f1398944352cMath IX Term 1 Solu

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CONTENTS

Unit-I : Number System

1. Real NumbersSummative Assessment Ø Worksheets 1 - 12 1 - 21

Formative Assessment Ø Worksheets 13 21 - 21

Unit-II : Algebra

2. PolynomialsSummative Assessment Ø Worksheets 14 - 27 42 - 42

Unit-III : Geometry

3. Introduction to Euclid's GeometrySummative Assessment Ø Worksheets 29 - 33 43 - 48

4. Line and Angles Summative Assessment Ø Worksheets 35 - 42 49 - 57

5. Triangle Summative Assessment Ø Worksheets 44 - 55 58 - 77

Unit-IV : Co-ordinate Geometry

6. C0-ordinate GeometrySummative Assessment Ø Worksheets 57 - 64 78 - 90

Unit-V : Mensuration

6. Areas Summative Assessment Ø Worksheets 66 - 71 91 - 101

1 REAL NUMBERS

SUMMATIVE ASSESSMENT WORKSHEET-1

Section A

1. 4 28 3 7÷ = 4 2 7 3 7× ÷ (1)

2.5x = 7P

⇒5 5

= 7P (1)

P = 25

7 7× = 257 (1)

Section B

3. x = 3 – 2 2

3 2 2( – ) ×

( )( )3 2 23 2 2

–3 2 2

= 3 + 2 2 (½)

+1 + 2 × x × 1

= 3 – 2 2 + 3 + 2 2 + 2

= 8 (½)

1= ± 2 2 (½)

13 41 1

3 35 8 27 +

13 41 1

3 33 3(2 ) (3 ) +

3 34 45(2 3) 5(5) + = (½+½)

4 14(5 ) 5 5.= = (½)

Section C

5. x = 1p

5 2 6+=

15 2 6( )+

× ( – )( – )5 2 65 2 6

LAER RBM EUN S

P-2 T I C S IX--M A T M AEH M - 1 SE RT

=5 2 625 24

––

= 5 – 2 6 (½)

p2 + x2 = (5 + 2 6 )2 + (5 – 2 6 )2

= (5)2 + (2 6 )2 + 2 × 5 × 2 6 + (5)2 + 2 62e j – 2 × 5 × 2 6 (½)

= 25 + 24 + 20 6 + 25 + 24 –20 6 (½)= 98. (½)

√2√3

–2 –1 O 1BA

P Q 2 3

Let AB = BC = 1 unit length.

Using Pythagoras theorem we see that

OC = 1 12 2+ = 2 (½)Construct CD = 1 unit length perpendicular to OC, then using Pythagoras theorem, we seethat

OD = 2 12 2e j + = 3 (½)

Using a compass with centre O and radius OD, draw an arc which intersects the number

line at the point Q, then corresponds to 3. (1)

Section D

7. Let x = 1·32 1·32222..............=10x = 13·222...........

100x = 132·222........... (½)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+ = + =119 3590 99

+ (½)

=119 11 35 10 1309 350 1659·

990 990 990× + × += = (½)

Section A

1. b2 = a ⇒ a b= . (1)

2. 2 2 2( )( ) ( ) ( )a b a b a b a b+ − = − = − (1)

3. 1 / 4 4 1 / 4 4 1 /4 1 14(81) 81 (3 ) (3 ) 3 3− − −× = × = × (1)

= 3º = 1

Section B

2 1+= 1 2 1

2 1 2 1−×

+ −(½)

= 2 1 2 12 1 1

− −=

−(½)

= 1·414 – 1

= 0·4145. The two irrational number between 0·5 and 0·55 are

0·51010010001000001............... (1)and 0·502002000200002............. (1)

Section C

6. 2x × (22)x = 3 1 / 3 5 1 / 5(2 ) (2 )× (½)2x · 22x = 21 × 21 (½)2x + 2x = 21 + 1 (½)

23x = 22 (½)Comparing powers on both sides, we get

3x = 2 (½)

x = 23 · (½)

7. Since LCM of 5 and 6 is 30,

∴1 1 56 6 5

= × = 5 5 25

30 5 150× = (½)

and1 1 65 5 6

= × = 6 5 30

30 5 150× = (½)

Hence, four rational numbers between 16 and

15 are :

26 27 28 29, , , ·150 150 150 150 (½+½+½+½)

Section D8. Given that xa = y, yb = z, zc = x

∴ xabc = (xa)bc (1)= (y)bc (½)

LAER RBM EUN S

= (yb)c (1)= (z)c (½)

xabc = x1 (½)Comparing powers on both sides, we get

abc = 1. (½)

9. x2 + y2 + xy =

2 23 1 3 1 3 1 3 13 1 3 1 3 1 3 1

+ − + −+ + × − + − + (1)

= 3 1 2 3 3 1 2 3 13 1 2 3 3 1 2 3

+ + + −+ ++ − + +

= ( )( )

( )( )

2 2 3 2 2 31

2 2 3 2 2 3

+ −+ +

− + (½)

= ( ) ( )

( )( )

2 22 3 2 3

12 3 2 3

+ + −+

− + (½)

= 4 3 4 3 4 3 4 3

+ + + + −+

−(½)

= 14 + 1 = 15. (½)

10. ( ) ( ) ( )2 2 22 2 5 3 2 3 2 1− + + − −

= ( ) ( ) ( )2 2 222 2 (5) 2 2 2 5 3 2 3 2 3 2 3+ − × × + + + × ×

( )2 22 (1) 2 2 1− − + × × (2)

= 8 25 20 2 18 3 6 6 2 1 2 2+ − + + + − − + (1)

= 51 18 2 6 6− + . (1)

Section A

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

2. Sometimes rational and sometimes irrtational. (1)3. x = 0·777..............

⇒ 10x = 7·777...... ⇒ 10x – x = 7·77........ – 0·777............

⇒ 9x = 7 ⇒ x = 79 (1)

Section B

4. (4 3 3 2) (4 3 3 2)+ × − (½)

= 2 2(4 3) (3 2)− (½)= 48 – 18 (½)= 30 (½)

5.3 / 4 3 / 281 25

− − ×

3 / 4 3 / 24 23 52 3

− − ×

= 3 33 5

− − ×

= 3 32 3

3 5 ×

1255= 1

Section C

6. (i) (xy + yx)–1 = (52 + 25)–1 (½)

= (25 + 32)–1 (½)

= (57)–1 = 157 (½)

(ii) (xx + yy)–1 = (55 + 22)–1 (½)= (3125 + 4)–1 (½)

= (3129)–1 = 13129 . (½)

7. x = ( 2 ) ( 2 ) 2 2 2

( 2 ) ( 2 )p q p q p q p q

p q p q+ + − + × + × −

+ − −1

x = 2 22 2 4

2 2p p qp q p q

+ −+ − +

⇒ x = 2 2( 4 )

p p qq

+ −1

⇒ 2qx = 2 24p p q+ − ⇒ 2qx – p = 2 24p q− 1

On squaring both sides,4q2x2 + p2 – 4pqx = p2 – 4q2

⇒ 4q (qx2 – px) = – 4q2

⇒ qx2 – px + q = 0.

Section D

8. a + b = 2 5 2 52 5 2 5

− +++ −

= ( ) ( )

( ) ( )

2 22 5 2 5

2 5 2 5

− + +

+ − (1)

= 4 5 4 5 4 5 4 54 5

+ − + + +− (1)

= 18 181 = −− (½)

∴ (a + b)3 = (– 18)3 = – 5832. (½)

LAER RBM EUN S

9.2 2 2

. .a b c c aa b b c

b c ax x xx x x

= ( ) ( ) ( )2 2 2 2 2 21 1 1

. .a b b c c a c aa b b cx x x− − − ++ + (1)

= 1 1 1( )( ) ( )( ) ( )( )( ) ( ) ( ). .

a b a b b c b c c a c aa b b c c ax x x− + × − + × − + ×+ + + (1)

= . .a b b c c ax x x− − − (½)

= a b b c c ax − + − + − (1)

= x0 = 1. (½)

10 20 40 5 80+ + − − = 15

10 2 5 2 10 5 4 5+ + − − (1)

3 10 3 5− (½)

= ( )( )( )

10 5153 10 5 10 5

− + (1)

= ( )5 10 510 5

× +−

= 5(3·2 2·2) 5·45+ = . (1)

Section A

1.3 / 4

1 / 416

16− = 3 /4 1 / 4 1(16) (16) 16+ = = 1

2. x = 2·9 2·99..........= ⇒ 10x = 29·99..... ⇒ 10x – x = 29·99.............. – 2·99.......... ½

⇒ 9x = 27 ⇒ x = 27 39

3. Sum of 2 5 and 3 7 = 2 5 3 7+ = 12·4 1

Section B

4. ( ) ( )+ +5 2 3 2 5 2 = ( )+ + +5 2 15 3 2 5 2 2 (1)

= ( )+5 2 17 8 2 (½)

= + ×85 2 40 2

= +85 2 80 . (½)

⇒ = 2157 3·4512625

Section C

6.5 25 2

⇒ 5 2 5 25 2 5 2

+ +×− ×

⇒ 5 2 2 5 2

5 2+ +

− ⇒

7 2 103

⇒ 7 2(3.162)

⇒ 7 6·324

⇒ 13·324

3 ⇒ 4·441 1

7. Mark the distance 9·3 units from a fixed point A on a givenline 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.

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 .

Section D

8.( )( )

39 3 27

−+ −× −

3 3(3 ) 3 (3 )

+ × −×

= 613 (1)

3 8n n n

+ + −× = 6

13 (1)

33 [3 1]

× = 613 (½)

3 3 83 8n m− × = 3– 6 (½)

LAER RBM EUN S

⋅625 ) 2157 (3 4512 1875 2820 2500 3200 3125 750 625 1250 1250 ×

Comparing powers on both sides, we get

3n – 3m = – 6 (½)

– 3(m – n) = – 6

m – n = 2. (½)

9. (i)0·714285...

7 5·000000491073028

2014605640355

∴57 = 0·714285 (1)

and0·81

11 9·0088

∴ 911 = 0·81 (½)

Thus, three different irrational numbers between 5 9and7 11 are 0·75075007500075000075.....,

0·767076700767000767.... and 0·80800800080000..... (½)(ii) Number system (½)(iii) Those who are irrational in there approach get fail in their efforts. (½)

Section A

1. 1 /2 1 /4 2[{(81) } ]− − = [{(81)–1/2 }]–1/2 = (81)1/4 = (34)1/4 = 3 1

2.150 =

1 1 2 2105 5 2 5 2 2

= × =× ×

So, rationalising factor is 2. ½

3.1500

2 15 =

1 1500 1 10100 52 15 2 2

= × = = 1

Section B4. LCM of 5 and 7 is 35,

∴ 35 = ×3 7

5 7 = 2135 (½)

and 57 = ×5 5

7 5 = 2535 (½)

so, < < < <21 22 23 24 2535 35 35 35 35 (½)

The required three rational numbers are 22 23 24, and35 35 35 . (½)

5. Given, a = 2 and b = 3.

∴ ab + ba = 23 + 32 (½)

= 8 + 9 (1)

= 17. (½)

Section C

6. Let x = 0·328 = 0·3282828......... (½)10x = 3·282828......... (½)

1000x = 328·282828......... (½)1000x – 10x = 328·2828........ – 3·2828........ (½)

990x = 325·00 (½)

x = 325990 = 65

198 . (½)

7.+ +−− +

5 3 5 37 4 3 7 4 3

= ( ) ( ) ( ) ( )

( ) ( )+ + − + −

5 3 7 4 3 5 3 7 4 3

7 4 3 7 4 3(1)

= + + + − + − +−

35 20 3 7 3 12 35 20 3 7 3 1249 48

= +40 3 241 (½)

= ( )+8 3 5 3 . (½)

Section D8. (1 + x + y–1)–1 + (1 + y + z–1)–1 + (1 + z + x–1)–1

= 1 1 11 1 11 1 1x y zy z x

+ ++ + + + + +

= + + = ⇒ =+ ++ + + +

∵1 1 1 1, ( 1 )1 1 111 1

xyz xyy xy zx

y xy x

= 1 1 11 1 1y xy y xy xy y

+ ++ + + + + + (1)

LAER RBM EUN S

1 1 1y xy

y xy y xy xy y+ ++ + + + + + (½)

= ( )( )

1y xyy xy

+ + . (½)

Value Based Question9. (i) Six rational number between 3 and 4 are 3·125, 3·225, 3·335, 3·445, 3·555, 3·625 1

(ii) Number system. 1(iii) Those who are rational in there approach get fails in their efforts. 1

Section A

1. We know that 5 2·236= the irrational number is 5. 1

2.1 / 2

4 3 2x

= ( ) ( )1 / 21 / 4 1 / 8

3 32 2x x

= 2 1 / 3 1 / 8 2 1 / 2 4 1/12[( ) ] ( )x x x= = ½

Section B

3. 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 = × × ×= =× × ×2196 2 3 3 122 122990 2 3 3 55 55 . (½)

4. Given, 17 = 0·142857142857......... (½)

27 = 0·285714285714........ (½)

Number can be 0·160160016000....... (1)

Section C

5.− −

−2 33 4

(216) (256)=

( ) ( )− −

2 33 43 4

6 4(1)

= − −−2 34 1

6 4 (½)

= 4 × 62 – 43 (½)= 144 – 64 (½)= 80. (½)

1 1a a

a b b aab ab

++ −

= b ba b b a

++ −

= ( ) ( )

( )( )b a b b b a

b a b a− + +

+ − 1

2 2b ab b ab

− + +

= – 2

a b−1

Section D

7.[ ] [ ] [ ]

2 2 2( ) ( ) ( )

4. .a b b c c a

= 2( ) 2( ) 2( )

4 4 4. .

a b b c c a

a b cx x x

= 2 2 2 2 2 2

4 4 4. .

a b b c c a

a b cx x x

= 2 2 2 2 2 2

a b b c c a

a b cx

x+ + + + +

+ + (1)

= 4 4 4

4 4 4 1a b c

a b cxx

+ + = . (1)

1 1a b b ax x− −+

1 1a b

b ax x

1 1b a a b

b ax x x x

++ + 1

b a a bx x

x x x x+

a bx x

9. ( ) ( )7 5

1 2 2 32 22 4 3 5

2 3 2 32 3 2 3

−− −

− −× ××

× ×= ( ) ( )

7 56 82 23 5

3 32 2

−× (1)

7 56 82 27 53 52 2

× ×−

× ×−× (1)

= 21 20

21 252 2

−× (½)

= 2 1 2 0 1

21 25 42 2 2

− −= (½ + ½)

= 3 × 22 = 12. (½)qq

LAER RBM EUN S

Section A

1.177 59 3 3413 59 7 7

×= =× 1

2. 0·3 0·4+ = 0·33... 0·44... 0·77+ = ½

Let, x = 0·7 ⇒ 10x = 7·77... ⇒ 10x – x = 7·77... – 0·77... ½

⇒ 9x = 7·0 ⇒ x = 79

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

Section B

4. ( )−

− ÷( )

ca b c b

ab a cx x

xx = −

− ÷ab ac bc

acba bcx x

xx (½)

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

5. − +50 98 162 = × × − × × + × × × ×5 5 2 7 7 2 3 3 3 3 2 (½)

= − + ×5 2 7 2 3 3 2 (½)

= − +2 2 9 2 (½)= 7 2 (½)

6. Given, a = 2 and b = 3.(A) (ab + ba)–1 = (23 + 32)–1 (½)

= (8 + 9)–1 = 17–1 = 117 (½)

(B) (aa + bb)–1` = (22 + 33)–1 (½)

= (4 + 27)–1 = 31–1 = 131 . (½)

7.12 = ×1 2

2 2 (rationalising) (½)

= = =2 1·414 0·7072 2

∴ + π12 = 0·707 + 3·141 = 3·848. (1)

Section C

8. x = +1 2

1x = ( )

( )( )

( )− −× = −+ −

2 1 2 112 12 1 2 1

= −2 1 (1)

1x x− = ( ) ( )+ − −1 2 2 1 (½)

= + − +1 2 2 1 = 2 (½)

∴ ( )−31x x = 23 = 8. (1)

9. + ++ + +1 1 2

1 2 2 3 3 5

( )( ) ( )

( )( )

− − −× + × + ×

+ − + − + −

2 1 3 2 5 31 1 22 1 2 1 3 2 3 2 5 3 5 3

= ( ) ( ) ( )− − −+ +− − −

2 1 3 2 2 5 32 1 3 2 5 3

= − + − + −2 1 3 2 5 3 (½)

= −5 1 . (½)

Section D

10. ( ) ( ) ( )11 1

. .a b cbc caab

c abx x x

x xx= ( ) ( ) ( )

1 1 1. .a b b c c aab bc cax x x− − − (1)

= . .a b b c c aab bc cax x x− − −

= a b b c c acaab bcx

− − −+ + (½)

= ac bc ab ca bc ababcx

− + − + − (1)

0abcx x= (½)

= 1. (½)qq

Section A1. n – 1 12. 1·011 1·011011011.....= ½

So, 1·011 is smallest ½

Section B3. 7x = 1

x = 17

(½)(7 10 0·14285773028

∴ x = 0·142857 . (½)

LAER RBM EUN S

4. (13 + 23 + 33)–3/2 = (1 + 8 + 27)–3/2 (½)

= (36)–3/2 (½)

= [(6)2]–3/2 = 6–3 (½)

= =31 1

2166 . (½)

5. 3 40 = 3 32 2 2 5 2 5× × × = (½)

3 320 = × × × × × × = × =3 3 32 2 2 2 2 2 5 2 2 5 4 5 (½)

∴ 3 3 40 – − = × − × −3 3 3 3 34 320 5 3 2 5 4 4 5 5 (½)

= − − = −3 3 3 36 5 16 5 5 11 5 . (½)

6.30 29 28

31 30 293 3 3

3 3 ( 1)(3 )

+ + −=

28 2 1

29 2 13 (3 3 1)

3 (3 3 1)

+ −1

= 9 3 1

3(9 3 1)+ ++ − ½

= 13 13

3 11 33=

Section C

7. Given, x = +3 8

∴ x2 = ( )+ = + + × ×2

3 8 9 8 2 3 8 (½)

= +17 12 2 (½)

⇒ 21

x = ( )

( )( )

17 12 2117 12 2 17 12 2

= −−

17 12 2289 288 = −17 12 2 (½)

Now, +22

1xx = + + −17 12 2 17 12 2 (½)

= 34. (½)

Section D8. 52x – 1 – 25x – 1 = 2500

⇒ 52x – 1 – [(5)2]x – 1 = 2500 ½⇒ 52x – 1 – 52x – 2 = 2500 ½⇒ 52x – 2 (5 – 1) = 2500 1⇒ 52x – 2 = 625 ⇒ 52x – 2 = 54 ½on comparing both sides, we get ½

2x – 2 = 4 ⇒ x = 3 19. Since L.C.M. of 3, 4, 6, 12 is 12,

∴1 4

3 3 42 2 ×= = 4 12 124122 2 16= = (1)

1 34 4 35 5 ×= =

3 12 123125 5 125= = (½)

1 26 6 27 7 ×= =

2 12 122127 7 49= = (½)

12 3 = 12 3 (½)

In descending order 12 12 12 12125, 49, 16, 3 (1)

i.e., 4 6 3 125, 7, 2, 3 . (½)

Section A

1.32 48 2( 8 12) 2( 2 3) 28 12 2( 2 3) ( 2 3)

+ + += = =+ + +

Section B

2. 32 3 3 2×⇒ 2(3)1/3 × 3(2)1/2 ½⇒ 21+1/2 × 31+1/3 ½⇒ 23/2 × 34/3 ½⇒ 6(72)1/6 ½

Section C

3. x = +9 4 5= + +5 4 4 5

= ( ) + + × ×2 25 (2) 2 2 5

x = ( )+2

x = +5 2 (1)

1x = ( )

( )( )

5 215 2 5 2

= − = −−5 2 5 25 4 (1)

∴ − 1xx = ( ) ( )+ − −5 2 5 2

= + − +5 2 5 2 = 4. (1)

4. ( ) ( )22 3·3 2x x

= ( )8116

( ) ( )−22 2.3 3x x

= ( )432 (1)

( ) −223

x x= ( )−42

LAER RBM EUN S

Comparing the power on both sides, we getx – 2x = – 4 (½)

– x = – 4⇒ x = 4. (½)

5.+−

5 35 3

= ( )( )

5 35 35 3 5 3 (1)

= + + × × +=−5 3 2 5 3 8 2 15

5 3 2 (1)

= ( )+= +

2 4 154 152

∴ + 15a b = +4 15Comparing on both sides, we get

a = 4, b = 1. (½)

6. +− +4 3

3 3 2 2 3 3 2 2=

( ) ( )( ) ( )

+ + −

4 3 3 2 2 3 3 3 2 2

3 3 2 2 3 3 2 2(½)

= + + −−

12 3 8 2 9 3 6 227 8

= +21 3 2 219 (½)

= × + ×21 1·732 2 1·41419 (½)

= +36·372 2·82819 (½)

= =39·2 2·063.19 (½)

Section D

7. 3 1 127 4 3

(2187) (256) (1331 )− − −− + = [ ]

3 1 127 4 34(2187) 5(256) 2 (1331)− + (½)

= ( ) ( ) ( )3 1 2

7 4 37 4 34 3 5 4 2 11− + (1)

= 4 × 33 – 5 × 4 + 2 × 112 (1)= 4 × 27 – 20 + 2 × 121 (1)= 108 – 20 + 242 = 330. (½)

Section A

1.4 3 2x

= 23 2 1 / 4 1 / 4 2 1 / 3 1 / 43( ) ( ) [( ) ]x x x = = ½

= (x2)1/12 = x1/6 ½

Section B

2. Let x = =0·237 0·237237237.....

1000x = 237·237237........ (½)

1000x – x = 237·237237........ – 0·237237..... (½)

999x = 237 (½)

x = 237999

. (½)

Section C

3. ( ) ( ) ( )+ − +

+93 1 1 12

3 12= ( ) ( )

( )( )

−− + − + ×

12 393 36 1 1212 3 12 3

= ( ) ( )−− + − + −

9 12 33 6 1 12 12 3

= − − + −3 5 12 12 3 (½)

= – 5. (½)4. (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 2a b c a b cx + + − − − 1= xº = 1 1

5. ( )− −÷2 1

43 2( )x y xy = ( )−

− ÷

2 11 1 13 242 2 2. ( )x y xy (½)

= − −÷1 1

23 4. ( )x y xy (½)

= − ×1 1

23 4. ( )x y xy (½)

= − + +1 1 123 4 4.x y (½)

= −1 912 4.x y (½)

x. (½)

6.−30

5 3 3 5= ( )

( )( )

5 3 3 5305 3 3 5 5 3 3 5

( ) ( )+

−2 2

30 5 3 3 5

5 3 3 5(1)

= ( )+−

30 5 3 3 575 45

= ( )+30 5 3 3 530

= +5 3 3 5 . (½)

LAER RBM EUN S

7. ( ) ( )−− −

− −× ××

7 51 2 2 32 2

2 4 3 55 7 5 75 7 5 7 = ( ) ( )−× ××

7 54 2 5 32 22 1 3 2

7 7 7 75 5 5 5

= ( ) ( )−×

7 56 82 23 5

7 75 5

= ( ) ( )− ×

1 17 52 26 8

3 57 75 5 (½)

= ( ) ( )−

−×1 1

42 402 221 25

7 75 5

= ( )×1

42 25 221 40

7 55 7

= ( )×1

2 4 27 5 (½)

= × 27 5 = 175. (½)

Section D

8. x = 3 2 2−

1x = ( )

( )( )3 2 21

3 2 2 3 2 2

− +(½)

( )3 2 29 8+

− = 3 2 2+ (½)

21x = ( )2

3 2 2 9 8 12 2+ = + + = 17 12 2+ (½)

x2 = ( )23 2 2 9 8 12 2− = + − = 17 12 2− (½)

Now, 44

− = ( ) ( )2 22 2

1 1x xx x

− + (½)

= ( ) ( ) ( ) ( )17 12 2 17 12 2 17 12 2 17 12 2 − − + − + + (½)

= ( ) ( )17 12 2 17 12 2 17 12 2 17 12 2− − − − + + (½)

= 24 2 34− × = 816 2− . (½)

Section A1. (16)1/4 – 6(343)1/3 + 18(243)1/5 – (196)1/2

⇒ (2)4/4 – 6(7)3/3 + 18 (3)5/5 – (14)2/2 ½⇒ 2 – 42 + 54 – 14 = 0 ½

Section B2. False.

Justification : 147 147 49 7 ,

75 25 575= = = 1

which is a rational number. 1

Section C

3. + −+ + +6 3 2 4 3

2 3 6 3 6 2

= ( )( )

( )( )

( )( ) ( )

( )( )

− − −× + × − ×

+ − + − + −

3 2 3 2 6 3 6 26 4 33 2 3 2 6 3 6 3 6 2 6 2 (1)

= − − −+ −− − −18 12 3 12 3 6 4 18 4 6

3 2 6 3 6 2 (1)

= ( ) ( )− −− + −

3 12 6 4 18 618 12 3 4

= − + − − +18 12 12 6 18 6 = 0. (½)

4. ( ) 1xab

−= ( )2 8xb

⇒ ( ) 1xab

−= ( ) (2 8)xa

b− −

Comparing powers on both sides, we getx – 1 = – (2x – 8) (½)

x – 1 + 2x – 8 = 0 (½)3x – 9 = 0 (½)

x = 3. (½)5. Since L.C.M. of 3, 2, 4 be 12,

∴ 3 44, 3, 6 = 1 1 13 2 44 , 3 , 6 (½)

= × × ×1 4 1 6 1 33 4 2 6 4 34 , 3 , 6 (½)

= × × ×1 1 14 6 312 12 124 , 3 , 6 (½)

= 1 1 112 12 12256 ,729 ,216 (½)

In ascending order = 1 1 112 12 12(216) ,(256) ,(729)

i.e. = 4 36, 4, 3 . (½ + ½)

Section D

6. x = ( )( )( )2 31

2 3 2 3

− + (½)

= 2 34 3+− = 2 3+ (½)

2x3 – 2x2 – 7x + 5 = ( ) ( ) ( )3 22 2 3 2 2 3 7 2 3 5+ − + − + + (½)

= ( ) ( ) ( )332 2 3 3 2 3 2 3 2 4 3 4 3 + + × × + − + + 14 7 3 5− − + .(1)

= 2 8 3 3 12 3 18 14 8 3 9 7 3 + + + − − − − (½)

= ( )2 26 15 3 23 15 3+ − − (½)

= 52 30 3 23 15 3+ − − = 29 15 3+ . (½)

LAER RBM EUN S

Section A

1. (4 3 3 2) (4 3 3 2)+ × − ½

2 2(4 3) (3 2) (16 3) (9 2) 48 18 30− = × − × = − = ½

Section B

2. 2(4 3 3 5)−

= (16 3) (9 5) 2 4 3 3 5× + × − × × × 1

= 48 45 24 15 93 24 15+ − = − 1

Section C3. (729)–1/6 ½

= 1 / 61

(729) ½

= 6 / 61

= 13 1

4. + + −5 2 6 8 2 15 = + + × + + − ×3 2 2 3 2 5 3 2 5 3 (½)

= ( ) ( ) ( ) ( )+ + × × + + − ×2 2 2 2

3 2 2 3 2 5 3 2 5 3 (1)

= ( ) ( )+ + −2 2

3 2 5 3 (½)

= + + −3 2 5 3 (½)

= +2 5 . (½)

3 / 4 3 / 2 381 2516 9

− − − 5 × ÷ 2 ½

3 / 4 3 / 2 316 981 25

2 × ÷ 5 ½

3 3 32 3 53 5 2

= 8 27 125

27 125 8 × ×

= 88 ½

= 1 ½

Section D

4 5+ = ( )( )( )

5 41 5 4 5 45 45 4 5 4

− −× = = −−+ − (½)

15 6+ = ( )

( )( )

6 51 6 5 6 56 56 5 6 5

− −× = = −−+ −(½)

16 7+ = ( )

( )( )

7 61 7 6 7 67 67 6 7 6

− −× = = −−+ − (½)

= ( )( )( )

( )8 7 8 71 8 78 78 7 8 7

− −× = = −−+ −

18 9+ = ( )

( )( )

9 81 9 8 9 89 89 8 9 8

− −× = = −−+ −(½)

L.H.S. = 5 4 6 5 7 6 8 7 9 8− + − + − + − + − (1)

= 4 9− + = – 2 + 3 = 1 = R.H.S. Proved (½)qq

FORMATIVE ASSESSMENT WORKSHEET-13

This worksheet contains Activities which are to be perform in the class itself.qq

LAER RBM EUN S

2 POLYNOMIALS

Section A1. p(x) = x2 + 11x + k

p(–4) = 0 ⇒ (– 4)2 + 11 × – 4 + k = 0 ½16 – 44 + k = 0 ⇒ k = 28 ½

2. No. of zeros of a cubic polynomial = 3 1

Section B

3. + +2 2 2a b c

bc ac ab=

+ +3 3 3a b cabc

= 3abcabc

+ + = ∴ + + =

∵3 3 3

a b ca b c abc (1)

= 3. (½)4. p(x) = kx2 – x – 4

(x + 1) is a factor of p(x), then p(– 1) = 0 (½)

∴ k (– 1)2 – (–1) – 4 = 0 (½)

k + 1 – 4 = 0 (½)

k = 3 (½)

Section C

5. + − −

3 5 3 5y yx x

= + − + + + − + + −

3 5 3 5 3 5 3 5 3 5 3 5y y y y y yx x x x x x

= + + × × + + − × × + −

2 2 22 2 222 2

5 9 25 3 5 9 25 3 5 9 25y y y y y yx x x x x

5 9 25y yx

2225 3 25y yx

6. x2 – y2 + xy = + − + −− + ×

− + − +

2 23 2 3 2 3 2 3 23 2 3 2 3 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)

Section D7. p(x) = x4 – 2x3 + 3x2 – px + 3p – 7

Put x + 1 = 0 or x = – 1 in p(x), we getp(– 1) = (– 1)4 – 2(– 1)3 + 3(– 1)2 – p(– 1) + 3p – 7 = 19 (½)

1 + 2 + 3 + p + 3p – 7 = 19 (½)4p – 1 = 19 (½)

4p = 20p = 5 (½)

∴ The polynomial p(x) = x4 – 2x3 + 3x2 – 5x + 15 – 7= x4 – 2x3 + 3x2 – 5x + 8 (½)

Put x + 2 = 0 or x = – 2 in p(x)p(– 2) = (– 2)4 – 2(– 2)3 + 3(– 2)2 – 5(– 2) + 8 (½)

= 16 + 16 + 12 + 10 + 8 (½)

= 62. (½)

Value Based Questions8. (i) (998)3 = (1000 – 2)3

We know that(a – b)3 = a3 – b3 – 3ab(a – b) (½)

Hence, (998)3 = (1000)3 – (2)3 – (3) (1000) (2) (1000 – 2) (½)= 1000000000 – 8 – 6000 × 998= 1000000000 – 5988008= 994011992. (1)

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

Section A1. x – 2 = 0 ⇒ x = 2

p(2) = 2 × (2)2 + 3 × 2 – k = 0 ½⇒ 8 + 6 – k = 0 ⇒ k = 14 ½

2. For zeros put p(x) = 0 ½x (x – 2) (x – 3) = 0,

then x = 0, 2, 3 ½

Section B

3. + − + +

2 42 4

1 1 1 1x x x x

x x x x = − + +

2 2 42 2 4

1 1 1x x x

x x x ,

as (a + b) (a – b) = a2 – b2 (1)

4 41 1

x xx x

− + (½)

= − 8

x. (½)

YLOP LIM AON S

4. − +

14 2a b

= + − +

14 2a b

= − + − + + × × − + × × + × ×

222(1) 2 2 (1) 2 (1)

4 2 4 2 2 4a b a b b a

= + + − − +2 2

1 .16 4 4 2a b ab a

b (½)

5. Let 14 = a, 13 = b, – 27 = ca + b + c = 14 + 13 – 27 = 0 (½)

∴ a3 + b3 + c3 = 3abc (1)(14)3 + (13)3 + (– 27)3 = 3 × 14 × 13 × (– 27)

= – 14742. (½)

Section C6. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) (1)

= 280 + 2 × 92

(a + b + c)2 = 280 + 9 = 289

a + b + c = =289 17 (½)

(a + b + c)3 = 173 = 4913. (1)7. p(x) = 3x2 – mx – nx

If (x – a) is a factor of p(x), thenp(a) = 0 (1)

3(a)2 – m × a – n × a = 0 (½)

0,a ≠ a[3a – m – n] = 0 (½)Since 3a – m – n = 0 (½)

∴ a = +3

m n. (½)

8. a7 + ab6 = a(a6 + b6) (½)= a[(a2)3 + (b2)3] (1)= a(a2 + b2) [(a2)2 + (b2)2 – a2 × b2] (1)= a(a2 + b2) (a4 + b4 – a2b2). (½)

Section D

= + + × ×22

1 12x x

x x(1)

= 7 + 2 = 9 (½)

= =9 3 (½)

33 1( )x

+ + × 2

1 1 1–x x x

x x x (1)

x= 3 × (7 – 1) (½)

= 3 × 6 = 18. (½)qq

Section A

1. Degree of a polynomial 3 is 0. 12. It is not defined. 1

Section B3. 1033 = (100 + 3)3 (½)

= 1003 + 33 + 3 × 100 × 3 (100 + 3) (½)

= 1000000 + 27 + 900 × 103

= 1000000 + 27 + 92700 (½)

= 1092727. (½)

4. f(x) = 2x3 – 3x2 + 7x – 6

Put x – 1 = 0 or x = 1 in f(x) (½)

f(1) = 2 × 13 – 3 × 12 + 7 × 1 – 6 (½)

= 2 – 3 + 7 – 6 (½)

= 0 (½)

Hence, (x – 1) is a factor of f(x).

Section C5. 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)] (1½)= (5x – 3x + z) [25x2 + 9y2 + z2 + 15xy + 3yz – 5xz] (½)

6. 3 – 12(a – b)2 = 3[1 – 4(a – b)2] (1)

= 3[(1)2 – {2(a – b)}2] (1)

= 3(1 + 2a – 2b) (1 – 2a + 2b). (1)7. 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)

Section D8. p(x) = x4 + 1

Put x – 1 = 0 or x = 1 in p(x)p(1) = (1)4 + 1 (2)

= 1 + 1 = 2Hence, we must be subtracted 2 from x4 + 1 so that it is exactly divisible by (x – 1). (1)Resultant polynomial for divisible by (x – 1)

= x4 + 1 – 2= x4 – 1. (1)

YLOP LIM AON S

Value Based Questions9. (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 (1)(ii) Factorisation of Polynomials (½)(iii) Expression of one’s desires and news is very necessary. (½)

Section A1. Let p(y) = 5y3 – 2y2 – 7y + 1 ½

then, remainder = p (0) = 0 – 0 – 0 + 1 = 1 ½

2.2 2x y

= – 1 ⇒ x2 + y2 = – xy ½

⇒ x2 + y2 + xy = 0 ½x3 – y3 = (x – y) (x2 + y2 + xy)

= (x – y) (0)= 0

Section B3. x2 – 9 = (97)2 – (3)2 ½

= (97 + 3) (97 – 3) ½= 100 × 94 ½= 9400 ½

4. 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.

Section C5. p(x) = x6 – ax5 + x4 – ax3 + 3x – a + 2

(x – a) is a factor of polynomial p(x), thenp(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. p(x) = 2x3 – 5x2 + px + bx = 2 and x = 0 are zeroes of polynomial, then

p(2) = 2 × 23 – 5 × 22 + p × 2 + b = 0 (½)

⇒ 16 – 20 + 2p + b = 0

⇒ 2p + b = 4 ...(i) (½)

and p(0) = 2 × 0 – 5 × 0 + p × 0 + b = 0 (½)

⇒ b = 0 ... (ii) (½)

From (i) and (ii), we get

2p + 0 = 4 (½)

p = 4/2 = 2.

then, p = 2 and b = 0. (½)

Section D7. Let q(x) = px2 + 5x + r

(x – 2) and ( )12x − are factors of q(x), then

q(2) = p(2)2 + 5 × 2 + r = 04p + 10 + r = 0 (1)

4p + r = – 10 ...(1) .

( )12q + = ( )21 15 02 2p r+ × + =

5 04 2p r+ + = (1)

54 2p r+ = − ...(2) .

Subtracting eqn. (1) from eqn. (2), we get

44p r p r+ − − = 5 102

p p−=

p−= 15

p = – 2From eqn. (1), 4 × (– 2) + r = – 10

r = – 10 + 8r = – 2 = p. (1)

Section A1. p(0) = 0 – 3 × 0 + 2 = 2 ½

p(2) = 22 – 3 × 2 + 2 = 4 – 6 + 2 = 0 ½p(0) + p(2) = 2 + 0 = 2

YLOP LIM AON S

2.1 32 2

x x + +

= 2 3 1 32 2 4

x x x+ + + ½

= 2 32

4x x+ + ½

Section B3. x4 – y4 = (x2)2 – (y2)2 (1)

= (x2 – y2) (x2 + y2) (½)

= (x – y) (x + y) (x2 + y2). (½)4. (x + 2y)2 = x2 + 4y2 + 2 × x × 2y (1)

= 17 + 4 × 2 (½)

= 17 + 8 = 25

(x + 2y) = =25 5 . (½)5. 249 × 251 = (250 – 1) (250 + 1) (½)

= (250)2 – (1)2 (½)= 62500 – 1 (½)= 62499. (½)

Section C

6. + − −

2 22 15 23 3

a a = + + − + − +

2 1 2 15 2 5 2

3 3 3 3a a a a (1)

= − + + +

2 1 2 17 3

3 3a a (1)

1 37 3

3 3a a (½)

17 (3 1)

3a a . (½)

7. a + b + c = 9 ½(a + b + c)2 = 92 ½

a2 + b2 + c2 + 2(ab + bc + ca) = 81 ½35 + 2(ab + bc + ca) = 81 ½

2(ab + bc + ca) = 81 – 35 = 46 ½ab + bc + ca = 46/2 ½ab + bc + ca = 23.

Section D8. (a + b + c)2 – (a – b – c)2 + 4b2 – 4c2

= (a + b + c + a – b – c)(a + b + c – a + b + c) + (2b)2 – (2c)2 (1)= 2a × (2b + 2c) + (2b – 2c)(2b + 2c) (1)= (2b + 2c)(2a + 2b – 2c) (1)= 2(b + c) × 2(a + b – c) (½)= 4(b + c)(a + b – c). (½)

9. Let p(x) = x3 + 2x2 – 5ax – 8

and g(x) = x3 + ax2 – 12x – 6

When p(x) and q(x) divided by (x – 2) and (x – 3) leave remainders p and q, then

p(2) = p and g(3) = q

p(2) = 23 + 2 × 22 – 5a × 2 – 8 (1½)

= 8 + 8 – 10a – 8

p(2) = 8 – 10a = p ...(1)

g(3) = 33 + a × 32 – 12 × 3 – 6

g(3) = 27 + 9a – 36 – 6

g(3) = – 15 + 9a = q (1½)

If q – p = 10

∴ – 15 + 9a – 8 + 10a = 10

19a – 23 = 10

19a = 33

a = 3319 . (1)

Section A

1. Coefficient of x in expression 2 72

+ − is 2π

2. Put x + 1 = 0 ⇒ x = – 1 in given expression, so ½f(– 1) = (– 1)11 + 101 = – 1 + 101 = 100 ½

Section B3. m (m – 1) – n (n – 1) = m2 – m – n2 + n (½)

= m2 – n2 – m + n= (m – n) (m + n) – (m – n) (1)= (m – n) (m + n – 1). (½)

4. −1

Squaring both sides, we get

= ( )23 (½)

+ − × ×22

1 12x x

x x= 3 (1)

x= 3 + 2 = 5. (½)

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

Put −12

x = 0 ⇒ =12

x in p(x) (½)

YLOP LIM AON S

Remainder =

p (½)

= + + +

3 21 1 1 12 2 2

= + + +1 1 1

18 4 2

= + + +1 2 4 8

. (½)

Section C6. p(x) = 3x3 – 5x2 + kx – 2

Remainder on dividing p(x) by (x + 2) is :p(– 2) = 3(– 2)3 – 5(– 2)2 + k(– 2) – 2 (½)

= – 24 – 20 – 2k – 2= – 46 – 2k ...(1) (½)

q(x) = – x3 – x2 + 7x + kRemainder on dividing q(x) by (x + 2) is :

q(– 2) = – (– 2)3 – (– 2)2 + 7(– 2) + k (½)= 8 – 4 – 14 + k= – 10 + k ...(2) (½)

Since remainders are same, then from eqs. (1) and (2), we have– 46 – 2k = – 10 + k (½)

3k = – 36k = – 12. (½)

7. x3 + y3 – 12xy + 64= (x)3 + (y)3 + (4)3 – 3 × (x) (y) (4) (1)= (x + y + 4) (x2 + y2 + 16 – xy – 4y – 4x) (1)

= 0 × (x2 + y2 + 16 – xy – 4y – 4x) = 0 (1)

8. a6 – b6 1

(a3)2 – (b3)2 1

⇒ (a3 + b3) (a3 – b3) 1

Section D9. 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)

10. (a) 4a2 – 9b2 – 2a – 3b= (2a)2 – (3b)2 – (2a + 3b) (½)

= (2a – 3b) (2a + 3b) – (2a + 3b) (1)

= (2a + 3b) (2a – 3b – 1) (½)

(b) a2 + b2 – 2(ab – ac + bc)= a2 + b2 – 2ab + 2ac – 2bc (½)= (a – b)2 + 2c(a – b) (½)= (a – b) [(a – b) + 2c]

= (a – b) (a – b + 2c). (1)qq

Section A1. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)

a3 + b3 + c3 – 3abc = 0, as a + b + c = 0 ½a3 + b3 + c3 = 3abc ½

2. (x – 2)3 = (x)3 – (2)3 – 3 × x × 2 (x – 2)= x3 – 8 – 6x2 + 12x ½

coefficient of x2 in the expension of (x – 2)3 = – 6. ½

Section B3. Let x2 + 7 = p and 2x – 1 = q, then given expression

= 12p2 – 8pq – 15q2

= 12p2 – 18pq + 10pq – 15q2 (½)= 6p(2p – 3q) + 5q(2p – 3q) (½)= (2p – 3q) (6p + 5q)= [2(x2 + 7) – 3(2x – 1)] [6(x2 + 7) + 5(2x – 1)] (½)= (2x2 + 14 – 6x + 3) (6x2 + 42 + 10x – 5)= (2x2 – 6x + 17) (6x2 + 10x + 37). (½)

Section C4. 2x3 + 2y3 + 2z3 – 6xyz= 2(x3 + y3 + z3 – 3xyz) (½)

= 2(x + y + z)(x2 + y2 + z2 – xy – yz – zx) (1)= (x + y + z)(2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx) (½)= (x + y + z)[x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx] (½)= (x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]. Hence Proved. (½)

5.( ) ( ) ( )

( ) ( ) ( )− + − + −

− + − + −

3 3 32 2 2 2 2 2

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 = 0Then a3 + b3 + c3 = 3abc (½)For Numerator a2 – b2 + b2 – c2 + c2 – a2 = 0For Denominator a – b + b – c + c – a = 0 (½)

∴( ) ( ) ( )

( ) ( ) ( )− + − + −

− + − + −

3 3 32 2 2 2 2 2

a b b c c a

a b b c c a=

( ) ( ) ( )( )

× − − −

− − −

2 2 2 2 2 233 ( ) ( )a b b c c a

a b b c b c (½)

= − + − + − +

− − −( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )a b a b b c b c c a c a

a b b c c a (½)

= (a + b) (b + c) (c + a). (½)

YLOP LIM AON S

6. p(x) = x3 + 3x2 – 2x + 4

p(2) = (2)3 + 3(2)2 – 2 × 2 + 4 (½)= 8 + 12 – 4 + 4 = 20 (½)

p(– 2) = (– 2)3 + 3(– 2)2 – 2(– 2) + 4 (½)= – 8 + 12 + 4 + 4 = 12 (½)

p(0) = 0 + 0 – 0 + 4 = 4 (½)p(2) + p(– 2) – p(0) = 20 + 12 – 4 = 28. (½)

Section D7. 2x3 + 2y3 + 2z3 – 6xyz= 2(x3 + y3 + z3 – 3xyz) (½)

= 2(x + y + z)(x2 + y2 + z2 – xy – yz – zx) (½)= (x + y + z)(2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx) (½)= (x + y + z)[x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx] (½)= (x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]. Hence Proved. (½)

Now, 2(13)3 + 2(14)3 + 2(15)3 – 6 × 13 × 14 × 15= (13 + 14 + 15)[(13 – 14)2 + (14 – 15)2 + (15 – 13)2] (½)= 42 × [1 + 1 + 4] (½)= 42 × 6 = 252. (½)

− + − + +

− +− + −

− +±∓

3 1(Quotient)3 2 3 8 3 2

3 2 – 3 2

0 (Remainder)

xx x x x x

x xx x (2)

Divisor × Quotient= (x2 – 3x + 2) (3x + 1) (1)= 3x3 – 8x2 + 3x + 2

∴ Dividend= Divisor × Quotient + Remainder (Division algorithm) (1)qq

Section A

+ = 21

= (4)2 – 2 = 16 – 2 = 14 1

2 283 17

83 83 17 17

− × +=

2 2(83 17)(83 83 17 17 )

(83 83 17 17 )

+ − × +

− × + = (83 + 17) = 100 1

Section B3. 2y3 + y2 – 2y – 1 = y2 (2y + 1) – 1 (2y + 1) 1

= (2y + 1) (y2 – 1) ½= (2y + 1) (y – 1) (y + 1) ½

4. x4 – 125xy3 = x(x3 – 125y3) (½)

= x[(x)3 – (5y)3]= x(x – 5y) [x2 + (5y)2 + x × 5y] (1)

= x(x – 5y) (x2 + 25y2 + 5xy). (½)

Section C

= (3)3 (1)

x3 + + × × +

31 1 13 x xx x x

= 27 ⇒ x3 + 31

x = 27 – 9 = 18 (2)

6. 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)

7. (a2 – 2a)2 – 23(a2 – 2a) + 120= (a2 – 2a)2 – 15(a2 – 2a) – 8(a2 – 2a) + 120 (½)

= (a2 – 2a)[a2 – 2a – 15] – 8[a2 – 2a – 15] (½)

= (a2 – 2a – 15)(a2 – 2a – 8) (½)

= (a2 – 5a + 3a – 15)(a2 – 4a + 2a – 8) (½)

= [a(a – 5) + 3(a – 5)][a(a – 4) + 2(a – 4)] (½)

= (a – 5)(a + 3)(a – 4)(a + 2). (½)

Section D

8. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) (1)(6)2 = a2 + b2 + c2 + 2 × 11 (½)

a2 + b2 + c2 = 36 – 22 = 14 (½)a3 + b3 + c3 – 3abc = (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)] (1)

= 6 × (14 – 11) = 6 × 3 = 18. (1)Alternative method.

a3 + b3 + c3 – 3abc = (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)] (1)= (a + b + c) [(a + b + c)2 – 3(ab + bc + ca)] (1)= 6[62 – 3(11)] = 6[36 – 33) = 6 × 3 = 18. (2)

9. (2x + 3y)3 = (2x)3 + (3y)3 + 3 × 2x × 3y(2x + 3y) (1)= 8x3 + 27y3 + 18(2x2y + 3xy2) (1)= 730 + 18 × 15 (½)= 730 + 270 (½)

(2x + 3y)3 = 1000 (½)

2x + 3y = 3 1000 = 10. (½)qq

YLOP LIM AON S

Section A1. Degree of x3 + 5 = 3, Degree of 4 – x5 = 5 ½

∴ Degree of (x3 + 5) (4 – x5) = 3 + 5 = 8 ½2. a = 3 + b ⇒ a – b = 3

(a – b)3 = 33 ⇒ a3 – b3 – 3ab (a – b) = 27 ½⇒ a3 – b3 – 9ab = 27 ½

Section B

3. +3 364 125x y = ( )+3

3(4 ) 5x y (½)

= ( ) ( ) + + − × 224 5 (4 ) 5 4 5x y x y x y (1)

= ( ) + + − 2 24 5 16 5 4 5x y x y xy . (½)

4. x4y4 – 256z4 = (x2y2)2 – (16z2)2 (½)

= ( ) ( )− +2 2 2 2 2 216 16x y z x y z (½)

= ( ) − + 2 2 2 2 2(4 ) 16xy z x y z (½)

= ( ) ( ) ( )− + +2 2 24 4 16xy z xy z x y z . (½)

Section C5. Let p(x) = 3x3 + x2 – 20x + 12

Given (3x – 2) is a factor of p(x).

3x – 2 ) 3x3 + x2 – 20x + 12 ( + −2 6x x 3x3 – 2x2

3x2 – 20x + 12 3x2 – 2x (1½) – 18x + 12 – 18x + 12 ×

x2 + x – 6 = x2 + 3x – 2x – 6 (½)= x(x + 3) – 2(x + 3) (½)= (x + 3) (x – 2) (½)

The other factors are (x – 2) and (x + 3).

Section D6.

3x + 4) 6x4 + 11x3 + 13x2 – 3x + 27 ( 2x3 + x2 + 3x – 5 quotient (½) 6x4 + 8x3

3x3 + 13x2 (½) 3x3 + 4x2 (½) 9x2 – 3x 9x2 + 12x (½)

– 15x + 27 – 15x – 20 (½)

47Thus, quotient = 2x3 + x2 + 3x – 5 and remainder = 47.

– –

7. R.H.S. = (x – y) (x2 + y2 + xy)= x3 + xy2 + x2y – x2y – y3 – xy2 (1)

= x3 – y3 = L.H.S. (1)

Now, 216x3 – 125y3 = (6x)3 – (5y)3 (½)

= (6x – 5y) [(6x)2 + (5y)2 + 6x × 5y] (1)

= (6x – 5y) (36x2 + 25y2 + 30xy]. (½)

Section A1. On expanding we get

⇒ 3 3 2 21 8 12 6

27 27 27 27x y xy x y− + − ½

3 2 28 4 227 27 9 9x y xy x y− + − ½

Section B2. (2x – 3y + z)2 = [2x + (– 3y) + z]2 (½)

= (2x)2 + (– 3y)2 + z2 + 2 × 2x × (– 3y) + 2 × (– 3y) × z + 2 × 2x × z (1)= 4x2 + 9y2 + z2 – 12xy – 6yz + 4xz. (½)

Section C

3. 3 21 9 127 216 2 4p p p− − + = ( ) ( )3 23 21 1 1(3 ) 3 (3 ) 3 36 6 6p p p− − × × + × (1½)

= − − × × −

33 1 1 1(3 ) 3 3 3

6 6 6p p p (½)

p . (1)

Section D4. Let p(x) = x2 + px + q and q(x) = x2 + mx + n

Since x + a is a factor of p(x) and q(x), then p(– a) = 0 and q( – a) = 0 (1)p(– a) = 0 ⇒ (– a)2 + p(– a) + q = 0

a2 – pa + q = 0 ...(1) (½)q(– a) = 0 ⇒ (– a)2 + m(– a) + n = 0

a2 – ma + n = 0 ...(2) (½)Subtracting eqn. (2) from eqn. (1), we get

a2 – pa + q – a2 + ma – n = 0 (1)

a(m – p) = n – q

a = n qm p

−−

YLOP LIM AON S

5. b = ( )

( )( )

−= ×

5 2 61 15 2 6 5 2 6a

= ( )−

= −−

5 2 65 2 6

25 24(½)

∴ a + b = + + − =5 2 6 5 2 5 10 (½)

Also b = 1 1aba

⇒ = (½)

a2 + b2 = (a + b)2 – 2ab (½)= (10)2 – 2 × 1= 100 – 2 = 98 (½)

(a3 + b3) = (a + b)3 – 3ab(a + b) (½)= (10)3 – 3 × 1 × 10= 1000 – 30 = 970. (½)

6. Volume of cuboid = 8x3 + 12x2 – 2x – 3= 4x2(2x + 3) – 1 (2x + 3) (½)= (2x + 3) (4x2 – 1) (½)= (2x + 3) (2x – 1) (2x + 1) (½)

The possible dimensions are (2x – 1), (2x + 1) and (2x + 3). (½)Verification : For x = 5Dimensions are (2 × 5 – 1), (2 × 5 + 1) and (2 × 5 + 3) i.e., 9, 11, 13 units. (½)

Volume = 9 × 11 × 13 = 1287 cubic units (½)Given volume = 8(5)3 + 12(5)2 – 2(5) – 3 (½)

= 1000 + 300 – 10 – 3= 1287 cubic units. (½)

7. x4 + 2x3y – 2xy3 – y4 = (x2)2 – (y2)2 + 2x3y – 2xy3 (½)= (x2 – y2) (x2 + y2) + 2xy(x2 – y2) (1)= (x2 – y2) (x2 + y2 + 2xy) (1)= (x – y) (x + y) (x + y)2 (1)= (x – y) (x + y)3. (½)

Section A1. 4x2 + 3x – 6

x + 1 ) 4x3 + 7x2 – 3x – 6 4x3 + 4x2

3x2 – 3x So, quotient is 4x2 + 3x – 6 1 3x2 + 3x – 6x – 6 – 6x – 6 ×

2.3 3 343

729p q + =

= + + −

2 2 77 499 81 9

pqpq p q ½

Section B

3.3 3 38 1 1

15 3 5 − −

⇒512 1 1

3375 27 125− − = 75

x½+½

⇒ on solving we get, x = 8 ½+½

Section C4. Factors of 12 = (+ 1, + 2, + 3, + 4, + 6, + 12) (½)

p(x) = 6x3 – 25x2 + 32x – 12p(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)

5. (i) + + − −22

1 22 2x x

+ + × × − −

22 1 1 2( ) 2 2x x x

x x x(1)

= + − +

21 12x xx x

= + + −

1 12x x

x x . (1)

(ii) x4 – x4 = (x2)2 – (y2)2

= (x2 + y2) (x2 – y2)= (x2 + y2) (x – y) (x + y)

Section D6. Let p(x) = ax4 + 2x3 – 3x2 + bx – 4

x2 – 4 or (x – 2) (x + 2) is a factor of p(x), then put x = 2 in p(x)p(2) = a(2)4 + 2(2)3 – 3(2)2 + b × 2 – 4 = 0

16a + 16 – 12 + 2b – 4 = 0 16a + 2b = 0 8a + b = 0 ...(1) (1)

Again put x = – 2 in p(x)p(– 2) = a(– 2)4 + 2(– 2)3 – 3(– 2)2 + b × – 2 – 4 = 0

16a – 16 – 12 – 2b – 4 = 016a – 2b = 32 (1)8a – b = 16 ...(2) .

Adding eqs. (1) and (2)16a = 16 ⇒ a = 1 (½)

− + − − −

−− +

+ − −−

−−

2 4 3 2

2 14 2 3 8 4

2 8 4 2 8

x xx x x x x

x x xx x

YLOP LIM AON S

By equation (1)8 × 1 + b = 0 ⇒ b = – 8 (½)

p(x) = x4 + 2x3 – 3x2 – 8x – 4= (x2 – 4) (x2 + 2x + 1) (1)

= (x – 2) (x + 2) (x + 1)2.7. Let f(x) = ax3 – 3x2 + 4

and g(x) = 2x3 – 5x + aWhen divided by (x – 2)

f(2) = p and g(2) = q (1½)f(2) = a × 23 – 3 × 22 + 4

p = 8a – 12 + 4p = 8a – 8 ...(1)

g(2) = 2 × 23 – 5 × 2 + aq = 16 – 10 + a (1½)q = 6 + a ...(2)

p – 2q = 4, (given)8a – 8 – 12 – 2a = 4

6a – 20 = 4a = 4. (1)

Section A1. The zero of a polynomial is

p(0) = 3(0)2 + 2(0) – 1 = – 1 12. 4x3 – x8 + 20 – 5x5

∴ degree = 8 1

Section B3. 8 – 27a3 – 36a + 54a2 1

(2)3 – (3a)3 – 3(2)2 (3a) + 3 (3a)2 (2) ½= (2 – 3a)3 ½

4. a2 + 4b2 + 9c2 + 4ab + 12bc + 6ac – a2 – 4b2 – 9c2 + 4ab – 12bc + 6ac – 6b2 – 9b 1= 8ab + 12ac – 6b2 – 9b 1

Section D5.

− −

+ + − −

+− −+

− −− −

3 61 4 3 10

3 3 10

3 36 106 6

x xx x x x

x xx xx x

= Remainder

Verification :Let f(x) = x3 + 4x2 – 3x – 10 (1)

By remainder theorem, remainder

f(– 1) = (– 1)3 + 4(– 1)2 – 3(– 1) – 10

= – 1 + 4 + 3 – 10 = – 4. (1)

6. Let p(x) = bx3 + 3x2 – 3Put x – 4 = 0 or x = 4 in p(x), we get

p(4) = b(4)3 + 3(4)2 – 3 = R1 (½)64b + 48 – 3 = R1

64b + 45 = R1 (½)Let q(x) = 2x3 – 5x + bAgain put x – 4 = 0 or x = 4 in q(x), we get

q(4) = 2(4)3 – 5(4) + b = R2 (½)= 128 – 20 + b = R2

108 + b = R2Given that 2R1 – R2 = 0 (½)⇒ 2(64b + 45) – (108 + b) = 0 (1)

128b + 90 – 108 – b = 0 (½)127b – 18 = 0

b = 18127

. (½)

7. (x + y)3 + (y + z)3 + (z + x)3 – 3(x + y)(y + z)(z + x)= (x + y + y + z + z + x) [(x + y)2 + (y + z)2 + (z + x)2

– (x + y)(y + z) – (y + z)(z + x) – (x + y)(z + x)]= 2(x + y + z)[x2 + y2 + 2xy + y2 + z2 + 2yz + z2 + x2 + 2zx – xy – xz – y2 – yz – yz – xy – z2 – zx – xz – x2 – yz – xy]= 2(x + y + z) [x2 + y2 + z2 – xy – yz – zx]= 2(x3 + y3 + z3 – 3xyz). (1+1+1+1)

Section A1. (x + y + z) (x2 + y2 + z2 + 2xy + 2yz + 2xz)

⇒ It is the expandable form of ½(x + y + z)3 ½

Section B2. (2x)3 + (3y)3 = (2x + 3y) (4x2 + 9y2 – 6xy)

= (8) (4x2 + 9y2 – 12) 1= 32x2 + 72y2 – 96 1

Section C3. (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)

YLOP LIM AON S

= (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)

Section D4. g(x) = x2 – 3x + 2

= x2 – 2x – x + 2= x(x – 2) – 1(x – 2)= (x – 2)(x – 1) (1)

Zero of x – 2 is 2Zero of x – 1 is 1

f(x) = 2x4 – 6x3 + 3x2 + 3x – 2f(2) = 2(24) – 6(23) + 3(22) + 3(2) – 2 (1)

= 32 – 48 + 12 + 6 – 2 = 0f(1) = 2(1)4 – 6(1)3 + 3(1)2 + 3(1) – 2 (1)

= 2 – 6 + 3 + 3 – 2 = 0⇒ (x – 1) and (x – 2) are factors of f(x).∴ f(x) is exactly divisible by g(x). (1)

5. (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)

6. x – 3 = 0 ⇒ x = 3 ½= a (3)3 + 4(3)2 + 3 (3) – 4 ½= 27a + 41 ...(1) 1= (3)3 – 4(3) + a= 15 + a ...(2) ½

As (1) = (2)∴ 27a + 41 = 15 + a ½

a ⇒ – 1 1qq

Section A1. ∴ Given x + 2 = 0 ⇒ x = – 2

3(– 2)3 – 5 (– 2)2 + k (– 2) – 2 = – (– 2)3 – (– 2)2 + 7 (– 2) + k– 24 – 20 – 2k – 2 = + 8 – 4 – 14 + k ½

k = – 12 ½

Section B2. (x + 2)2 + p2 + 2p(x + 2) = (x + 2)2 + (p)2 + 2 × (x + 2) × p (1)

= (x + 2 + p)2, [∵ a2 + b2 + 2ab = (a + b)2] (1)

Section C3. 250x3 – 432y3 = 2[125x3 – 216y3] (½)

= 2[(5x)3 – (6y)3] (½)

= 2(5x – 6y) [(5x)2 + (6y)2 + 5x × 6y] (1½)

= 2(5x – 6y) (25x2 + 36y2 + 30xy). (½)

4. p(x) = x3 – 3x2 – 9x – 5Put x = – 1

p(– 1) = (– 1)3 – 3(– 1)2 – 9(– 1) – 5 (½)= – 1 – 3 + 9 – 5= 0 (½)

Hence, (x + 1) is a factor of p(x), thenx3 – 3x2 – 9x – 5 = x2(x + 1) – 4x(x + 1) – 5(x + 1) (½)

= (x + 1) (x2 – 4x – 5) (½)= (x + 1) (x2 – 5x + x – 5)= (x + 1) [x(x – 5) + 1(x – 5)] (½)= (x + 1) (x – 5) (x + 1). (½)

Section D5. (a + b)3 + (a – b)3 + 6a(a2 – b2) = (a + b)3 + (a – b)3 + 3 × 2a (a – b)(a + b) (1)

= (a + b)3 + (a – b)3 + 3(a + b)(a – b)[(a + b) + (a – b)] (1)= [(a + b) + (a – b)]3 (1)= (2a)3

= 8a3 (1)

6. p3 – q3 = (p – q)(p2 + q2 + pq) (1)

= (p – q)[(p – q)2 + 3pq] (1)

= + ×

210 10 539 9 3 (1)

10 1005

10 100 4059 81

= 5050729

7. Let p(x) = 2x3 – 5x2 + x + 2Put x = 1 in p(x)

p(1) = 2(1)3 – 5(1)2 + 1 + 2= 2 – 5 + 1 + 2 = 0 (1)

Hence, (x – 1) is a factor of p(x).

YLOP LIM AON S

Put x = 2 in p(x)

p(2) = 2(2)3 – 5(2)2 + 2 + 2

= 16 – 20 + 2 + 2 = 0 (1)Hence, (x – 2) is a factor of p(x).

Put x = 12

− in p(x) (1)

p − = 2 × − − − + − +

3 21 1 15 22 2 2

= − × − × − +1 1 1

2 5 28 4 2

= − − − +1 5 1

24 4 2

= − − − +=

1 5 2 80

Hence, (2x + 1) is a factor of p (x).

8. p(x) = x3 – 2x2 – 5x + 6Put x = 1 in p(x)

p(1) = (1)3 – 2(1)2 – 5 × 1 + 6 (1)= 1 – 2 – 5 + 6= 0

Hence, (x – 1) is a factor of p(x), thenx3 – 2x2 – 5x + 6 = x2(x – 1) – x(x – 1) – 6(x – 1) (1)

= (x – 1)(x2 – x – 6) (½)= (x – 1)(x2 – 3x + 2x – 6) (½)= (x – 1)[x(x – 3) + 2(x – 3)] (½)

= (x – 1)(x – 3)(x + 2). (½)

Section AThis worksheet contains Activities which are to be perform in the class itself.

3 INTRODUCTION TO EUCLID’S GEOMETRY

Section A1. AB + BC = AC

x + 3 + 2x = 4x – 5⇒ 3x + 3 = 4x – 5 ½⇒ 4x – 3x = 3 + 5 ⇒ x = 8 ½

Section B2. AC = BD (given)

AB + BC = BC + CD 1AB = CD 1

3. (a) Infinite if they are collinear. 1(b) Only one if they are non-collinear. 1

Section C4. Given,

AC = DC and CB = CE 1AC + CB = CD + CE 1

⇒ AB = DE 15. Since, x + y = 10 and x = z,

∵ x + y = z + y 1⇒ 10 = z + y 1Hence, z + y = 10. 1

Section D6. To prove : ∠A > ∠C 1

Cons. : Join ACProof : In ∆ DAC, CD > AD

∠1 > ∠3 ...(i) 1In ∆ ABC : BC > AB

∠2 > ∠4 ...(ii) 1Adding equations (1) and (2), we get

∠1 + ∠2 > ∠3 + ∠4∠A > ∠C 1

Section A1. Dimensions of surface = Length and Breadth which is 2. 1

RTNI ICU TDO O EOTN ’IL DCU S .. .

Section B2. AC = DC

CB = CE (given)Adding AC + CB = DC + CE ½

AB = DE ½If equals are added to equals, the wholes also equal. 1

P BLet AB be a ⊥ to a line L and AP is any other line segment.In right ∆ ABP, ∠B > ∠P (∵ ∠B = 90º) 1⇒ AP > AB or AB < AP 1

Section C4. Since, ∠1 = ∠3 and ∠2 = ∠4, therefore

⇒ ∠1 + ∠2 = ∠3 + ∠4 1⇒ ∠BAD = ∠BCD 1⇒ ∠A = ∠C 1

Section D5. (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 fourthpostulate).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 equalto the same thing are equal to one another.’’) (2)

Section A1. Only one line can pass through two distinct point : 1

Section B2. 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 betweenthem. (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)

3. AB = BC (given)∵ AX + BX = BY + CYSince, BX = BY ½∴ AX + BY = BY + CY ½

AX = CYAxiom : If equals are subtracted from the equals, the remainders are equals. 1

Section C

4. Here, OX = 12 XY, PX = 1

2 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)

Section D5. (i) False : Because infinitely many lines can pass through a single point. (½)

(ii) False : Because only one line can pass through twodistinct points.

A B (1)(iii) True : Because a terminated line i.e., line segment AB can

be produced indefinitely on both the sides (Euclid’spostulate 2).

A B (1)(iv) True : Two circles are equal if :

(a) their circumferences are equal, or (1)(b) their radii are equal.

(v) True : Things which are equal to the same thing are equal to one another (Euclid’s axioms).(½)

Value based question6. (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 fourthpostulate).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 areequal to the same thing are equal to one another.’’) (1)

(ii) Introduction to Euclid’s Geometry. (½)

(iii) Equality leads to democracy. (½)qq

Section A1. To form a straight line. 1

Section B

2. Axion : If C be mid-point of line AB, then AC = 12

AC = 12

AB ⇒ AD = 12

⇒ AD = 1 12 2

⇒ AD = 14

3. x – 15 = 25On adding 15 to both sides, we have

x – 15 + 15 = 25 + 15 (½)x = 40 (½)

Euclid’s Axiom : If equals are added to equals, the wholes are equal. (1)4.

Let AB has 2 mid points, say X and Y, then (½)

2AB = AX and 2

AB = AYY (½)

∴ AX= AY (Things which are equal to the same thing are equal to one another) (½)

Hence, X and Y coincide. Thus, every line segment has one and only one mid point. (½)

Section C5. Given, ∠ABC = ∠ACB

⇒ ∠1 + ∠4 = ∠2 + ∠3 (1)⇒ ∠1 + ∠4 – ∠4 = ∠2 + ∠3 – ∠3 (As ∠3 = ∠4) (1)⇒ ∠1 = ∠2. (1)

6. AB = 2AE (E is the mid point of AB) (1)CD = 2DF (F is the mid point of CD) (1)

Also AE = DF (Given)Therefore AB = CD (Things which are double of the same thing are equal to one another) (2)

Value based question7. (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 two right angles then by Euclid’s fifth postulate the lines will not meet on thisside of l. Next, we know that the sum of the interior angles on the other side of line l will also betwo right angles, Therefore, they will not meet on the other side also. So, the lines m and n nevermeet and are, therefore, parallel. (2)

(ii) Introduction to Euclid’s Geometry. (½)

(iii) Universal truth. (½)qq

Section A1. Only one line can pass through two distinct points. 1

Section B2. AB = AD

AC = AD ½AB = AC ½

Axiom : Things which are equal to the same thing are equal. 13. Given : ∠1 = ∠4 and ∠3 = ∠2

Again ∠2 = ∠4 ½Axiom : Things which are equal to the same thing are equal to one another, 1∴ ∠1 = ∠3 ½

4. 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)

Section C5. Given, C is the mid-point of AB.

∴ AC = CB …(1) (1)If possible, let D be another mid-point of AB.∴ AD = DB …(2) (1)A B CD

Now, subtracting (1) from (2), we getAD – 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)

6. Given,: AC = BC (1)

A B C (1)

So, AC + AC = AC + BC (Equals are added to equals)⇒ 2AC = AB, (... AC + CB coincides with AB)

∴ AC = 12

AB. (1)

Section D7. Given : Three lines l, m and n in a plane such that l || m and m || n. (1)

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.

8. ∠1 = ∠2 ...(1) (½)∠3 = ∠4 ...(2) (½)

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)

(B) Match1. D2. E3. A4. F5. C6. H7. G8. B

4 LINES AND ANGLES

Section A1. Let ∠a = 2x, ∠b = 3x

Then 2x + 3x = 5x = 180° (½)x = 36°

∠a = 2x = 72°∠h + ∠a = 180° (Corresponding exterior angles) (½)

∠h = 180° – 72° = 108°.

Section B

2.43 of a right angle =

4 90º 120º3

× = 1

Supplement of 120º = 180º – 120º = 60º. 1

Section C3. ∠BMN + ∠DNM = 180°, (Interior angles) (1)

⇒ 2 2BMN DNM∠ ∠+ = 180

2° ⇒ ∠1 + ∠2 = 90° (1)

∠1 + ∠2 + ∠3 = 180° ⇒ ∠3 = 90°.

PM and PN intersect at right angle. (1)

Section D4. 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° – yz = 175 – 65°

z = 110°. (½)

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Value based question

5. (i) ∠TOP = 12 ∠TOB (½)

∠ORS = 12 ∠ORD (½)

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)

(ii) Lines and angles. (½)

(iii) Bisectoring among human beings gives rise to deterioration in the society. (½)qq

Section A1. In ∆ABC, ∠A + ∠B + ∠C = 180º

Also, in ∆ DEF, ∠D + ∠E + ∠F = 180º ½∴ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360º

= 4 × 90º ⇒ k = 4 ½

Section B2. In ∆ ABC, ∠A + ∠B + ∠C = 180º

2x – 5º + 5x + 5º + 3x + 50º = 180º10x + 50º = 180º ⇒ x = 13º 1

∠A = 2x – 5º = 26º – 5º = 21∠B = 5x + 5º = 65º + 5º = 70º 1

∠C = 3x + 50º = 39º + 50º = 89º

Section C3. Given, ∠POR : ∠ROQ = 5 : 7,

Let ∠POR = 5x∠ROQ = 7x ⇒ 5x + 7x = 180º ⇒ 12x = 180º 1

x = 15º∠POR = d = b = 5x = 5 × 15º = 75º 1∠ROQ = c = a = 7x = 7 × 25º = 105º 1

Section D4. 4b + 75º + b = 180º

5b = 180º – 75º = 105º (Linear pair) 1

b = 105º 21º

4b = 4 × 21º = 84º ⇒ a = 4b = 84º 12c + a = 180º

⇒ 2c = 180º – 84º = 96º ⇒ c = 48º 1

Value based question5. (i) Two plane mirrors PQ and RS are placed parallel to each other i.e. PQ || RS. An incident ray

AB after reflection takes the path BC and CD.BN and CM are the normals to the plane mirrors PQ and RS respectively. (½)

Since BN ⊥ PQ, CM ⊥ RS and PQ || RS (½)∴ BN ⊥ RS ⇒ BN || CM (½)Thus BN and CM are two parallel lines and transversal BC cuts them at B and C respectively.∴ ∠2 = ∠3But ∠1 = ∠2 and ∠3 = ∠4∴ ∠1 + ∠2 = ∠2 + ∠2 and ∠3 + ∠4 = ∠3 + ∠3 (½)⇒ ∠1 + ∠2 = 2 (∠2) and ∠3 + ∠4 = 2(∠3)⇒ ∠1 + ∠2 = ∠3 + ∠4 (½)⇒ ∠ABC = ∠BCD.Thus, lines AB and CD are intersected by transversal BC, such that

∠ABC = ∠BCDi.e. alternate interior angles are equal.Therefore, AB || CD (½)(ii) Lines and angles (½)(iii) Similarity leads to unanimity. (½)

Section A1. No, The following statement is not correct because ∠2 and ∠8 are not supplementary. 1

Section B2. ∠1 = ∠x = 75º (Alternate angles) ½

∠2 = 180º – x = 180º – 75º = 105º 1

∠2 = 105º = 75º + 30º = 175º 90º3

+ × ½

ENIL NDN AAS G SEL

Section C3. Since AC || DE, ∠ACB = ∠DEC ½

∴ y = 55º (Corresponding angles) 1

∠ABC = 180º – (70º + 55º) = 55º ½

∴ x = ∠ABC = 55º (Corresponding angles) 1

Section D4. y + 20º = 58º (Cross angle)

y = 58º – 20º = 38º 1½

∠PRQ = 180º – (58º + 22º) (Linear pair)

= 180º – 80º = 100º (Angle sum property) 1½

x = 180º – (100º + 38º)

= 180º – 138º = 42º 1qq

Section A1. x + 10º + x + x + 20º = 180º

3x = 180º – 30º = 150º ½

x = 150º 50º

Section B2. ∠FEC + ∠DCE = 155º + 25º = 180º ½

(Angle on the same side of transversal)⇒ EF || CD ½

∠BCD = ∠BCE + ∠ECD

= 35º + 25º = 60º = ∠ABC (Alternate angles) ½∴ AB || CD

So, AB || CD || EF

Thus, AB || EF ½

Section C3. 5x – 20º + 2x – 10º = 180º

7x = 180º + 30º = 210º

x = 30º 1

y = 180º – (5x – 20º) = 180º – (150º – 20º)

y = 180º – 130º = 50º 1

z = 2x – 10º

= 60º – 10º = 50º 1

Section D4. ∠ACD = ∠ABC = x (say) ½

∠BCD = ∠CPD = y (say) ½

∠APC = x + y (Exterior angle) 1½

∠ACP = x + y 1

∠ACP = ∠APC ½

Section A

1. x + 50º + 90º = 180º ⇒ x = 40º ½

q = x = 40º ½

Section B

2. ∠a = 2x and ∠b = 3x

∠a + ∠b = 90º

⇒ 2x + 3x = 90º

x = 18º

a = 2 × 18º = 36º ½

b = 3 × 18º = 54º ½

c = 180º – b = 180º – 54º ½

= 126º ½

Section C3. Draw AS ⊥ to PQ

Draw ⊥ from A to M

In ∆ PSA

∠ PSA + ∠SAP + ∠ APS = 180°

90° + ∠SAP + 15° = 180° (1)

∠ SAP = 75°

∠ BAS = 75° + 35°

= 110°

∠ NMA = 90°

∠ BMA = 90° – 10°

= 80° (1)

(Exterior angle) ∠ BAS = ∠ ABM + ∠ AMB

110° = x + 80°

x = 30° (1)

ENIL NDN AAS G SEL

Section D

4. x = 42° (Alternate angles) (1)∠1 = 24° + 42° = 66° (Exterior Angle) (½)

y = ∠1 = 66° (∵ CE = AC) (1)z = 180° – (x + y) = 180° – 108° (1)

= 72°. (½)qq

Section A1. AB = AC

∴ ∠C = ∠BThen, ∠A + ∠B + ∠C = 180º ½

50º + ∠B + ∠B = 180º ½2∠B = 130º ⇒ ∠B = 65

Section B2. Let x = 3k, y = 2k

Then, x + y = 3k + 2k = 180º ½(Angle on the same side of transversed)

5k = 180ºk = 36º

∴ x = 3k = 108º (½)y = 2k = 72º (½)

Thus, ∠z = ∠x = 108°. (½)

Section C3.

∠ACD = ∠A + ∠B (Exterior angle is the sum of opp. interior angle) (½)

12 ext. ∠ACD = 2 2

∠ ∠+A B

∠2 = ∠1 + 12 ∠A ...(1) (½)

Also, ∠2 = ∠1 + ∠E(Exterior angle is the sum of opp. interior angle) ...(2) (½)From eqs. (1) and (2), we get (½)

∴ ∠1 + 2A∠ = ∠1 + ∠E (1)

∠E = 2A∠ . Proved.

Section D4. Since, AE || DC

∠D + ∠1 = 180° (Angles on the same side of transversal)∠1 = 180° – 110° = 70° (1)

∠4 = ∠1 = 70° (Alternate angles) (½)Again, 97° + ∠2 = 180° (Angle on the same side of transversal) (½)

∠2 = 180° – 97° = 83°∠3 = ∠2 = 83° (Vertically opp. angles) (½)

In ∆BEF,∠3 + ∠4 + ∠EBF = 180° (Angle sum property) (1)

83° + 70° + ∠EBF = 180°∠EBF = 180° – 153°∠EBF = 27°. (½)

Section A1. AB = AC ⇒ ∠ACB = ∠B ⇒ 80º + ∠ACB + ∠B = 180º ½

2 ∠ACB = 100 ⇒ ∠ACB = 50º, Again, ∠ACB + x = 180º ½50º + x = 180º ⇒ x = 130º

Section B2. 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°x + y = 180°

⇒ AOB is a straight line. (½)

Section C3. EF ⊥ CD ⇒ ∠CFE = 90° (1)

90° + z = ∠CFGz = 130° – 90°

= 40° (1)x = ∠CFG (Alt. int. angles)

= 130°x + y = 180° (Linear pair) (1)

130° + y = 180°y = 50°.

4. In ∆ABC,∠A + ∠B + ∠C = 180° (½)

∠A + ∠C = 180° – ∠B = 180° – 90° = 90° (½)

12 (∠A + ∠C) = 45° (½)

ENIL NDN AAS G SEL

In ∆AOC,12 ∠A + 1

2 ∠C + ∠AOC = 180° (½)

45° + ∠AOC = 180° (½)∠AOC = 180° – 45° = 135°. (½)

Section D5. 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°. Hence proved.qq

Section A1. Since, 110º + x + 120º = 360º ½

∴ x = 360º – 230º = 130º ½

Section B2. 3z – 42° = 2z + 13° (Alternate interior angles)

z = 42° + 13° (½)z = 55° (½)

∠DNM = 2z + 13° = 110° + 13° = 123° (½)∠CNM = 180° – ∠DNM

= 180° – 123° = 57°. (½)

Section C3. y = 88° (Corresponding angles) (½)

a = 180° – 88° = 92° (Linear Pair) (½)

b = 180° – 110° = 70° (Linear Pair) (½)

x = 180° – (92 + 70)°, (Angle sum property) (½)

= 180° – 162° (½)

= 18°. (½)

4. Given AB || CD, CD || EF, EA⊥AB. (½)∠BEF = 55°

y = 180° – 55° = 125° (Co-interior angles) (½)x = y = 125° (Corresponding angles) (½)

EA ⊥ AB∴ z = 90° – 55° = 35°. (½)

Section D5. Ext. angle ∠PBC = x + z

2∠OBC = x + z (½)Ext. angle ∠QCB = x + y

2∠OCB = x + y (½)∠BOC + ∠OCB + ∠OBC = 180° (Angle sum property) (1)

2∠BOC + 2∠OCB + 2∠OBC = 360°2∠BOC + x + y + x + z = 360°

2∠BOC + 180° + x = 360°2∠BOC = 180° – x° (1)

∠BOC = 90° – 12 x. (1)

This worksheet contains activities which due to be perform in the class itself.qq

ENIL NDN AAS G SEL

5 TRIANGLES

Section A1. ∠ABC = 180º – 125º = 55º ½

AB = AC ½So, ∠ABC = ∠ACB = 55°In ∆ABC ∠A + ∠B + ∠C = 180°

x + 55° + 55° = 180°x = 70°

Section B2. In ∆PQS and ∆PRS

PQ = PR (given) ½

PS = PS (Common)

∠PSQ = ∠PSR = 90º ½

By R.H.S. rule, ∆PQS ≅ ∆PRS ½

⇒ ∆QPS = ∆RPS

Hence, PS bisects ∠P (by c.p.c.t) ½

3. In ∆ABP and ∆ACP, AB = AC (Given) (½)AP = AP (Comman)

∠APB = ∠APC = 90º (AP ⊥ BC) (½)By RHS rule, ∆ABP ≅ ∆ACP (½)⇒ ∠B = ∠C (½)

Section C4. AB = AC (½)

∴ ∠B = ∠C = xBA = BC

∴ ∠A = ∠C = xAlso, AC = BC (½)

∠A = ∠B = xBut ∠A + ∠B + ∠C = 180° (1)

3x = 180°x = 60°

∴ ∠A = ∠B = ∠C = 60°. (1)

5. 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 + SRQS + SR < PQ + PR. (½)

Section D6. ∠XAK + ∠KAH = 180° (Linear pair) .

∠KAH = 180° – 137° = 43°AB = AC (1)

∴ ∠ABC = ∠ACB∠ABC + ∠ACB = 137°, (ext. angle)

∴ ∠ABC = ∠ACB = 1372

° = 68·5° (½)

CH = CB (Given) .⇒ ∠CBA = ∠CHB = 68·5° (½)∴ ∠HCB = 180° – 137° = 43° (1)

∠CHK = ∠HCB = 43°. (Alternate angles) (1)

Value based questions7. (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 ABHence, CD bisects AB. (1)(ii) Congruency of triangles. (½)(iii) Equality is the sign of democracy. (½)

Section A1. By SAS or side Angle side congruency. 1

Section B2. AX || BY

⇒ ∠BAX = ∠ABY (Alternate angles) ...(1) (½)∠AXY = ∠BYX (Alternate angles) ...(2) (½)

In ∆PAX and ∆PBY, AX = BY (Given) ...(3) (½)From (1), (2) and (3), we get

∆APX ≅ ∆BPY, (By ASA) Proved. (½)

AIRT EL SGN

3. 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)

Section C4. In ∆ABC, AB > BC ⇒ ∠C > ∠A ...(1) .

AB > AC ⇒ ∠C > ∠B ...(2) (1)On adding (1) and (2), we get

2∠C > ∠A + ∠B2∠C + ∠C > ∠A + ∠B + ∠C (1)

3∠C > 180°∠C > 60°. (1)

5. Since AB||DC ∠ABP = ∠CDP (Alternate interior angles) (1)∠APB = ∠CPD (Vertical opposite angles) .

PD = PB (Given) (1)∆APB ≅ ∆CPD (By ASA) .

AP = PC (By c.p.c.t) (1)

⇒ P is also the mid-point of AC.

Section D6. Construction : Take BD = AB. Join AD (1)

Proof : In ∆ABD,AB = BD ⇒ ∠1 = ∠3 and ∠4 > ∠1 ⇒ ∠4 > ∠3 (1)In ∆ADC, ∠3 > ∠2∴ ∠4 > ∠3 > ∠2⇒ ∠4 > ∠2. (1)⇒ AC > DC (1)⇒ AC > BC – BD

AC > BC – AB∴ BC – AB < AC

Similarly, AC – AB < BC

and BC – AC < AB.

Value based question7. (i) l and m are two parallel lines intersected by another pair of parallel lines p and q.

AD || BC

and AB || CD. (½)

⇒ ABCD is a parallelogram.i.e. AB = CDand BC = AD (½)

Now in ∆ABC and ∆CDA, we haveAB = CDBC = AD

and AC = AC∴ We have ∆ABC ≅ ∆CDA (By SSS criterion of congruence) (1)(ii) Congruency of triangles. (½)

(iii) Equality is the sign of democracy. (½)qq

Section A1. AD = BC (given)

∠BAD = ∠ABC (given)AB = AB (common) ½

∴ ∆DAB ≅ ∆CBA (by SAS)∴ ∠BDA = ∠ACB (by c.p.c.t) ½

Section B2. In ∆PAB and ∆PDC, PA = PD

AB = CD 1½by RHS ∆PAB ≅ ∆PCD ½

PB = PC ⇒ ∠PCB = ∠PBC Proved3. In ∆ABY and ∆ACX

AB = AC (P is mid point of AD)AY = AX (Given) 1∠A = ∠A (Common)

∴ By SAS, ∆ABY ∆ ≅ ACX Proved 1

Section C4. PQ = PR ⇒ ∠PQR = ∠PRQ (Angles opp. to equal sides) (½)

ST||QR ⇒ ∠PST = ∠PQR (Corresponding angles) (1)∠PTS = ∠PRQ (Corresponding angles) (½)

∴ ∠PST = ∠PTS (½)⇒ PS = PT. (Sides opp. to equal angles) (½)

5. ∠DBC = ∠DCB (Given)

∴ DC = DB(Sides opp. to equal angles are equal) ...(i) (½)

In ∆ABD and ∆ACD,AB = AC (Given)

BD = CD [from (i)] (½)

AD = AD (Common) .∆ABD ≅ ∆ACD (SSS) (1)

∠BAD = ∠CAD (By c.p.c.t.) (½)

Hence, AD is bisector of ∠BAC. Proved. (½)

AIRT EL SGN

Section D6. AD = BD

⇒ ∠ABD = ∠DAB = 59°(Angles opp. to equal sides are equal) (1)

In ∆ABD,

59° + 59° + ∠ADB = 180°

∠ADB = 180° – 118° = 62° (1)

∠ACD = 62° – 32° = 30° (Exterior angle is equal to 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)7. Given, ∠BAB = ∠EAC

Adding ∠CAD on both sides, we get

∠BAD + ∠DAC = ∠EAC + ∠DAC (1)

∠BAC = ∠EAD

In ∆ABC and ∆ADE,

AB = AD (Given)

∠BAC = ∠EAD (Proved)

AC = AE (given)

∴ ∆ABC ≅ ∆ADE (By SAS) (2)

⇒ BC = DE (By PCS) (1)qq

Section A1. OA = OB

OP = OP ½∠AOP = ∠BOP ½∆OAP ≅ ∆OBP

Section B2. In ∆BED and ∆CFD,

∠DEB = ∠DFC = 90° (½)

BD = DC (D is the mid point)

ED = FD (Given)

∴ ∆BED ≅ ∆CEF (By RHS) (1)

∠B = ∠C. (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. Proved. (By c.p.c.t.) (½)

Section C4. PQRS is a square.

(i) SRT is an equilateral triangle.

∴ ∠PSR = 90°, ∠TSR = 60° (1)

⇒ ∠PSR + ∠TSR = 150°. Similarly ∠QRT = 150°

∆PST and ∆QRT, we have PS = QR (S)

∠PST = ∠QRT = 150° (A) and ST = RT (S) (1)

By SAS, ∆PST ≅ ∆QRT

⇒ PT = QT (By c.p.c.t.) Proved.(ii) In ∆TQR, QR = RT (Square and equilateral are ∆ on same base) (½)

⇒ ∠TQR = ∠QTR = x (½)

∴ x + x + ∠QRT = 180°2x + 150°= 180° ⇒ 2x = 30° ∴ x = 15°. Proved.

5. ∠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 (½)

∴ ∆BAE ≅ ∆BCD (By AAS) (½)∴ AE = CD. (By c.p.c.t.) Proved. (½)

Section D6. Join OD and OB.

In ∆AOB and ∆COB,

AIRT EL SGN

AO = CO (Given) .OB = OB (Common) .AB = BC (∵ ABCD is a rhombus) .

∆AOB ≅ ∆COB (SSS Congruence.) .∠AOB = ∠COB (By c.p.c.t.) ...(1) (1)

Similarly, ∆AOD ≅ ∆COD (½)∠AOD = ∠COD (By c.p.c.t.) ...(2) (½)

But, ∠AOD + ∠COD + ∠COB + ∠AOB = 360° (½)2(∠AOD + ∠AOB) = 360° (½)

∠AOD + ∠AOB = 180° (½)⇒ D, O, B are collinear. (½)

Section A1. ∠ADB = ∠ACB = 90º (given)

AD = BC (given) (½)AB = AB (common)

∴ ∆ABD ≅ ∆BAC (by R.H.S.) (½)

Section B2. Let ∠DAC be ∠3,

∠1 = ∠2 (Given) (½)

∠1 + ∠3 = ∠2 + ∠3

∠BAC = ∠EAD ...(i) (½)Given that, BD = CE

BD + DC = CE + DC (½)BC = DE ...(ii) .

∠B = ∠E (Given) ...(iii) .

From (i), (ii) and (iii), we get∆ABC ≅ ∆AED. (By AAS rule) (½)

Section C3. Given, ∠B > ∠A and ∠C > ∠D,

∴ OA > OB ...(1)

OD> OC (Side opposite to greater angle is longer) ...(2) (1½) Adding (1) and (2), we get

OA + OD > OB + OCAD > BC. (1½)

4. AB = AC 2 2AB AC⇒ = (½)

⇒ AE = AF,Since E and F are 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)

Section D5. Proof : In ∆ADE, we have

AD = AE (½)∠ADE = ∠AED

180° – ∠ADE = 180° – ∠AED (½)∠ADB = ∠AEC ...(i) (½)

Consider ∆ABD and ∆ACEAD = AE

∠ADB = ∠AEC From (i) .BD = EC (1½)

By SAS congruence, ∆ABD ≅ ∆ACEBy c.p.c.t., AB = AC (1)∴ ∆ABC is an isosceles triangle.

6. ∠FEC = ∠ECD = 10° (Alternate angles are equal) (½)∠BCA = 90°

∠BCD + ∠DCA = 90°10° + ∠DCA = 90° (1)

∠DCA = 80°AC = DC

∠DAC = ∠ADC = 180 802

° − ° = 50° (½)

In ∆BAC,∠A + ∠B + ∠BCA = 180°

50° + ∠B + 90° = 180°∠B = 40° (1)

In ∆BDE,∠B + ∠BED + ∠BDE = 180°

40° + 90° + ∠BDE = 180°∠BDE = 50°. (1)

Section A1. AB = AC

⇒ ∠C = ∠B ½∠A + ∠B + ∠C = 180°

⇒ 90° + ∠B + ∠B = 180° ⇒ 2∠B = 90° ∴ ∠B = 45°. ½

AIRT EL SGN

Section B2. (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 pair ofalternate angles.Hence, AD || BC. (1)

Section C

3. ∠1 + ∠5 = 180° = ∠2 + ∠6 ⇒ ∠1 = ∠2 (∵ ∠5 = ∠6) (1)

In ∆CAP and ∆BAP,

∠1 = ∠2 (Proved)

∠3 = ∠4 (AD is bisector of ∠BAC)

AP = AP

∴ ∆CAP ≅ ∆BAP (By AAS) (1)

⇒ CP = BP. (By c.p.c.t.) Proved. (1)

4. BC = DE (Given)Add CD both side

BC + CD = CD + DEBD = CD (1)

In ∆ABD and ∆FECAB = EF (Given)BD = CD (1)

∠ABD = ∠FEC = 90°So, By RHS esiteria

∆ABD ≅ ∠FEC (Proved) (1)

Section D

5. In ∆ALB and ∆NCMAL = CN (given)BL = CN (given)

∴ ∆ALB ≅ ∆NCM (By SAS)AB = NM∠A = ∠N (c.p.c.t.) (½)

∴ In ∆ABC and ∆NML,Now, AL = CN⇒ AL + LC = LC + CN

AC = LN (½)

AB = NM (Proved)AC = LN (Proved)

LA = LN (Proved) (1)∴ ∆ABC ≅ ∆NML. (SSS) (½)

6. (i) AB = AC, BD = CD, AD = DA (1)∆ABD ≅ ∆ACD (By SSS) ...(i)

∴ ∠BAD = ∠CAD (By c.p.c.t.) .(ii) AB = AC, ∠BAP = ∠CAP, AP = AP

∆ABP ≅ ∆ACP (By SAS) (1)∴ BP = CP (By c.p.c.t.) .(iii) From (i), ∠BAD = ∠CAD⇒ AP is the bisector of ∠A (1)

BD = CD, BP = CP, DP = DP∆BDP ≅ ∆CDP ⇒ ∠BDP = ∠CDP

DP is the bisector of ∠D. (1)qq

Section A

1. Let the angles of triangle are 5x, 3x and 7x, then5x + 3x + 7x = 180°

15x = 180°x = 12° (½)

∴ Angles are 60°, 36°, 84° ∵ Each angle is less than 90°

∴ The triangle is an acute angled triangle. (½)

Section B

2. AB = AD ⇒ ∠ABD = ∠ADB ...(i) (½)BC = CD ⇒ ∠CBD = ∠CDB ...(ii) (½)

Adding eq. (i) and (ii), we get∠ABD + ∠CBD = ∠ADB + ∠CDB (½)

⇒ ∠ABC = ∠ADC. Proved. (½)

Section C3. In ∆PAM and ∆QBM, (½)

AM = BM (Given)

∠1 = ∠2 (V.O.A.) (½)

∠3 = ∠4 = 90°

∴ ∆PAM ≅ ∆QBM (By ASA Cong.) (1½)

∴ PA = BQ. (By c.p.c.t.) Proved. (½)

AIRT EL SGN

4. (i) In ∆ABC and ∆DCB,AB = DC (Given)

∠ABC = ∠DCB (Given) (1)BC = CB (Common)

∴ ∆ABC ≅ ∆DCB (By SAS) (1)(ii) ⇒ AC = DB. (By c.p.c.t.) (1)

Section D5. (i) In ∆ABD and ∆BAC

∠DAB = ∠CBA (Given) .AB = BA (Comman) (1)AD = BC (Given) .

∴ ∆ABD ≅ ∆BAC (By SAS) Proved. (1)(ii) ∴ BD = AC (By c.p.c.t.) Proved. (1)(iii) ∠ABD = ∠BAC. (By c.p.c.t.) Proved. (1)

6. Proof : We are given two triangles ABC and PQR in which :∠B = ∠Q, ∠C = ∠R

and BC = QRWe need to prove that ∆ABC ≅ ∠PQR (1)There are three cases.

Case I : Let AB = PQIn ∆ABC and ∆PQR,

∠B = ∠Q (Given) .BC = QR (Given) .AB = PQ (Assumed) .

∴ ∆ABC ≅ ∆PQR (By SAS rule) (1)Case II : Suppose AB ≠ PQ and AB < PQTake a point S on PQ such that QS = ABJoin 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 = ∠QRSWhich is impossible unless ray RS coincides with RP.∴ AB must be equal to PQ. (½)So, ∆ABC ≅ ∆PQRCase III : If AB > PQ.We can choose a point T on AB such that TB = PQ and repeating the arrangements as given inCase II, we can conclude that AB = PQ and so,

∆ABC ≅ ∆PQR (½)qq

Section A1. Let the angle of triangle are 3x, 4x and 3x

then 3x + 4x + 3x = 180º10x = 180º

x = 180º 18º10

So, Angles are 3 × 18 = 54º4 × 18 = 72º and 54º ½

Section B2. RP = RQ (Given) .

⇒ ∠RQP = ∠RPQ (Proved) (½)⇒ ∠MQP = ∠NPQ

In ∆PQN and ∆QPM,

∠NPQ = ∠MQP (½)

PN = QM (Given) .

PQ = PQ (Common) .∴ ∆PQN ≅ ∆QPM (By SAS) (½)

⇒ ∠PQN = ∠QPM (By c.p.c.t.) .

⇒ ∠PQO = ∠QPO (½)

⇒ OP = OQ. (Sides opposite to equal angles are equal) Proved.

Section C3. Proof : In ∆ABC and ∆BAD,

BC = AD (Given)∠CBA = ∠BAD (Given)

AB = AB (2)∴ ∆ABC ≅ ∆BAD (By SAS cong.)

∴ BD = AC (By c.p.c.t.) (1)4. In ∆ABC,

AB + BC > AC ...(1) (1)

AIRT EL SGN

In ∆ACD,CD + DA > AC ...(2) (1)

Adding (1) and (2), we getCD + DA + AB + BC > 2AC. (1)

Section D5. In ∆RTS, RT = ST

∠TSR = ∠TRS (Angles opposite to equal sides are equal) ...(1) (½)∠1 = ∠4 (V.O.A) .

⇒ 2∠2 = 2∠3⇒ ∠2 = ∠3 ...(2) (½)Subtract eqn. (2) from eqn. (1), we get

∠TRS – ∠2 = ∠TSR – ∠3∠TRB = ∠TSA (1)

In ∆RBT and ∆SAT,∠RTB = ∠STA (Common angles) .

RT = ST (Given) .∠TRB = ∠TSA (Proved) (1)

∴ ∆RBT ≅ ∆SAT (By ASA congruence) .⇒ RB = AS (By c.p.c.t.) (1)

6. ∠PQT = ∠RQU and ∠TQS = ∠UQS∴ ∠PQT + ∠TQS = ∠RQU + ∠UQS (1)⇒ ∠PQS = ∠RQS

PQ = RQ, QS = QS (1)∴ ∆PQS ≅ ∆RQS (By SAS) (½)∴ ∠P = ∠R (By c.p.c.t.) .Again, in ∆PQT and ∆RQU,

∠P = ∠R∠PQT = ∠RQU

PQ = QR∴ ∆PQT ≅ ∆RQU (By ASA) (1)

∴ QT = QU. (By c.p.c.t.) (½)qq

Section A1. Then, if AC = BC (given an isosceles triangle)

So, Acc. to SAS propertythe angles B and C are also equal.i.e., ∠B = ∠CSo, Let x be the angle,

x + x + 90º = 180º ½x = 45º

So, ∠B = 45º ½A

Section B2. ∠AOB = ∠COD (given)

∠AOB – ∠COB = ∠COD – ∠COB 1In ∆ AOC and ∆BOD

AO = OBOC = OD

∠AOC = ∠BOD∴ ∆AOC ≅ ∆BOD (SAS) (½)

AC = BD (By CPCS) (½)

Section C3. Proof : In ∆PQR,

PQ = PR∠PQR = ∠PRQ (Angles opp. to equal sides are equal) ...(i) (1)

In ∆PQS, ∠PQR > ∠PSQ (Ext. angle of a ∆ is greater than .each of interior opp. angles) .(½)

∴ ∠PRQ > ∠PSQ, using (i) (½)⇒ ∠PRS > ∠PSR ⇒ PS > PR (½)

PS > PQ (∵ PR = PQ) (Side opp. to greater angle is larger) (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)∠DAB = ∠CBA. (By c.p.c.t.).

Section D5. AL = CN (Given) .

AL + LC = CN + LCi.e. AC = LNIn ∆ALB and ∆NCM, (1)

AL = CN (Given) .∠ALB = ∠NCM = 90°

BL = CM (Given) .∴ ∆ALB ≅ ∆NCM (SAS) .⇒ AB = NM (c.p.c.t.) (1)

In ∆BLC and ∆MCL,BL = MC (Given) .

∠BLC = ∠MCL = 90°

LC = CL∴ ∆BLC ≅ ∆MCL (By SAS) .

AIRT EL SGN

⇒ BC = ML (c.p.c.t.) (1)∴ In ∆ABC and ∆NML,

AB = NM (Proved) .AC = NL (Proved) .BC = ML (Proved) .

∴ ∆ABC ≅ ∆NML. (SSS) (1)6. ∵ ∠1 + ∠3 = 180° (Linear pair)

∵ ∠2 + ∠4 = 180° (Linear pair)∵ ∠1 + ∠3 = ∠2 + ∠4

But ∠1 = ∠2⇒ ∠3 = ∠4

In ∆OAC and ∆ODB, (1)∠3 = ∠4

∠AOC = ∠DOB (V.O.A.) (½)

OA = OD (Given) (½)

∴ ∆OAC ≅ ∆ODB (ASA) (1)

OC = OB (c.p.c.t.) (½)

⇒ ∆OCB is an isosceles triangle (½)qq

Section A1. ∠ABC = 180° – 125° = 55° (½)

AB = AC

⇒ ∠ACB = ∠ABC = 55°

∴ x = 180° – 55° – 55° = 70°. (½)

Section B2. 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.) (½)∴ BD = AE. Proved. (½)

Section C3. In ∆ABD and ∆CDB,

AB = CD (Given) (1)∠ABD = ∠CDB (Given)

BD = DB (Common) (1)∴ ∆ABD ≅ ∆CDB (By SAS)⇒ AD = CB. (By c.p.c.t.) (1)

4. In ∆COD and ∆BOA,OD = OA (Given)OC = OB (Given) (1)

∠DOC = ∠AOB = 90°∴ ∆COD ≅ ∆BOA (By SAS) (1)⇒ CD = AB. (By c.p.c.t.) (1)

Section D5. In ∆ABC,

AB = BC (Given) (1)∴ ∠1 = ∠2 (Angles opposite to equal sides)In ∆ABD and ∆CBD,

AB = BC∠1 = ∠2AD = CD (1)

∴ ∆ABD ≅ ∆CBD (By SAS cong.) .∠3 = ∠4 (By c.p.c.t.) (½)

∠3 + ∠4 = 180°∴ ∠4 = 90°∴ ∠ADE = ∠4 (V.O.A) .∴ ∠ADE = 90° Proved. (½)

In ∆EAD and ∆ECD, ∠A = ∠CAD = CD (Given)

∠ADE = ∠CDE = 90° (½)

and DE = DE (Common)So, ∆EAD = ∆ECD . (By AAS)

⇒ AE = CE. (By cpct) (½)

6. In ∆AMB and ∆PNQ,

AB = PQ (Given) .

AM = PN (Given) .

∠1 = ∠2 = 90°

⇒ ∆AMB ≅ ∆PNQ (By RHS) .

⇒ ∠3 = ∠4 (By c.p.c.t.) (2)

Similarly, ∆AMC ≅ ∆PNR

⇒ ∠5 = ∠6

∴ in ∆ABC and ∆PQR, AB = PQ (Given) .

AC = PR (Given) .

∠A = ∠P (∠3 + ∠5 = ∠4 + ∠6) (Proved) .

⇒ ∆ABC ≅ ∆PQR. (By SAS) (2)qq

AIRT EL SGN

Section A1. AD = BC

∠BAD = ∠ABCAB = AB

∴ ∆DAB ≅ ∆CBA (By cpct) ½∴ ∠BDA = ∠ACB ½

Section B

2. Given, ∠BAD = ∠EAC

Adding ∠CAD on both sides, we get

∠BAD + ∠CAD = ∠EAC + ∠CAD (½)

∠BAC = ∠DAE

AC = AE (Given) (½)

AB = AD (Given)

∆BAC ≅ ∆DAE (By SAS) (½)

∴ BC = ED (By c.p.c.t.) (½)

Section C

3. 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)

4. In ∆PQS,

PQ + QS > PS (Sum of a any two sides is greater than the third side) ...(1) (2)

In ∆PSR,

PR + SR > PS ...(2) (1)

∴ On adding (1) & (2)

PQ + PR + QS + SR > PS + PS

PQ + QR + RP > 2 PS

Section D5. In ∆BCE and ∆CBF, we have

∠BEC = ∠CFB (each 90°)BC = BC (common)

BE = CF (given)By RHS congluence.,

∆BCE ≅ ∆CBF∠BCE = ∠CBF (2)

In ∆ABC∠C = ∠B ⇒ AB = AC

Similarly, we prove AB = BC (1)Then, AB = BC = ACSo, ∆ABC is equilateral triangle. (1)

6. By triangle inequality property,

In ∆ABC, AB + BC > AC ...(1) (½)

In ∆BCD, BC + CD > BD ...(2) (½)

In ∆CDA, CD + DA > AC ...(3) (½)

In ∆DAB, DA + AB > BD ...(4) (½)Adding (1), (2), (3) and (4), we getAB + BC + BC + CD + CD + DA + DA + AB > AC + BD + AC + BD (1)

2(AB + BC + CD + DA) > 2(AC + BD)

Hence, Perimeter > Sum of its diagonals. (1)qq

Section A

1. ∠ADB = ∠ACB = 90º

AD = BC (½)

AB = AB

∴ ∆ABD ≅ ∆BAC (½)

Section B2. In ∆ABC,

∠A + ∠B + ∠C = 180°∠A = 180° – 20 – 70 = 90° (½)

So, ∠BAN = ∠NAC = 45°In ∆ABM

∠BAM = 180 – 90 – 70 = 20 (½)∠BAN = ∠BAM + ∠MAN

45 = 20 + ∠MAN (½)∠MAN = 25°

Section C3. Given, BR = CQ

BR + BQ = CQ + BQQR = BC (½)

AIRT EL SGN

In ∆ABC and ∆PQR,QR = BC (Proved above) .AB = PQ (Given) (1)

∠BAC = ∠QPR = 90° (Given) .∴ ∆ABC ≅ ∆PQR (By RHS) (1)

⇒ AC = PR. (By c.p.c.t.) (½)4. In ∆ABE and ∆ACD,

AB = AC (Given) (½)∠C = ∠B (Angle opp. to equal sides) (½)BE = CD (Given) (½)

∴ ∆ABE ≅ ∆ACD (By SAS) (1)⇒ AD = AE. (By c.p.c.t.) (½)

Section D

5. In ∆ABM and ∆PQN, AB = PQ (Given)

BC = QR (Given)

⇒ 12 BC = 1

i.e. BM = QN (½)

AM = PN (Given) .

∴ ∆ABM ≅ ∆PQN (SSS) (½)

⇒ ∠ABM = ∠PQN (c.p.c.t.) .

i.e. ∠ABC = ∠PQR (1)

Now, in ∆ABC and ∆PQR,AB = PQ (Given) .

∠ABC = ∠PQR (Proved) .(½)

BC = QR (Given) .∴ ∆ABC ≅ ∆PQR. (SAS) (1½)

6. Produce AD to E such thatAD = 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 + DEAC + AB > AD + AD (∵DE = AD) (½)

AC + AB > 2AD. Proved. (1)qq

This worksheet contain activities that are to be perform in the class itself.qq

AIRT EL SGN

6 COORDINATE GEOMETRY

Section A1. Clearly, (– 2, 5) i.e., x coordinate is – ve of y coordinate is + ve, so it lies the IInd quadrant.

12. So, clearly, from y-axis, the ⊥ distance is 7. 1

Section B

3. On x-axis : C (– 3, 0), F (6, 0), G (3, 0) 1

On y-axis : A (0, 2), D (0, – 3), E (0, 4) 1

4. (i) Since the point lies on x-axis at a distance of 9 units from y-axis. Hence its co-ordinates are(9, 0) (1)

(ii) According to question, the required co-odinates are (0, –9). (1)

5. The point on y-axis has x-cordinate 0.

Since it lies 4 units distance in negative direction of y-axis. (1)

∴ The point is (0, – 4). (1)

Section C

6. According to question, plot the points A(0, 3), B(5, 3), C(4, 0) and D(–1, 0) on graph paper.

From figure, ABCD is a parallelogram. The point (2, 2) will be inside the figure. (1)

Section D

7. (i) E (–1, 2) (1)

(ii) D (–4, – 1) (1)

(iii) 2 – 4 = – 2 (1)

(iv) 3 + (– 4) = – 1 (1)

Value based questions8. (i) Draw X´ OX and Y´ OY

as the co-ordinate axes andmark their point of intersection Oas 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 obtainthe 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 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)

In order to plot (3, – 1), we take 3 units on OX and then 1 unit below x-axis parallel to OY ́toobtain the point E(3, – 1) (½)

(ii) Co-ordinate geometry. (½)

(iii) Co-ordination among people is good for progress. (½)

Section A1. Mirror image is (1, 2) 12. The co-ordinater of a point whose ordinate is 6 of lies on y-axis is (0, 6). 1

Section B3. Plotting of P(2, – 6)

Coordinates of M(2, 0) (½)

Coordinates of N(0, – 6). (½)

ROOC AN TID OEGE RT YEM

-1-2-3-4-5-6

2 3 4 5 6 7 8

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-2-3-4-5-6-7

(2,-6)N P

Section C4.

-1-2-3-4-5-6

2 3 4 5 6 7 8

A (4, 0)

B (0, 4)

From figure, OA = 4 unitsOB = 4 units

Area of ∆OAB = 12 × OA × OB

= 12 × 4 × 4 = 8 square unit. (1)

Section D5. According to question (– 2, 9) = (1 + x, y2)

⇒ – 2 = 1 + x or x = – 3 (1)and 9 = y2 or y = 3, as y > 0 (1)So, P(y, x) = P(3, – 3),which will be in IV quadrant.

Q(2, x) = Q(2, – 3),which will be in IV quadrant.

R(x2, y – 1) = R(9, 2),which will be in I quadrant

S(2x, – 3y) = S(– 6, – 9),which will be in III quadrant. (½+½+½+½)

6. Since ∆ABC is equilateral triangle with side 2a units, thereforeAB = BC = CA = 2a units

O is mid point of AC, then (1)

OA = 12 AC

= 12 (2a) = a units (1)

Now, in right ∆AOB

OB = 2 2–AB OA

= 2 2(2 ) –a a

= 2 24 –a a

= 23 3 unitsa a= (1)

Thus, co-ordinates of B are (0, 3).a (1)

Value based question7. (i) Let OB = a

⇒ AB = 2a⇒ AC = 2 × a = 2aUsing Pythagoras theorem in ∆COA (1)

(– , 0)a ( , 0)a

(2a)2 = a2 + (OC)2

⇒ (OC)2 = 3a2

⇒ OC = 3a

So co-ordinates of C are (0, 3)a

Similarly, co-ordinates of D are −(0, 3)a (1)

(ii) Co-ordinate geometry. (½)(iii) Co-ordinate among people is the first condition of success in any society. (½)

Section A1. It is (0, – 7). 12. It lies on IV and II quadrant. 1

Section B

3. (A) P(0, 5) (½)

(–2, 0)

(B) Q(0, – 3) (½)

(C) R(5, 0) (½)

(D) S(–2, 0) (½)

Section C4. Drawing the points, A(3, 2), B(2, 3), C(–4, 5) and D(5, – 3). Joining AB, BC, CA, AD, we get a

quadrilateral ABCD.

-1-2-3-4-5-6

2 3 4 5 6 7 8

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-2-3-4-5-6-7

(5,-3)

(-4,5)

Section D5. From figure the co-ordinate of vertices of a rectangle are :

O(0, 0) (1)

A(– p, 0) (1)

B(– p, – q) (1)

C(0, – q) (1)

A (–, , 0)p (0, )q

C (0, – )qB p q(– , – )

(O, O)

–5 –4 –3 –2 –1–1–2

–3 –4

O1 2 3 4 5

(–3, 5)

(–3, –5) (2, –5)

6. (i) (– 3, – 5) lies in III quadrant, as x < 0 and y < 0.

(ii) (2, – 5) lies in IV quadrant, as x > 0 and y < 0.(iii) (– 3, 5) lies in II quadrant, as x < 0 and y > 0. (½+½+½)Verification : The points (– 3, – 5), (2, – 5) and

(– 3, 5) are plotted as shown in figure :

Result is verified : (i) (– 3, – 5) lies in III quadrant.(ii) (2, – 5) lies in IV quadrant.(iii) (– 3, 5) lies in II quadrant. (1)

Section A1. It lies on the IV quadrant. 12. It lies on the x-axis. 1

Section B3. Co-ordinates of A = (5, 0) (½)

Co-ordinates of B = (5, 3) (½)

Co-ordinates of C = (–2, 4) (½)

Co-ordinates of D = (0, –3). (½)

Section C4.

-1-2-3-4-5-6-7-8

2 3 4 5 6 7 8

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-2-3-4-5-6-7

(-3,-3) (3,-4)

(5,-2)

(0,7)(-1,7)

(-8,0)3

Section D5. (i) Required points are A(0, 3) and L(0, – 4). (2)

(ii) Required points are G(5, 0) and I(– 2, 0). (1)(iii) Required points are D(– 5, 1) and H(– 5, – 3). (1)

6. The points A(– 3, – 4), B(– 2, 0),C(– 1, 4) and D(1, 0)

are plotted as shown below :(i) A(– 3, – 4) lies in III quadrant.(ii) B(– 2, 0) lies on x-axis.(iii) C(– 1, 4) lies in II quadrant.

(iv) D(1, 0) lies on x-axis. (½+½+½+½)qq

Section A1. In II and IV quadrant. 1

2. AB = 2 2(0 0) (4 0) 0 16 4− + − = + =

BC = 2 2(0 4) (0 0) 16 0 4− + − = + =

Area of triangle = 12 × base × height (½)

= 12 × 4 × 4

= 8 sq. units (½)

Section B3. (a) IV quadrant (b) II quadrant (½+½)

(c) III quadrant (d) I quadrant (½+½)

Section C4. From the graph,

QR = 2 + 2 = 4

So, PQ = QR = RP = 4 (1)

Now OR = 2

So, from right triangle OPR,

OP2 = PR2 – OR2 (1)

OP2 = 42 – 22 = 12

So. OP = 2 3 (½)

Therefore, co-ordinates of P are (0, 2 3 ). (½)

Section D5.

–5 –4 –3 –2 –5 1 2 3 4 5

(I quadrant)

(IV quadrant)(III quadrant)

(II quadrant)

(T)(B)

(E)(S)

Section A1. (x + 2, 4) = (5, y – 2) (½)

⇒ x + 2 = 5 and 4 = y – 2⇒ x = 5 – 2 = 3 and y = 4 + 2 = 6 (½)The co-ordinates (x, y) = (3, 6)

2. The co-ordinates of the origin is (0, 0). 1

Section B3. (A) (–3, – 2) (½) (B) 0 (½)

(C) 0 (½) (D) Q and T. (½)

Section C4. (i)

(ii) Rhombus ABCD. (1)(iii) AB lies in II quadrant and AD lies in I quadrant. (1)

Section D5. The points A(– 3, 0), B(5, 0) and C(0, 4) can be plotted as shown below :

Figure formed is a triangle ABC.

1 2 3 4 5 –1

–3 –2 –1

C(0, 4)

(–3, 0)A

(5, 0)

Area of ∆ABC = 12 × AB × OC

= 12 × 8 × 4;

(as AB = 8 and OC = 4)= 16 sq. units. (2) + (2)

-1-2-3-4-5

2 3 4 5

-2-3-4-5

A (0,4)

D (3,0)

C (0,-4)

B (-3,0)

6. (i) Plot the points M(4, 3), N(4, 0), O(0, 0) and P(0, 3) on graph paper.

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2 3 4 5 6 7 8

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-2-3-4

O (0,0)

M (4,3)

N (4,0)

P (0,3)

(ii) By joining MNOP, the figure will be rectangle. (1)(iii) Perimeter of rectangle = 2(l + b) = 2(4 + 3) = 14 units. (1)

Section A1. At the origin. 12. Point C (3, a – 3) lie on x-axis

∴ a – 3 = 0 ½a = 3 ½

Section B

3. OA = OB = 6 32 = units and BC = AD = 3 units.

Co-ordinates of A be (–3, 0) (½)Co-ordinates of B be (3, 0) (½)Co-ordinates of C be (3, 3) (½)Co-ordinates of D be (–3, 3). (½)

Section C4. In ∆POQ,

∠POQ = 90º

QR = 8

OQ = 4 (1)

OP2 + OQ2 = PQ2 (½)

OP2 + 42 = 82, (PQ = QR = RP) (½)

OP2 = 48

OP = 48 or 4 3 (½)

The co-ordinates of P are ( 48,0) or (4 3,0). (½)

Section D5. (i) Plot the point M(5, – 3) and N(–3, –3) on graph paper.

(ii) Length of MN = 3 + 5 = 8 units(iii) From figure, (½)

A(3, – 3)B(1, – 3)C(–1, – 3) (½+½+½)

Plot the three vertices of the rectangle as A(3, 2),B(– 4, 2), C(– 4, 5). (2)

We have to find the co-ordinates of the fourth vertex Dso that ABCD is a rectangle. Since the opposite sides of arectangle are equal, so the abscissa of D should be equal toabscissa of A, i.e., 3 and the ordinate of D should be equal tothe ordinate of C, i.e., 5.

So, the co-ordinates of D are (3, 5). (2)

Section A1. On comparing we get, b = 2 ½

So, from 2 – a + b = 6, we get ½a = – 2 (1)

2. We get square.

Section B3. In, II quadrant, III quadrant, ½+½+½+½

IV quadrant and I quadrant.

–5 –4 –3 –2 –1–1

–3 –4 –5

O1 2 3 4 5

(–4, 5)C D

A(3, 2)(–4, 2)B

Section C4. (i)

(ii) Graph is a straight line. (1)5. OB = a, OA = a

(–a, 0) (a, 0)

⇒ AB = 2a⇒ AC = 2aUsing Pythagoras Theorem in ∆COA, (1)

(2a)2 = a2 + (OC)2

⇒ (OC)2 = 3a2

⇒ OC = 3a (1)So, co-ordinates of C are (0, 3)a (½)

Similarly, co-ordinates of D are (0, 3)a− . (½)

Section D

6. Points Quadrants

(–2, 4) II

(1, 2) I

(–3, –5) III

(3, –1) IV

(–1, 0) X-axis

– 1– 2– 3– 4– 5

1 2 3 4 5–4 –3 –2 –1

(–2, 4)

(–1, 0)(1, 2)

(3, –1)

(–3, –5)

–4 –3 –2 –1–1–2 –3 –4

O1 2 3 4 5 6

D(2, 6)

C(2, 0)

(–2, 3)A

B(–2, 0)

7. The given points A(– 2, 3), B(– 2, 0), C( 2, 0) and D(2, 6) can be plotted as under :

On joining AB, BC, CD and DA, we get a trapezium ABCD.

From figure, BC = 2 – (– 2)

= 4 units (2)

and AB = 3 – 0

= 3 units (1)

Now, area of ∆ABC = 12 × BC × AB

= 12 × 4 × 3

= 6 sq. units. (1)qq

Section AThis worksheet contains activities that are to be performed in the class itself.

7 AREAS

Section A

1. x = 2 225 15 40 10− = × (½)

x = 20 cm.

Area of right triangle = 12 × base × height

= 12 × 15 × 20 = 150 cm2. (½)

2. Area of equilateral triangle= 23 16 34 a× =

a2 = 4 × 16 ⇒ a = 8 m (½)Perimeter = 3a = 3 × 8 = 24 m. (½)

Section B3. Perimeter of an equilateral triangle = 3a = 60 cm

a = 20 cm (1)

Area = 23 3 20 20 100 34 4a = × × = cm2. (1)

4. Sides of an isosceles triangle are a, a and b, then

a + a + b = 32 (½)2a + 12 = 32

a = 10 cm

s = 32 162 = cm (½)

Area = 16 (16 12)(16 10)(16 10)× − − − (½)

= 16 4 6 6× × × = 48 cm2. (½)

Section C5. Draw DE || BC and DF ⊥ AB

AE = 25 – 10 = 15 m

For ∆AED, s = 13 14 15 212+ + = m

Area of ∆AED = 21 8 7 6 84× × × = m2

Also, area of ∆AED = 1 152 × × DF = 84 (1)

⇒ DF = 2 84 5615 5× = (½)

Area of parallelogram EBCD = 5610 1125× = m2 (1)

Area of trapezium = 84 + 112 = 196 m2. (½)

AERA S

Section D6. Perimeter of the triangle = 84 cm

Ratio of its sides = 13 : 14 : 15Sum 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.

Section A

1. ( )( )( )S S a S b S c− − − 1

2. Perimeter = 4x + 5x + 6x = 150⇒ 15x = 150⇒ x = 10 ½Sides are 4 × 10 = 40 cm, 5 × 10 = 50 cm and 6 × 10 = 60 cm ½

Section B3. Let the sides are 5x, 12x and 13x then,

Perimeter = 5x + 12x + 13x = 12030x = 120

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

s = 120 602 = cm (½)

Area = ( )( )( )s s a s b s c− − −

= 60(60 20)(60 48)(60 52)− − − (½)

= 60 40 12 8× × × = 10 6 10 4 6 2 4 2× × × × × × ×= 10 × 6 × 4 × 2 = 480 cm2. (½)

4. Let third side = x

Then 24 + 10 + x = 60

x = 60 – 34 = 26 cm (½)

s = 60 302 = cm (½)

Area = ( )( )( )s s a s b s c− − −

= 30(30-24)(30-10)(30-26) (½)

= 30 6 20 4× × × = 10 3 3 2 10 2 4× × × × × ×

= 10 × 3 × 4 = 120 cm2. (½)

Section C

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

s = 6 6 4 16 82 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)

Section D6. 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)

Using Heron’s formula,

Area of one triangular leaf = ( – )( – )( – )s s a s b s c

= 42(42–28)(42–15)(42–41) (1)

= 42 14 27 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)qq

AERA S

Section A

1. Clearly, the length of the altitude of an equilateral triangle of side ‘a’ is 3

2. x = 2 210 6 8− = cm

Area of right triangle = 12 × base × height (½)

= 12 × 6 × 8 = 24 cm2. (½)

Section B3. Let the longest side be 90 cm and other side BC be 54 cm

AB = 2 290 54 72− = cm (½)

∴ Area of the triangle = 12 b h× ×

= 1 54 722 × × (1)

= 1944 cm2. (½)4. Perimeter of rhombus = 4 × side = 200

side = 50 m (½)

For ∆ABC, s = 50 50 80 902+ + = m (½)

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

= 90 (90 50)(90 50)(90 80)× − − −

= 90 40 40 10× × ×= 1200 m2. (½)

∴ Area of rhombus = 2 × Area of ∆ABC= 2 × 1200 = 2400 m2. (½)

Section C

5. For ∆ABC, s = 96 52 52 200 1002 2+ + = = m (½)

Area = ( )( )( )s s a s b s c− − −

= 100 4 48 48× × × (½)= 10 × 2 × 48 = 960 m2 (½)

Area of rhombus = 960 × 2 = 1920 m2 (½)

Area to be grazed by 20 cows = 1920 m2

Area to be grazed by 1 cow = 192020 (½)

= 96 m2. (½)

Section D6. a = 120 m, b = 170 m, c = 250 m (½)

s = 120 170 250 540 2702 2 2a b c+ + + += = =

Area = ( ) ( ) ( )s s a s b s c− − − (½)

= 270 (270 120)(270 170)(270 250)× − − −

= 270 150 100 20× × × (½)

= 81000000 (½)= 9000 m2 (½)

Area = 12 × base × height (½)

9000 = 12 × 250 × h (½)

h = 72 m. (½)

Value based question7. (i) Let the sides of the triangle be a, b and c.

a : b : c = 12 : 17 : 25

12 17a b= = 25

c k= (say)

⇒ a = 12k, b = 17k, c = 25k (½)Perimeter = 540 cma + b + c = 540

12k + 17k + 25k = 54054k = 540

k = 540 1054 = (½)

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

Now s = 1 5402 × = 270 cm

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

AERA S

∴ 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)

Section A

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

a = 20 m

Area = 23 3 20 20 100 34 4a = × × = m2. 1

2. S = + + + +

= =8 7 15

15 cm2 2

a b c1

Section B

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

s = 15 15 12 212 2a b c+ + + += = cm (½)

Area = ( )( )( )s s a s b s c− − − (½)

= 21 (21 15)(21 15)(21 12)× − − −

= 21 6 6 9× × × (½)

= 18 21 cm2. (½)

Section C

4. s = 20 50 50 1202 2

+ + = = 60 cm (½)

Area of one triangular piece = ( )( )( )s s a s b s c− − − (½)

= 60(60 20)(60 50)(60 50)− − −

= 60 40 10 10× × × (½)

= 200 6 cm2 (½)

Required cloth = 10 × 200 6 = 2000 × 2·45 (½)= 4900 cm2. (½)

Section D5. Rhombus,

Perimeter = 52 cm (1)

Side = 52 134 = cm

Diagonal = 24 cm (1)OB = OD = 12 cm

OA = 2 213 12 169 144 25 5− = − = = cm (1)

Area of rhombus = 14 5 122× × × (1)

= 120 cm2

Value based question

6. (i) If a, b, c be the lengths of sides BC, CA and AB of ∆ABC and if s = 1( ),2 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) = 3

2a (½)

s – a = 32 2a aa− =

s – b = 32 2a aa− =

s – c = 32 2a aa− = (½)

area = ( ) ( ) ( ) ( )32 2 2 2a a a a = 32 2

a a× ×

4a . ...(1) (1)

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

Here, a + a + a = 180⇒ 3a = 180

⇒ a = 180 603 =

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 life. (½)

Section A

1. Let the side of an isosceles right angle triangle be x, x and 5 2.

So, x2 + x2 = 2(5 2) ½2x2 = 25 × 2

x = 5 cm ½

AERA S

2. x2 + x2 = ( )212 2

2x2 = 144 × 2

x = 12 cm (½) Given 3a = 4x (a, be a side of triangle)

3a = 4 × 12a = 16 cm

Area of equilateral triangle = 23 3 16 164 4a = × ×

= 64 3 cm2. (½)

Section B

3. Diagonals bisect each other at O.

Draw LOM ⊥ AB.OL = OM = 3 cm

Area of shaded region OAB = 1 8 32 × × (2)

= 12 cm2.

Section C

4. 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 cm2 (½)

No. of rose beds = Total areaoftriangularfield

Areaoccupiedbyeachrosebed (½)

= 306 516 = . (½)

5. s = 26 28 30 842 2

+ + = = 42 cm (½)

Area of ∆ = ( )( )( )s s a s b s c− − − (½)

= 42(42 26)(42 28)(42 30)− − − (½)

= 42 16 14 12× × × (½)

= 336 cm2

Again, area of parallelogram = Area of triangle (1)b × h = 336 (½)

28 × h = 336 (½)

h = 12 cm (½)

Section D

6. ar (∆BCD) = × × =1

12 5 302

m2 (½)

BD = 2 212 5 13+ = m2 (½)

For ∆ABD, s = 13 8 9 152+ + = cm

Area of ∆ABD = 15 2 7 6× × ×

= 6 35 m2 (1½)

ar (quad. ABCD) = ( )+30 6 35 m2. (½)

7. Let the equal sides be x, then

32x x x+ + = 42

72x = 42 (1)

x = 12 cm

Then sides are 12 cm, 12 cm and 18 cm

s = 12+12+18

2 = 42 212 = cm (1)

Area = ( )( )( )s s a s b s c− − −

= 21(21 12)(21 12)(21 18)− − − (1)

= 21 9 9 3× × ×

= 27 7 cm2. (1)

Section A

1. Sides are 3x, 4x, 5x, then,

3x + 4x + 5x = 36

⇒ 12x = 36 ⇒ x = 3 ½

∴ Sides are a, b, c are 9 cm, 12 cm, 15 cm. ½

2. Area of ∆ABC = 214 4 8 cm

2× × = 1

Section B

3. Let a = 20 cm, b = 30 cm, c = 40 cm

Then s = 20 30 402 2

a b c+ + + += = 45 cm

AERA S

Now, using Heron’s formula,

Area of ∆ = ( – )( – )( – )s s a s b s c (1)

= 245(45–20)(45–30)(45–40) cm

= 245 25 15 5 cm× × ×

= 23 3 5 5 5 3 5 5 cm× × × × × × ×

= 23 5 5 15 cm× × ×

= 75 × 3·873 cm2

= 290·48 cm2. (1)

Section C

4. Area of the equilateral triangle = 23 3 10 104 4a = × × (½)

= 25 3 sq. cm = 25 × 1·732

= 43·3 sq. cm (1)

In ∆BDC, DC = 2 210 8 6− = cm (½)

∴ Area of ∆BDC = 1 6 8 242 × × = sq. cm (½)

∴ Area of unshaded portion = 43·3 – 24

= 19·3 sq. cm. (½)

Section D

5. Let the equal sides be ‘a’ units, then

s = 24 2 242 2

a a a+ + += (½)

= a + 12

Area = ( )( )( )s s a s b s c− − − (½)

60 = ( 12)( 12 )( 12 )( 12 24)a a a a a a+ + − + − + −

= ( 12)(12)(12)( 12)a a+ − (½)

60 = 2( 144) 12a − ×

5 = 2 144a − (½)

On squaring both sides, we get25 = a2 – 144a2 = 25 + 144 = 169a = 13 cm (½)

Perimeter = 13 + 13 + 24 = 50 cm (½)

6. In ∆ABC, AC2 = AB2 + BC2

AC2 = 92 + 402

= 81 + 1600 = 1681 (½)

AC = 41 m

Area of ∆ABC = 12 × b × h (½)

= 12 × 9 × 40

= 180 m2

For ∆ACD, s = 28 41 15 84 422 2+ + = = m (½)

Area = − − −( )( )( )s s a s b s c

= − − −42(42 28)(42 41)(42 15) (½)

= 42 14 1 27× × × (½)

= 126 m2 (½)Area of quadrilateral = Area of ∆ABC + Area of ∆ACD (½)

= 180 + 126 = 306 m2. (½)qq

Section AThis worksheet contains activities that are to be performed in the class itself.

AERA S

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