3 x  12 + = + =2 x 8 3 2 2 3 x  12

In these type of Qu. Ans. is always 2

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4a  1  1 2

a  a  a  .......  =

e.g. 12  12  12  ......... 

77  1 2

e.g. 72  72  ..........

[a =72]

-3

4  72  1  1 17  1 9 = 2 2

= Q.

[a=19]

p 4- e t 35 it 91 io 5, n 86 k 85 in g 9

4  19  1  1 = 2

m

12  4  1  1 =3 2

e.g. 19  19  19  .......... =

[a= 2]

59 15

=

[This tricx applies to all]

.co



52 3 = a + b 3 , then find the value of a + b 74 3

If

Solution:-

52 3 74 3 =a+b 3  74 3 74 3

m



52 3 = a+b 3 74 3

w Ph .c o .

 35 - 6 3 - 24 = a + b 3  11- 6 3 = a + b 3

28

97

w w

Comparing coefficient both sides we get a = 11 , b = - 6  a+ b = 5

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2017

CIRCLES

CIRCLE (o`Ùk)

A circle is a locus i.e. path of a point in a plane which moves in such a way that its distance from P a fIxed point always remains constant. In fgure, 'O' is the fixed point and P is a moving point in the same plane. The path traced by P is called a circle. Fixed point O is the centre of the circle and the constant distance OP is called radius of the circle.

Unless stated otherwise, by an arc AB we shall mean the minor arc.

A

B

P

Major Arc

circumference (ifjf/)

Terminology Related to CIRCLES Radius (f=kT;k) : Line segment joining centre and any point of circle.

Radius (f=kT;k)

2)

A piece of a circle between two points is called an arc. Consider two points M and N on the circle. We fnd that there are two pieces of circle between A and B. One is longer and other is smaller. The longer piece is called major arc and smaller piece is called minor arc

OR A circle is a simple closed curve, all the points of which are at the same distance from a given fIxed point. Circumference Length of a complete circle is called its circumference.

1)

Arc of a circle (pki ) :

4)

A

B

 Major arc is denoted by APB and minor arc is denoted by  AQN

Q

Note:-When M and N are ends of a diameter then both the arcs are equal and both are called semicircle. 5)

Segment of a circle (o`Ùk[k.M) :

The region between a chord and an arc of a circle is called a segment. There are two segments corresponding to two arcs, major segment and minor segment. Major segment is the segment enclosed by major arc. Centre of the circle lies in the major segment. Minor segment is the segment enclosed by minor arc.

Chord (thok) : A line segment joining any two points on a circle is called chord of the circle.

Major segment Centre P

Q

Chord (thok)

3)

Diameter (O;kl) : A chord passing through the centre of a circle is known as its diameter. Diameter is longest chord.

Minor segment Major segment Minor segment

6) Sector of a circle(f=kT;[k.M) : The part of a circle enclosed by an arc and two radii is called a sector. Major Sector

Diameter (O;kl)

MATHEMATICS BY- RAHUL RJ

Minor Sector

GEOMETRY & MENSURATION

7)

Semicircle (v¼Zo`Ùk) : A diameter of a circle divides the circumference of the circle into two equal arcs and each of these arcs is known as a semicircle.

8) Tangent of a circle ( LiZ'kjs[kk) : A tangent is a straight line which touches the circumference of a circle at only one point. A tangent does not intersect the circumference, if produced infinitely on either sides. A O P

B

9)

Secant of a circle ( izfrPNsnhjs[kk ) : A straight line intersecting the circle at two points, is called a secant. A O

B

10) Concentric Circles (ldsfUæ;o`Ùk) : Circles having the same centre are said to be concentric circles.

Important formulas of CIRCLE

π r2

1.

Area of circle =

2.

Perimetre of circle = 2 π r

3.

Area of semicircle =

4.

Perimeter of semicircle = ( π +2)r

5.

Area of a quadrant of a circle =

6.

Perimeter of a quadrant of a circle = 

7.

Area of a sector of a circle =

8.

Length of arc =

1 2 πr 2 1 2 πr 4

π  + 2 r 2 

θ × 2πr 360

θ × πr 2  360

Properties related to Circle

5).

1). One and only one circle can pass through given three non- collinear points.

If two circles intersect at two points then the line seg- ment joining their centre is perpendicular bisector of common chord. A

In case, the three given points are collinear, then a single circle cannot pass through these 3 points

O

2).The perpendicular from the centre of a circle to a chord bisects the chord.

B

AC = BC and OC ⊥ AB

B

P



O A

O'

C

# In a circle with radius 13 cm, a chord is drawn at the distance 12 cm from the centre. Find the length of the chord. Conversely, the line joining the centre of a circle Solution:- Distance of the chord from the centre is 12 cm to the midpoint of a chord is perpendicular tothe chord. obviously, OC AB. C is the midpoint of AB. If AP = PB then OPAB .

If OPAB then AP = PB

3).Equal chords of a circle are equidistant from the centre. A L

2

O 13

B A

O D M

C

Chords AB and CD are equidistant from the centre O, ie OL = OM if AB = CD. Conversely, chords of a circle which are equidistant from the centre are equal.

4).The perpendicular bisectors of two chords of a circle intersect at its centre. A

cm

12 cm

B

C

By the Pythagoras Theorem, 2

2

OC2  AC2  OA 2  12  AC  13  Length of the chord = AC + CB = 2AC = 10 cm # The radius of the circle is 13 cm. The lengths of two parallel chords on same sides of centre of the circle are 24 cm and 10 cm respectively. The distance between the chords is 1) 4 cm 2) 7 cm 3) 6 cm 4) 8 cm Solution:- AB and CD are parallel chords. The perpendicular from the centre bisects the chord.

E

C

F C

D

cm

O E

D

13

13

cm

O B

A

5 cm

F

B

By the Pythagoras Theorem: The perpendicular bisectors of two chords AB and CD of a circle intersect at its centre O.

OF  132  5 2 =

144 = 12 cm

OE = 132  122  5 cm  EF is the distance between the chords and EF = OF – OE = 12 cm – 5 cm = 7 cm GEOMETRY & MENSURATION

TRICK:- CE will become OF & AF will become OE OF=CE=12 cm OE=AF=5 cm

and EF = OF – OE = 12 cm – 5 cm = 7 cm # In a circle of radius 17 cm, two parallel chords are drawn on opposite sides of a diameter. The distance between the chords is 23 cm. If length of one chord is 16 cm, then the length of the other one is : (a) 15 cm (b) 23 cm (c) 30 cm (d) 34 cm Solution: (c) Let PQ and RS be two parallel chords of the circle on the opposite sides of the diameter AB = 16 cm

# Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and both chords are on the same side of its centre. If the distance between AB and CD is 3 cm, then find the radius of the circle. Solution:- The perpendicular bisector OL of AB and OM of CD are in the same line because AB || CD. Let OM = x cm and radius of the circle be r cm.

Then, from ΔOMD 2

 11  121 x2     r 2  x2   r 2 .... (1) 4  2

From ΔOLB , 2

x  3   5   r 2  x 2  6x  61  r 2 .... (2) 4  2 From (1) and (2), we get 2

60 5 121 61  6x  x  x 2  6x  4 2 4 4 Substituting x into (1), we get

x2 

Now, PN = 8 (Since ON is the perpendicular bisector) In DPON ON 2 = OP 2 – PN 2 = (17)2 – (8)2 = 289 – 64 = 225 or ON = 15 ⇒ ∴ OM = 23 – 15 = 8 In DORM, RM 2 = OR2 – OM 2 2 17 – 82 = 289 – 64 = 225 or RM = 15 ⇒ RS = 15 × 2 = 30 cm

TRICK:-

NP becomes OM =8 In DORM, RM 2 = OR2 – OM 2 2 17 – 82 = 289 – 64 = 225 or RM = 15 ⇒ RS = 15 × 2 = 30 cm

2

2

25 121 146 5  11  1 r2         r  146 4 4 4 2  2  2

TRICK:-

# What is the distance in cm between two parallel chords of lengths 32 cm and 24 cm in a circle of radius 20 cm? 1) 1 or 7 2) 2 or 14 3) 3 or 21 4) 4 or 28

# Two circles of radii 10 cm and 8 cm intersect each other and length of the common chord is 12 cm. Find the distance between their centres. (a) 13.8 cm (b) 13.29 cm (c) 13.2 cm (d) 12.19 cm

Solution:-The two parallel chords lie either on the same side of the diameter or on the oppo- site sides of the diameter. If we find the distance between parallel chords of any one of the above two cases, then it serves our purpose.

Solution: (b) Here, OP = 10 cm; O′P = 8 cm P 10 O

8

Q

12 20

PQ = 12 cm

x O y

20

O

L

PL = 1/2 PQ ⇒ PL =

∴

In rt. DOLP, OP2 = OL2 + LP2

16

Distance between chords = x  y  20 2  12 2  20 2  16 2  16  12  28 cm Only one option contains 28. Hence theywill definitely be 4 cm apart in the othercase.

TRICK:- 16 will become x & 12 will become y

1 × 12 ⇒ PL = 6 cm 2

(using Pythagoras theorem) ⇒ (10)2 = OL2 + (6)2 ⇒ OL2 = 64; OL = 8 In DO′LP, (O′L)2 = O′P2 – LP2 = 64 – 36 = 28 O′L2 = 28 ⇒ O′L = 28 ∴

O′L = 5.29 cm OO′ = OL + O′L = 8 + 5.29 OO′ = 13.29 cm

Distance between chords x+y=28 or x-y=4 cm

# Two equal circles pass through each other's centre. If the radius of each circle is 5 cm, what is the length of the common chord?

# A chord of length 14 cm is at a distance of 6 cm from the centre of a circle. Find the length of another chord at a distance of 2 cm from the centre of the circle. (a) 18 cm

(b) 16 cm

(c) 10 cm (d) 12 cm Solution:- (a) Let AB be the chord of length 14 cm. at a distance of 6 cm from the centre O. Draw OE ^ AB. Then, BE = 7 cm and OE = 6 cm. \ OB2 = OE2 + BE2 = (62 + 72) = 85. Let CD be the chord at a distance of 2 cm from O. Now, OF = 2 cm, OD2 = OB2 = 85. A C

E 7

7 F

B D

O

(a) 5 3

(b) 10 3

5 3 2 Solution: (a) (c)

(d) 5 A

O

M

O

B

Given, distance between the centres of two circle = 5 cm OO' = 5 cm 5 ∴ OM = cm 2

\ FD = 9 cm. \ CD = 2 ´ FD = 18 cm.

MATHEMATICS BY- RAHUL RAJ ∴ The length of common chord, AB = 2 × AM

In DOAM, OA2 = OM2 + AM2

Properties based on ANGLES of a Circle 1. Equal chord of a circle subtend equal angles at the centre. A

2

5 (5)2 =   + AM2 2 25 5 3 = cm 4 2 ∴ The length of common chord, AB = 2 × AM AM =

= 2×

P

O

25 −

B

5 3 = 5 3 cm 2

Q

If AB = PQ, then ∠AOB = ∠POQ

The converse is also true. If the angles subtended by two chords of a circle at the centre are equal, the chords are equal.

2. The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle. R

R

R P

O P

Q

O

P

Q

O

Q

In each of the above figures,

POQ  2PRQ 3. Angles in the same segment of a circle are equal.

x0

x0

∴ ∠ACB = ∠ADB = ∠AEB. 4. Angle in a semicircle is a right angle. P

B

A O

In figure, AOB is a diameter, hence AOBPA is a semicircle, therefore ∠APB = 90°. Conversely, the arc of a circle subtending a right angle at any point of the circle in its alternate segment is a semi-circle.

5.

Angle made by a chord in minor segment is obtuse and in major segment is acute. Acute Angle

Obtuse Angle

# Find the value of x in the following circle with centre O. A Solution:- We know that the angle

subtended by a chord at the centre is twice that at the circumference.  2x  112

x° O 112°

C

B

 x  56

# In the given figure, chord ED is parallel to the diameter AC of the circle. If ,CBE  65 then whatis the value of DEC? 1) 25° 3) 45°

B

2) 65° 4) 35°

O

A E

C D

Solution:- Draw the line from point A to point E. EAC  CBEAngles made by the same chord on the circumference are equal  = 65° Also, AEC  90 Angle made by diameter on the circumference   ACE  180  90  65  25 And ACE  DEC  25 (Alternate angles)

GEOMETRY & MENSURATION

TRICK:- If two chords intersect each other then angle made by them on point of intersection – A

Cyclic Quadrilateral Properties Cyclic quadrilateral (pfØ;prqHkqZt) : A cyclic quadrilateral is called cyclic quadrilateral if its all vertices lie on a circle.

B x0

C

D

D C

x=

B

1 × angle by 2

A

( arc AC – arc BD) at the centre A

1.

The sum of either pair or opposite angles of a cyclic quadrilateral is 1800.

D x

0

C

D

A

B

C

1 x = × angle by 2

B

∠ A + ∠ C = 1800 and ∠ B + ∠ D = 1800

(arc AC + arc BD) at centre

If the sum of any pair of opposite angles of a quadrilateral is 1800, then the quadrilateral is cyclic.

3.

If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to interior opposite angle.



2.

A

D

E

C

B

∠ CDE = ∠ A 4.

The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. A

D

P Q B

S R

C

PQRS is cyclic quadrilateral

2017

5.

If two sides of a cyclic quadrilateral are parallel then the remaining two sides are equal and diagonal are also equal.

6.

If two opposite sides of a cyclic quadrilateral are equal, then the other two sides are parallel.

7.

If the bisectors of the opposite angles ∠ P and ∠ R of a cyclic quadrilateral PQRS intersect the corresponding circle at A and B respectively, then AB is a diameter of the circle. P

#

In the given figure, ABCD is a cyclic quadrilateral; O is the centre of the circle. If BOD  160 , find the measure of BPD 1) 20° 3) 80°

2) 100° 4) None of these A

O 160°

D

B

S

Q

A

P

B C

Consider the arc BCD of the circle. This arc make angle BOD  160 at the centre of the circle and BAD

R

at a point A on the circumference. 8.

The sum of the angles in the four segments exterior to a cyclic quadrilateral is equal to 6 right angles. A

S

P

 BAD  BPD  180  80  BPD  180

B

D

 BPD  180  80  100

R

Q

#

C

∠ A + ∠ B + ∠ C + ∠ D = 900 x 6 = 5400 9.

1 BOD  80 2 Now, ABPD is a cyclic quadrilateral.  BAD 

I n t h e a d joi n i n g fi g u r e , A B C D i s a c yc l i c quadrilateral. If AB is a diameter, BC = CD and ÐABD = 40°, find the measure of ÐDBC. D

ABCD is a cyclic quadrilateral. AB and DC are produced to meet in E, then ∆ EBC ~ ∆ EDA .

d

X

C X

A

D

a

40°

B

C

E B

(a) 65 (c) 45

A

10. AB is diameter of a circle. Chord CD is equal to radius. If AC and BD when produced intersect at P, then ∠ APB = 600 A B 600 D C

(b) 25 (d) 60

SOLUTION:- (b) In DBCD, BC = CD, ÐBDC = ÐCBD = x In cyclic quadrilateral ABCD, ÐABC + ÐADC = 180°40° + x + 90° + x = 180° Þ x = 25°.

# In the following figure, O is the centre of the circle andÐABO = 30°, find ÐACB. C B

(a) 60° (c) 75°

A

(b) 120° (c) 90°

30° O

C

B A

Þ ÐABC + 110° = 180° ( ABCD is a cyclic quadrilateral ) Þ ÐABC = 180 – 110 Þ ÐABC = 70° Q AD || BC \ ÐABC + ÐBAD = 180° (Sum of the interior angles on the same side of transversal is 180°) 70° + Ð BAD = 180° Þ ÐBAD = 180° – 70° = 110° Þ ÐBAC + ÐDAC = 110° Þ 50° + ÐDAC = 110° Þ ÐDAC = 110° – 50 ° = 60° # If two sides of a cyclic quadrilateral are parallel then the remaining two sides are equal and the or diagonals are also equal. A cyclic trapezium is isosceles and its diagonals are equal.Conversely, If two nonparallel sides of a trapezium are equal it is cyclic. or An isosceles trapezium is always cyclic.

30°

O

D

OB = OA = radius of the circle ÐAOB = 180 – (30 + 30) {Sum of angles of triangle = 180°) Þ 120° 120 = 60°, because the angle subtended 2 by a chord at the centre is twice of what it can subtend at the circumference. Again, ABCD is a cyclic quadrilateral; So ÐACB = 180° – 60° = 120° (because opposite angles of cyclic quadrilateral are supplementary).

Then ÐADB =

# ABCD is a cyclic quadrilateral in which BC || AD, ÐADC = 110° and ÐBAC = 50° find ÐDAC (a) 60° (b) 45° (c) 90° (d) 120°

If there is a cyclic quadrilateral ABCD in which AB || CD then (i) AD = BC (ii) AC = DB # The bisectors of the angles formed by producing the opposite sides of a cyclic quadrilateral(provided that they are not parallel), intersect at right angle. Q

SOLUTION:-(a) ÐABC + ÐADC = 180° (sum of opposites angles of cyclic quadrilateral is 180°)

A 50° B

D

D

C

110° C

R A B

P

In a cyclic quadrilateral ABCD, AB and DC when produced meet at P and AD and BC whenproduced meet at Q. Bisectors of P and Q meet at a point R. Then PRQ  90.

TANGENT AND ITS PROPERTIES 1)

A tangent to a circle is perpendicular to the radius through the point of contact.

(ii)

∆PAO ≅ ∆PBO

(iii)

∠ P + ∠ O = 1800 ∠P & ∠O

(iv) PO is a angle bisector of

(v) OP is perpendicular bisector of AB (vi)

O

 < BA  AB

8.ALternate segment THEOREM

2) 3) 4)

OP ⊥ AB

B

P

A

A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle.

If a chord is drawn through the point of contact of a tangent to a circle, then the angle which this chord makes with the given tangent is equal to the angle formed in the alternate segment.

C

D

The perpendicular to a tangent through its point of a contact passes through the centre of the circle. P

5)

B

One and only one tangent can be drawn to a circle at a given point on the circumference.

If two circles touches each-other, the point of contact lies on the straight line joining their centers.

A

Q

In the figure, AB is a chord of a circle. PQ is a tangent at an end point A of the chord to the circle. C is any point on arc AB and D is any point on arc BA. ∠BAQ and ∠ACB are angles in the alternate segments ∠BAP and ∠ADB are angles in the alternate segments. Angles in the alternate segments of a circle are equal i.e. ∠BAQ = ∠ACB and ∠BAP = ∠ADB 9. If two chords AB and CD of a circle intersect inside a

6)

From any point outside a circle two tangents can be drawn to it and they are equal in length.

circle (or outside a circle when produced) at point E, then AE × EB = CE × ED. A

D

A B

A

E E D

B P

C

P

A

P

Q

O

B

(i) PA = PB

T

PA = PB

B

7)

C

R

PR x RQ = TR2 10) Two circles touch externally and through the point of contact a straight line is drawn, touches the circumference of both circle, then the tangent at its extremities are parallel.

A

B

E

A

B

O C

C

D

F

If AB | | CD than

AB | | CD 11) If a circle touches all the four sides of a quadrilateral then the sum of opposite pair of sides are equal.

D

∠ EOF = 900

15) Tangents at the end point of a diameter of a circle are parallel. R

S

A d

c

B

C

a

b

PQ | | RS

D

AB + DC = BC + DA

Q

16) Common tangents to two circle

s(s − a)(s − b)(s − c)(s − d)

Area =

12) A circle touching the side BC of ∆ABC at P and touching AB and AC produced at Q and R respectively then AQ =

P

1 (perimeter of ∆ABC) 2

No common tangent

A

B

P

C R

Q

13) Two tangents TP and TQ are drawn to a circle with centre O from an external point T, then ∠ PTQ = 2 ∠ OPQ .

One common tangent

Two common tangent

P

T

O

Q

Three common tangent

# IMPORTANT RESULT A

E

B

G

Four common tangent # Direct common tangent: A tangent to two circles are such that the two circles lies on the same side of the tangent, then the tangent is called direct tangent to the two circles. Q

F

C

D

(i) AB = CD = EF (ii) AE = EB = EG = GF = CF = FD # IMPORTANT RESULT

A B

P r1

C

2

O

O

∠ ACB = 900

R S

In the figure, PQ and RS are two direct common tangent # The point of intersection of direct common tangents to the same two circles. Length of these two common and indirect (transverse) common tan-gents to two circles tangents to the same two circles are equal. divide the line segment joining the two centres, both PQ = RS

Also

PQ = RS =

externally and inter-nally, in the ratio of their radii.

(OO′)2 + (r2 − r1 ) 2

A E

Here O, O′ are the centres and r1, r2 2

# Indirect or Transverse Common Tangent: If a tangent to two circles is such that the two circles lie on opposite sides of the tangent, then the tangent is called indirect tangent. Length of two indirect tangents to two circles is equal.

P

S

r1 O O

O

1

Q

In the figure, PQ and RS are two indirect common tangents to the same two circles. ∴ PQ = RS 2 2 Also PQ = RS = (OO′) − (r1 + r2 ) Here O, O′ are centres r1, r2 are radii of the two circles respectively.

Q

F

O´

P

D

G C

In the above figure, there are two non-intersecting circles. AB and CD are their direct commontangents, which when produced meet at P. Also, EF and GH are the transverse common tan-gents intersecting at Q. r is the radius of the circle having centre O and s is the radius of circle having centre O´.Then

(i)

P divides OO´ externally in the ratio r : s.

(ii)

OP r  O´P s Q divides OO´ internally in the ratio r : s OQ r  ie O´Q s

r2 R

B

H

ie



i.e.

# In the given figure, chords AB and CD of a circle intersect externally at P. If AB = 6 cm, CD = 3 cm and PD = 5 cm, then PB = ?

#In the given fig. PQ is a chord of a circle and PT is the tangent at P such that ÐQPT = 60°. Then ÐPRQ is equal to Q

O R (a) 5 cm (b) 6.25 cm (c) 6 cm (d) 4 cm Solution: (d) PA × PB = PC × PD (According to property of circle) ⇒ (x + 6) × x = 8 × 5 ⇒ x2 + 6x – 40 = 0 ⇒ (x + 10) (x – 4) = 0 ⇒ x = 4 ∴ PB = 4 cm # In the given figure, PAB is a secant and PT is a tangent to the circle from P. If PT = 5 cm, PA = 4 cm and AB = x cm, then x is equal to

P

(a) 135° (c) 120°

# (a) 2.5 cm (c) 2.25 cm

(b) 2.6 cm (d) 2.75 cm

Solution: (c) PA × PB = PT2 ⇒ 4 × (4 + x) = 25 25 ⇒ 4+x= = 6.25 ⇒ x = 2.25 cm 4 # In the given figure, two chords AB and CD intersect each other at O. If AO = 8 cm, CO = 6 cm and OD= 4 cm, find OB.

1) 3 cm 2) 4 cm 3) 5 cm 4) 6 cm

T

(b)

(b) 150° (d) 110°

If four sides of a quadrilateral ABCD are tangential to a circle, then. (a) AC + AD = BD + CD (b) AD + BC = AB + CD (c) AB + CD = AC + BC (d) AC + AD = BC + DB Since ABCD is a quadrilateral Again AP, AQ are tangents to the circle from the point A. D

D

P

B

A

S

C

R

O C

We know that, OA × OB = OC × OD  8  OB  6  4 OB=3 CM



A

Q

B

\ AP = AQ Similarly BR = BQ CR = CS DP = DS \ (AP + DP) + (BR + CR) = AQ + DS + BQ + CS = (AQ + BQ) + (CS + DS) Þ AD + BC = AB + CD

#

R

Find ÐBOA.

S

C

B

50°

O

r1

30°

A

(a) 100° (c) 80°

F 1.

(b) 150° (d) Indeterminate

ÐCAF = 100°. Hence ÐBAC = 80° Also, ÐOCA = (90–ACF) = 90 – 50 = 40° = ÐOAC (Since the triangle OCA is isosceles) Hence ÐOAB = 40° In isosceles DOAB, ÐOBA will also be 40° Hence, ÐBOA = 180 – 40 – 40 = 100°

(a)

r2

P

Q

I

II

O

(b) In DSOQ and DROP ÐO is common ÐS = ÐR = 90º (tangent at circle) \ D SOQ ~ DROP

Þ

RP OP PQ + OQ PQ = = = +1 SQ OQ OQ OQ

4 PQ PQ 4 1 = + 1 or = -1 = 3 OQ OQ 3 3 Þ PQ = 7 and OQ = 21 Þ

Directions for Questions 1–3: Answer the questions on the basis of the information given below.

\ Required ratio =

In the adjoining figure, I and II are circles with centers P and Q respectively. The two circle touch each other and have a common 2. tangent that touches them at points R and S respectively. This common tangent meets the line joining P and Q at O. The diameters of I and II are in the rartio 4 : 3. It is also known 3. thatthe length of PO is 28 cm.

(b) PQ = r1 + r2 = 7 As the ratio of radii is 4 : 3. So, the only value which satisfies the radii of circle II = 3 (c) In DSOQ, Þ SO2 + SQ2 = OQ2 Þ SO2 = 212 – 32 = (21 – 3) (21 + 3) = 18 × 24 = 432

R

Þ SO = 12 3

S

# P

7 1 = 21 3

O

Q

In the given diagram O is the centre of the circle and CD is a tangent. ÐCAB and ÐACD are supplementary to each other ÐOAC 30°. Find the value of ÐOCB:

A I

1.

(b) 1 : 3 (d) 3 : 4

C

What is the radius of the circle II?

(a) 2 cm (c) 4 cm

3..

O

What is the ratio of the length of PQ to that of QO?

(a) 1 : 4 (d) 3 : 8

2.

B

II

(b) 3 cm (d) 5 cm

The length of SO is

(a) 8 3 cm

(b) 10 3 cm

(c) 12 3 cm

(d) 14 3 cm

(a) (c) (a)

D

30° (b) 20° 60° (d) None of these ÐOCD = 90° ÐOAC = ÐOCA = 30° ÐACD = ÐACO + ÐOCD = 30° + 90° = 120° \ ÐBAC = 180° – 120° = 60° Þ ÐBCD = 60° (ÐBCD = ÐBAC) Þ ÐOCB = ÐOCB – ÐBCD = 90° – 60° = 30°

#.

Two circles touch each other internally. Their radii #A smaller circle touches internally to a larger circle at A and are 2 cm and 3 cm. The biggest chord of the outer circle passes through the centre of the larger circle. O is the centre whichis outside the inner circle is of length of the larger circle and BA, OA are of the diameters of the larger and smaller circles respectively. Chord AC intersects (a) 2 2 cm (b) 3 2 cm the smaller circle at a point D. If AC = 12 cm, then AD is: (c) 2 3 cm (d) 4 2 cm B (d) C 2 cm

AB = \

2

2

O'

3 - 1 = 2 2 cm 1 cm

AC = 4 2 cm C

B

O

O

3 cm

2 2 cm

D

A

A

(a) (c)

#

In the given figure, AB is chord of the circle with centre O, BT is tangent to the circle. The values of x and y are

y O

B X

(c)

(b) 6 cm (d) Data insufficient

(b) ÐADO is a right angle (angle of semicircle) Again when OD is perpendicular on the chord AC and OD passes through the centre of circle ABC, then it must bisect the chord AC at D. \ AD = CD = 6 cm

P

(a) 52º, 52º (c) 58º, 58º

4 cm 5.6 cm

O

32º

D

90°

(b) 58º, 52º (d) 60º, 64º

Given AB is a circle and BT is a tangent, ÐBAO = 32º # Here, Ð OBT = 90º [Q Tangent is ^ to the radius at the point of contact] OA = OB [Radii of the same circle] \Ð OBA = Ð OAB = 32º [Angles opposite to equal side are equal] \Ð OBT = Ð OBA + Ð ABT = 90º or 32º + x = 90º . Ð x = 90º – 32º = 58º . Also, Ð AOB = 180º – ÐOAB – ÐOBA = 180º – 32º – 32º = 116º 1 Now Y = AOB 2 [Angle formed at the center of a circle is double the angle formed in the remaining part of the circle] 1 = × 116º = 58º . 2



A The figure below shows two concentric circles with centre O. PQRS is a square inscribed in the outer circle. It also circumscribes the inner circle, touching it at point B, C, D and A. What is the ratio of the perimeter of the outer circle to that of polygon ABCD?

P

D

S

A

Q

B

O

R

C

(a)

p 3

(b) p

(c)

p 2

(d)

p 4

(c)

Let the diagonal of PQRS be 2r.

#

Therefore, side = r 2 .

r 2 ´ 2 = r. Now, ABCD is a square. And side 2 Perimeter of ABCD = 4r. Circumference of bigger circle = 2pr. p Therefore, required ratio = 2



#

In the adjoining figure ‘O’ is the centre of the circle and PQ, PR and ST are the three tangents. ÐQPR = 50°, then the value of ÐSOT is:

Q

M

O

A circle with radius 2 is placed against a right angle. Another smaller circle is also placed as shown in the adjoining figure. What is the radius of the smaller circle?

R (a) (c)

S

30° 65°

50°

P

T (b) 75° (d) can’t be determined

(c) ÐROQ = 180° – 50 = 130° æQ ÐOQP + ÐORP + ÐQOR + ÐQPR = 360°ö çè and ÐOQP = ORP = 90° ÷ø

Now, since RT = TM and QS = SM \ also OR = OM = OQ \ ÐROT = ÐTOM and ÐMOS = ÐSOQ

(a)

3-2 2

(b)

(c)

7-4 2

(d) 6 - 4 2

4- 2 2

\ ÐSOT =

1 ÐROQ 2

130 = 65° 2 The diagram below represents three circular garbage cans, each of diameter 2 m. The three cans are touching as shown. Find, in metres, the perimeter of the rope encompassing the three cans.

\ ÐSOT =

SOL:-. (d)

# O

2

2

A D r r

O' B OABC is square with side = 2

C

\ OB = 22 + 22 = 2 2

OB = 2 2 = OD + r + O'B = 2 + r + r 2 Þ r ( 2 +1) = 2( 2 –1) Þr=

2( 2 - 1)

( 2 + 1)

= 6- 4 2

=

2( 2 - 1) 2 = 2(2 + 1 - 2 2) 2 -1

(a) 2p + 6 (c) 4p + 6 (a)

\

(b) 3p + 4 (d) 6p + 6

ÐAOB = ÐCO¢D = ÐFO¢¢E = 120° Distance between 2 centres = 2 m BC = DE = FA = 2 m Perimeter of the figure = BC + DE + FA + circumference of sectors AOB, CO¢D and FO¢¢E. But three equal sectors of 120° = 1 full circle of same radius.

B

A

Therefore, perimeter of surface = 2pr + BC + DE + FA = (2p + 6)m

C O¢ 120°

O 120°

D

O¢¢ F

120° E

# In the given figure (not drawn to scale), A, B and C are three points on a circle with centre O . The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If ATC  30 and ACT  50 then BOA is C 50° O B

30° A

1) 100° 3) 40

T

2) 80° 4) 65°

1; From the alternate segment theorem ABC  ACT  ABC  50 Also, CAB  ACT  ATC = 50° + 30° = 80°

 BCA  180  ABC  CAB  180  50  80  50Now, BOA  2.BCA  2  50  100



[Angle made by chord at the centre is double that at the circumference]

MATHEMATICS BY- RAHUL RAJ



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