SECTION CHAPTER B1 REAL NUMBERS · 2016-05-21 · REAL NUMBERS SECTION B CHAPTER 1 TOPIC-1 Rational Numbers SUMMATIVE ASSESSMENT WORKSHEET-1 Solutions 1. 58 1000 ... 2157 625 = 3.4512

P-1S O L U T I O N S

REAL NUMBERS

SECTION

BSECTIONCHAPTER

1TOPIC-1Rational Numbers

SUMMATIVE ASSESSMENT WORKSHEET-1Solutions

1. 581000

= 0.058 (Decimal point is shifted three places

to the left) 1

2.

256

= 2.04124 ..... . It is non-repeating, so it is not a

rational number.

204

= 2.2360 ..... . It is non-repeating, so it is not a

rational number.

2. 27 = 2.2727 ..... . As, 27 is repeating continously, so it is a rational number.

2 . 3 = 6 = 2.4494 .... . It is non-terminating, non-Repeating So it is not a rational number.

So, 2. 27 is a rational number. 1

3. Let, x = 0.999.... ⇒ 10x = 9.999.... ⇒ 10x–x = (9.999.....) (– 0.999.......) ½ ⇒ 9x = 9 ½ ⇒ x = 1

4.

982

= 982

= 49 = 7

So, it is a rational number. 1 5. Yes, zero is a rational number.

Zero can be expressed as 0 0 0

, ,5 26 100

etc,

which are in the form of, pq

where p and q are

integers and q ≠ 0. 2

6. Let, x = 0. 6 x = 0.6666 ....(i) multiplying 10 on both the sides, we get, 10 x = 6.6666 ....(ii) 1 From (ii) – (i), we get 9 x = 6.0

⇒

x =

69

=

23

.

1

7.

x =

17

0.1428577 10

7 30 28 20 14 60 56 40 35 50 49 1

\ x = 0. 142857 [CBSE Marking Scheme, 2012] 2

8. Since LCM of 7 and 11 is 77,

\ 9

11 =

911

× 77

= 6377

× 33

= 441539

1

and

57

=57

× 1111

= 5577

× 33

= 385539

1

Hence, three rational numbers between57

and

911

are :

386539

,

387539

,

388539

1

9. Any example & verification of example : Let m = 4/5, n = 9/2

Sum = + =4 9 535 2 10

(Rational Number) 1

product =

3610

(Rational Number) 1

diffrence = =

9 4 37–

2 5 10 (Rational Number) 1

division =

9 4 452 5 8

÷ = (Rational Number) 1

P-2 M A T H E M A T I C S - I X T E R M - 1


1.

177413

= 59 × 359 × 7

= 37

.

1

2.

78

= 0.875.

1

3. 1. 011 = 1.011011011....... . 1 4. n – 1. 1

5. 1000x = 237. 237

⇒ x =

237999

[CBSE Marking Scheme, 2012]

Alternative Method : Let

x = 0. 237 = 0.237237237237.......

⇒ 1000x = 237.237237........ ½

⇒ 1000x – x = (237.237237.....) – (0.237237....) ½

⇒ 999x = 237 ½

\

x =

237999

. ½

6.

2157625

= 3.4512 1½

Terminating ½ [CBSE Marking Scheme, 2012]

Alternative Method : 3.4512

625) 2157 1

187528202500

32003125

750 625

1250 1250

×

2157625

= 3.4512 (Terminating) 1

7. Alternative Method : LCM of 5 and 7 is 35

\

35

= 35

× 77

= 2135

½

and

57

=

57

×

55

=

2535

½

so,

2135

< 2235

< 2335

< 2435

< 2535

½

The required three rational numbers are 22

35, 23

35

and

2435

.

½

[CBSE Marking Scheme, 2012] 8. Let a = 3 & b = 4 Here, we find six rational numbers, i.e., n = 6

So d =

– + 1

b an

= 4 – 36 + 1

=

17

1st rational number = a + d = 3 +

17

=227

½

2nd rational number = a + 2d = 3 +

27

=237

½

3rd rational number = a + 3d = 3 +

37

=247

½

4th rational number = a + 4d = 3 +

47

=257

½


57

=267

½


67

=277

½

So, six rational numbers are

227

,

237

,

247

,

257

,

267

&

277

.

9. LCM of 3, 4, 6, and 12 is 12

1 43 123 12

1 34 124 12

1 2126 6 12

112 1212

2 = 2 = 2 = 16

5 = 5 = 5 = 125

7 =7 =7 = 49

3 = 3 = 3

2

Descending order is

12 12 12 12125, 49 , 16 , 3 1

i.e., 4 3 1265 , 7 , 2 , 3 . 1

[CBSE Marking Scheme, 2012] Alternative Method : Since, LCM of 3, 4, 6, 12 is 12

\1 4 4

12 43 123 4 122 = 2 = 2 = 2 = 16×

1

1 3 3

12 34 124 3 125 = 5 = 5 = 5 = 125×

½

1 2 2

12 2 126 6 2 127 =7 =7 = 7 = 49×

½

12 3 = 12 3


In Descending order 12 12 12 12125, 49 , 16 , 3 1

i.e., 4 3 1265 , 7 , 2 , 3 . 1

10. 1. 32 = 119

90 1½

0. 35 =

3599

1½

1. 32 + 0. 35 =

1659990

1

[CBSE Marking Scheme, 2012] Alternative Method : Let,

x = 1. 32 = 1.32222.......

⇒ 10x = 13.222..........

⇒ 100x = 132.222......

⇒ 100x – 10x = (132.222.......) – (13.222.....) ½

⇒ 90x = 119.00

⇒

x =

11990 ½

Again, let y = 0. 35 = 0.353535.......

⇒ 100y = 35.3535........ ½

⇒ 100y – y = (35.3535......) – (0.3535.......) ½

⇒ 99y = 35

⇒

y =

3599

½

\ 1. 32 + 0. 35 = x + y =

11990

+ 3599

½

=

119 11 + 35 10990

× ×

=

1309+350990

=

1659990

1

TOPIC-2Irrational Numbers


1. 0.13 is a terminating number. So, it is not an irrational number.

0.13 15 = 0.131515......, 15 is repeating continuously, so it is not an irrational number.

0.1315 = 0.13151315...., 1315 is repeating continuously, so it is not an irrational number.

0.3013001300013...., non-terminating and non-recurring decimal. Hence, it is an irrational number.

So, 0.3013001300013 is an irrational number. 1 2. No, it may be rational or irrational. 1 3. Let, x = 0.777.... ⇒ 10x = 7.777...... ⇒ 10x – x = (7.777......) – (0.777.....) ⇒ 9x = 7

⇒ x =

79

.

1

4. Sum of 2 5 and 3 7 = 2 5 +3 7 . 1

5. Required two irrational number are : 0.1 = 0.10 (i) .10100100010000......... 1 (ii) .1020020002000........ 1

6. 0.5101001000100001....... and 0.502002000200002 ..... . [CBSE Marking Scheme, 2012]

7.

14.5

D

AA O B C E

14.5

4.5

Mark the distance 4.5 units from a fixed point A on a given line to obtain a point B such that AB = 4.5 units. From B, mark a distance of 1 unit and mark the new point as C. Find the mid-point of AC and mark that point as O. Draw a semi-circle with centre O and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semi-circle at D.

Then, BD = 4.5

To represent 4.5 on the number line, let us treat the line BC as the number line, with B as zero, C as 1, and so on. 1

Draw an arc with centre B and radius BD, which intersects the number line at E. 1

\E represents 4.5 1 8. Let two irrational numbers are :

6 and 3 ,

(i) 6 – 3 = Difference is an irrational number.


(ii) +6 3 = sum is an irrational number.

(iii) × = =6 3 18 3 2

= product is an irrational number.

(iv) =6 / 3 2 = division is an irrational number. 1 × 4 = 4


1. Since, 5 = 2.236

Hence, the irrational number between 2 and 2.5 is 5 . 1

2. 0.3 0.4+ = (0.333.....) + (0.444.....)

= 0.777...... Let, x = 0.777....... 10x = 7.777....... ⇒ 10 x – x = (7.777.......) – (0.777.......) ⇒ 9x = 7.0

⇒ x = 7

9.

3. 2 is non-terminating, non-recurring. 1

4. 2( 2 + 5) =

2( 2 ) + 2( 5) + 2 × 2 × 5

= 2 + 5 + 2 10 .

= 7 + 2 10 (Irrational Number). 1

5. Let x = 2.218 = 2.218181818........ ⇒ 10x = 22.18181818........ ½ ⇒ 1000x = 2218.181818..... ½ ⇒ 1000x – 10x = (2218.181818......) –

(22.181818.....) ⇒ 990x = 2196.00 ½

⇒ x = 2196990

= 2 3 3 1222 3 3 55× × ×× × ×

= 12255

½


6. Given,

17

= 0.142857142857......... ½

27

= 0.285714285714......

½

Hence, required number can be 0.160160016000... [CBSE Marking Scheme, 2012] 1

7.

–2 –1 –1

2OA

1B

P Q 2 3

3

3

2

D

C1

1

�–�

Let AB = BC =1 unit length

Using Pythagoras theorem, we see that

OC = 2 21 + 1

= 2 ½

Construct CD =1 unit length perpendicular to OC, then using Pythagoras theorem, we see that

OD = 2 2( 2 ) 1+

=3

½

Using a compass with centre O and radius OD, draw an arc which intersects the number line at the point Q, then Q corresponds to 3 . 1

8.

19.3

D

AA O B C E

19.39.3

Mark the distance 9.3 units from a fixed point A on a given line to obtain a point B such that AB = 9.3 units. From B, mark a distance of 1 unit and mark the new point as C. Find the mid-point of AC and mark that point as O. Draw a semi-circle with centre O and radius OC.

Draw a line perpendicular to AC passing through B and intersecting the semi-circle at D. 1

Then, BD = 9.3 1

To represent 9.3 on the number line,let us treat the line BC as the number line, with B as zero, C as 1, and so on.

Draw an arc with centre B and radius BD, which intersects the number line at E. 1

\E represents 9.3


9. 10x = 3.178178 ...(1) 1 ⇒ 10000x = 3178.178178 ...(2) 1

Subtracting (1) from e.q. (2)

9990x = (3178.178 ...... – 3.178....) 1

⇒ x = =3175 6359990 1998

1


TOPIC-3nth Root of a Real Number


1.

( )5+ 5 ( )5 – 5

=

( ){ }225 – 5

½

= {25 – 5} = 20 ½ [CBSE Marking Scheme, 2014]

2. 12 8× = 2 3 ×2 2 = 4 3 2× = 4 6 . 1

3. 4 28 ÷ 3 7 = 4 ×2 7 ÷ 3 7 = 83

1

4. +8 3 – 2 3 4 3 = +3(8 – 2 4)

= 10 3 2

5. 2 50 × 3 32 × 4 18

⇒10 2 ×12 2 ×12 2

= 2880 2 2

6. x = 3 – 2 2

⇒ x = 1 + 2 – 2 2 ½

⇒ x =

( )21 – 2

⇒ x = 1– 2 ½

⇒

1x

=

11 – 2

×

1 21 2

++

⇒ =

1 21 – 2+

= –

( )1 2+

½

\ x +

1x

= 1 – 2 – 1 – 2

= – 2 2 ½

7. 4 16 =

4 2 2 2 2× × × = 2

3 343 =

3 7 7 7× × = 7

5 243 =

5 3 3 3 3 3× × × × = 3

196 = 14 1

\ 4 16 – 6 3 343 + 18 5 243 – 196

= 2 – 6 × 7 + 18 ×3 – 14 = 2 – 42 + 54 –14 = 56 –56 = 0. 1

8. ( )4 3 3 2+

×( )4 3 – 3 2

=

( )24 3

–

( )23 2

1

= 48 – 18 ½ = 30. ½ [CBSE Marking Scheme, 2013, 2012]

9. 3 45 – 125 200 – 50+

= 9 5 – 5 5 +10 2 – 5 2 1½

= 4 5 5 2+ 1½

10. (i)

( )( )

2 – 1

2 1+

is an irrational number

( )( )

2 – 1

2 1+ =( )( )

( )( )

2 – 1 2 – 1

2 1 2 – 1×

+

= ( )2

2 – 1

2 – 1

=

( )22 – 1

1 = 2 – 1

which is an irrational number. 1 Let, there is a number x such that x3 is an irrational

number but x5 is a rational number.

Let, x = 5 7 be the number

⇒

x3 = ( )35 7 = 35(7) ½

is an irrational number

But, x5 = ( )55 7 = (7)5/5 = 7

= 7, is a rational number ½ (ii) Accepting own mistakes gracefully, co-operative

learning among the classmates. 1


1.

5x

= p 7 ⇒ 5 57

×

= p 7

⇒

p =

257 7×

= 257

. 1

2. b2 = a ⇒

a = b. 1


3. ( )a b+ ( )–a b = (a)2 – ( )2

b = a2 – b. 1

4. False 1

Justification :

14775

=

14775

=

4925

= 75

1

which is a rational number. [CBSE Marking Scheme, 2012]

5.

( )24 3 – 3 5

=

( )24 3

+

( )23 5 – 2 × 4 3

× 3 5

[Using (a–b)2 = a2 + b2 –2ab]

= 48 + 45 – 24 15 ½

= 93 – 24 15

= 3(31 – 8 15 ). ½


6. 50 – 98 + 162 1

= 5 5 2× × – 7 7 2× × + 3 3 3 3 2× × × × ½

= 5 2 – 7 2 + 9 2 ½

= 7 2 [CBSE Marking Scheme, 2012]

Alternative Method :

50 – 98 + 162 ½

= 5 5 2× × – 7 7 2× × + 3 3 3 3 2× × × ×

= 5 2 – 7 2 + 3 × 3 2 ½

= –2 2 + 9 2 ½

= 7 2 ½

7. 3 40 =

3 32 5× = 2 3 5 ½

3 320 = 4 3 5 ½

\ 3 3 40 – 4 3 320 –

3 5 = 3 × 2 3 5 – 4 × 4 3 5 –

3 5

= –11 3 5 1

[CBSE Marking Scheme, 2012] Alternative Method :

3 40 =

3 2 2 2 5× × × = 2 3 5 ½

3 320 = 3 2 2 2 2 2 2 5× × × × × ×

= 2 × 2 3 5 = 4 3 5 ½

3 40 – 4

3 320 – 3 5

= 3 × 2 3 5 – 4 × 4 3 5 – 3 5 ½

= 6 3 5 – 16 3 5 – 3 5 = – 11 3 5 . ½

8. LCM of 2 and 3 is 6 1

32 3 3 2× = 2 × 3

6 2 33 2× ½

= 6 6 72 ½


Alternative Method : LCM of 2 and 3 is 6

\ 2 3 3 = 2 ( )

1 23 23 ×

= 2 6 23 ½

and 3 2 = 3 ( )1 32 32 ×

= 3 6 32 ½

Now, 2 3 3 × 3 2 = 26 9 × 3

6 8 ½

= 6 6 72 ½

9. ( )–1 / 6729 = ( )–1 / 663 =

–13 2

=

13

1

[CBSE Marking Scheme, 2012] Alternative Method : 3 729 3 243 3 81 3 27 3 9 3 3 1 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36 2

( )

–16729

= ( )–1

6 63 ½

= ( )–13

=

13

½

10. Justify, 2

5 3+ =

( )( )

5 – 325 3 5 – 3

×+

½

=

( ) ( )2 210 – 6 10 – 6

5 – 35 – 3=

½

=

10 – 62

Again, 28 + 98 + 147

= 2 2 7× × + 2 7 7× × + 3 7 7× × ½

= 2 7 + 7 2 + 7 3 ½

Value : Co-operative learning among classmates without any gender and religious bias . 1


TOPIC-4Laws of Exponents with Integral Powers


1.

1–2 –2 2 2 4

4 4 4132 143 13 11 13143 132 11 12 12

× = = = ×

=

1213

12

=

132 3 1

2. [(16)½]½

= ( )1

1 22 24

= [ ]124 ½

= 2 ½

3. ( )( )

( – )

( – )

a b cc

b

b a c a

xxxx

÷

= 1 1

–

–

ab ac bc

ba bc ac

x xx x

÷ = (xab–ac–ba+bc) ÷ (xbc–ac) = x0 1

= 1 [CBSE Marking Scheme, 2014]


( – )

( – )

ca b c b

b a c a

x xx x

÷

= –

–

ab ac bc

ba bc ac

x xx x

÷

½

= (xab–ac–ba+bc) ÷ (xbc–ac) ½ = (xbc–ac) ÷ (xbc–ac) ½ = 1 ½

4.

30 29 28

31 30 29

3 3 33 3 – 3

+ ++

=

28 2 1

29 2 1

3 (3 3 1)3 (3 3 – 1)

+ ++

1

=

(9 3 1)3(9 3 – 1)

+ ++

½

=

133 11×

=

1333

½


5.

( ) +

1/ 431/3 1/35 8 27

=

( ) +

1/ 433 1/3 3 1/35 (2 ) (3 )

1

= +

1/ 435 (2 3)

1

= ( ) = =

1/ 41/ 43 45 (5) 5 5

1

6. ( )2

2 2 – 5 + ( )2

3 2 3+ – ( )2

2 – 1

= ( )22 2 – 2( 2 )(5) + (5)2 + ( )2

3 2 + 2(3 2 )

( 3 ) + ( 3 )2 – ( )2

2 – (1)2 +2( 2 ) (1) 1

= 8 + 25 – 20 2 + 18 + 3 + 6 6 – 2 – 1 + 2 2 1

= 51 – 18 2 + 6 6 1


7.

2 3 1– – –

3 4 5

4 1 2

(216) (256) (243)

+ +

=

–2 –3 –13 4 53 4 5

4 1 2

(6 ) (4 ) (3 )

+ +

1

=

–2 –3 –14 1 2

6 4 3+ +

1

=

–1 –1 –1

4 1 236 64 3

+ +

1

= 4 × 36 + 1 × 64 + 2 × 3 ½ = 214 ½

8.

–3 3–3

4 281 9 516 25 2

× ÷

=

3 33

4 216 9 281 25 5

× ÷

1

=

3 334 24 2

4 22 3 2

53 5

× ÷

½


=

3 34 2 34 22 3 5

3 5 2

× ×

1

=

3 3 32 3 53 5 2

× × ½

=

3 3 3

3 3 32 3 53 5 2

× ×

=

3 3

3 32 33 2

×

½

= 1 ½


1. ( )1

– 4481 81× = ( ) ( )1 1

–4 44 43 3×

= 3–1 × 31 = 30 = 1. 1

2.

( ){ }2–1

–1 4281

=

( ){ }–1

–1 2281

= ( )1481

= ( )1

4 43 = 3. 1

3. = 1

3 45(2 3) + 1

= 1

4 4(5 ) 1

= 5 [CBSE Marking Scheme, 2012] Alternative Method :

13 41 1

3 35 8 27

+

=

13 41 1

3 33 35 (2 ) (3 )

+

½

=

13 45(2 3) +

=

13 45(5) 1

=

14 4(5 ) = 51= 5. ½

4.

3 3– –

4 281 2516 9

× =

3 3– –4 24 23 5

2 3

× ½

=

–3 –33 52 3

× ½

=

3 32 33 5

× ½

=

3

3

25

=

8125

½


5.

3 3– – –3

4 281 25 516 9 2

× ÷

½

=

31 33

4 216 9 281 25 5

× ÷

1

=

3 3 32 3 23 5 5

× ÷ ½

=

38 3 527 5 2

× × =

8 27×

27 8 = 1 1

[CBSE Marking Scheme, 2012, 2013] Alternative Method :

3 3– – –3

4 281 25 516 9 2

× ÷

=

3 3– –4 2 –34 23 5 5

2 3 2

× ÷

1

=

–3 –3 –33 5 52 3 2

× ÷ ½

=

3 3 32 3 23 5 5

× ÷ ½

=

8 27 125× ×

27 125 8

½

=

8 27×

27 8 = 1

½

6.

7 5––1 2 –2 32 2

2 – 4 3 –5

5 7 5 75 7 5 7

× ×× × × ½

=

7 5–4 2 5 32 2

2 1 3 2

7 7 7 75 5 5 5

× ×× × ×

1

=

7 5–6 82 2

3 5

7 75 5

×

=

7 56 52 2

3 8

7 55 7

×

½


=

1 142 252 2

21 40

7 55 7

×

=

142 25 2

21 40

7 55 7

×

½

= ( )1

2 4 27 5×

= 7 × 52 = 7 × 25 = 175 [CBSE Marking Scheme, 2012]


7 5––1 2 –2 32 2

2 – 4 3 –5

5 7 5 75 7 5 7

× ×× × ×

=

7 5–4 2 5 32 2

2 1 3 2

7 7 7 75 5 5 5

× ×× × ×

½

=

7 5–6 82 2

3 5

7 75 5

×

½

=

1 17 –52 26 8

3 5

7 75 5

×

½

=

1 142 252 2

21 40

7 55 7

×

½

=

142 25 2

21 40

7 55 7

× ×

= ( )1

2 4 27 5×

½

= 7 × 52 = 175 ½

7.

( ) 2 2 2( ) ( )

4

. .

( )

a b b c c a

a b c

x x x

x x x

+ + +

=

2 2 2 2 2 2

4 4 4. .

. .

a b b c c a

a b cx x x

x x x

+ + +

2

=

4 4 4

4 4 4

a b c

a b cxx

+ +

+ +

½

= 1 1½ [CBSE Marking Scheme, 2012]


( )

( )4

2 2 2a+b (b+c) (c+a)

a b c

x . x . x

x x x

=

2( ) 2( ) 2( )

4 4 4. .

. .

a b b c c a

a b cx x x

x x x

+ + +

1

=

2 2 2 2 2 2

4 4 4. .

. .

a b b c c a

a b cx x x

x x x

+ + +

1

=

2 2 2 2 2 2

4 4 4

a b b c c a

a b cx

x

+ + + + +

+ +

1

= 4 4 4

4 4 4

a b c

a b cxx

+ +

+ + = 1 1

8.

7 5––1 2 –2 32 2

2 – 4 3 –52 3 2 32 3 2 3

× ×× × ×

=

7 5–6 82 2

3 53 32 2

×

1½

=

7 56 8 –

2 2

7 53 5 –

2 2

3 3

2 2

× ×

× ××

1

=

2521 2

21 202

3 23

2

×

= 3 × 22 = 3 × 4 = 12 1½ [CBSE Marking Scheme, 2012]


7 5––1 2 –2 32 2

2 – 4 3 –52 3 2 32 3 2 3

× ×× × ×

=

7 5–6 82 2

3 53 32 2

×

1

=

7 56 8 –

2 2

7 53 5 –

2 2

3 3

2 2

× ×

× ××

1

=

21 –20

21 25–

2 2

3 3

2 2

×

1

= 21–20

21 25–

2 2

3

2

= 1

4–

2

3

2

½

= 3 × 22 = 12. ½


1.

15002 15

=

1 15002 15

×

=

1100

2×

=

102

= 5.

1

2.

3 / 4

–1 / 4

1616

= ( )

3 14 416 +

= (16)1 = 16.

1


3. (13 + 23 + 33)–3/2 = (1 + 8 + 27)–3/2 ½ = (36)–3/2 ½ = [(6)2]–3/2 = 6–3 ½

=

3

16

=

1216

½


4. Given, a = 2 and b = 3. ab + ba = 23 + 32 ½ = 8 + 9 1 = 17 ½ [CBSE Marking Scheme, 2012]

5. (xa–b)a+b. (xb–c)b+c. (xc–a)c+a

= 2 2 2 2 2 2– – –. .a b b c c ax x x 1

= 2 2 2 2 2 2– – –a b b c c ax + +

1

= x0 = 1 (any number to the power 0 is 1) 1 [CBSE Marking Scheme, 2012]

6. 2–

3

4

(216)

–

3–

4

1

(256) =

–2 –3

3 43 4

4 1–

(6 ) (4 )

1

= 4 × 62 – 43 1 = 144 – 64 = 80 1 [CBSE Marking Scheme, 2011, 2012]


2–

3

4

(216)

–

3–

4

1

(256)

= –2

3 3

4

(6)

– –3

4 4

1

(4)

1

=

–2

46

–

–3

14

½

= 4 × 62 – 43 ½ = 144 – 64 ½ = 80 ½

7.

1 1.–

ab aba b a a b a

++

1

= 2 2

2 2

––

b ab b abb a

+ +

1

=

2

2 2

–2–b

a b 1


–1 –1

–1 –1 –1 –1

a a+

a +b a – b =

1

1 1a

a b+

+

1

1 1–

a

a b

1

=

1a

a bab+

+

1

–a

b aab ½

=

ba b+

+ –

bb a ½

=

( ) ( )( )( )–

–b a b b b a

b a b a+ +

+ ½

=

2 2

2 2

––

b ab b abb a

+ +

½

=

( )2

2 2

2– –

ba b

=

2

2 2

–2–b

a b ½

8. LHS = 1 1 1

1 1 11 1y xy y xy

y xy x

+ ++ + + + + +

1

( xyz = 1 ⇒

1z

= xy) ½

=

11 1 1

y xyxy y xy y xy x

+ ++ + + + + +

1

=

11

y xyxy y

+ ++ +

= 1 = RHS 1½

Hence proved.[CBSE Marking Scheme, 2012]

Alternative Method : (1 + x + y–1)–1 + (1 + y + z–1)–1 + (1 + z + x–1)–1

=1 1 1

1 1 11 1 1x y z

y z x

+ ++ + + + + +

1

=

1 1 11 1 111 1y xyxy xy x

+ ++ ++ + + +

( xyz = 1 ⇒

1z

= xy)

=

1 1 11 11y xy xy yy xy

y xy

+ ++ + + ++ +

1

=

11 1 1

y xyy xy y xy xy y

+ ++ + + + + +

1½

=

( )( )

11

y xyy xy

+ ++ +

= 1.

½

9. 52x–1 – (52)x–1 = 2500 ½ ⇒ 52x–1 – 52x–2 = 2500 ½ ⇒ 52x–2 (5 – 1) = 2500

⇒

52x–2 =

25004

= 625 = 54 1

⇒ 2x – 2 = 4 ⇒ 2x = 6 \ x = 3 1 [CBSE Marking Scheme, 2012]


Alternative Method : Given, 52x–1 – (25)x–1 = 2500 ⇒ 52x–1 – [(52)]x–1 = 2500 ½ ⇒ 52x–1 – 52x–2 = 2500 ½ ⇒ 52x–2 (5 – 1) = 2500 1

⇒

52x–2 =

25004

= 625 ½

⇒ 52x–2 =54 ½

comparing the powers of both sides, we get 2x – 2 = 4 ½ 2x = 6 ⇒x = 3 ½

TOPIC-5Rationalisatoin of Real Numbers


1.

150

=

15 5 2× ×

=

1 25 2 2

× =

210 1

So, rationalising factor is 2 .

2.

6 – 4 3 6 – 4 36+4 3 6 – 4 3

×

½

=

( )26 – 4 3

36 – 48 ½

=

36 48 – 48 3–12

+

½

=

84 – 48 3–12

= – (7 – 4 3 ) = 4 3 – 7

½

3. x = 3 – 2 2 ⇒

1x

= 3 + 2 2 . ½

21

xx

+ = 8 1

⇒ x +1x

= +2 2 ½


x = 3 – 2

⇒

1x

=

( )

( )( )3 2 21

3 – 2 2 3 2 2

+×

+

=

( )3 2 2

9 – 8

+

= 3 + 2 2 ½

21

xx

+ =

1 12x x

x x+ + × ×

½

= 3 – 2 2 3 2 2 2+ + + ½

⇒ 21

xx

+ = 8

\ 1

+xx

= +2 2 ½

4. x = 2 + 3

1x

= 2 – 3 1

x +

1x

= 4 1

Squaring, both sides, we get

22

1+x

x = 14 1

5. 3 2

5 2+

+ =

3 2 5 – 25 2 5 – 2

+×

+ 1

= 5 3 5 2 – 6 – 225 – 2

+

1

= 5 3 5 2 – 6 – 223

+ 1

6.

( )

( )( )

2 3 412 3 – 4 2 3 4

+ +×

+ + +

1

=

( ) ( )2

2 3 4 2 3 42 3 2 6 – 42 3 – 4

+ + + +=

+ ++

=

2 3 4 1 – 2 61 2 6 1 – 2 6+ +

×+

1

=

2 22 3 4 – 2 12 – 2 18 – 4 6

1 – (2 6 )+ +

½

=

2 3 4 – 4 3 – 6 2 – 4 61 – 24

+ +

½


=

–5 2 – 3 3 2 – 4 6–23

+

½

=

5 2 3 3 4 6 – 223

+ + +

½

7. LHS

=

1 1 1 1 1– –

3 – 8 8 – 7 7 – 6 6 – 5 5 – 2+ +

=

( ) ( )( )

( ) ( )( )

( ) ( )2 2 2 2 22

8 7 7 63 8–

3 – 8 8 – 7 7 – 6

+ +++

–

( )( ) ( )

( )( ) ( )2 2 2 2

6 – 5 5 2

6 – 5 5 – 2

++

1

=

( ) ( ) ( )8 7 7 6 6 53 8– –

9 – 8 8 – 7 7 – 6 6 – 5

+ + +++

1

+

( )5 +2

5 – 4 [ a2 – b2 = (a – b) (a + b)] 1

= 3 + 8 – 8 – 7 + 7 + 6 – 6 – 5 + 5 + 2

= 3 + 2 = 5 = RHS 1


1.

1 ( 2 – 1)( 2 1) ( 2 – 1)

×+

=

2 –1

= 1.414 – 1 1

= 0.414 1 [CBSE Marking Scheme, 2012]


12 1+

= 1 ( 2 – 1)

( 2 1) ( 2 – 1)×

+

=

( )( ) ( )2 2

2 – 1

2 – 1

1

=

2 – 12 – 1

= 1.414 – 1

1 = 0.414. 1

2.

12

=

12

×

22

(rationalising)

=

22

= 1.414

2 = 0.707 1

12

+ p = 0.707 + 3.141

= 3.848. 1


3.

5 65 – 6

+

=

5 65 – 6

+

×

5 65 6

++

½

=

( )( )

2

2 2

5 6

5 – ( 6 )

+

½

=

25 6 10 625 – 6

+ +

½

=

31 10 619

+

½

\ a + b 6 =

3119

+

1019

6

½

Comparing the rational and irrational parts of both sides, we get

a =

3119

, b =

1019

½

4. x =

13 – 2 2

×

3 2 23 2 2

++

x =

3 2 29 – 8+

= 3 + 2 2

1

and

y =

13 2 2+

×

3 – 2 23 – 2 2

=

3 2 29 – 8+

= 3 – 2 2 1

x + y+ xy = 3+2 2 + 3 – 2 2 + (3+2 2 )

(3 – 2 2 ) ½

= 6 + 9 – 8

= 7 ½

5. (i) We know that between two rational numbers x and y, such that x < y there is a rational number

+2

x y

.

i.e.,

3 <

72

< 4

Now, a rational number between 3 and

72

< 4 is :

1 7

32 2

+ =

1 6 72 2

+ × = 13

4 ½

A rational number between

72

and 4 is :


1 7

42 2

+ =

1 7 82 2

+ × = 154

½

\

3 < 134

<

72

< 154

< 4

½

Further a rational between 3 and

134

is :

1 13

32 4

+ =

1 12 132 4

+ × = 25

8


154

and 4 is :

1 15

42 4

+ = 1 15 16

2 4+

× = 31

8 ½


318

and 4 is :

1 31

42 8

+ =

1 31 322 8

+ × = 63

16 ½

\

3 < 25

8 < 13

4<

72

< 154

< 318

< 6316

< 4

Hence, six rational numbers between 3 and 4 are

25 13 7 15 31, , , , ,

8 4 2 4 8 and 63

16 1

(ii) Number system. ½

(iii) Rationality is always welcomed.

6. x =

3 23 – 2

+

and

y =

3 – 23 2+

⇒

x2 + y2 =

23 23 – 2

+

+

23 – 23 2

+

1

=

3 2 2 63 2 – 2 6

+ + +

+ 3 2 – 2 63 2 2 6

+ + +

1

=

5 2 65 – 2 6

+

+

5 – 2 65 2 6+

½

=

( ) ( )( ) ( )

2 2

22

5 2 6 5 – 2 6

5 – 2 6

+ +

½

=

25 24 20 6 25 24 – 20 625 – 24

+ + + +

= 98 1

7. LHS =

( ) ( )( )( )

2 22 5 3 2 5 – 3

2 5 – 3 2 5 3

+ +

+

1

( ) ( )2 24 5 3 2 2 5 3 4 5 3 – 2 2 5 3

2 5 – 3

× + + × × + × + × ×

1

=

20 3 4 15 20 3 – 4 1520 – 3

+ + + +

=

4617

=

4617

+

15 ×

(0)

1

\

4617

+

15 (0) = a + 15 b

= RHS

comparing both sides, we get

a =

4617

, b = 0 1



1.

2 2 8– +

5 +2 10 – 2 2 2

=

( )

2 ( 5 – 2) 2–

( 5 2) ( 10 – 2 2 )5 – 2×

+

×

( 10 2 2 ) 8 2( 10 2 2 ) 2 2

++ ×

+ 1

= 10 – 2 2 2( 10 2 2 ) 8 2–

5 – 4 10 – 8 2+

+

1

=

10 – 2 2 – 2( 10 2 2 )2+

+ 4 2

=

10 – 2 2 – 10 – 2 2 + 4 2 = 0 . 1

2.

12 7 3 3+

=

1(2 7 3 3)

×+

(2 7 – 3 3)(2 7 – 3 3)

=

2 2

2 7 – 3 3(2 7 ) – (3 3)

1

= 2 7 – 3 34 7 – 9 3× ×

1

= 2 7 – 3 3

28 – 27 1

=

2 7 – 3 31

= 2 7 – 3 3 .


3. x =

2[ 2 – 2 ]2 – 2

p q p qp q p q+ ++ +

1

=

14q

( )2 22 2 – 4p p q+ 1

⇒ 2qx – p = 2 2– 4p q

⇒ 4q (qx2 – px) = – 4q2 ⇒ qx2 – px + q = 0. 1 [CBSE Marking Scheme, 2012]


x =

2 – 2 2 – 22 – – 2 2 – 2

p q p q p q p qp q p q p q p q

+ + + +×

+ + +

½

= ( ) ( )

( ) ( )2 – 2 2 2 – 2

2 – – 2p q p q p q p q

p q p q+ + + × + ×

+

½

= 2 22 2 – 4

2 – 2p p qp q p q

++ +

½

= ( )2 22 – 4

4

p p q

q

+

⇒ 2qx = p + 2 2– 4p q

⇒

2qx – p = 2 2– 4p q Squaring both sides, we get 4q2x2 + p2 – 4pqx = p2 – 4q2 ½ 4q (qx2 – px) = – 4q2 ½ qx2 – px + q = 0. ½

4. (i)

0.7142857 5.000000

– 4910–7

30–28

20–14

60–56

40–355

\

57

=0. 714285 1

and

0.8111 9.00

– 8820

–119

\

911

=0. 81 1

Thus, three different irrational numbers between 57

and

911

are 0.75075007500075000075.......,

0.767076700767000767.... and 0.80800800080000...... (ii) Number system. (iii) Those who are irrational in their approach

fail in their efforts. 1

5. x =

12 – 3

⇒

x =

1 (2 3)2 – 3 (2 3)

+×

+

⇒

=

2 34 – 3+

= 2+ 3 1

⇒ (x – 2) = 3

⇒ ( )2– 2x = ( )23 = 3 1

x2 – 4x + 4 = 3 x2 – 4x + 4 – 3 = 0 ½ \ x2 – 4x + 1 = 0

\ x3 – 2x2 – 7x + 5

x(x2 – 4x + 1) + 2(x2 – 4x + 1) + 3 1

= x × 0 + 2 × 0 + 3 = 3 ½ [CBSE Marking Scheme, 2012]

FORMATIVE ASSESSMENT WORKSHEET-13 & 14 Note : Students should do these activities themselves.

qqq


TOPIC-1Polynomials


1. p (y) = y2 – y + 1 p (0) = 02 – 0 + 1 = 1 1 2. x + 4 is a factor of x2 + 3x + m ⇒ p (– 4) = 0 ⇒ (– 4)2 + 3(– 4) + m = 0 ⇒ 16 – 12 + m = 0 ⇒ m = – 4 1

3. 3 x2 – x – 1. 1 4. Not defined. 1 5. Linear polynomial. 1 6. Degree of x3 + 5 = 3 Degree of 4 – x5 = 5 Degree of (x3 – 5) (4 – x5) = 3 + 5 = 8. 1 7. Not a polynomial. 1 8. (x + 3) is a factor of x3 + 3x2 – kx – 3 ⇒ p (–3) = 0 ⇒ (–3)3 + 3(–3)2 – k (–3) –3 = 0 1 – 27 + 27 + 3k – 3 = 0 k = 1 1

9. (x + 2) is a factor of p(x) = 2x3 – kx2 + 3x + 10 ½ ⇒ p (–2) = 0 ⇒ 2 (–2)3 – k (–2)2 + 3(–2) +10 = 0 1 ⇒ – 16 – 4k – 6 + 10 = 0 k = – 3 ½ [CBSE Marking Scheme, 2014]

10. p(x) = 2x2 + kx + 2 (x – 1) is a factor of p(x), then p(1) = 0 ½

\ 2 (1)2 + k(1) + 2 = 0 ½

2 + k + 2 = 0 ½

k = –2 – 2 ½ 11. f(x) = 3x + 5 then, f(7) = 3 × 7 + 5 = 26 1 then, f(5) = 3 × 5 + 5 = 20 1 \ f(7) – f(5) = 26 – 20 = 6. 1 12. f(x) = x2 – 5x + 7 Then, f(2) = 22 – 5 × 2 + 7 = 1 1 f(–1) = (– 1)2 – 5 (–1) + 7 = 13 1

and f13

=

213

– 513

+ 7 =

499

1

f(2) – f(–1) + f13

= 1 – 13 + 499

=

–599

1


1. Constant polynomial is 7. 1 2. Binomial. 1

3. Degree of a polynomial 3 is 0. 1 4. Every real number is a zero of the zero polynomial. 1 5. No. of zeroes of cubic polynomial = 3. 1 6. (x + 1)3 – (x + 1) = (x + 1) [(x+1)2 – 1] One factor of (x + 1)3 – (x + 1) is (x + 1). 1 7. Given, p(x) = x2 + 11x + k Since – 4 is a zero of polynomial p(– 4) = 0 ⇒ (– 4)2 + 11 × (– 4) + k = 0

⇒ 16 – 44 + k = 0 \ k = 28. 1

8. 2 x2 + 3x or 3 y2 + 3y. 1

9. Coefficient of x in expression x2 + 2π

x – 7 is 2π

. 1

10. p(x) = x3 – 3x2 – 2x + 6 ½

Then, p ( 2 ) = ( )32 –3 2( 2 ) – 2 ( )2 + 6 ½

= 2 2 – 6 – 2 2 + 6 ½ = 0 ½ 11. Linear polynomial → 1 + x, degree = 1 ½ Quadratic polynomial → x2 + x, degree = 2 ½ Cubic polynomial → x – x3, 7x3, degree = 3 1

POLYNOMIALS

SECTION

BSECTIONCHAPTER

2


12. p(x) = x2 – 3x + 6

(i) When x = 2

then, p ( )2 = ( )22 – 3 × 2 + 6 ½

= 2 – 3 2 + 6 ½

= 8 – 3 2 ½ (ii) When x = 3 then, p(3) = 32 – 3 × 3 + 6 ½ = 9 – 9 + 6 ½ = 6 ½ 13. (i) (I) Possible length and breadth of the rectangle

are the factors of its given area. Area = 25a2 – 35a + 12

= 25a2 – 15a – 20a + 12 = 5a(5a – 3) – 4(5a – 3) = (5a – 4)(5a – 3) So, possible length and breadth are (5a – 3) and (5a – 4)

units, respectively. 1 (II) Area = 35y2 + 13y – 12 = 35y2 + 28y – 15y – 12 = 7y(5y + 4) – 3(5y + 4) = (7y – 3)(5y + 4) So, possible length and breadth are (7y – 3) and

(5y + 4) units respectively. 1 (ii) Factorisation of Polynomials. ½ (iii) Expression of one’s desires and news is very

necessary. ½

TOPIC-2Remainder Theorem

SUMMATIVE ASSESSMENT WORKSHEET-17Solutions 1. Let, p(y) = 5y3– 2y2 – 7y + 1, then remainder = p(0) = 0 – 0 – 0 + 1 = 1. 1 2. Put, x + 1 = 0 ⇒ x = – 1 Then remainder is : f(– 1) = (– 1)11 + 101 = –1 + 101 = 100. 1 3. p(x) = 2x3+ 3x2 – 9x + 4 So, x2 + 2x – 3

2x – 1) 2x3 + 3x2 – 9x + 4 2x3 – x2

(–) (+) 4x2 – 9x 4x2 – 2x (–) (+) – 7x + 4 – 6x + 3 (+) (–)

– x + 1 (Remainder) Quotient = x2 + 2x – 3 Remainder = – x + 1 2

4. p(x) = x3 + x2 + x + 1

Put, x – 12

= 0 ⇒x = 12

in p(x) ½

Remainder = p 12

½

= 31

2

+ 21

2

+ 12

+ 1 ½

= 18

+ 14

+ 12

+ 1

= 1+2+4+8

8=

158

. ½

[CBSE Marking Scheme, 2012] 5. Factors of 12 = (±1, ± 2, ± 3, ± 4, ± 6, ± 12 ) ½ p(x) = 6x3 – 25x2 + 32x – 12 p(2) = 6(2)3 – 25(2)2 + 32 × 2 – 12 = 48 – 100 + 64 – 12 = 112 – 112 = 0 ½ \ x = 2 is a zero of p(x) or (x – 2) is a factor of p(x) ½ 6x3 – 25x2 + 32x – 12 = 6x2(x – 2) – 13x(x – 2) + 6(x – 2) ½ = (x – 2) (6x2– 13x + 6) = (x – 2) (6x2– 9x – 4x + 6) ½ = (x – 2) [3x(2x – 3) –2(2x – 3)] = (x – 2) (2x – 3) (3x – 2). ½ 6. 4y2 – 4y + 14 y2 + 4y + 2) 4y4 + 12y3 + 6y2 + 50y + 26 1 4y4 + 16y3 + 8y2

– – – – 4y3 – 2y2 + 50y 1 – 4y3 – 16y2 – 8y + + + 14y2 + 58y + 26 1 14y2 + 56y + 28 – – 2y – 2 1 So, 2y – 2 must be subtracted.


7. Let p(x) = ax3 + 3x2 – 13 and g(x) = 2x3 – 5x + a R1 = p(2) R2 = g(2) 1½ + 1½ p(2) = a(2)3 + 3(2)2 – 13 R1 = 8a + 12 – 13 = 8a – 1 R2 = 2(2)3 – 5(2) + a = 16 – 10 + a ⇒ R2 = 6 + a R1 = R2 ⇒ 8a – 1 = 6 + a ½ ⇒ 7a = 7 \ a = 1 ½ [CBSE Marking Scheme, 2014]

Alternative Method : Let p(x) = ax3 + 3x2 – 13 and g(x) = 2x3 – 5x + a 1 R1 = p(2) R2 = g(2) p(2) = a(2)3 + 3(2)2 – 13 = 8a + 12 – 13 1 = 8a – 1 So, R1 = 8a – 1 g(2) = 2(2)3 – 5(2) + a = 16 – 10 + a 1 = 6 + a

So, R2 = 6 + a According to the question, R1 = R2 then, 8a – 1 = 6 + a ⇒ 8a – a = 6 + 1 ⇒ 7a = 7 1 \ a = 1 8. (4x2 + 3x – 12) quotient x2 + 2x + 2 ) 4x4 + 11x3 + 2x2 – 11x + 6 ½ 4x4 + 8x3 + 8x2

– – – 3x3 – 6x2 – 11x + 6 ½ 3x3 + 6x2 + 6x – – – – 12x2 – 17x + 6 ½ – 12x2 – 24x – 24 + + + 7x + 30 ½ Thus, quotient = 4x2 + 3x – 12 and remainder = 7x + 30 By Remainder theorem, Divide nd = (Divisor × Quotient) + Remainder ½ \ 4x4 + 11x3 + 2x2 – 11x +6 = [(x2 + 2x + 2) × (4x2 + 3x – 12)] + (7x + 30) ½ = [4x4 + 8x3 + 8x2 + 3x3 – 6x2 + 6x – 12x2 – 24x – 24]

+ 7x + 30 = 4x4 + 11x3 + 2x2 – 11x + 6 1


1. For zeroes, put p(x) = 0 x (x – 2) (x – 3) = 0, then, x = 0, 2, 3. 1 2. Putting, x = 0 p(0) = 0 – 3 × 0 + 2 = 2 Putting, x = 2 p(2) = 22 – 3 × 2 + 2 = 4 – 6 + 2 = 0 Thus, p(0) + p(2) = 2 + 0 = 2. 1 3. Here, p (x) = x4 + x3 – 2x2 + x + 1, and the zero of x – 1 is 1. ½ So, p (1) = (1)4 + (1)3 – 2 (1)2 + 1 + 1 = 2 1 So, by the Remainder Theorem, 2 is the remainder

when x4 + x3 – 2x2 + x + 1 is divided by x – 1. ½ 4. Here, p(x) = x3 – ax2 + 6x – a, and the zero of x – a is a. ½ So, p(a) = a3 – a.a2 + 6a – a = 5a. 1 So, by the remainder theorem 5a is the remainder

when x3 – ax2 + 6x – a, is divided by x – a. ½ 5. Let, p(x) = 3x3 – 5x2 + kx – 2, and q(x) = – x3 – x2 + 7x + k

Put, x + 2 = 0 or x = – 2 in p (x) and q (x) P(– 2) = 3 (– 2)3 – 5(– 2)2 + k(– 2) – 2 = – 24 – 20 – 2k – 2 1 = –2k – 46 q(– 2) = – (– 2)3 – (– 2)2 + 7 (– 2) +k = 8 – 4 – 14 +k 1 = – 10 +k \ p(x) and q(x) leave the same remainder when

divided by x + 2. ⇒ –2k – 46 = k – 10 ⇒ –3k = 36 \ k = – 12 1 6. p(x) = x3 + 8x2 + 17x + a p(x) leave the same remainder when divided by

(x + 2) and (x +1). ½ p(–2) = (–2)3 + 8(–2)2 + 17(–2) + a = – 8 + 32 – 34 + a = – 10 + a 1 p(–1) = (–1)3 + 8(–1)2 + 17(–1) + a(–1) = – 1 + 8 – 17 + a = – 10 – a 1 Remainders are equal


So, – 10 + a = – 10 – a ½ ⇒ 2a = 0

\ a = 0 1

7. R1 = f(1) ⇒ 14 – 2(1)3 + 3(1)2 – a(1) + b = 5 ⇒ 1 – 2 + 3 – a + b = 5 ⇒ b – a = 3 ...(1) 1 R2 = f(–1) ⇒ (–1)4 – 2(–1)3 + 3(–1)2 –a(–1) + b = 19 1 + 2 + 3 + a + b = 19 ⇒ a + b = 13 ...(2) 1 Solving (1) and (2), we get b = 8 and a= 5 1 R = f(2) = (2)4 – 2(2)3 + 3(2)2 – 5 × 2 + 8 = 12 – 10 + 8 = 10 1 [CBSE Marking Scheme, 2013]

Alternative Method : f(x) = x4 – 2x3 + 3x2 – ax + b Put, x – 1 = 0 or x = 1 in f(x), we get ⇒ f(1) = (1)4 – 2(1)3 + 3(1)2 – a × 1 + b

⇒ 5 = 1 – 2 + 3 – a + b ⇒ 5 = 2 – a + b ⇒ a – b = 2 – 5 1 ⇒ a – b = – 3 ...(1) Again put, x + 1 = 0 or x = – 1 in f(x), we get f(– 1) = (– 1)4 – 2 (– 1)3 + 3 (– 1)2 – a (– 1) + b ⇒ 19 = 1 + 2 + 3 + a + b ⇒ 19 – 6 = a + b ⇒ a + b = 13 ...(2) 1 Adding equations (1) and (2), 2a = 10 a = 5 By equation (2), 5 + b = 13 \ b = 8 \ f(x) = x4 – 2x3 + 3x2 – 5x + 8 1 Again put, x – 2 = 0 or x = 2 in f(x) f(2) = (2)4 – 2 (2)3 + 3 (2)2 – 5 × 2 + 8 = 16 – 16 + 12 – 10 + 8 = 20 – 10 = 10 1

TOPIC-3Factor Theorem


1. x + 4 is a factor of x2 + 3x + m = p(x) ⇒ p(– 4) = 0 ⇒ 16 – 12 + m = 0 ⇒ m = – 4 1 2. 2x – 1 is a factor of p(x) = 6x2 + kx – 2

⇒

12

p = 0

⇒

1 16. . – 2

4 2+ k

= 0

⇒ k = 1 1 3. 8x3 – (2x – y)3 = (2x)3 – (2x – y)3

= [2x – (2x – y)][(2x)2 + (2x – y)2

+ 2x(2x – y)] Since (a3 – b3) = (a – b) (a2 + b2 – ab) = y[4x2 + 4x2 + y2 – 4xy

+ 4x2 – 2xy] = y[12x2 + y2 – 6xy] 2 4. 9x2 + 6xy + y2 = (3x)2 + 2 × (3x) × y + y2 1 = (3x + y)2 [ a2 + 2ab + b2 = (a + b)2] 1 5. Give Exp = (x – y)2 – 7(x + y) (x – y) + 12(x + y)2

= (x – y)2 – 4(x + y) (x – y) – 3(x + y) (x – y) + 12(x + y)2

= (x – y) [x – y – 4x – 4y] – 3(x + y) [x – y – 4x – 4y] 1 = (x – y) [– 5y – 3x] – 3(x + y) [– 3x – 5y] = (– 5y – 3x) [x – y – 3 (x + y)] 1 = – (5y + 3x) (– 2x – 4y) = (5y + 3x) (2x + 4y) = 2(x + 2y) (5y + 3x) 1

6.

p q3 3 343

729+

=

pq

33 7

( )9

+ 1

=

pq

79

+

pq pq2

2 7 7( )

9 9

+ − × 1

a3 + b3 = (a + b) (a2 + b2 – ab)

=

pq

79

+

pqp q2 2 49 7

81 9 + −

1

7. a6 – b6 = (a3)2 – (b3)2 1 = (a3 – b3) (a3 + b3) 1 = (a – b) (a2 + b2 + ab) (a + b) (a2 + b2 – ab) 1 = (a – b) (a + b) (a2 + b2 + ab) (a2 + b2 – ab). [CBSE Marking Scheme, 2012]

8. p(x) = 2x3 – 3x2 – 11x + 6 p(–2) = – 16 – 12 + 22 + 6 = 0 ⇒ – 2 is a zero of p(x)


p(3) = 54 – 27 – 33 + 6 = 0 ⇒3 is a zero of p(x) (x + 2) (x – 3) = x2 – x – 6 is a factor of p(x)

2

( )– – 6p x

x x = 2x – 1

\ p(x) = (x + 2) (x – 3) (2x – 1) 4 9. Factor theorem – statement Let p(x) = x3 – 3x2 – x + 3

The factors of the constant term 3 are ±1, ±3 p(1) = 13 – 3(1)2 – 1 + 3 = 0 (x – 1) is a factor p(–1) = (–1)3 – 3(–1)2 – (–1) + 3 = 0 (x + 1) is a factor p(3) = 33 – 3(3)2 – 3 + 3 = 0 (x – 3) is a factor (x – 1) (x + 1) (x – 3) are the factors of p(x). 4


1. x2 – 3x = x(x – 3) 1 2. 12a2b – 6ab2 = 6ab(2a – b) 1 3. 8a3 + 8b3 = (2a)3 + (2b)3

= (2a + 2b) [(2a)2 + (2b)2 – (2a) × (2b)] 1 [ a3 + b3 = (a + b) (a2 + b2 – ab)] = 2(a + b) × 4(a2 + b2 – ab) 1 = 8(a + b) (a2 + b2 – ab) 4. (x + 3y)3 + (3x – y)3 = [x3 + (3y)3 + 3 × x × 3y (x +

3y)] +[(3x)3 – y3 – 3 × 3x × y (3x – y)]

( ) ( )

a b a b ab a b

and a b a b ab a b

3 3 3

3 3 3

+ = + +3 +

( – ) = – – 3 ( – )

= x3 + 27y3 + 9x2y + 27xy2 + 27x3 – y3 – 27x2y + 9xy2 1 = 28x3 + 26y3 – 18x2y + 36xy2. 1

5. To factorise,

a a2 22 1

5 23 3

+ − − ½

a2 – b2 = (a + b) (a – b) 1½

=

a a

2 15 + +2

3 3 −

a a2 1

5 + 2 +3 3

− 1

=

a

17 +

3

(3a + 1)

[CBSE Marking Scheme, 2012] 6. 9x2 + y2 + z2 – 6xy + 2yz – 6xz = (– 3x)2 + (y)2 + (z)2 + 2 × (– 3x)(y) + 2 × (y) (z) +

2 (– 3x) (z) 1 = (– 3x + y + z)2 1 If x = 1, y = 2, z = – 1, then (– 3x + y + z)2 = (– 3 × 1 + 2 – 1)2 ½

= (– 3 + 2 – 1)2

= 4. ½

7. 125x3 – 27y3 + z3 + 45xyz = (5x)3 + (– 3y)3 + (z)3 – 3 × (5x) (– 3y) (z) 1 = (5x – 3y + z) [(5x)2 + (– 3y)2 + (z)2 – (5x)(– 3y) –

(– 3y) (z) – (5x) (z)] 1 a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca ) = (5x – 3y + z) [25x2 + 9y2 + z2 + 15xy + 3yz – 5xz] [CBSE Marking Scheme, 2012] 1

8. We have, x2 – y2 + y2 – z2 + z2 – x2 = 0 (x2 – y2)3 + (y2 – z2)3 +(z2 – x2) = 3(x2 – y2) (y2 – z2)

(z2 – x2) a + b + c = 0 ⇒a3 + b3 + c3 = 3abc (x2 – y2)3 + (y2 + z2)3 + (z2 – x2)3

= 3(x + y) (x – y) (y + z) (y + z) (z – x) (z + x) Similarly, x – y + y – z + z – x = 0 (x – y)3 + (y – z)3 + (z – x)3 = 3(x – y) (y – z) (z – x)

\

2 2 3 2 2 3 2 2 3

3 3 3( – ) ( – ) ( – )

( – ) ( – ) ( – )+ ++ +

x y y z z xx y y x z x

=

3( )( – )( – )( )( )( – )3( – )( – )( – )

+ + +x y x y y z z x y z z xx y y z z x

= (x + y) (y + z) (z + x) 4 9. Let p = m(n2 – p2) + n(p2 – m2 + p(m2 – n2) p(m = n) = n(n2 – p2) + n(p2 – n2) + p(n2 – n2) = n(n2 – p2) – n(n2 – p2) + 0 = 0 m – n is a factor of p Similarly p(n = p) = 0 & p(p = m) n – p is a factor of p. and p – m is a factor of p. 4


1. x – 2 is a factor of p(x), then p(2) = 0 p(2) = 2 × (2)2 + 3 × 2 – k = 0 ⇒ 8 + 6 – k = 0

\ k = 14. 1

2. Since,

1–

3f

= 0


\

1–

3 is a zero of polynomial f(x)

So,

x

1+

3 or 3x + 1 is a factor of f(x). 1

3. 64a3 – 27b3 – 144a2b + 108ab2

= (4a)3 – (3b)3 – 3 × (4a)2 × (3b) + 3 × (4a) × (3b)2 ½ = (4a)3 – (3b)3 – 3 × 4a × 3b (4a – 3b) ½ = (4a – 3b)3 1 4. a9 + b9 + 3a6b3 + 3a3b6

= (a3)3 + (b3)3 + 3 × (a3)2 (b3) + 3 (a3) × (b3)2 ½ = (a3)3 + (b3)3 + 3 × a3 b3 × (a3 + b3) ½ = (a3 + b3)3 1 5. Let, p(x) = x6 – ax5 + x4 – ax3 + 3x – a + 2 (x – a) is a factor of the polynomial p(x), then p(a) = 0 1 ⇒ a6 – a × a5 + a4 – a × a3 + 3 × a – a + 2 = 0 1 ⇒ a6 – a6 + a4 – a4 + 3a – a + 2 = 0 ½ ⇒ 2a = – 2 ⇒ a = – 1. ½

6. a7 + ab6 = a(a6 + b6) 1 = a[(a2)3 + (b2)3] = a(a2 + b2) [(a2)2 + (b2)2 – a2 × b2] 1 = a(a2 + b2) (a4 + b4 – a2b2). 1 [CBSE Marking Scheme, 2012]

7. Let, x2 – 4x = a \ Expression = a (a – 1) – 20 ½ = a2 – a – 20 = (a – 5) (a + 4) ½ = (x2 – 4x – 5) (x2 – 4x + 4) = (x – 5) (x + 1) (x – 2)2 2 [CBSE Marking Scheme, 2012]

Alternative Method : (x2 – 4x)(x2 – 4x – 1) – 20 = (x2 – 4x)2 – (x2 – 4x) – 20

½ = (x2 – 4x)2 – 5(x2 – 4x) + 4(x2 – 4x) – 20 ½ = (x2 – 4x)[x2 – 4x – 5] + 4[x2 – 4x – 5] ½ = (x2 – 4x – 5) (x2 – 4x + 4) = (x2 – 5x + x – 5)[(x)2 – 2 × 2 × x + (2)2] ½ = [x(x – 5) + 1(x – 5)][x – 2]2 ½ = (x – 5)(x + 1) (x – 2)(x – 2). ½ 8. p(x) = 3x3 + 5x2 – 11x + 3 p(1) = 3 + 5 – 11 + 3 = 0 \1 is a zero of p(x) p(–3) = – 81 + 45 + 33 + 3 = 0 – 3 is a zero of p(x) (x – 1) (x + 3) = x2 + 2x – 3 is a factor of p(x)

2

( )2 – 3+

p xx x

= 3x – 1, when we divide physically

Hence p(x) = (x – 1) (x + 3) (3x – 1) 4 9. Factor of 60 = (±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12,

±15, ±30, ±60) ½ p(x) = x3 – 12x2 + 47x – 60 p(3) = (3)3 – 12(3)2 + 47(3) – 60 = 27 – 108 + 141 – 60 = 168 – 168 = 0 ½ \ x = 3 is a zero of p(x) or (x – 3) is a factor of p(x) ½ x3 – 12x2 + 47x – 60 = x2(x – 3) – 9x(x – 3) + 20(x – 3) ½ = (x – 3) (x2 – 9x + 20) ½ = (x – 3) (x2 – 5x – 4x + 20) ½ = (x – 3) [x(x – 5) – 4(x – 5)] ½ = (x – 3) (x – 4) (x – 5) ½


1. a3 – 1 = (a – 1) (a2 + 1 + a × 1) Then factors of (a3 – 1) are (a – 1) and (a2 + 1 + a). 1

2. x2 + 5 2x + 12 = x2 + 3 2x + 2 2x + 12

= x

( )x+3 2

+

( )x2 2 +3 2

=

( )( )x x+3 2 +2 2 . 1

Factors are 3 2 and 2 2x x+ + .

3. (x + 2)2 + p2 + 2p(x + 2) = (x + 2)2 + (p)2 + 2 × (x + 2) × p 1 = (x + 2 + p)2, [ a

2 + b2 + 2ab = (a + b)2] 1 4. 8 – 27a3 – 36a + 54a2

= (2)3 – (3a)3 – 18a(2 – 3a) 1

= (2)3 – (3a)3 – 3 × 2 × 3a(2 – 3a) = (2 – 3a)3. 1

5. p(x) = 3x2 – mx – nx If (x – a) is a factor of p(x), then p(a) = 0 1 ⇒ 3(a)2 – m × a – n × a = 0 ½ ⇒ a[3a – m – n] = 0, a ≠ 0 ½ ⇒ 3a – m – n = 0 ½

\ a =

m n3+

. ½

[CBSE Marking Scheme, 2012] 6. 250x3 – 432y3 = 2[125x3 – 216y3] ½ = 2[(5x)3 – (6y)3] ½ = 2(5x – 6y) [(5x)2 + (6y)2 + 5x × 6y]

1½ a3 – b3 = (a – b) (a2 + b2 + ab)


= 2(5x – 6y) (25x2 + 36y2 + 30xy). ½

7. Let, p(x) = x3 – 3x2 – 9x – 5, p (– 1) = – 1 – 3 × 9 – 5 = 0 Since, (x + 1) is a factor of x3 – 3x2 – 9x – 5 ½ \ (x3 – 3x2 – 9x – 5) = (x + 1) (x2 – 4x – 5) 1 x2 – 4x – 5 = x2 – 5x + x – 5 = x(x – 5) + 1(x + 5) = (x + 1) (x – 5) 1 Factors are : (x + 1) (x + 1) (x – 5) ½ [CBSE Marking Scheme, 2012]

8. p(x) = 2x4 – ax3 + 4x2 – x + 2

If (2x + 1) is a factor of p(x), then 2x + 1 = 0, x =

–12

is a zero of the polynomial p(x) 1

So,

p

–12

= 0

½

p

–12

= 2 ×

4–12

– a

3–12

+

2–14

2

–

–12

2 = 0 1

⇒

2 ×

116

– a

–18

+

14

4

–

–12

+ 2 = 0 ½

⇒

18

+ a8

+ 1 +

12

+ 2 = 0 ½

⇒

298

+ 8

a

= 0

\ a = – 29 ½ 9. (x – 3) is a factor of p(x) = 2x4 + 3x3 – 26x2 – 5x + 6,

then it completely divide p(x) 2x3 + 9x2 + x – 2 (quotient) 1 x – 3) 2x4 + 3x3 – 26x2 – 5x + 6 2x4 – 6x3

– + 9x3 – 26x2

9x2 – 27x2

– + x2 – 5x x2 – 3x – + – 2x + 6 – 2x + 6 + – 0 2 \ Remainder = 0. So, (x – 3) is a factor of p(x). 1

TOPIC-4Algebraic Identities


1.

x12

+ x

32

+ = x2 +

x

32

+

x12

+

34

= x2 + 2x +

34

. 1

2. Given,

x yy x

+ = – 1

⇒

x yyx

2 2+

= – 1

⇒ x2 + y2 = – xy ⇒ x2 + y2 + xy = 0 ⇒ x3 – y3 = (x – y) (x2 + y2 + xy) = (x – y) × 0 = 0. 1

3. (2x – 3y + z)2

= 4x2 + 9y2 + z2 – 12xy – 6yz + 4xz 2[CBSE Marking Scheme, 2014]

Alternative Method : By using the identity, (a + b + c)2

= a2 + b2 + c2 + 2ab + 2bc + 2ca

(2x + (–y) + z)2 = (2x)2 + (–y)2 + z2 + 2(2x)(–y) + 2(–y) (z) + 2(z)(2x) 1

= 4x2 + y2 + z2 – 4xy – 2yz + 4xz 1

4.

x y

31 23 3

−

=

x

313

–

y32

3

– 3 ×

x

13

×

y23

yx

1 23 3

− 1

=

x3

27 –

y3827

–

xy23

x y23 3

−

=

x3

27 –

y3827

–

x y229

+

xy249

1

5. x + y + z = 0 1 ⇒ x + y = – z ⇒ (x + y)3 = (– z)3

⇒ x3 + y3 + 3xy (–z) = – z3

⇒ x3 + y3 – 3xyz = – z3 1 ⇒ x3 + y3 + z3 = 3xyz 1 6. (2a + 3b)3 – (2a – 3b)3 = x3 – y3 where 2a + 3b =x

and 2a – 3b = y


= (x –y) (x2 + xy + y2) = [(2a + 3b) – (2a – 3b)][(2a + 3b)2 + (2a + 3b) (2a – 3b) + (2a – 3b)2] 1 = 6b[(4a2 + 12ab + 9b2) + (4a2 – 9b2) + (4a2 – 12ab + 9b2)] = 6b(12a2 + 9b2) 1 = 6b × 3 × (4a2 + 3b2) = 18b(4a2 + 3b2) 1 7. 1113 = (100 + 11)3 1 = (100)3 + 3(100)2 (11) + 3(100) (11)2 + (11)3

1 = 1367631 1

8. 22

1+x

x =

21–

x

x+ 2 1

= 4 + 2 = 6

22

21 +

xx

=

44

1+x

x + 2 1

⇒ (6)2 =

44

1+x

x + 2

36 – 2 = 4

41

+xx

1

44

1+x

x = 34 1

9. RHS. =

12

(x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]

=

12

(x + y + z) [x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx]

=

12

(x + y + z) [2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx] 2

=

12

(x + y + z). 2[x2 + y2 + z2 – xy – yz – zx]

=

12

x3 + y3 + z3 – 3xyz [By identity] 2

= LHS.


1. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 –ab – bc – ca) a3 + b3 + c3 – 3abc = 0, as a + b + c = 0 a3 + b3 + c3 = 3abc. 1 2. (x – 2)3 = (x)3 – (2)3 – 3 × x × 2 (x – 2) = x3 – 8 – 6x2 + 12x Coefficient of x2 in the expansion of (x – 2)3 = – 6. 1

3. x2 – 9 = (97)2 – (3)2 1 = (97 + 3) (97 – 3) = 100 × 94 ½

= 9400. ½ [CBSE Marking Scheme, 2012]

4. (x + 2y)2 = x2 + 4y2 + 2 × x × 2y 1 = 17 + 4 × 2 ½ = 17 + 8 = 25

⇒ (x + 2y) = ± 25 = ±5. ½

5. + + −2 2( 2 3) ( 5 2 ) = 2 2( 2 ) ( 3) 2 2+ + ×

2 2 3 ( 5) ( 2 ) 2 5 2× + + − × ×

1

= 2 + 3 + 2 6 + 5 + 2 – 2 10

= + −12 2 6 2 10 1

= + −2(6 6 10 ) 1

6. (x + y + 2z) (x2 + y2 + 4z2 – xy – 2yz – 2zx) = (x + y + 2z) [x2 + y2 + (2z)2 – x × y – y × 2z – x × 2z] 1

We know that (a + b + c) (a2 + b2 + c2 – ab – bc – ca)

= a3 + b3 + c3 – 3abc 1 \ (x + y + 2z) (x2 + y2 + (2z)2 – x × y – y × 2z – x × 2z) = (x)3 + (y)3 + (2z)3 – 3 × x × y × 2z = x3 + y3 + 8z3 – 6xyz. 1 7. Given, a + b + c = 3x or 3x – a – b – c = 0 ½ \ (x – a)3 + (x – b)3 + (x – c)3 – 3(x – a) (x – b) (x – c) = [x – a + x – b + x – c][(x – a)2 + (x – b)2 + (x – c)2

– (x – a)(x – b) – (x – b)(x – c) – (x – a)(x – c)] 1½ = [3x – a – b – c][(x – a)2 + (x – b)2 + (x – c)2 – (x – a)

(x – b) – (x – b)(x – c) – (x – a) (x – c)] ½ = 0 × [(x – a)2 + (x – b)2 + (x – c)2 – (x – a)(x – b)

– (x – b)(x – c) – (x – c)(x – a)] = 0. ½

8.

1x

x+

= 5

On squaring both sides, we get 1

21+x

x

= 52

x2 + x

21+ 2 × x + 1

x = 25

[ (a + b)2 = a2 + b2 + 2ab)] 1

x2 +

2

1x

+ 2 = 25 1

x2 +

21x

= 25 – 2


x2 +

2

1x

= 23 1

9. (i) 103 × 107 = (100 + 3) (100 + 7) = 1002 + (3 + 7) × 100 + 3 × 7 1 = 10000 + 1000 + 21

= 11021 1 (ii) (102)3 = (100 + 2)3

= (100)3 + 23 + 3 × 100 × 2(100 + 2) = 1000000 + 8 + 600 × 102 = 1000000 + 8 + 61200 1 = 1061208 1


1. x2 +x21

=

x

x

21 + – 2 (x)

1x

= (4)2 – 2 = 16 – 2 = 14. 1 2. Area of rectangle = 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 1 = 5a(5a – 4) – 3(5a – 4) ½ = (5a – 3) (5a – 4) = length × breadth ½ \ Length and breadth are (5a – 3) and (5a – 4)

respectively. 3. 2x + 3y = 8 ⇒ (2x + 3y)3 = 83 ½ ⇒8x3 + 27y3 + 3 × 2x × 3y (2x + 3y) ⇒ = 512 ½ 8x3 + 27y3 + 18 × 2 × 8 = 512 ½ ⇒ 8x3 + 27y3 = 512 – 288 ⇒ 8x3 + 27y3 = 224. ½ 4. Given, x + y = 5, then x3 + y3 + 15xy – 125 = x3 + y3 – 125 + 15xy = (x)3 + (y)3 + (– 5)3 – 3 × x × y × (– 5) ½ = (x + y – 5) [(x)2 + (y)2 + (– 5)2

– x × (– 5) – y × (– 5) – x × y)] 1 = (5 – 5) (x2 + y2 + 25 + 5x + 5y – xy) ½ = 0 (x2 + y2 + 25 + 5x + 5y – xy) ½ = 0 ½

5.

14,x

x+ =

(given)

On squaring both sides, we get

21x

x +

= (4)2 1

⇒

22 1 1

2x xx x

+ + × × = 16 1

⇒

22

12x

x+ +

= 16

⇒

22

1x

x+

= 14. 1

6. Given, a + b + c = 9 ½ ⇒ (a + b + c)2 = 92 ½ ⇒ a2 + b2 + c2 + 2(ab + bc + ca) = 81 ⇒ [as, (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc+ca)] ½ ⇒ 35 + 2(ab + bc + ca) = 81 ½ ⇒ 2(ab + bc + ca) = 81 – 35 = 46 ½ ⇒ ab + bc + ca = 46/2 ⇒ ab + bc + ca = 23. ½ 7. a – b = 7 and a2 + b2 = 85 ½ ⇒ (a – b)2 = (7)2 1 ⇒ a2 – 2ab + b2 = 49 ⇒ 85 – 2ab = 49 ⇒ 2ab = 36 ⇒ ab = 18 1 Then, a3 – b3 = (a – b) (a2 + b2 + ab) ½ = (7) (85 + 18) = (7) (103)

= 721 1


1.

3 3

2 283 17

83 83 17 17+

− × +

=( )

( )2 2

2 2

(83+17) 83 – 83×17+17

83 – 83×17+17

a3 + b3 = (a+ b) (a2 + b2 – ab)

= 83 + 17 = 100. 1

2. 14 = a, 13 = b, – 27 = c

a + b + c = 14 + 13 – 27 = 0 ½

\ a3 + b3 + c3 = 3abc 1

(14)3 + (13)3 + (– 27)3 = 3 × 14 × 13 × (– 27)

= – 14742. ½

3. 249 × 251 = (250 – 1) (250 + 1) 1 = (250)2 – (1)2 = 62500 – 1 ½ = 62499. ½ [CBSE Marking Scheme, 2012]


4. x2 – y2 + xy = + −

− − +

2 23 2 3 23 2 3 2

+ −+ × − +

3 2 3 23 2 3 2

½

=

+ −− +

− +5 2 6 5 2 6

15 2 6 5 2 6

½

= ( ) ( )

( )( )+ − −

+− +

2 25 2 6 5 2 6

15 2 6 5 2 6

½

= + + − − +

+−

25 24 20 6 25 24 20 61

25 24 ½

= +40 6 1

= 40 × 2·4 + 1 = 96 + 1 = 97. 1

5.

( ) ( ) ( )( ) ( ) ( )− + − + −

− + − + −

3 3 32 2 2 2 2 2

3 3 3

a b b c c a

a b b c c a

Both Numerator and Denominator are of the form a3 + b3 + c3 ½

We know that when a + b + c = 0 then a3 + b3 + c3 = 3abc ½ For Numerator, a2 – b2 + b2 – c2 + c2 – a2 = 0 For Denominator , a – b + b – c + c – a = 0

\

( ) ( ) ( )( ) ( ) ( )− + − + −

− + − + −

3 3 32 2 2 2 2 2

3 3 3

a b b c c a

a b b c c a

=

( ) ( ) ( )( )

× − − −

− − −

2 2 2 2 2 23

3 ( ) ( )

a b b c c a

a b b c b c 1

=

− + − + − +− − −

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )

a b a b b c b c c a c aa b b c c a ½

= (a + b) (b + c) (c + a). ½[CBSE Marking Scheme, 2012]

6. (i) (998)3 = (1000 – 2)3

We know that

(a – b)3 = a3 – b3 – 3ab(a – b) 1

Hence, (998)3 = (1000)3 – (2)3 – (3) (1000) (2) (1000 – 2)

= 1000000000 – 8 – 6000 × 998

= 1000000000 – 5988008

= 994011992. 1

(ii) Algebraic Identities ½

(iii) Satisfaction solves the identity crisis among people. ½

7. Expansion of (x + y + z)2

= x2 + y2 + z2 + 2(xy + yz + zx) 1

⇒ 100 = 40 + 2(xy + yz + zx) 1 ⇒ (xy + yz + zx) = 30 1 x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy

– yz – zx) 1 = 10(40 – 30) = 100


(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) 1

⇒ (10)2 = 40 + 2(xy + yz + zx)

⇒ 100 – 40 = 2(xy + yz + zx) 1

⇒

xy + yz + zx = 60

302

= ½

and x3 + y3 + z3 – 3xyz

= (x + y + z) [x2 + y2 + z2 – (xy + yz + zx)] 1

= 10 [40 – 30]

= 10 × 10 = 100. ½

8. x3 – 8y3 – 36xy – 216

= (x)3 + (– 2y)3 + (– 6)3 – 3(x)(– 2y)(– 6) 1

= [x + (– 2y) + (– 6)][x2 + (– 2y)2 + (– 6)2 – (x)(– 2y) – (– 2y)(– 6) – (x)(– 6)] 1

= (x – 2y – 6) (x2 + 4y2 + 36 + 2xy – 12y + 6x) ½

= 0 × [x2 + 4y2 + 36 + 2xy – 12y + 6x],

( x = 2y + 6 or x – 2y – 6 = 0) 1

\ x3 – 8y3 – 36xy – 216 = 0. ½

FORMATIVE ASSESSMENT WORKSHEET-27 Note : Students should do this activity themselves.

qqq


TOPIC-1Euclid’s Geometry


1. A system of axioms is called consistent, when it is impossible to deduce from these axioms, a statement that contradicts any axiom or previously proved statement. 1

2. AB + BC = AC ⇒ x + 3 + 2x = 4x – 5 ⇒ 3x + 3 = 4x – 5 ⇒ 4x – 3x = 3 + 5 \ x = 8 1 3. Theorem requires a proof. 1 4. Let, First thing = x Second thing = y then, x = 2y 1

5. A B C D \ AC = BD (given) ⇒ AB + BC = BC + CD 1 ⇒ AB = CD. 1

[CBSE Marking Scheme I, 2012 , 2014]

6. Euclid’s axioms (1) Things which are equal to the same thing are

equal to one another. (2) If equals are added to equals, the wholes are

equal. 1 + 1[CBSE Marking Scheme, 2012, 2014]

7. AB = CD (Given) ⇒ AB + BC = BC + CD ½+½ ⇒ AC = BD Euclid’s axiom used : If equals are added to

equals, the wholes are equal. 1 [CBSE Marking Scheme I, 2012, 2014]

8. AC = DC (Given) CB = CE Adding, AC + CB = DC + CE ½ AB = DE ½ If equals are added to equals, the wholes are

equal. 1 [CBSE Marking Scheme, 2012]

9. Given, AC = BC

A C B 1 So, AC + AC = AC + BC 1 (Equals are added to equals) ⇒ 2AC = AB, (\ AC + CB coincides with AB)

\

AC =

12

AB. 1

10. Given, C is the mid-point of AB. \ AC = CB …(1) 1 If possible, let D be the another mid-point of AB. \ AD = DB …(2) 1

A C BD Now, subtracting (1) from (2), we get AD – AC = DB – CB ⇒ – CD = CD ⇒ 2CD = 0 ⇒ CD = 0 \ C and D coincide. Hence, every line segment has one and only one

mid-point. 1 11. Concermed, Caring. Things equal to same things

are equal to one another. Rehman contributed ` 500. All right angles are, equal to one another (OR) Any

1 postulate of Euclid can be stated.


1. A surface is that which has length and breadth. 1

y

xSurface breadth ( )ylength ( )x

INTRODUCTION TO EUCLID’S GEOMETRY

SECTION

BSECTIONCHAPTER

3


2. Dimension of Surface = Length and Breadth (which is 2). 1

3. Lines are parallel if they do not intersect on being extended.

For example :

A

B

or

A B Lines A and B are parallel lines.

1

4.

D

CB

A

AB = AD AC = AD AB = AC 1 Things which are equal to the same thing are

equal to one another. 1 [CBSE Marking Scheme I, 2012]

5. Euclid’s axiom : If C be the mid-point of a line segment AB, then AC = AB.

AC =

12

AB ½

and

AD =

12

AC

½

⇒

AD =

1 12 2

AB

½

⇒

AD =

14

AB. ½

[CBSE Marking Scheme, 2012] 6. Given : Three lines l, m and n in a plane such that

l || m and m || n. 1

l

m

n

To prove : l || n. Proof : If possible, let l be not parallel to n, then l and

n should intersect in a unique point, say A. Thus, through a point A, outside m, there are two

lines l and n, both parallel to m. 1 This contradicts the parallel line axiom. So, our assumption is wrong. Hence, l || n. 1 7. (i) If a straight line l falls on two straight lines m

and n such that the sum of the interior angles on one side of l is equals to two right angles then by Euclid’s fifth postulate the lines will not meet on this side of l. Next, we know that the sum of the interior angles on the other side of line l will also be two right angles. Therefore, they will not meet on the other side also. So, the lines m and n never meet and are therefore, parallel. 2

(ii) Introduction to Euclid’s Geometry. ½ (iii) Universal truth. ½ 8. (i) The terms need to be defined are : Polygon : A simple closed figure made up of three

or more line segments. 1

Line segment : Part of a line with two end points.

Line : Undefined term.

Point : Undefined term. Angle : A figure formed by two rays with a common

initial point.

Ray : Part of a line with one end point.

Right angle : Angle whose measure is 90°. Undefined terms used are : Line, Point. 1 Euclid’s fourth postulate says that ‘‘all right angles

are equal to one another.’’ In a square, all angles are right angles, therefore, all

angles are equal (From Euclid’s fourth postulate). Three line segments are equal to fourth line

segment. (Given) Therefore, all the four sides of a square are equal (by

Euclid’s first axiom ‘‘things which are equal to the same thing are equal to one another.’’) 1

(ii) Introduction to Euclid’s geometry. ½ (iii) Equality leads to democracy. ½


1. ∠1 = ∠2 ...(1) ∠3 = ∠4 ...(2) Adding (1) + (2), we get

∠1 + ∠3 = ∠2 + ∠4 ½ ⇒ ∠ABC = ∠DBC ½ Euclid’s axiom used : If equals are added to

equals, wholes are equal. 1[CBSE Marking Scheme, 2012]


2. In a circle having centre at P, we have PR = PQ = radius ½ In a circle having centre at Q, we have QR = QP = radius ½ Euclid’s first axiom : Things which are equal to the

same thing are equal to one another. ½ \ PR = PQ = QR. ½

3. x – 15 = 25 x – 15 + 15 = 25 + 15 ⇒ x = 40 1 If equals are added to equals, the wholes are

equal. 1 [CBSE Marking Scheme, 2012]

4. Here, OX =

12

XY, PX =

12

XZ 1

⇒ XY = 2(OX), XZ = 2(PX) Also, OX = PX, (Given)1 XY = XZ, (Because things which are double of the same

things are equal to one another.) 1 5. Here, ∠3 = ∠4 and ∠1 = ∠3 and ∠2 = ∠4. Euclid’s

first axiom says, the things which are equal to same things are equal to one another. 2

1

34

2

So, ∠1 = ∠2. 1

TOPIC-2Euclid’s Postulates


1. If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles. 1

2. Meeting place of two walls. 1

3. Only one line passes through two distinct points.

A

B

1

4. Playfair’s Axiom (statement) : For every line l and for every point P not lying on l, there exists a unique line m passing through P and parallel to l. It is equivalent to Euclid’s fifth postulate. 2

5. (a) Infinite, if they are collinear.

(b) Only one, if they are non-collinear. 1 + 1

6. AB = 2AE

(E is the mid-point of AB) ½

CD = 2DF

(F is the mid-point of CD) ½

Also, AE = DF (Given)

Therefore, AB = CD (Things which are double of the same thing are equal to one another) 1

7. There are two undefined terms, line and point. They are consistent, because they deal with two different situations.

(i) Says that given two points be A and B, there is a point C, lying on the line which is in between them. 1

(ii) Says that given A and B, we can take C not lying on the line passing through A and B.

These ‘Postulates’ do not follow from Euclid’s postulates.

However, (ii) follow from given postulate (i). 1

8. AB = BC (given) BX = BY (given) 1 If equals are subtracted from equals, then remains

are also equal. AB – BX = BC – BY ⇒ AX = CY 2

9. Since x + y = 10 and x = z, therefore, x + y = z + y 1

⇒ 10 = z + y 1

Hence, z + y = 10. 1

10. Their sales in July will also be equal as things which are double of the same things are equal to one another.

Two other axioms are :

(i) The whole is greater than the Part. (ii) Things which are halves of the same thing are equal

to one another. 4



1. Two planes intersect each other to form a straight line. 1

2. M NC P

MN = MC + CN 1

3. A BX Y

Let AB has 2 mid-points, say X and Y,

then,

2AB

= AX and

2AB

= AY ½ + ½

\ AX = AY ½ ⇒X and Y coincides ½ [CBSE Marking Scheme 2013, 12, 11]

4. Let AB be perpendicular to a line l and AP is any other line segment.

In right ∆ABP, ∠B > ∠P, (\ ∠B = 90°) 1 ⇒ AP > AB or AB < AP. 1

5. Given, ∠ABC = ∠ACB ⇒ ∠1 + ∠4 = ∠2 + ∠3 1 ⇒ ∠1 + ∠4 – ∠4 = ∠2 + ∠3 – ∠3 (As, ∠3 = ∠4) 1 ⇒ ∠1 = ∠2. 1 6. Since ∠1 = ∠3 and ∠2 = ∠4, therefore adding both

equations ∠1 + ∠2 = ∠3 + ∠4 1 ⇒ ∠BAD = ∠BCD 1 ⇒ ∠A = ∠C. 1 7. Axiom : If equals be subtracted from equals, the

remainders are equal. Two more axiom are : (i) Things which are halves of the same thing are equal

to one other. (ii) The whole is greater than the part OR any of Eulid’s

Axioms can be stated. 4

FORMATIVE ASSESSMENT WORKSHEET-33Note : Students should do this activity themselves.

qqq


TOPIC-1Different Types of Angles


1. x + 20° + 2x – 20° + 60° = 180° (\Straight line makes an angle of 180°) ⇒ 3x = 180° – 60° = 120° ⇒ x = 40° Thus, ∠COD = 2x – 20° = 80° – 20° = 60°. 1 2. Let the angle be x, then Angle x = Complement of x ⇒ x = 90° – x ⇒ x = 45°. 1 3. Complementary angle of 65° = 90° – 65° = 25°. 1 (As sum of complemetary angles is 90°) 4. (3x – 15°) + (x + 5°) = 90° ⇒ x = 25° Angles are 60° and 30°. 2

5. Let, ∠a = 2x and ∠b = 3x ½ Then, ∠a + ∠b = 90° ⇒ 2x + 3x = 90° \ x = 18° ½ a = 2 × 18° = 36° b = 3 × 18° = 54° \ c = 180° – b = 180° – 54° = 126°. 1 [CBSE Marking Scheme, 2012]

6. OP ⊥ AB ⇒ ∠POA = 90° ½ Let, ∠POQ = a \ ∠QOR = 3a ∠ROA = 5a ½ ⇒ a + 3a + 5a = 90° ⇒ 9a = 90° ½ ⇒ a = 10° \ x = 10° ½ and y = 3 × 10° = 30° ½ z = 5 × 10° = 50°. ½ [CBSE Marking Scheme, 2012, 2014]

7. 4b + 75° + b = 180° (Linear pair) 1 ⇒ 5b = 180° – 75° = 105°

⇒

b =

1055

°

= 21°

4b = 4 × 21° = 84° 1 ⇒ a = 4b = 84° (V.O.A.) 1 2c + a = 180° (Linear pair) ⇒ 2c = 180 – 84° = 96°

⇒ c = 48°. 1

SUMMATIVE ASSESSMENT WORKSHEET-35Solutions 1. Angles (30° – a) and (125° + 2a) are supplementary

of each other, then 30° – a + 125° + 2a = 180° ⇒ a = 180° – 155° = 25°. 1 2. The complement of (90° – a) = 90° – (90° – a) (As sum of complementary angles is 90°) = 90° – 90° + a = a. 1 3. Let the angle be x, then

By given condition, x =

15

(90° –x) ⇒ 6x = 90°

⇒ x = 15°. 1

4. ∠POC = 2y (Vertically opp. angles)

\ 5y + 2y + 5y = 180° 1 ⇒ 12y = 180° ½ ⇒ y = 15°. ½ [CBSE Marking Scheme, 2012]

5. x + y + z + w = 360° 1 ⇒ (x + y) + (z + w) = 360° ⇒ x + y + x + y = 360°

(Since, z + w = x + y) ½ ⇒ 2(x + y) = 360 °

LINES AND ANGLES

SECTION

BSECTIONCHAPTER

4


\ x + y = 180° ⇒ AOB is a straight line, as straight line makes an

angle of 180°. ½ [CBSE Marking Scheme, 2012]

6. Let, ∠AOB = 2x and ∠BOC = 3x ∠AOB + ∠BOC = ∠AOC ⇒ 2x + 3x = 75°

⇒

x =

755

°

= 15°

\ ∠AOB = 2x = 30° 1 and ∠BOC = 3x = 45°. 1 [CBSE Marking Scheme, 2012]

7. Join AC, 1D

C3

1

2

A B

4

In ∆ABC Ext. ∠4 = ∠ACB + ∠CAB ...(i) Again, in ∆ACD Ext. ∠3 = ∠DAC + ∠DCA ...(ii) ½ Adding (i) and (i), we get ∠3 + ∠4 = (∠ACB + ∠DCA)

+ (∠CAB + ∠DAC) 1 = ∠1 + ∠2 \ ∠3 + ∠4 = ∠1 + ∠2. ½ [CBSE Marking Scheme, 2012]

SUMMATIVE ASSESSMENT WORKSHEET-36Solutions 1. Angle (55° + 3a) and (115° – 2a) are supplement of

each other, then 55° + 3a + 115° – 2a = 180° ⇒ a = 180° – 170° = 10°. 1 2. Here, x + 10° + x + x + 20° = 180° ⇒ 3x = 180° – 30° = 150°

\

x =

1503

°

= 50°. 1

3. Here, 5x + 4x = 180° (\Straight line makes an angle of 180°)

⇒ x =

1809

°

= 20°.

1

4. Given, a – b =

16

×180° ½

⇒ a – b = 30° ...(1) a + b = 180° (Linear Pair) ...(2) ½ Adding (1) & (2) 2a = 210° ⇒ a = 105° ½ ⇒ b = 180° – a = 180° – 105° = 75°. ½ [CBSE Marking Scheme, 2012]

5. Let the two supplementary angles are 2x and 3x, then

2x + 3x = 180°

⇒

x =

1805

°

= 36°

1 Hence, the angles are 2x and 3x or 72° and 108°. 1 [CBSE Marking Scheme, 2012]

6.

A C

QP O

BD

12

34

½

OP and OQ are bisectors of ∠AOD and ∠BOC. 1 \ ∠1 = ∠2 and ∠3 = ∠4 ∠AOC = ∠BOD (Vertically opp. angles) \ ∠1 + ∠AOC + ∠3 = ∠2 + ∠BOD + ∠4 But ∠1 + ∠AOC + ∠3 + ∠2 + ∠BOD + ∠4 = 360° 1 \ ∠1 + ∠AOC + ∠3 = ∠2 + ∠BOD + ∠4 = 180° ½ \ OP and OQ are in the same line.

7. Given,

∠POR : ∠ROQ = 5 : 7

Let, ∠POR = 5x

and ∠ROQ = 7x

5x + 7x = 180°

⇒ 12x = 180°

⇒

x =

18012

= 15° 2

∠POR = d = b = 5x = 5 × 15° = 75°

∠ROQ = c = a = 7x = 7 × 15°

= 105°. 1



TOPIC-2Transversal Line


1. Let, l || AB || CD (by construction)A B

120°

C D

140° ExF

O

O'

I

then, ∠OEF + ∠BOE = 180° ∠OÉF + ∠DOE = 180° [Corresponding interior angles] ∠OEF + ∠OÉF + 120° + 140° = 360° ⇒ ∠OEO´ = 360° – 260° \ x = 100°. 1 2. ∠3 = ∠1 = 60° (corr ∠S) ∠2 + ∠3 = 180° ∠2 + 60° = 180° (∠3 = 60°) ∠2 = 120° = 2 × 60° = 2 ∠1 2 3.

A BD E

100°140°

C21l

Draw l|| AB \ l|| DE as AB || DE

∠1 + 140° = 180 (Co. int. angles) \ ∠1 = 40° ½ Similarly, ∠2 = 80°. ½ ∠1 + ∠BCD + ∠2 = 180 (St. angle) ½ 40° + ∠BCD + 80° = 180°. \ ∠BCD = 60°. ½ [CBSE Marking Scheme, 2012, 2013]

4.

O

D

B

FE

C

5 °y

A5 °y

2 °y

∠COD = 5y° (VOA) 5y + 5y + 2y = 180° ⇒ 12y = 180° ⇒ y = 15° 3

5.

O

QR

SP

∠POR + ∠ROQ = 180° (linear pair) 2 ⇒ 5x + 7x = 180°

⇒ x = 15 \ ∠POR = 75° = ∠QOS, ∠ROQ = 105° = ∠POS 2


1.

m

n

1

2x

∠1 = ∠x = 75° (Alternate angles) ½ ∠2 = 180° – x = 180 – 75° = 105° 1 ∠2 = 105° = 75° + 30°

= 75° + 1

903

× ° . ½


2.

x

yy

l3 l4

l1

l221

x = ∠1 (Corresponding angles) ½ ∠2 = ∠1 (Corresponding angles) ½ ∠2 + 2y = 180° ⇒ x + 2y = 180° ½


⇒ 2y = 180° – x

\

y = 90° – 2

x

½

3.

A

D

BC

Oy°

x°

w°

z°

x + y + w + z = 360° ⇒ 2(x + y) = 360° ( x + y = w + z) ⇒ x + y = 180° \ AOB is a straight line. 3

4.

B

O

C

A

E

D

84°

2 °x

z°

y°75°

84° + 2x° = 180° (linear pair) ⇒ 2x = 96° 2 ⇒ x = 48° y + 75 = 2x (VOA) ⇒ y = 2 × 48° – 75° = 96° – 75° = 21°

z = 84° (VOA) 2


1. ∠PQS + ∠QSF = 180° (Angles on the same side of transversal) ⇒ ∠PQS + ∠RFE = 180°, as ∠QSF = ∠EFR

⇒ 60° + ∠RFE = 180° (Corresponding ∠S)

\ ∠RFE = 120°. 1 2. ∠70° + ∠PRS = 180° ⇒ ∠PRS = 110° = ∠TRO (Vertically opposite angles) In ∆TRO, x + 110° + 20° = 180° ⇒ x = 50°. 1

3.

PL

Q

M N ∠M = ∠N (Given) Exterior ∠NLQ = ∠M + ∠N = 2∠N 1 2∠NLP = 2∠N ⇒ ∠NLP = ∠N (Alternate interior angles) \ LP||MN. 1 [CBSE Marking Scheme, 2012]

4. EG is the bisector of ∠AEF \ ∠AEG = ∠GEF = a Similarly, ∠EFH = ∠HFD = b 1

\ ∠GEF = ∠EFH ( a = b) But these are alternate interior angles \ EG || FH ½ Again, ∠AEF = 2a and ∠EFD = 2b \ ∠AEF = ∠EFD = 2a or 2b 1 But these are alternate angles. \ AB || CD. ½

5. AB ||DC ∠CDB = ∠ABD = x = 35° [alternate angles] 1 x + y + 80° = 180° 1 ∠ADB = y = 65° [angle sum property] 1 ∠DCB = z = 180° – [35° + 35°] = 110° 1 [CBSE Marking Scheme, 2012]

Alternative method : AB || DC y + 35° + 80° = 180° (Corresponding interior angles) 1 ⇒ y = 180° – 115° ⇒ y = 65° ½ ∠ABD = ∠CDB ⇒ x = 35°

(Alternate angles) 1 In ∆BCD, 35° + y – 30° + z = 180° 1 ⇒ z = 180° – 5° – y ⇒ z = 175 – 65° ⇒ z = 110°. ½


TOPIC-3Angle Sum Property of a Triangle


1. We know that x + 60° = 100° (Exterior angle is the sum of the two interior opposite angles) \ x = 40°. 1 2. Since sum of all the exterior angles formed by

producing the sides of a polygon is 360°. \ x° + y° + z° = 360°. 1 3. ∠ACD = ∠A + ∠B 1 [Exterior angle is the sum of

the two interior opposite angles]

= 60 + 70

∠ACD = 130° 1

4. Given, ∠A + ∠B = 65° Given, ∠B + ∠C = 140° ∠A + ∠B + ∠B + ∠C = 65° + 140 = 205° ½ But, ∠A + ∠B + ∠C = 180

(Angle sum prop. of ∆) ⇒ 180° + ∠B = 205° 1 ⇒ ∠B = 25° ½ \ ∠C = 140° – 25° = 115° [CBSE Marking Scheme, 2012, 2013]

Alternative Method : We know that, ∠A + ∠B + ∠C = 180° ...(i)

(Angle Sum prop. of ∆) ⇒ 65° + ∠C = 180° 1 ⇒ ∠C = 115° Again by (i), ∠A + 140° = 180° ⇒ ∠A = 40° ½ Again by (i), 40° + ∠B + 115° = 180° (Angle Sum prop. of ∆) ⇒ ∠B = 25°. ½

5.

E

A G B

H125°

x

C F D

135°

1

Draw EH||AB. Then ∠AGE = ∠GEH and ∠CFE = ∠FEH (Alternate int ∠S) x = ∠GEH + ∠FEH = ∠AGE + ∠CFE ...(i) 1 But ∠AGE + 135° = 180°, ∠CFE + 125° = 180 (linear pair) ∠AGE = 45°, ∠CFE = 55° 1 Hence from (i) x = 45° + 55° = 100°.


1. In ∆ABC,

∠A + ∠B + ∠C = 180°

Also, in ∆DEF,

∠D + ∠E + ∠F = 180°

\ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F

= 360° = 4 × 90°

Hence, k = 4. 1

2. x = ∠APQ = 40° (Alternate angles) ½ x + y = 118° ½ (Exterior angle is the sum of

the two opposite interior angles)

⇒ 40° + y = 118° ½ \ y = 118° – 40° = 78°. ½

3. In ∆ABC, ∠A + ∠B + ∠C = 180° 1

(Angle Sum prop. of ∆) ⇒ 2x – 5° + 5x + 5° + 3x + 50° = 180° 10x = 130° \ x = 13° ∠A = 2x – 5° = 21° ∠B = 5x + 5° = 70° ∠C = 3x + 50° = 89° 1 [CBSE Marking Scheme, 2012]

Alternative Method : In ∆ABC, ∠A + ∠B + ∠C = 180° (Angle Sum prop. of ∆) ⇒ 2x – 5° + 5x + 5° + 3x + 50°


= 180° (By given values) ⇒ 10x + 50° = 180° ⇒ 10x = 130° ½ ⇒ x = 13° ∠A = 2x – 5° = 26° – 5° = 21° ½ ∠B = 5x + 5° = 65° + 5° = 70° ½ ∠C = 3x + 50° = 39° + 50° = 89° ½ 4. In ∆BDE, ∠B + ∠D + ∠DEB = 180° (Angle sum property of a triangle) ⇒ 40° + x + 90° = 180° ⇒ x = 50° 1

In ∆DCF, ∠D + ∠FCD = ∠AFD 1 (Exterior angle is the sum of the two interior opposite angles) ⇒ 50° + y = 110° ⇒ y = 60° 1

5. (i) ∠TOP =

12∠TOB

½

∠ORS =

12∠ORD

But, ∠TOB = ∠ORD (l || m and corresponding angles) ½ \ ∠TOP = ∠ORS ½ But, they are corresponding angles w.r.t. transversal

TR and lines OP and RS. Hence, OP || RS. ½

T

P BA

C D

S

R

Ol

m

(ii) Lines and angles. ½ (iii) Bisectoring among human beings gives rise to

deterioration in the society.


1. ∠QPR = 75° (Vertically opposite angles) Again, ∠PQR + ∠QPR = 105° (Exterior angle) ⇒ ∠PQR + 75° = 105° ⇒ ∠PQR = 30°. 1

2. A

CB40°

60°

∠C = 180° – (60° + 40°) = 180° – 100° ½ = 80° ⇒ AC is the smallest side ½ Reason : Side opposite to smaller angle is shortest. 1 [CBSE Marking Scheme, 2012]

3.

P Q

L

y

x

38°

75°R T

∠1

Given : PQ ⊥ PR PQ || RL ∠RQT = 38° ∠QTL = 75° To find : x & y ∠1 = ∠y ...(1) [alt. interior angles] Now, In ∆QRT ∠QTL = ∠TQR + ∠QRT 1 [by exterior property of triangles] ⇒ 75° = 38° + ∠1 ⇒ ∠1 = 37 ° ⇒ ∠y = 37° ...(2) [from eq.(1)] 1 Now, In ∆QTR ∠QPR + ∠x + ∠y = 180° [by ASPT] ⇒ 90° + ∠x + 37° = 180° ∠x = 53° [from eq (2)] 1 4. (i) In ∠ABC, ∠BCD = ∠BAC + ∠ABC [Exterior angle equal to the sum of opposite two interior angles] 6x + 2 = 3x + 15 + 2x – 1 6x + 2 = 5x + 14 6x – 5x = 14 – 2 x = 12° 2


(i) Exterior angle property of a triangle is used in the above problem. 1

(ii) By doing so, students exhibit the importance of water. 1

5. ∠PBC = x + z ½

⇒ 2∠OBC = x + z ∠QCB = x + y ⇒ 2∠OCB = x + y ½

∠BOC + ∠OCB + ∠OBC = 180° (Angle sum prop. of ∆)

⇒ 2∠BOC + 2∠OCB + 2∠OBC = 360° (Multiply by 2 both sides) ⇒ 2∠BOC + x + y + x + z = 360° ⇒ 2∠BOC + 180° + x = 360° ⇒ 2∠BOC = 180° – x

⇒

2∠BOC = 90° –12

x

3



1. In ∆ABC, ∠A + ∠B + ∠C = 180° (By given conditions) ⇒ ∠A + 2∠A + 6∠A = 180° ⇒ 9∠A = 180° ⇒ ∠A = 20°. 1 2. AB = AC \ ∠C = ∠B Then, ∠A + ∠B + ∠C = 180° (Prop. of isosceles ∆) ⇒ 50° + ∠B + ∠B = 180° ⇒ 2∠B = 130° ⇒ ∠B = 65° 1 3. Exterior angle = Sum of opposite two interior angle

½ other angle = 180 – 110 = 70° 110 = 30 + x 1 ⇒ x = 80°. ½

4.

A

B D C

21

45° 55°

∠1 = ∠2 = x, ∠B = 45° ∠A + ∠B + ∠C = 180°, ∠C = 55° (Angle sum prop. of ∆) ⇒ 2x + 45° + 55° = 180° ⇒ 2x = 80° \ x = 40° ∠ADB = ∠2 + ∠C

(Exterior angle is the sum of the two interior opposite angles) 1 = 40° + 55° = 95° ½ Similarly, ∠ADC = ∠1 + ∠B = 45° + 40° = 85°. ½

5.

A

B C

x

y z

12l

To prove : Sum of all the angles of ∆ABC is 180°. 1 Construction : Draw a line l parallel to BC. Proof : Since l||BC, we have ∠2 = ∠y (Alternate angles are equal) ... (i) 1 Similarly, l||BC ∠1 = ∠z (Alternate angles are equal) ...(ii) Also, sum of angles at a point A on line l is 180°. ½ \ ∠2 + ∠x + ∠1 = 180° i.e., ∠y + ∠x + ∠z = 180°

(from (i) and (ii)) \ ∠x + ∠y + ∠z = 180° ½ ⇒ ∠A + ∠B + ∠C = 180° Sum of all angles of a ∆ is 180° 1 \ 5x + 6x + 7x = 180° ⇒ x = 10° Angles are 50°, 60° and 70° respectively.

Hence proved. [CBSE Marking Scheme, 2012]

FORMATIVE ASSESSMENT WORKSHEET-44, 45 & 46 Note : Students should do these activities themselves.

qqq


TOPIC-1Criteria for Congruence of Triangles


1. ASA congruence : Two triangles are congruent, if two angles and the included side of one triangle are equal to two angles and the included side of other triangle. 1

2. Since AB = DE, ∠A = ∠D and ∆ABC ≅ ∆DEF by SAS.

Therefore AC = DF. 1A

B C

D

E F

3.

A

B C

D

Join AD.

In ∆ABC and ∆ACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common) 1

By using SSS Congruency Rule,

∆ABD ≅ ∆ACD ½

\ ∠ABD = ∠ACD (By c.p.c.t.) ½

4. In ∆DCB,

∠4 + ∠C + ∠2 = 180° ...(1)

(Angle sum prop. of ∆) ½

In ∆ADB,

∠3 + ∠A + ∠1 = 180° ...(2)

(Angle sum prop. of ∆) ½

Adding eq. (1) & (2),

⇒∠4 + ∠3 + ∠C + ∠A + ∠2 + ∠1 = 360°

∠A + ∠C + ∠B + ∠D = 360°

⇒ ∠A + ∠C + ∠B + ∠D = 360° Hence proved. 1

5. Given : D is the mid-point of side AC

To Prove :

BD =

12

AC

Const : Draw, DE || BC 1

Proof : In ∆ABC,

D is the mid–point of AC & ED || BC

\ By mid point theorem,

AE = EB ...(i)

Now, ED || BC (By const)

∠AED = ∠ABC

(Corresponding ∠’s)

∠AED = 90° 1

\ ∠AED = ∠DEB = 90° ...(2)

Now, In ∆ADE & ∆BDE,

AE = EB (from eq. ...(1))

∠AED = ∠DEB (from eq. ...(2))

ED = ED (Common)

\ ∆ADE ≅ ∆BDE (By SAS rule)

\ AD = BD ...(3) (By cpct)

But, AD = DC =

12

BC ...(4) 1

(\ D is the mid–point of side AC)

From eq. (3) & (4),

BD =

12

AC Hence Proved. 1

TRIANGLES

SECTION

BSECTIONCHAPTER

5



1. OA = OB (Given) OP = OP (Common) ∠AOP = ∠BOP (Given) ∆OAP ≅ ∠OBP (By SAS) 1 2. In ∆ABY and ∆ACX, AB = AC (Given) AY = AX (Given) 1 ∠A = ∠A (Common) \ By SAS, ∆ABY ≅ ∆ACX. Proved. 1 3. In ∆PAB and ∆PDC, PA = PD (Given) (P is the mid-point of AD) AB = CD (Side of a square) ∠PAB = ∠PDC = 90° By R.H.S., ∆PAB ≅ ∆PDC 1½ \ PB = PC (By c.p.c.t.)

(Angles opp. to equal sides are equal) ⇒ ∠PCB = ∠PBC. Proved. ½

4. Proof : In ∆ABD and ∆BAC, AD = BC (Given) BD = AC (Given) AB = AB (Common side) 2 By SSS congruence axiom, ∆ABD ≅ ∆BAC ⇒ ∠ADB = ∠BCA (By c.p.c.t.) 1 and ∠DAB = ∠CBA. (By c.p.c.t.)

5. Given, ∠DCA = ∠ECB Adding ∠DCE to both sides, we get ∠DCA + ∠DCE = ∠ECB + ∠DCE ⇒ ∠ECA = ∠DCB 1 In ∆ACE and ∆BCD, AC = BC

(Given) ∠ECA = ∠DCB

(Proved) ∠EAC = ∠DBC

(Given) \ ∆ACE ≅ ∆BCD (By AAS cong.) 2 \ BD = AE. Proved. 1 [CBSE Marking Scheme, 2012]


1. AD = BC (Given) ∠BAD = ∠ABC (Given) AB = AB (Common) \ ∆DAB ≅ ∆CBA (By SAS) \ ∠BDA = ∠ACB (By c.p.c.t.) 1 2. OA = OB (O is the mid-point of AB) ∠AOC = ∠BOD

(Vertically opposite angles) ½ OC = OD

(O is the mid-point of CD) ½ ∆AOC ≅ ∆BOC (By SAS) ½ ⇒ AC = BD. (By c.p.c.t.) Proved ½ 3. In ∆ADC and ∆ABC, AC is common ∠DAC = ∠BAC ½ ∠DCA = ∠BCA (Given) ½ Hence, ∆ADC ≅ ∆ABC (By AAS rule) ½ ⇒ CD = BC. (By c.p.c.t.) Proved. ½

4. ∠BDC + ∠CDA = 180° (Linear pair) ⇒ ∠BDC + x = 180° ⇒ ∠BDC = 180° – x ½

Similarly, ∠BEA = 180° – y° ½ Since, x = y (Given) \ ∠BDC = ∠BEA ½ ∠B = ∠B (Common) AB = BC 1 \ ∆BAE ≅ BCD (ASA) ½ \ AE = CD (c.p.c.t.) [CBSE Marking Scheme, 2012]

5. Given, AD = BC

∠1 = ∠2

∠3 = ∠4

Let us consider ∆ABC and ∆ABD

AB = AB (Common side) 1

∠1 = ∠2

∠3 = ∠4

So, ∠1 + ∠3 = ∠2 + ∠4 1

∠DAB = ∠CBA

and, AD = BC 1

Then, from the SAS Congruence Rule

∆ABC ≅ ∆ABD 1

\ BD = AC Hence Proved



1. Let the angles of triangle are 5x, 3x and 7x, then 5x + 3x + 7x = 180° ⇒ 15x = 180° Thus, x = 12° \ Angles are 60°, 36°, 84° Each angle is less than 90° \ The triangle is an acute angled triangle. 1 2. ∠BDA = ∠ACB = 90° (Given) AD = BC (Given) AB = AB (Common) \ ∆ABD ≅ ∆BAC (By RHS) 1 3. (i) In ∆AOD and ∆BOC, OA = OB (Given) OD = OC (Given) ∠AOD = ∠BOC

(Vertically opposite angles) So, by SAS criteria, ∆AOD ≅ ∆BOC 1 (ii) ∠CBA = ∠DAB (By c.p.c.t.) AD and BC are two lines intersected by AB such that

∠CBA = ∠DAB and they form a pair of alternate angles.

Hence, AD || BC. 1

4.

C

B

A2 D

134

5

P6

∠1 + ∠5 = 180° = ∠2 + ∠6 (linear pair)

⇒ ∠1 = ∠2 ( ∠5 = ∠6) 1 In ∆CAP and ∆BAP, ∠1 = ∠2 (Proved) ∠3 = ∠4

(AD is the bisector of ∠BAC)

AP = AP \ ∆CAP ≅ ∆BAP (By AAS) 1 ⇒ CP = BP. (By c.p.c.t.) Proved. 1 [CBSE Marking Scheme I, 2012]

5.

P

Q R

O

1 Proof : QO is bisector of ∠PQR

∠OQR =

12∠PQR =

12

∠Q

RO is bisector ∠PRQ

\

∠ORQ = 12∠PRQ

12

∠R

In ∠OQR ∠QOR + ∠OQR + ∠ORD = 180° (Angle sum property)

∠QOR +

12∠Q +

12∠R = 180°

\ ∠QOR = 180° –

12

(∠Q + ∠R)

But in ∠PQR ∠P + ∠Q + ∠R = 180° ∠Q + ∠R = 180° – ∠P

∠QOR = 180° –

12

(180° – ∠P) 2

= 180° – 90° +

12∠P

= 90° +

12

∠P Hence Proved.

These type of rallies spread awareness among people for not to kill girl child and helping in equalising sex ratio. 1


1. ∆CBA ≅ ∆PRQ 1 2. Interior opposite angles. ∠BCD = ∠A + B 1

3.

A

E F

B CD

In ∆BED and ∆CFD,

∠DEB = ∠DFC = 90° ½

BD = DC (D is the mid–point)

ED = FD (Given)

\ ∆BED ≅ ∆CFD (By RHS) 1

⇒ ∠B = ∠C. (By c.p.c.t.) Proved. ½


4. PQRS is a square. (given)

T

RS

QP

x

x

(i) SRT is an equilateral triangle. (given) \ ∠PSR = 90°, ∠TSR = 60° ½ ⇒ ∠PSR + ∠TSR = 150°. Similarly, ∠QRT = 150° In ∆PST and ∆QRT, we have PS = QR (S) ∠PST = ∠QRT = 150° (A) and ST = RT (S) ½ By SAS, ∆PST ≅ ∆QRT ⇒ PT = QT (By c.p.c.t.) Proved. (ii) In ∆TQR, QR = RT (Square and equilateral ∆ on same base) ½ ⇒ ∠TQR = ∠QTR = x ½ \ x + x + ∠QRT = 180° ⇒ 2x + 150° = 180° ⇒ 2x = 30° \ x = 15°. Proved. 1

5. Proof : We are given two triangles ABC and PQR in which

∠B = ∠Q, ∠C = ∠R and BC = QR We need to prove that ∆ABC ≅ ∠PQR 1 There are three cases. Case I : Let AB = PQ In ∆ABC and ∆PQR, ∠B = ∠Q (Given) BC = QR (Given)

AB = PQ (Assumed) \ ∆ABC ≅ ∆PQR (By SAS rule) 1

A

CB

P

RQ

Case II : Suppose AB ≠ PQ and AB < PQ Take a point S on PQ such that QS = AB

A

CB

P

RQ

S

Join RS. In ∆ABC and ∆SQR, AB = SQ (By construction) BC = QR (Given) ∠B = ∠Q (Given) 1 \ ∆ABC ≅ ∆SQR (By SAS rule) ⇒ ∠ACB = ∠QRS (By c.p.c.t.) But, ∠QRP = ∠ACB ⇒ ∠QRP = ∠QRS Which is impossible unless ray RS coincides with

RP. \ AB must be equal to PQ. ½ So, ∆ABC ≅ ∆PQR Case III : If AB > PQ. We can choose a point T on AB such that TB = PQ

and repeating the arguments as given in Case II, we can conclude that AB = PQ and so,

∆ABC ≅ ∆PQR ½ [CBSE Marking Scheme, 2013, 2012]

TOPIC-2Some Properties of Triangles


1. In ∆PEQ and ∆PER, ∠PEQ = ∠PER = 90° ∠QPE = ∠RPE (Given) PE = PE (Common) \ ∆PEQ ≅ ∆PER (by ASA)

\ PQ = PR. (By c.p.c.t.)

P

RQ E90°90°

1


2. AB = AD ⇒ ∠ABD = ∠ADB ...(i) ½ BC = CD ⇒ ∠CBD = ∠CDB ...(ii) ½ Adding eqns. (i) and (ii), we get ∠ABD + ∠CBD = ∠ADB + ∠CDB ½ ⇒ ∠ABC = ∠ADC. Proved. ½ [CBSE Marking Scheme, 2011, 2012]

3.

A

CB P In ∆ABP and ∆ACP, AB = AC (Given) ½ AP = AP (Common) ∠APB = ∠APC = 90°, (AP ⊥ BC) ½ By RHS rule, ∆ABP ≅ ∆ACP ½ ⇒ ∠B = ∠C. (By c.p.c.t.) ½

4.

A

B CY

X

1 AB = BC

⇒

12

AB =

12

BC

But

12

AB = BX

\

12

BC = BX (ii) ½

But

12

BC = BY (i) ½

From (i) and (ii) BX = BY 1

5.

B D C

E

A

O

32

5

61 4

2∠3 + 2 ∠4 = 180° – ∠ABC = 90° \ ∠3 + ∠4 = 45° ∠1 = 90° + ∠3 (External angle) ∠2 = 90° + ∠4 (External angle) Adding, ∠1 + ∠2 = 180° + ∠3 + ∠4 = 180° + 45° (External angle) = 225° 4


1. AB = AC ⇒ ∠C = ∠B By angle sum property,

90°BA

C

∠A + ∠B + ∠C = 180° ⇒ 90° + ∠B + ∠B = 180° ⇒ 2B = 90° \ B = 45°. 1 2. In ∆PQS and ∆PRS, PQ = PR (Given) ½ PS = PS (Common) ∠PSQ = ∠PSR = 90° (PS is altitude) ½ By R.H.S. rule,

∆PQS ≅ ∆PRS ½ ⇒ ∠QPS = ∠RPS (By c.p.c.t.) ½ Hence, S bisects ∠P.

P

RQ S

3. From ∆DCF, we have x + y = ∠AFD = 110° ....(1) Also ∠AFE = 180° – 110° = 70° \ ∠FAE = 90° – 70° = 20° From ∆ACB y = ∠FAE + ∠ABC = 20° + 40° = 60° Then (1) gives x = 50°. 3


4.

D

CB

A

1 23

4

In ∆ABC, AB = AC ⇒ ∠1 = ∠2 ...(1) Angles opp. to equal sides are equal.

In ∆ADC., AB = AD \ AC = AD ½ In ∆BCD, ∠3 = ∠4 ...(2) ½ ∠1 + (∠2 + ∠3) + ∠4 = 180° ½ ⇒ ∠2 + ∠2 + ∠3 + ∠3 = 180° ½ ⇒ 2(∠2 + ∠3) = 180° ⇒ ∠2 + ∠3 = 90° ½ ⇒ ∠BCD is a right angle. ½ [CBSE Marking Scheme, 2014, 12]


1. Proof : In ∆ABF and ∆ACE, AB = AC (Given) 1 ∠A = ∠A (Common) 1 AF =

AE \ ∆ABF ≅ ∆ACE (by SAS cong.) \ By c.p.c.t., BF = CE. 1 [CBSE Marking Scheme, 2012]

Alternative method :

AB = AC ⇒

AB2

=

AC2

½

⇒ AE = AF, Since E and F are the mid-points of AB and AC. In ∆ABF and ∆ACE, AB = AC (Given) ½ ∠A = ∠A (Common) ½ AF = AE (Proved) ½ \ ∆ABF ≅ ∆ACE (By SSA cong.) ½ \ BF = CE. (By c.p.c.t) ½ 2.

A

CB ½ AB = AC ½ \ ∠B = ∠C = x BA = BC \ ∠A = ∠C = x Also, AC = BC ½ ∠A = ∠B = x But, ∠A + ∠B + ∠C = 180° ½ ⇒ 3x = 180° ⇒ x = 60° \ ∠A = ∠B = ∠C = 60°. 1 [CBSE Marking Scheme, 2012]


A

CB

Let ∆ABC be an equilateral triangle, so that AB = AC = BC.

Now, AB = AC ⇒ ∠B = ∠C ....(1) ( Angles opp. to equal sides are equal) CB = CA ½ ⇒ ∠A = ∠B ....(2) ( Angles opp. to equal sides are equal) From (1) and (2), we have ∠A = ∠B = ∠C 1 Also, ∠A + ∠B + ∠C = 180°, (Angle sum property) \ ∠A + ∠A + ∠A = 180° ½ ⇒ 3∠A = 180° ⇒ ∠A = 60° \ ∠A = ∠B = ∠C = 60°. Thus, each angle of an equilateral triangle is 60°. 1

3.

A

C D E B We have here ∠CAE = ∠EAB (given) .....(1) Also ∠CAD + ∠DAE = ∠CAE = ∠EAB

[Use (1)].....(2) Now ∠CAD + ∠C = 90° = ∠DAE + ∠EAB + ∠B ∠CAD + ∠C = ∠DAE + ∠CAD + ∠DAE + ∠B [Use (2)] = 2 ∠DAE + ∠CAD + ∠B \ 2 ∠DAE = ∠C – ∠B


⇒ ∠DAE = 12

(∠C – ∠B) 3

4. Proof AB = AC ⇒ ∠ABC = ∠ACB and BD = CD ⇒ ∠DBC = ∠DCB ∠ABC – ∠DBC = ∠ACB – ∠DCB ⇒ ∠ABD = ∠ACD 1 ∆ABD ≅ ∆ACD by SAS 1

⇒ ∠BAP = ∠CAP ∆ABP ≅ ∆ACP by SAS 1

\ BP = PC, ∠APB = ∠APC \ ∠APB = ∠ADC = 90° AP is perpendicular to BC AP is perpendicular bisector of BC 1

D

B P C

A

TOPIC-3Inequalities of a Triangle


1. If D is a point on the side BC of a ∆ABC such that AD bisect ∠BAC, then AB > BD. 1

2. In ∆ABC, AC > AB (Given) \ ∠ABC > ∠ACB (Angles opposite to larger side is greater) ½ \ ∠ABC + ∠1 > ∠ACB + ∠1

(Adding ∠1 on both sides) ½ \ ∠ABC + ∠1 > ∠ACB + ∠2

(AD bisects ∠A, ∠1 = ∠2) ½ \ ∠ADC > ∠ADB.

(Exterior angle property of triangle) ½

3. Proof : In ∆AOB, ∠B > ∠A ∠A < ∠B 1 OB < OA ...(1)

In ∆COD, ∠C > ∠D ∠D > ∠C 1 OC < OD ...(2) Adding (1) + (2), OB + OC > OA + OD BC > AD or, AD < BC [CBSE Marking Scheme, 2012]

4. In ∆PQS, PQ + QS > PS ...(1) 1 (Sum of any two sides is greater than the third side) In ∆PSR, PR + SR > PS ...(2) Adding (1) & (2), 1 PQ + QS + PR + SR > 2PS (QR = QS + SR) PQ + QR + RP > 2PS. 1 [CBSE Marking Scheme, 2012]

5. AD = BD ⇒ ∠ABD = ∠DAB = 59° (Angles opp. to equal sides are equal) 1 In ∆ABD, 59° + 59° + ∠ADB = 180° ⇒ ∠ADB = 180° – 118° = 62° 1 and ∠ACD = 62° – 32° = 30° (Exterior angle is equal to the sum

of interior opposite angles) In ∆ABD, AB > BD

(Side opp. to greatest angle is the longest) 1 Also in ∆ABC, AB < AC

⇒ BD < AC. 1


1. No, Because, 2.3 + 3.1 = 5.4 cm (third side) \ Not possible to construct a triangle. 1 2. Since ∠C > ∠A > ∠B, \ AB > BC > AC.

A

B C

70°

30° 80° 1


3.

P

TS

Q R

Construction : Produce QS to meet PR in T ½

In ∆PQT, PQ + PT > QT

⇒ PQ + PT > QS + ST ...(1) ½

In ∆SRT, TR + ST > SR ...(2) ½

Adding (1) and (2), we get

PQ + PT + TR + ST > QS + ST + SR 1

⇒ PQ + PR > QS + SR

⇒ QS + SR < PQ + PR. ½


4.

A

C B

In ∆ABC, as AB is the greatest side

⇒ AB > BC ⇒ ∠C > ∠A ...(1)

AB > AC ⇒ ∠C > ∠B ...(2) 1

On adding (1) and (2), we get

2∠C > ∠A + ∠B ⇒ 2∠C + ∠C > ∠A + ∠B + ∠C 1

⇒ 3∠C > 180°

\ ∠C > 60°. 1

[CBSE Marking Scheme, 2012] 5. (i) AB and CD intersect at O \ ∠AOD = ∠BOC (Vertically opp. angles) ...(i) In ∆AOD and ∆BOC, we have ½ ∠AOD = ∠BOC From ...(ii)

∠DAO = ∠CBO = 90° (Given) ½ and AD = BC (Given) \ ∆AOD ≅ ∆BOC (By AAS congruence criterion) ⇒ OA = OB (By c.p.c.t.) i.e., O is the mid-point of AB Hence, CD bisects AB. 1 (ii) Congruency of triangles. 1 (iii) Equality is the sign of democracy. 1 6.

A

B CD

E

Construction : Produce AD to E such that AD = DE. Join EC. ½ In triangles ADB and EDC, AD = DE (Const) ½ BD = DC (Given) ∠ADB = ∠EDC (V.O.A) ½ \ ∆ADB ≅ ∆EDC (SAS congruence axiom) ½ ⇒ AB = EC (By c.p.c.t.) In ∆AEC, AC + EC > AE [Triangle Inequality Property] ½ \ AC + AB > AE (EC = AB)

⇒ AC + AB > AD + DE ⇒ AC + AB > AD + AD ( DE = AD) ½

⇒ AC + AB > 2AD. Proved. 1 [CBSE Marking Scheme, 2013, 12]


qqq


TOPIC-1Cartesian System


1. P (2, – 3) and Q (–3, 2) lie in IV and II quadrants. 1 2. (3, –2) 1 3. In II quadrant, x < 0 Points = (–1, 0), (–2, 0) 1 4. The distance of a point from the y-axis is called its

x-co-ordinate, or abscissa. 1 5. The P is on x-axis \ y = 0 P is at a distance of 4 units from y-axis to its left. \ In second quadrant, the co-ordinates of the point

P = (– 4, 0). 1 6. II Quadrant. 1

7. (A) P(0, 5) ½ (B) Q(0, – 3) ½ (C) R(5, 0) ½ (D) S(–2, 0) ½

y

xx'

y'

654321

–1–2–3–4

Q(0, – 3)

R(5, 0)(–2, 0)O

P(0, 5)

–1 –5 –4 –3 –2 –1 1 2 3 4 5 6

8. (A) II quadrant ½

(B) III quadrant ½ (C) IV quadrant ½ (D) I quadrant. ½

9.

y

y'

x'x

D A

BC–1

–2

–3

(–1, 3) (1, 3)

0–2–3 2

(–1, –1) (1, – 1)

1

3

2

1

–1

3

Co-ordinates of D are (–1, 3). 10. In a point (– 5, 3), x < 0 and y > 0 \Point (– 5, 3) lies in II quadrant. ½ In a point ( 4, – 3), x > 0 and y < 0 \ Point (4, – 3) lies in IV quadrant ½ In a point (5, 0), x > 0 and y = 0 \ Point (5, 0) lies on x-axis ½ In a point (6, 6), x > 0 and y > 0 \ Point (6, 6) lies in I quadrant ½ In a point (– 5, – 4), x < 0 and y < 0 \Point (– 5, – 4) lies in III quadrant 1


1. (x, y) = (y, x) 1 2. 7 1

3. The co-ordinate of x-axis = (x, 0) 1

4. Negative ordinate i.e., (x, –y) 1

COORDINATE GEOMETRY

SECTION

BSECTIONCHAPTER

6


5. The point on y-axis has x-co-ordinate 0.

Since it lies at a distance of 4 units in the negative direction of y-axis. 1

\ The point is (0, – 4). 1

6. (a) A point which lies on x and y-axes is (0, 0) i.e., origin ½

(b) A point whose abscissa is 5 and ordinate is 6 i.e., x = 5 and y = 6 is (5, 6) ½

(c) A point whose ordinate is 6 i.e., y = 6 and lies on y-axis is (0, 6) ½

(d) A point whose ordinate is 3 and abscissa is – 7 i.e., y = 3 and x = – 7 is (–7, 3) ½

(e) A point whose abscissa is 3 i.e., x = 3 and lies on x-axis is (3, 0) ½

(f) A point whose abscissa is 4 and ordinate is 4 i.e., x = 4 and y = 4 is (4, 4) ½

7. Firstly, we plot all the points i.e.,

A(3, 0), B(5, 0), C(5, 3) and D(3, 3) on a graph paper and join all these points. ½

–4–5 –3 –2 –1 0 1 2 3 4 5 6

5

4

3

2

1

–1

–2

–3

–4

–5

Y

Y'

XX'

D(3, 3) C(5, 3)

B(5, 0)(3, 0)A

1½

The obtained figure ABCD is a rectangle, since AB = CD and BC = DA and all lines are perpendicular to each other. 1

8. (i) E(– 1, 2) (ii) D(– 4, – 1) (iii) Co-ordinates of A = (2, – 2) Co-ordinates of B = (4, 1) \The abscissa of A – abscissa of B = 2 – 4 = – 2 (iv) Co-ordinates of C = (3, 3) Co-ordinates of F = (– 1, – 4) \ The ordinate of C + ordinate of F = 3 + (– 4) = – 1

1 + 1 + ½+½ + ½+½


1. (0, 0) 1 2. Origin 1 3. Point C(3, a – 3) lie on x-axis \ a – 3 = 0

⇒ a = 3. 1

4. (A) II quadrant ½ (B) III quadrant ½ (C) I quadrant ½ (D) II quadrant. ½ 5. 4, 0, 4, –3. 2 6. The vertices of the rectangle OABC are O(0, 0),

A(– 6, 0), B(– 6, – 3), C(0, – 3) 3 7. (i) Draw XÓX and YÓY as the co-ordinate axes

and mark their point of intersection O as the origin (0, 0).

In order to plot the points (– 2, 8), we take 2 units on OX´ and then 8 units parallel to OY to obtain the point A(– 2, 8).

Similarly, we plot the point B(– 1, 7). ½ In order to plot (0, – 1.25), we take 1.25 units below

the x-axis on the y-axis to obtain C(0, – 1.25). ½

In order to plot (1, 3), we take 1 unit on OX and then 3 units parallel to OY to obtain the point D(1, 3) ½

X' X

Y'

Y

12

45678

–1–2–3–4–5

–1–2–3–4–5 0 1 2 4 5 6 7 8E(3, –1)

(0, –1.25)C

D(1, 3)

A(–2, 8)

B

3

3

(–1, 7)

1 In order to plot (3, – 1), we take 3 units on OX and

then 1 unit below x-axis parallel to OY´ to obtain the point E(3, – 1) ½

(ii) Co-ordinate geometry. ½ (iii) Co-ordination among people is good for

progress. ½


TOPIC-2Plotting a Point in a Plane


1.

3

2

1

1

2

3

3 2 1 1 2 3

(0, 0) (3, 0)

(0, 3) (3, 3)

Y

XX'

Y'

Vertices are (0, 0), (3, 0), (3, 3), & (0, 3). 2 2. For plotting a point A(5, –3), we will take a distance

of – 3 units in the negative direction of y-axis and a distance of 5 units in the positive direction of x-axis, which is shown in the figure given below. Similarly, we plot all the points i.e., B(– 6, 0), C(– 2, – 3) and D(– 4, 3). 1

–6 –5 –4 –3 –1 1 2 3 4 6O

54

21

–1–2

–4–5

–2 5

3

–3

Y

Y'

A(5,–3)

XX'

D(–4, 3)

B(–6, 0)

C(–2,–3)

1

3.

–1

–2

–2 1 2

2

1(–1,1)

–1

(0,2)

(1,0)(–1,0)

A Pentagon

Y

Y'

XX'

(1,1)

4. (i) The co-ordinate of B = (–2, 3) 1 (ii) E is the point which is identified by the coordinates

(–3, –2) 1 (iii) The co-ordinate of the point D is (6, 2) 1 \ Abscissa is 6. (iv) The co-ordinate of the point C is (3, –1) 1 \ Ordinate is –1.


1. Square

Y

XX'

Y'

(0, 2) B C (2, 2)

D (2, 0)(0, 0) A

1 2. The co-ordinate of point Q = (– 3, – 3·5) 1 3. (2 – a + b, b) = (6, 2) \ b = 2 and 2 – a + b = 6 ⇒ 2 – a + 2 = 6 ⇒ – a = 6 – 4 = 2 \ a = – 2. 1

4.

–25 –20 –15 –10 –5 5 10 15 20 25O

Y

XX'

Y'

–5

–10

–15

–20

–25

5

10

15

20

25 B(3,23)

A(23,3)

1

After plotting these two points AB line segment is formed.


Mid-point =

3+23 23+3,

2 2

= (13, 13) 1

5.

–6 –5 –4 –3 1 2 3 4 5 6 7

1098

654321

–1–2–3–4–5–6–7

XX'

Y'

YA(3,10)

C(–1,–6)

–2 –1

7

O

B(–3,5)

After joining the points, a triangle is formed. 2

6.

y

y'

xx'

54321

–l–2–3

–l–2–3–4–5–6 O1 2 3 4 5 6 7 8

B (0, 4)

A (4, 0)

2

From figure, OA = 4 units OB = 4 units

Area of ∆OAB =

12

× OA × OB

=

12

× 4 × 4 = 8 square unit. 1

7. (i) Plotting of points M(5, – 3) and N(–3, –3) on the graph paper is shown in the diagram.

(ii) Length of MN = 3 + 5 = 8 units (iii) From figure, ½ A(3, – 3), B(1, – 3), C(–1, – 3) ½ + ½ + ½

y

x' x

y'

4321 1 2 3 4 5 6 7 8

–6–5–4

–4–5

B A M(5,–3)N(–3,–3)C

–3–2–1 –1–2–3

2

SUMMATIVE ASSESSMENT WORKSHEET-64Solutions 1. Position of thief = (5, 0) or (– 5, 0)

y

y'

x' x

(0, 5) Police

(5, 0)Thief

(–5, 0)Thief

1

2. The given points are : (–2, – 4) = A (2, 4) = B (0, –2) = C (4, – 6) = D The point A will lie in III quadrant. ½ The point B will lie in I quadrant. ½ The point C will lie O in y-axis (i.e., x = 0) ½ The point D will lie in IV quadrant. ½ Verification :

–6 –5 –4 –3 –1 1 2 3 5 6O

Y

XX'

Y'

6

5

4321

–1

–2

–3

–5

–6

B(2, 4)

C(0, –2)

A(–2, –4)

–2 4

–4

After plotting these points on cartesian plane, we find that point A is lying in III quadrant, B is lying in I quadrant, C is in the negative direction of y-axis and D is in IV quadrant.

Thus, result is verified. 3. According to the question, plot the points A(0, 3),

B(5, 3), C(4, 0) and D(–1, 0) on the graph paper.


1 2 3 4 5–1–2–3–4–1

–2

–3

–4

1

2

3

4B(5, 3)

0

D

C(4, 0)

A(0, 3)

1 From figure, ABCD is a parallelogram. The point (2,

2) will lie inside the figure. 1

4.

–1

–2

–3

–4

–5

–6

1 2 3 4 5 6

6

5

4

3

2

1

(–2,5)

(–4, 3)

–6 –5 –4 –3 –2 –1

(0,2.5)

(2,5)

(4,–3)

(6,–1)

TOPIC-3Graph of a Linear Equations


1. Given equation 2x + y – 4 = 0 \ y = – 2x + 4 1 2. Given equation, 3x + 6y – 4 = 0 ... Line intersect the x-axis ... y = 0 ½ 3x + 6 × (0) – 4 = 0

⇒ x =

43

So, the point where line intersect the x-axis is

(43

, 0). ½

3. x-axis. 1 4. Given, y = 3x + 6 Put x = 0, we get y = 3 × 0 + 6 = 6 Similarly at x = 1, y = 9 \ Points (0, 6), (1, 9) lies on the line y = 3x + 6. 1 5. The given equation is : y = x Putting x = 0, we get y = 0 ½ Putting x = 1, we get y = 1 ½ Putting x = 2, we get y = 2 ½ Thus, we have the following table :

x 0 1 2 3

y 0 1 2 3

Now, plot the points R(0, 0), P(1, 1) and a(2, 2) on the graph paper.

–4 –3 –2 –1 4321

–4

–3

–2

–1

4

3

2

1

XX'

Y'

Y

a (2, 2)

P(1, 1)(0, 0) R

½ 6. The given equation is : y = 2x + 1 Putting x = 0, we get y = 0 ½ Putting x = 1, we get y = 3 ½ Putting x = 2, we get y = 5 ½ Thus, we have the following table :

x 0 1 2 3

y 1 3 5 7 Now Let plot the points A(0, 1), B(1, 3), C(2, 5) and

D(3, 7) on the graph paper.


Y

Y'

X' XA(0, 1)

–1–2–3–4–5 1 2 3 4 5 6 7

7654321

–1

–2

–3

–4

D(3, 7)

C(2, 5)

B(1, 3)

½ 7. The given equation is: y = 3x – 2 Putting x = 0, we get : y = (–2) ½ Putting x = 1, we get y = 1 ½ Putting x = 2, we get y = 4 ½ Thus, we have the following table

½

x 0 1 2 3

y –2 1 4 7 Now plot the points A(0, –2), B(1, 1), C(2, 4) and

D(3, 7) on the graph paper. Join A, B, C, D and extend it in both the directions. Then, line AD is the graph of the equation y = 3x – 2.

y

y'

x' x–5 –4 –3 –2 –1 1 2 3 4 5 6

–1

–2

–3

–4

1234567

0

A(0, –2)

B(1, 1)

C(2, 4)

D(3, 7)

1


qqq


TOPIC-1Area of Triangle


1.

s =

2a b c+ +

=

13 14 152

+ + = 21 cm ½

Area = ( – )( – )( – )s s a s b s c

= 21 (21 13)(21 14)(21 15)× − − −

= 21 8 7 6 21 4 84× × × = × = cm2. ½

2. Area of equilateral triangle =

2316 3

4a× =

a2 = 4 × 16 ⇒ a = 8 cm

Thus, Perimeter = 3a = 3 × 8 = 24 cm. 1

3.

A

x

B C15

25

x =

2 225 15 40 10− = ×

x = 20 cm. ½

Area of right triangle =

12

× base × height

=

12

× 15 × 20 = 150 cm2. ½

4. Perimeter = 4x + 5x + 6x = 150 15x = 150 ⇒ x = 10 Sides are 40 cm, 50 cm and 60 cm. 1

5.

32

a. 1

6. Perimeter of an equilateral triangle = 3a = 60 m

a = 20 m

Area =

34

a2 =

34

× 20 × 20

= 100 3 m2. 1

7.

A

x

B C6

10

x = 2 210 6 8− = cm ½

Area of right triangle =

12

× base × height

=

12

× 6 × 8 = 24 cm2 ½

8.

A

x

B C

5 2

x

x2 + x2 = ( )25 2

2x2 = 25 × 2 x = 5 cm. 1

9. Perimeter of an equilateral triangle = 3a = 60 ⇒ a = 20 cm ½

Area =

23 320 20

4 4a = × × ½

= 100 3 cm2. 1 [CBSE Marking Scheme, 2012]

10. In right angled ∆ ABC, A

B Ca

a

AREAS

SECTION

BSECTIONCHAPTER

7


By Pythagoras theorem, AC2 = AB2+ BC2

=a2 + a2 = 2a2 1

⇒ AC = 2 a ½ Perimeter of ∆ABC = AB + BC + CA

= a + a + 2 a

= 2a + 2 a

= a (2 + 2 ) 1½


1.

12 2

xSquare

x

x2 + x2 = ( )212 2

⇒ 2x2 = 144×2 \ x = 12 cm Given, 3a = 4x

(a, be a side of triangle) ⇒ 3a = 4 ×12 \ a = 16 cm

Area of equilateral triangle =

23 316 16

4 4a = × ×

= 64 3 cm2. 1

2. Sides are 3x, 4x, 5x, then 3x + 4x + 5x = 36 ⇒ 12x = 36 ⇒ x = 3 \ Sides a, b, c are 9 cm, 12 cm, 15 cm. 1 3.

A

B C

4

4

90°

Area of ∆ABC = 12

×4×4 = 8 cm2. 1

4. Area of an equilateral triangle

=

234 4 3

4× =

cm2. 1

5. Area of an equilateral triangle

=

234

a

½

\

23

4a

= 81 3

1

⇒ a2 = 81 × 4 a = 9 × 2 = 18 cm Perimeter of equilateral triangle = 3a = 3 × 18 = 54 cm. ½ [CBSE Marking Scheme, 2012]

6. Using Pythagoras theorem, ½ (Hypotenuse)2 = (Base)2 + (Perpendicular)2 ⇒ 102 = (8)2 + (x)2

⇒ 100 – 64 = x2

x = 6 cm 1

\ Area =

12

× base × height ½

=

12

× 8 × 6

= 24 cm2 1 7. (i) Let the sides of the triangle be a, b and c. a : b : c = 12 : 17 : 25

12 17a b=

=

25c

k= (say)

⇒ a = 12k, b = 17k, c = 25k ½ Perimeter = 540 cm ⇒ a + b + c = 540 ⇒ 12k + 17k + 25k = 540 ⇒ 54k = 540

\

k =

54010

54= ½

Hence a = 12 × 10 = 120 cm, b = 17 × 10 = 170 cm, c = 20 × 10 = 250 cm

Now, s =

12

×540 = 270 cm

(s – a) = (270 – 120) cm = 150 cm (s – b) = (270 – 170) cm = 100 cm (s – c) = (270 – 250) cm = 20 cm ½

\ Area of triangle = ( – )( – )( – )s s a s b s c

= 270 150 100 20× × × cm2

= 100 27 15 10 2× × × cm2

= 100 3 3 3 3 5 2 5 2× × × × × × × cm2

= 100 × 3 × 3 × 5 × 2 = 9000 cm2 ½ (ii) Heron’s formula. ½ (iii) A balanced ratio leads to good result. ½



1.

A B

CD

aa

a

a

16

For ∆BCD, s =

168

2a a

a+ + = +

\ Area of ∆BCD

= ( 8)( 8 16)( 8 )( 8 )a a a a a a+ + − + − + −

= ( 8)( 8)(8)(8)a a+ −

= 28 64a − ½

\ Area of rhombus = 2 × Area of ∆BCD

⇒ 96 = 2 × 28 64a −

⇒ 6 = 2 64a −

⇒ 36 = a2 – 64 ⇒ a2 = 64 + 36 = 100 \ a = 10 cm. ½ 2.

A

x

B Cx

y

Area of an isosceles right triangle

=

12

× base × height

8 =

12

× x × x

⇒ x = 4 cm

\ y = 2 24 4 32+ = cm.

= 4 2 cm. 1

3. Let ‘b’ be the equal sides length and ‘a’ be the base.A

B CDa

b bx

a/2 a/2

x =

2 2 22 4

2 2a b a

b− − =

Area of triangle =

12

× base × height

=

12

× a × x

=

12

× a × 2 242

b a−

=

2 244a

b a−

1

4. (i) If a, b, c be the lengths of sides BC, CA and AB of

∆ABC and if s =12

(a + b + c) then,

Area of ∆ABC = ( – )( – )( – )s s a s b s c \ Here, a = b = c (= a) and

s =

12

(a + a + a) = 32a

½

s – a =

32 2a a

a− =

s – b =

32 2a a

a− =

s – c =

32 2a a

a− =

½

Area =

32 2 2 2a a a a

=

3

2 2a a

× ×

=

234a

. ...(1) 1

(ii) To find the area, when its perimeter = 180 cm2

Here, a + a + a = 180 ⇒ 3a = 180

⇒ a =

18060

3=

Hence, required area =

23(60)

4×

= 900 3 cm2. 1

(iii) Heron’s formula. ½ (iv) We should follow the traffic rules to save our

lives. ½


TOPIC-2Heron’s Formula


1. ∆ = ( )( )( )s s a s b s c− − −

where, 2s = a + b + c. 1

2.

34

.a2

1

3. s =

7 24 25

2

+ +

= 28 cm

Area = ( )( )( )28 28 - 7 28 - 24 28 - 25

= 28×21×4×3 = 84 cm2. 1

4. s =

18

2

a b c+ +⇒

=

13 14

2

c+ +

⇒ c = 36 – 27 = 9 cm. 1

5. s

=

8 7 5

2 2

a b c+ + + +=

= 10 cm. 1

6. s =

6+8+10= 12

2 cm.

Area = 12(12 - 6)(12 - 8)(12 - 10)

= 12×6×4×2 = 24 cm2

Cost of painting =

24×9= 2.16

100` . 1

7. a = 8 cm, b = 11 cm, Perimeter = 32 cm 1 \ c = 32 – (8 + 11) = 13 cm, s = 16

\ Area = 16(16 8)(16 11)(16 13)− − −

= 8 30 cm2 1

8. Third side = ( ) ( )2 2125 100− = 75 2

s = 150

Area of ∆ = 150 25 50 75× × × = 3750 m2

9.

5 12 16

20

B13A

D

C

3 For ∆ABD, s = 15

Area of ∆ABD = 15 2 3 10× × × = 30 cm2

For ∆BCD, s = 24

Area of ∆BCD = 24 4 12 8× × ×

= 96 cm2

Area of quadrilateral = 30 + 96 = 126 cm2. 10. Area of cloth required = 10 × Area of cloth for one

piece ½ Area of one piece of cloth is made by sides 60 cm, 60

cm and 20 cm is,

s =

60+60+202

= 70 cm ½

Area = 70(70 – 60)(70 – 60)(70 – 20)

= 70 10 10 50× × ×

= 7 10 10 10 5 10× × × × ×

= 10 × 10 × 35

= 100 35 cm2 1

Area of cloth required

= 10 × 100 × 35

= 1000 35 cm2 1


1. ( )( )( )s s a s b s c− − − 2. Heron’s Formula. 1 3. Sides are 16 cm, 16 cm, 16 cm

Semi, perimeter =

16 16 16

2 2

a b c+ + + +=

=

48= 24 cm

2 ½

\ Area of triangle = ( )( )( )s s a s b s c− − −


= 24(24 – 16)(24 – 16)(24 – 16)

= × × ×24 8 8 8

= × × × ×3 8 8 8 8

= 64 3 cm2

A

B C

16 cm 16 cm

16 cm ½

4.

s = 2

a b c+ +,

here a, b, c are the sides of a triangle.

s = 12 16 20 48

24cm2 2

+ += = ½

Area = ( – )( – )( – )s s a s b s c ½

=

24(24 – 12)(24 – 16)(24 – 20)

= 2 3 4 12 8 4× × × × ×

= 2 12 12 2 4 4× × × × ×

= 2×4×12 = 96 cm2 1

5. s =

2a b c+ +

=

70+80+902

= 120 cm ½

Area = ( )( )( )s s a s b s c− − − ½

= 120(120 – 70)(120 – 80)(120 – 90)

= 120 50 40 30× × ×

= 40 3 5 10 40 3 10× × × × × ×

= 40 × 10 × 3 × 5

= 1200 × 2.23

= 2676 cm2 1

6. Sides of a triangle are a = 6 cm, b = 6 cm, c = 4 cm

s =

6 6 4 168

2 2 2a b c+ + + += = = cm

½

A = ( )( )( )s s a s b s c− − − 1

= 8 2 2 4× × ×

= 8 2 cm2 ½

Area of two black triangles

= 8 2 ×2 = 16 2 cm2 ½

Area of two white triangles

= 16 2 cm2. ½


7. Area = ( )( )( )s s a s b s c− − − 1

A = 54 3 17 34 306× × × =

= 306 m2 1

No. of rose beds =

3066

= 51 1


s =

51 37 20 1082 2

+ + =

= 54 cm ½

Area = ( )( )( )s s a s b s c− − − ½

= 54(54 51) (54 37) (54 20)− − −

= 54 3 17 34× × × ½ = 306 m2 ½ No. of rose beds

=

Total area of triangular fieldArea occupied by each rose bed

½

=

30651

6= ½


1.

s =

120= 60

2

Sides = 5x + 12x + 13x = 120 30x = 120 x = 4 cm 5x = 20 cm, 12x = 48 cm, 13x = 52 cm.

Area of ∆ = ( )( )( )s s a s b s c− − − cm2

= 60 40 12 8 230400× × × =

= 480 cm2 2[CBSE Marking Scheme, 2012]


Let the sides are 5x, 12x and 13x, then Perimeter = 5x + 127 + 13x = 120 ⇒ 30x = 120


⇒ x = 4 cm ½ Then sides are 20 cm, 48 cm, 52 cm

s =

120= 60

2 cm

Area = ( )( )( )s s a s b s c− − − ½

= 60(60 20)(60 48)(60 52)− − −

= 60 40 12 8× × × ½

= 10 6 10 4 6 2 4 2× × × × × × ×

= 10 × 6 × 4 × 2 = 480 cm2. ½

2. a = 15 cm, b = 15 cm, c = 12 cm

s =

15 15 1221

2 2a b c+ + + += = cm

½

Area = ( )( )( )s s a s b s c− − −

= 21 (21 15)(21 15)(21 12)× − − − 1

= 21 6 6 9× × ×

= 18 21 cm2. ½

[CBSE Marking Scheme, 2012] 3. Perimeter of the triangle = 84 cm Ratio of its sides = 13 : 14 : 15 Sum of the ratio = 13 + 14 + 15 = 42 cm 1

\ a =

1342

× 84 = 13 × 2 = 26 cm

b =

1442

× 84 = 14 × 2 = 28 cm

c =

1542

× 84 = 15 × 2 = 30 cm

s =

26 28 302

+ +

s =

842

= 42 cm 1

Using Heron’s formula,

Area of ∆ = ( – )( – )( – )s s a s b s c

=

42(42 – 26) (42 – 28) (42 – 30) cm2 1

= 42 16 14 12× × × cm2

= 2 3 7 16 2 7 3 4× × × × × × × cm2

= 2 × 3 × 7 × 4 × 2 cm2 1

= 336 cm2. 4. For one identical triangular leaf, let a = 28 cm, b = 15 cm and c = 41 cm

Also, s =

28 15 412 2

a b c+ + + +=

=

842

= 42 cm 1

41 cm

28 cm

15 cm

Using Heron’s formula, Area of one triangular leaf

=

( – )( – )( – )s s a s b s c

= 42(42 – 28) (42 – 15) (42 – 41)

= 42 14 27 1× × × 1

= 3 14 14 3 9× × × × = 9 × 14 = 126 cm2 1 There are 6 leaves in a circle. So, total number of leaves in 2 circles = 2 × 6 = 12 \ Area of 12 leaves = (12 × 126) cm2 = 1512 cm2

Hence, total area to be painted red = 1512 cm2. 1

TOPIC-3Application of Heron’s Formula in finding Area of Quadrilaterals


1. In ∆ABC, s = 5 7 8

102

+ + = cm

Area of || gm ABCD

= 2×Area of triangle ABC

= 2 10(10 8)(10 7)(10 5)− − −

= 2 10 2 3 5× × ×

= 20 3 cm2 1


2.

50 mD C

BA

50 m50 m

50 m

80 m

Perimeter of rhombus = 4 × sides = 200 ⇒ side = 50 m ½

Area of ∆ABC = ( )( )( )s s a s b s c− − −

=

90 (90 50)(90 50)(90 80)× − − −

= 90 40 40 10× × × 1

= 1200 m2. ½ Area of rhombus = 2 × Area of ∆ABC = 2×1200 = 2400 m2. ½ [CBSE Marking Scheme, 2012]

3.

6 cm

D L C

BA M

O

8 cm

Diagonals bisect each other at O Draw, LOM⊥ AB and CD OL = OM = 3 cm Area of shaded region

OAB =

12

× 8 × 3

= 12 cm2. 2 [CBSE Marking Scheme, 2012]

Alternative Method : Diagonals bisect each other at O. Draw, LOM ⊥AB. OL = OM = 3 cm Area of shaded region OAB

=

12

× 8 × 3 = 12 cm2. 2

4. Rhombus,

D C

BA 13 cm

13 cm

13 cm13 cmO 12 cm

Perimeter = 52 cm 1

Side =

5213

4= cm

Diagonal = 24 cm 1 OB = OD = 12 cm

OA = 2 213 12 169 144 25 5− = − = = ½

Area of rhombus = 4 ×

12

× 5 × 12 ½

Area = 120 cm2 [CBSE Marking Scheme, 2012]

5. Let ABCD be the given field in the form of trapezium in which AB = 35 m, CD = 10 m, BC = 13 m, AD = 14 m and DC || AB ½

35 m

10 m

14 m 13 m

D C

BA E F

Through C, draw CE || DA and let it meet AB at E. Let h metres (CP) be the height of the trapezium. DC || AE and CE || DA \AECD is a parallelogram. AE = DC = 10 m and CE = DA = 14 m In ∆CEB, CB = 13 m, CE = 14 m ½ and BE = AB – AE = 35 – 10 = 25 m Let a = 14 m, b = 13 m and c = 25 m

Then, s =

2a b c+ +

=

14 13 252

+ +

= 26 m ½

Area of ∆CEB = ( – )( – )( – )s s a s b s c

= 26(26 – 14)(26 – 13)(26 – 25)

= 26 12 13 1× × ×

= 13 2 2 6 13× × × ×

= 226 6 m 1

Also, Area of ∆CEB =

12

× base × height

=

12

× BE × h

⇒ 12

× 25 × h = 226 6 m


\ h =

526

25m ½

Area of trapezium =

12

(DC+AB) × h

=

12

(10 + 35) ×

526

25

=

12

× 45 ×

526

25

=

926 6

5 ×

= 114.66 m2 (approx) 1


1. Area of trapezium ABCD = Area of rectangle AOCD + Area of ∆OBC In ∆OBC, by Phythagoras theorem BC2 = OC2 + OB2

⇒ 172 = OC2 + 82

⇒ OC2 = 289 – 64 = 225 \ OC = 15 cm ½ Area of rectangle AOCD = 6 × 15 = 90 cm2 ½

Area of ∆OBC =

12

× 8 × 15 = 60 cm2 ½

Area of Trapezium ABCD = 90 + 60 = 150 cm2 ½

2.

20 mD C

BA

20 m

20 m

24 m20 m

Semi-perimeter of triangle ABC

s =

2a b c+ +

=

20 20 24 642 2

+ + = = 32 cm ½

Area of ∆ABC = ( )( )( )s s a s b s c− − −

= 32(32 24)(32 20)(32 20)− − −

= 32 8 12 12× × ×

= 8 4 8 12 12× × × × = 8 × 2 × 12 = 192 cm2 1 \ Area of rhombus ABCD = 2 × Area of ∆ABC = 2 × 192 = 384 cm2. ½ 3. Draw a quadrilateral ABCD.

D C

A B

18 m

82 m

50 m

50 m

½ Here, ∠CBD = 90° Area of quadrilateral ABCD = Area of ∆BCD + Area

of ∆ABD ½ In ∆BCD, by phythogoras theorem, BC2 + BD2 = CD2

⇒ 182 + BD2 = 822

⇒ BD2 =822–182

= 6400 BD = 80 m

Area of ∆BCD =

12

× BC × BD ½

=

12

× 18 × 80 = 720 m2 ½

Semi-perimeter of ABD,

s =

50 50 802

+ +

= 90 m

Area = 90(90 – 50)(90 – 50)(90 – 80)

= 9 10 40 40 10× × × ×

= 3×10×40 = 1200 m2 ½ Area of quadrilateral ABCD = 720+1200 = 1920 m2 ½ 4. From Fig; Area of paper of each colour

=

12

× Area of kite.

Area of kite = Area of square + Area of

isosceles triangle. Area of square = (side)2

= (16)2

= 256 cm2

Area of isosceles triangle

= 2 24 –

4a

b a

1


Here, b = length of equal side, a = base of triangle.

=

2 244(6) – 4

4

=

4144 – 416

4

= 128

= 64 2 8 2× = cm2 1

Area of paper of each color

=

12

[Area of kite]

=

12

[256 + 8 2 ]

= 128 + 4 2

= 128 + 4×1.41 = 128 + 5.64 = 133.64 cm2 1

FORMATIVE ASSESSMENT WORKSHEET-76 Note : Students should do this activity themselves.

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SECTION CHAPTER B1 REAL NUMBERS · 2016-05-21 · REAL NUMBERS SECTION B CHAPTER 1 TOPIC-1 Rational Numbers SUMMATIVE ASSESSMENT WORKSHEET-1 Solutions 1. 58 1000 ... 2157 625 = 3.4512