Areas

Question 1:

Exercise 7A

Question 2:

Question 3:

Question 4:

Question 5:

Question 6:

Question 7:

Question 8:

Question 9:

Question 10:

Question 11:

Question 12:

Question 13:

Question 14:

Question 15:

(i) Area of an equilateral triangle =

Where a is the side of the equilateral triangle

Question 16:

Question 17:

Question 18:

Question 19:

Question 20:

Question 21:

Question 22:

Question 23:

Question 24:

Angles, Linesand Triangles

Question 1:

Exercise 4A

1. Angle: Two rays having a common end point form an angle.
2. Interior of an angle: The interior of ∠AOB is the set of all points in its plane, which lie on the same side of OA as B and also on same side of OB as A.
3. Obtuse angle: An angle whose measure is more than 90° but less than 180°, is called an obtuse angle.
4. Reflex angle: An angle whose measure is more than 180° but less than 360° is called a reflex angle.
5. Complementary angles: Two angles are said to be complementary, if the sum of their measures is 90o.
6. Supplementary angles: Two angles are said to be supplementary, if the sum of their measures is 180°.

Question 2:

∠A = 36° 27′ 46″ and ∠B = 28° 43′ 39″

∴ Their sum = (36° 27′ 46″) + (28° 43′ 39″)

Therefore, the sum ∠A + ∠B = 65° 11′ 25″

Question 3:

Let ∠A = 36° and ∠B = 24° 28′ 30″ Their difference = 36° – 24° 28′ 30″

Thus the difference between two angles is ∠A – ∠B = 11° 31′ 30″

Question 4:

• 1. Complement of 58o = 90o – 58o = 32o
  2. Complement of 16o = 90 – 16o = 74o
  3. 
     of a right angle =
     
     × 90o = 60o Complement of 60o = 90o – 60o = 30o
  4. 1o = 60′

⇒ 90o = 89o 60′

Complement of 46o 30′ = 90o – 46o 30′ = 43o 30′ (v) 90o = 89o 59′ 60″

Complement of 52o 43′ 20″ = 90o – 52o 43′ 20″

= 37o 16′ 40″

(vi) 90o = 89o 59′ 60″

∴ Complement of (68o 35′ 45″)

= 90o – (68o 35′ 45″)

= 89o 59′ 60″ – (68o 35′ 45″)

= 21o 24′ 15″

Question 5:

1. Supplement of 63o = 180o – 63o = 117o
2. Supplement of 138o = 180o – 138o = 42o
3. 
   of a right angle =
   
   × 90o = 54o

∴ Supplement of 54o = 180o – 54o = 126o

1. 1o = 60′

⇒ 180o = 179o 60′

Supplement of 75o 36′ = 180o – 75o 36′ = 104o 24′ (v) 1o = 60′, 1′ = 60″

⇒ 180o = 179o 59′ 60″

Supplement of 124o 20′ 40″ = 180o – 124o 20′ 40″

= 55o 39′ 20″

(vi) 1o = 60′, 1′ = 60″

⇒ 180o = 179o 59′ 60″

∴ Supplement of 108o 48′ 32″ = 180o – 108o 48′ 32″

= 71o 11′ 28″.

Question 6:

• 1. 
     Let the required angle be xo Then, its complement = 90o – xo

∴ The measure of an angle which is equal to its complement is 45o.

• 1. 
     Let the required angle be xo Then, its supplement = 180o – xo

∴ The measure of an angle which is equal to its supplement is 90o.

Question 7:

Let the required angle be xo Then its complement is 90o – xo

∴ The measure of an angle which is 36o more than its complement is 63o.

Question 8:

Let the required angle be xo Then its supplement is 180o – xo

∴ The measure of an angle which is 25o less than its supplement is

Question 9:

Let the required angle be xo Then, its complement = 90o – xo

∴ The required angle is 72o.

Question 10:

Let the required angle be xo Then, its supplement is 180o – xo

∴ The required angle is 150o.

Question 11:

Let the required angle be xo

Then, its complement is 90o – xo and its supplement is 180o – xo That is we have,

∴ The required angle is 60o.

Question 12:

Let the required angle be xo

Then, its complement is 90o – xo and its supplement is 180o – xo

∴ The required angle is 45o.

Question 13:

Let the two required angles be xo and 180o – xo. Then,

⇒ 2x = 3(180 – x)

⇒ 2x = 540 – 3x

⇒ 3x + 2x = 540

⇒ 5x = 540

⇒ x = 108

Thus, the required angles are 108o and 180o – xo = 180 o – 108o = 72o.

Question 14:

Let the two required angles be xo and 90o – xo. Then

⇒ 5x = 4(90 – x)

⇒ 5x = 360 – 4x

⇒ 5x + 4x = 360

⇒ 9x = 360

⇒ x =

= 40

Thus, the required angles are 40o and 90o – xo = 90 o – 40o = 50o.

Question 15:

Let the required angle be xo.

Then, its complementary and supplementary angles are (90o – x) and (180o – x) respectively.

Then, 7(90o – x) = 3 (180o – x) – 10o

⇒ 630o – 7x = 540o – 3x – 10o

⇒ 7x – 3x = 630o – 530o

⇒ 4x = 100o

⇒ x = 25o

Thus, the required angle is 25o.

Question 1:

Exercise 4B

Since ∠BOC and ∠COA form a linear pair of angles, we have

∠BOC + ∠COA = 180o

⇒ xo + 62o = 180o

⇒ x = 180 – 62

∴ x = 118o

Question 2:

Since, ∠BOD and ∠DOA form a linear pair.

∠BOD + ∠DOA = 180o

∴ ∠BOD + ∠DOC + ∠COA = 180o

⇒ (x + 20)o + 55o + (3x – 5)o = 180o

⇒ x + 20 + 55 + 3x – 5 = 180

⇒ 4x + 70 = 180

⇒ 4x = 180 – 70 = 110

⇒ x =

= 27.5

∴ ∠AOC = (3 × 27.5 – 5)o = 82.5-5 = 77.5o And, ∠BOD = (x + 20)o = 27.5o + 20o = 47.5o.

Question 3:

Since ∠BOD and ∠DOA from a linear pair of angles.

⇒ ∠BOD + ∠DOA = 180o

⇒ ∠BOD + ∠DOC + ∠COA = 180o

⇒ xo + (2x – 19)o + (3x + 7)o = 180o

⇒ 6x – 12 = 180

⇒ 6x = 180 + 12 = 192

⇒ x =

= 32

⇒ x = 32

⇒ ∠AOC = (3x + 7)o = (3 32 + 7)o = 103o

⇒ ∠COD = (2x – 19)o = (2 32 – 19)o = 45o

and ∠BOD = xo = 32o

Question 4:

x: y: z = 5: 4: 6

The sum of their ratios = 5 + 4 + 6 = 15 But x + y + z = 180o

[Since, XOY is a straight line]

So, if the total sum of the measures is 15, then the measure of x is 5. If the sum of angles is 180o, then, measure of x =

× 180 = 60

And, if the total sum of the measures is 15, then the measure of y is 4. If the sum of the angles is 180o, then, measure of y =

× 180 = 48 And ∠z = 180o – ∠x – ∠y

= 180o – 60o – 48o

= 180o – 108o = 72o

∴ x = 60, y = 48 and z = 72.

Question 5:

AOB will be a straight line, if two adjacent angles form a linear pair.

∴ ∠BOC + ∠AOC = 180o

⇒ (4x – 36)o + (3x + 20)o = 180o

⇒ 4x – 36 + 3x + 20 = 180

⇒ 7x – 16 = 180o

⇒ 7x = 180 + 16 = 196

⇒ x =

= 28

∴ The value of x = 28.

Question 6:

Since ∠AOC and ∠AOD form a linear pair.

∴ ∠AOC + ∠AOD = 180o

⇒ 50o + ∠AOD = 180o

⇒ ∠AOD = 180o – 50o = 130o

∠AOD and ∠BOC are vertically opposite angles.

∠AOD = ∠BOC

⇒ ∠BOC = 130o

∠BOD and ∠AOC are vertically opposite angles.

∴ ∠BOD = ∠AOC

⇒ ∠BOD = 50o

Question 7:

Since ∠COE and ∠DOF are vertically opposite angles, we have,

∠COE = ∠DOF

⇒ ∠z = 50o

Also ∠BOD and ∠COA are vertically opposite angles. So, ∠BOD = ∠COA

⇒ ∠t = 90o

As ∠COA and ∠AOD form a linear pair,

∠COA + ∠AOD = 180o

⇒ ∠COA + ∠AOF + ∠FOD = 180o [∠t = 90o]

⇒ t + x + 50o = 180o

⇒ 90o + xo + 50o = 180o

⇒ x + 140 = 180

⇒ x = 180 – 140 = 40

Since ∠EOB and ∠AOF are vertically opposite angles So, ∠EOB = ∠AOF

⇒ y = x = 40

Thus, x = 40 = y = 40, z = 50 and t = 90

Question 8:

Since ∠COE and ∠EOD form a linear pair of angles.

⇒ ∠COE + ∠EOD = 180o

⇒ ∠COE + ∠EOA + ∠AOD = 180o

⇒ 5x + ∠EOA + 2x = 180

⇒ 5x + ∠BOF + 2x = 180

[∴ ∠EOA and BOF are vertically opposite angles so, ∠EOA = ∠BOF]

⇒ 5x + 3x + 2x = 180

⇒ 10x = 180

⇒ x = 18

Now ∠AOD = 2xo = 2 × 18o = 36o

∠COE = 5xo = 5 × 18o = 90o

and, ∠EOA = ∠BOF = 3xo = 3 × 18o = 54o

Question 9:

Let the two adjacent angles be 5x and 4x. Now, since these angles form a linear pair. So, 5x + 4x = 180o

⇒ 9x = 180o

⇒ x =

= 20

∴ The required angles are 5x = 5x = 5 20o = 100o and 4x = 4 × 20o = 80o

Question 10:

Let two straight lines AB and CD intersect at O and let ∠AOC = 90o.

Now, ∠AOC = ∠BOD [Vertically opposite angles]

⇒ ∠BOD = 90o

Also, as ∠AOC and ∠AOD form a linear pair.

⇒ 90o + ∠AOD = 180o

⇒ ∠AOD = 180o – 90o = 90o

Since, ∠BOC = ∠AOD [Verticallty opposite angles]

⇒ ∠BOC = 90o

Thus, each of the remaining angles is 90o.

Question 11:

Since, ∠AOD and ∠BOC are vertically opposite angles.

∴ ∠AOD = ∠BOC

Now, ∠AOD + ∠BOC = 280o [Given]

⇒ ∠AOD + ∠AOD = 280o

⇒ 2∠AOD = 280o

⇒ ∠AOD =

= 140o

⇒ ∠BOC = ∠AOD = 140o

As, ∠AOC and ∠AOD form a linear pair. So, ∠AOC + ∠AOD = 180o

⇒ ∠AOC + 140o = 180o

⇒ ∠AOC = 180o – 140o = 40o

Since, ∠AOC and ∠BOD are vertically opposite angles.

∴ ∠AOC = ∠BOD

⇒ ∠BOD = 40o

∴ ∠BOC = 140o, ∠AOC = 40o , ∠AOD = 140o and ∠BOD = 40o.

Question 12:

Since ∠COB and ∠BOD form a linear pair So, ∠COB + ∠BOD = 180o

⇒ ∠BOD = 180o – ∠COB …. (1)

Also, as ∠COA and ∠AOD form a linear pair. So, ∠COA + ∠AOD = 180o

⇒ ∠AOD = 180o – ∠COA

⇒ ∠AOD = 180o – ∠COB …. (2)

[Since, OC is the bisector of ∠AOB, ∠BOC = ∠AOC] From (1) and (2), we get,

∠AOD = ∠BOD (Proved)

Question 13:

Let QS be a perpendicular to AB. Now, ∠PQS = ∠SQR

Because angle of incident = angle of reflection

⇒ ∠PQS = ∠SQR =

= 56o

Since QS is perpendicular to AB, ∠PQA and ∠PQS are complementary angles. Thus, ∠PQA + ∠PQS = 90o

⇒ ∠PQA + 56o = 90o

⇒ ∠PQA = 90o – 56o = 34o

Question 14:

Given : AB and CD are two lines which are intersecting at O. OE is a ray bisecting the

∠BOD. OF is a ray opposite to ray OE.

To Prove: ∠AOF = ∠COF

Proof : Since

and

are two opposite rays,

is a straight line passing through O.

∴ ∠AOF = ∠BOE and ∠COF = ∠DOE

[Vertically opposite angles] But ∠BOE = ∠DOE (Given)

∴ ∠AOF = ∠COF

Hence, proved.

Question 15:

Given:

is the bisector of ∠BCD and

is the bisector of ∠ACD. To Prove: ∠ECF = 90o

Proof: Since ∠ACD and ∠BCD forms a linear pair.

∠ACD + ∠BCD = 180o

∠ACE + ∠ECD + ∠DCF + ∠FCB = 180o

∠ECD + ∠ECD + ∠DCF + ∠DCF = 180o

because ∠ACE = ∠ECD and ∠DCF = ∠FCB

2(∠ECD) + 2 (∠CDF) = 180o

2(∠ECD + ∠DCF) = 180o

∠ECD + ∠DCF =

= 90o

∠ECF = 90o (Proved)

Question 1:

Exercise 4C

Since AB and CD are given to be parallel lines and t is a transversal. So, ∠5 = ∠1 = 70o [Corresponding angles are equal]

∠3 = ∠1 = 70o [Vertically opp. Angles]

∠3 + ∠6 = 180o [Co-interior angles on same side]

∴ ∠6 = 180o – ∠3

= 180o – 70o = 110o

∠6 = ∠8 [Vertically opp. Angles]

⇒ ∠8 = 110o

⇒ ∠4 + ∠5 = 180o [Co-interior angles on same side]

∠4 = 180o – 70o = 110o

∠2 = ∠4 = 110o [ Vertically opposite angles]

∠5 = ∠7 [Vertically opposite angles] So, ∠7 = 70o

∴ ∠2 = 110o, ∠3 = 70o , ∠4 = 110o, ∠5 = 70o, ∠6 = 110o, ∠7 = 70o and ∠8 = 110o.

Question 2:

Since ∠2 : ∠1 = 5 : 4.

Let ∠2 and ∠1 be 5x and 4x respectively.

Now, ∠2 + ∠1 = 180o , because ∠2 and ∠1 form a linear pair. So, 5x + 4x = 180o

⇒ 9x = 180o

⇒ x = 20o

∴ ∠1 = 4x = 4 × 20o = 80o And ∠2 = 5x = 5 × 20o = 100o

∠3 = ∠1 = 80o [Vertically opposite angles]

And ∠4 = ∠2 = 100o [Vertically opposite angles]

∠1 = ∠5 and ∠2 = ∠6 [Corresponding angles] So, ∠5 = 80o and ∠6 = 100o

∠8 = ∠6 = 100o [Vertically opposite angles] And ∠7 = ∠5 = 80o [Vertically opposite angles]

Thus, ∠1 = 80o, ∠2 = 100o, ∠3 = ∠80o, ∠4 = 100o, ∠5 = 80o, ∠6 = 100o, ∠7 = 80o and

∠8 = 100o.

Question 3:

Given: AB || CD and AD || BC To Prove: ∠ADC = ∠ABC

Proof: Since AB || CD and AD is a transversal. So sum of consecutive interior angles is

180o.

⇒ ∠BAD + ∠ADC = 180o ….(i)

Also, AD || BC and AB is transversal. So, ∠BAD + ∠ABC = 180o ….(ii) From (i) and (ii) we get:

∠BAD + ∠ADC = ∠BAD + ∠ABC

⇒ ∠ADC = ∠ABC (Proved)

Question 4:

1. 
   Through E draw EG || CD. Now since EG||CD and ED is a transversal.

So, ∠GED = ∠EDC = 65o [Alternate interior angles] Since EG || CD and AB || CD,

EG||AB and EB is transversal.

So, ∠BEG = ∠ABE = 35o [Alternate interior angles] So, ∠DEB = xo

⇒ ∠BEG + ∠GED = 35o + 65o = 100o.

Hence, x = 100.

1. 
   Through O draw OF||CD.

Now since OF || CD and OD is transversal.

∠CDO + ∠FOD = 180o

[sum of consecutive interior angles is 180o]

⇒ 25o + ∠FOD = 180o

⇒ ∠FOD = 180o – 25o = 155o

As OF || CD and AB || CD [Given] Thus, OF || AB and OB is a transversal.

So, ∠ABO + ∠FOB = 180o [sum of consecutive interior angles is 180o]

⇒ 55o + ∠FOB = 180o

⇒ ∠FOB = 180o – 55o = 125o

Now, xo = ∠FOB + ∠FOD = 125o + 155o = 280o.

Hence, x = 280.

1. 
   Through E, draw EF || CD.

Now since EF || CD and EC is transversal.

∠FEC + ∠ECD = 180o

[sum of consecutive interior angles is 180o]

⇒ ∠FEC + 124o = 180o

⇒ ∠FEC = 180o – 124o = 56o

Since EF || CD and AB ||CD

So, EF || AB and AE is a trasveral.

So, ∠BAE + ∠FEA = 180o

[sum of consecutive interior angles is 180o]

∴ 116o + ∠FEA = 180o

⇒ ∠FEA = 180o – 116o = 64o

Thus, xo = ∠FEA + ∠FEC

= 64o + 56o = 120o.

Hence, x = 120.

Question 5:

Since AB || CD and BC is a transversal.

So, ∠ABC = ∠BCD [atternate interior angles]

⇒ 70o = xo + ∠ECD ….(i)

Now, CD || EF and CE is transversal.

So, ∠ECD + ∠CEF = 180o

∴ ∠ECD + 130o = 180o

[sum of consecutive interior angles is 180o]

⇒ ∠ECD = 180o – 130o = 50o

Putting ∠ECD = 50o in (i) we get, 70o = xo + 50o

⇒ x = 70 – 50 = 20

Question 6:

Through C draw FG || AE

Now, since CG || BE and CE is a transversal.

So, ∠GCE = ∠CEA = 20o

∴ ∠DCG = 130o – ∠GCE

= 130o – 20o = 110o

[Alternate angles]

Also, we have AB || CD and FG is a transversal.

So, ∠BFC = ∠DCG = 110o

As, FG || AE, AF is a transversal.

[Corresponding angles]

∠BFG = ∠FAE [Corresponding angles]

∴ xo = ∠FAE = 110o. Hence, x = 110

Question 7:

Given: AB || CD

To Prove: ∠BAE – ∠DCE = ∠AEC

Construction : Through E draw EF || AB Proof : Since EF || AB, AE is a transversal. So, ∠BAE + ∠AEF = 180O ….(i)

[sum of consecutive interior angles is 180o] As EF || AB and AB || CD [Given]

So, EF || CD and EC is a transversal.

So, ∠FEC + ∠DCE = 180o ….(ii)

[sum of consecutive interior angles is 180o] From (i) and (ii) we get,

∠BAE + ∠AEF = ∠FEC + ∠DCE

⇒ ∠BAE – ∠DCE = ∠FEC – ∠AEF = ∠AEC [Proved]

Question 8:

Since AB || CD and BC is a transversal.

So, ∠BCD = ∠ABC = xo

[Alternate angles]

As BC || ED and CD is a transversal.

∠BCD + ∠EDC = 180o

⇒ ∠BCD + 75o =180o

⇒ ∠BCD = 180o – 75o = 105o

∠ABC = 105o

∴ xo = ∠ABC = 105o Hence, x = 105.

[since ∠BCD = ∠ABC]

Question 9:

Through F, draw KH || AB || CD

Now, KF || CD and FG is a transversal.

⇒ ∠KFG = ∠FGD = ro …. (i) [alternate angles]

Again AE || KF, and EF is a transversal.

So, ∠AEF + ∠KFE = 180o

∠KFE = 180o – po …. (ii) Adding (i) and (ii) we get,

∠KFG + ∠KFE = 180 – p + r

⇒ ∠EFG = 180 – p + r

⇒ q = 180 – p + r i.e., p + q – r = 180

Question 10:

Since AB || PQ and EF is a transversal.

So, ∠CEB = ∠EFQ [Corresponding angles]

⇒ ∠EFQ = 75o

⇒ ∠EFG + ∠GFQ = 75o

⇒ 25o + yo = 75o

⇒ y = 75 – 25 = 50

Also, ∠BEF + ∠EFQ = 180o

∠BEF = 180o – ∠EFQ

= 180o – 75o

∠BEF = 105o

[sum of consecutive interior angles is 180o]

∴ ∠FEG + ∠GEB = ∠BEF = 105o

⇒ ∠FEG = 105o – ∠GEB = 105o – 20o = 85o

In ∆EFG we have,

xo + 25o + ∠FEG = 180o

Hence, x = 70.

Question 11:

Since AB || CD and AC is a transversal.

So, ∠BAC + ∠ACD = 180o

⇒ ∠ACD = 180o – ∠BAC

= 180o – 75o = 105o

[sum of consecutive interior angles is 180o]

⇒ ∠ECF = ∠ACD [Vertically opposite angles]

∠ECF = 105o

Now in ∆CEF,

∠ECF + ∠CEF + ∠EFC =180o

⇒ 105o + xo + 30o = 180o

⇒ x = 180 – 30 – 105 = 45

Hence, x = 45.

Question 12:

Since AB || CD and PQ a transversal.

So, ∠PEF = ∠EGH [Corresponding angles]

⇒ ∠EGH = 85o

∠EGH and ∠QGH form a linear pair. So, ∠EGH + ∠QGH = 180o

⇒ ∠QGH = 180o – 85o = 95o

Similarly, ∠GHQ + 115o = 180o

⇒ ∠GHQ = 180o – 115o = 65o

In ∆GHQ, we have, xo + 65o + 95o = 180o

⇒ x = 180 – 65 – 95 = 180 – 160

∴ x = 20

Question 13:

Since AB || CD and BC is a transversal. So, ∠ABC = ∠BCD

⇒ x = 35

Also, AB || CD and AD is a transversal. So, ∠BAD = ∠ADC

⇒ z = 75

In ∆ABO, we have,

∠AOB + ∠BAO + ∠BOA = 180o

⇒ xo + 75o + yo = 180o

⇒ 35 + 75 + y = 180

⇒ y = 180 – 110 = 70

∴ x = 35, y = 70 and z = 75.

Question 14:

Since AB || CD and PQ is a transversal. So, y = 75 [Alternate angle]

Since PQ is a transversal and AB || CD, so x + APQ = 180o

[Sum of consecutive interior angles]

⇒ xo = 180o – APQ

⇒ x = 180 – 75 = 105

Also, AB || CD and PR is a transversal.

So, ∠APR = ∠PRD [Alternate angle]

⇒ ∠APQ + ∠QPR = ∠PRD [Since ∠APR = ∠APQ + ∠QPR]

⇒ 75o + zo = 125o

⇒ z = 125 – 75 = 50

∴ x = 105, y = 75 and z = 50.

Question 15:

∠PRQ = xo = 60o

[vertically opposite angles]

Since EF || GH, and RQ is a transversal. So, ∠x = ∠y [Alternate angles]

⇒ y = 60

AB || CD and PR is a transversal.

So, ∠PRD = ∠APR [Alternate angles]

⇒ ∠PRQ + ∠QRD = ∠APR [since ∠PRD = ∠PRQ + ∠QRD]

⇒ x + ∠QRD = 110o

⇒ ∠QRD = 110o – 60o = 50o

In ∆QRS, we have,

∠QRD + to + yo = 180o

⇒ 50 + t + 60 = 180

⇒ t = 180 – 110 = 70

Since, AB || CD and GH is a transversal So, zo = to = 70o [Alternate angles]

∴ x = 60 , y = 60, z = 70 and t = 70

Question 16:

1. Lines l and m will be parallel if 3x – 20 = 2x + 10

[Since, if corresponding angles are equal, lines are parallel]

⇒ 3x – 2x = 10 + 20

⇒ x = 30

1. Lines will be parallel if (3x + 5)o + 4xo = 180o

[if sum of pairs of consecutive interior angles is 180o, the lines are parallel] So, (3x + 5) + 4x = 180

⇒ 3x + 5 + 4x = 180

⇒ 7x = 180 – 5 = 175

⇒ x =

= 25

Question 17:

Given: Two lines m and n are perpendicular to a given line l.

To Prove: m || n Proof : Since m ⊥ l So, ∠1 = 90o Again, since n ⊥ l

∠2 = 90o

∴ ∠1 = ∠2 = 90o

But ∠1 and ∠2 are the corresponding angles made by the transversal l with lines m and n and they are proved to be equal.

Thus, m || n.

Question 1:

Exercise 4D

Since, sum of the angles of a triangle is 180o

∠A + ∠B + ∠C = 180o

⇒ ∠A + 76o + 48o = 180o

⇒ ∠A = 180o – 124o = 56o

∴ ∠A = 56o

Question 2:

Let the measures of the angles of a triangle are (2x)o, (3x)o and (4x)o. Then, 2x + 3x + 4x = 180 [sum of the angles of a triangle is 180o ]

⇒ 9x = 180

⇒ x =

= 20

∴ The measures of the required angles are: 2x = (2 × 20)o = 40o

3x = (3 × 20)o = 60o

4x = (4 × 20)o = 80o

Question 3:

Let 3∠A = 4∠B = 6∠C = x (say) Then, 3∠A = x

⇒ ∠A =

4∠B = x

⇒ ∠B =

and 6∠C = x

⇒ ∠C =

As ∠A + ∠B + ∠C = 180o

Question 4:

∠A + ∠B = 108o [Given]

But as ∠A, ∠B and ∠C are the angles of a triangle,

∠A + ∠B + ∠C = 180o

⇒ 108o + ∠C = 180o

⇒ C = 180o – 108o = 72o

Also, ∠B + ∠C = 130o [Given]

⇒ ∠B + 72o = 130o

⇒ ∠B = 130o – 72o = 58o Now as, ∠A + ∠B = 108o

⇒ ∠A + 58o = 108o

⇒ ∠A = 108o – 58o = 50o

∴ ∠A = 50o, ∠B = 58o and ∠C = 72o.

Question 5:

Since. ∠A , ∠B and ∠C are the angles of a triangle . So, ∠A + ∠B + ∠C = 180o

Now, ∠A + ∠B = 125o [Given]

∴ 125o + ∠C = 180o

⇒ ∠C = 180o – 125o = 55o

Also, ∠A + ∠C = 113o [Given]

⇒ ∠A + 55o = 113o

⇒ ∠A = 113o – 55o = 58o Now as ∠A + ∠B = 125o

⇒ 58o + ∠B = 125o

⇒ ∠B = 125o – 58o = 67o

∴ ∠A = 58o, ∠B = 67o and ∠C = 55o.

Question 6:

Since, ∠P, ∠Q and ∠R are the angles of a triangle. So, ∠P + ∠Q + ∠R = 180o ….(i)

Now, ∠P – ∠Q = 42o [Given]

⇒ ∠P = 42o + ∠Q ….(ii)

and ∠Q – ∠R = 21o [Given]

⇒ ∠R = ∠Q – 21o ….(iii)

Substituting the value of ∠P and ∠R from (ii) and (iii) in (i), we get,

⇒ 42o + ∠Q + ∠Q + ∠Q – 21o = 180o

⇒ 3∠Q + 21o = 180o

⇒ 3∠Q = 180o – 21o = 159o

∠Q =

= 53o

∴ ∠P = 42o + ∠Q

= 42o + 53o = 95o

∠R = ∠Q – 21o

= 53o – 21o = 32o

∴ ∠P = 95o, ∠Q = 53o and ∠R = 32o.

Question 7:

Given that the sum of the angles A and B of a ABC is 116o, i.e., ∠A + ∠B = 116o. Since, ∠A + ∠B + ∠C = 180o

So, 116o + ∠C = 180o

⇒ ∠C = 180o – 116o = 64o

Also, it is given that:

∠A – ∠B = 24o

⇒ ∠A = 24o + ∠B

Putting, ∠A = 24o + ∠B in ∠A + ∠B = 116o, we get,

⇒ 24o + ∠B + ∠B = 116o

⇒ 2∠B + 24o = 116o

⇒ 2∠B = 116o – 24o = 92o

∠B =

= 46o

Therefore, ∠A = 24o + 46o = 70o

∴ ∠A = 70o, ∠B = 46o and ∠C = 64o.

Question 8:

Let the two equal angles, A and B, of the triangle be xo each. We know,

∠A + ∠B + ∠C = 180o

⇒ xo + xo + ∠C = 180o

⇒ 2xo + ∠C = 180o ….(i) Also, it is given that,

∠C = xo + 18o ….(ii)

Substituting ∠C from (ii) in (i), we get,

⇒ 2xo + xo + 18o = 180o

⇒ 3xo = 180o – 18o = 162o x =

= 54o

Thus, the required angles of the triangle are 54o, 54o and xo + 18o = 54o + 18o = 72o.

Question 9:

Let ∠C be the smallest angle of ABC. Then, ∠A = 2∠C and B = 3∠C

Also, ∠A + ∠B + ∠C = 180o

⇒ 2∠C + 3∠C + ∠C = 180o

⇒ 6∠C = 180o

⇒ ∠C = 30o

So, ∠A = 2∠C = 2 (30o) = 60o

∠B = 3∠C = 3 (30o) = 90o

∴ The required angles of the triangle are 60o, 90o, 30o.

Question 10:

Let ABC be a right angled triangle and ∠C = 90o Since, ∠A + ∠B + ∠C = 180o

⇒ ∠A + ∠B = 180o – ∠C = 180o – 90o = 90o

Suppose ∠A = 53o

Then, 53o + ∠B = 90o

⇒ ∠B = 90o – 53o = 37o

∴ The required angles are 53o, 37o and 90o.

Question 11:

Let ABC be a triangle. Given, ∠A + ∠B = ∠C

We know, ∠A + ∠B + ∠C = 180o

⇒ ∠C + ∠C = 180o

⇒ 2∠C = 180o

⇒ ∠C =

= 90o

So, we find that ABC is a right triangle, right angled at C.

Question 12:

Given : ∆ABC in which ∠A = 90o, AL ⊥ BC To Prove: ∠BAL = ∠ACB

Proof :

In right triangle ∆ABC,

⇒ ∠ABC + ∠BAC + ∠ACB = 180o

⇒ ∠ABC + 90o + ∠ACB = 180o

⇒ ∠ABC + ∠ACB = 180o – 90o

∴ ∠ABC + ∠ACB = 90o

⇒ ∠ ACB = 90o – ∠ABC ….(1)

Similarly since ∆ABL is a right triangle, we find that,

∠BAL = 90o – ∠ABC …(2)

Thus from (1) and (2), we have

∴ ∠BAL = ∠ACB (Proved)

Question 13:

Let ABC be a triangle. So, ∠A < ∠B + ∠C

Adding A to both sides of the inequality,

⇒ 2∠A < ∠A + ∠B + ∠C

⇒ 2∠A < 180o

⇒ ∠A <

= 90o

[Since ∠A + ∠B + ∠C = 180o]

Similarly, ∠B < ∠A + ∠C

⇒ ∠B < 90o

and ∠C < ∠A + ∠B

⇒ ∠C < 90o

∆ABC is an acute angled triangle.

Question 14:

Let ABC be a triangle and ∠B > ∠A + ∠C Since, ∠A + ∠B + ∠C = 180o

⇒ ∠A + ∠C = 180o – ∠B Therefore, we get

∠B > 180o – ∠B

Adding ∠B on both sides of the inequality, we get,

⇒ ∠B + ∠B > 180o – ∠B + ∠B

⇒ 2∠B > 180o

⇒ ∠B >

= 90o

i.e., ∠B > 90o which means ∠B is an obtuse angle.

∆ABC is an obtuse angled triangle.

Question 15:

Since ∠ACB and ∠ACD form a linear pair. So, ∠ACB + ∠ACD = 180o

⇒ ∠ACB + 128o = 180o

⇒ ∠ACB = 180o – 128 = 52o

Also, ∠ABC + ∠ACB + ∠BAC = 180o

⇒ 43o + 52o + ∠BAC = 180o

⇒ 95o + ∠BAC = 180o

⇒ ∠BAC = 180o – 95o = 85o

∴ ∠ACB = 52o and ∠BAC = 85o.

Question 16:

As ∠DBA and ∠ABC form a linear pair. So, ∠DBA + ∠ABC = 180o

⇒ 106o + ∠ABC = 180o

⇒ ∠ABC = 180o – 106o = 74o

Also, ∠ACB and ∠ACE form a linear pair. So, ∠ACB + ∠ACE = 180o

⇒ ∠ACB + 118o = 180o

⇒ ∠ACB = 180o – 118o = 62o

In ∠ABC, we have,

∠ABC + ∠ACB + ∠BAC = 180o 74o + 62o + ∠BAC = 180o

⇒ 136o + ∠BAC = 180o

⇒ ∠BAC = 180o – 136o = 44o

∴ In triangle ABC, ∠A = 44o, ∠B = 74o and ∠C = 62o

Question 17:

(i) ∠EAB + ∠BAC = 180o [Linear pair angles]

110o + ∠BAC = 180o

⇒ ∠BAC = 180o – 110o = 70o

Again, ∠BCA + ∠ACD = 180o [Linear pair angles]

⇒ ∠BCA + 120o = 180o

⇒ ∠BCA = 180o – 120o = 60o

Now, in ∆ABC,

∠ABC + ∠BAC + ∠ACB = 180o

xo + 70o + 60o = 180o

⇒ x + 130o = 180o

⇒ x = 180o – 130o = 50o

∴ x = 50 (ii)

In ∆ABC,

∠A + ∠B + ∠C = 180o

⇒ 30o + 40o + ∠C = 180o

⇒ 70o + ∠C = 180o

⇒ ∠C = 180o – 70o = 110o

Now ∠BCA + ∠ACD = 180o [Linear pair]

⇒ 110o + ∠ACD = 180o

⇒ ∠ACD = 180o – 110o = 70o In ∆ECD,

⇒ ∠ECD + ∠CDE + ∠CED = 180o

⇒ 70o + 50o + ∠CED = 180o

⇒ 120o + ∠CED = 180o

∠CED = 180o – 120o = 60o

Since ∠AED and ∠CED from a linear pair So, ∠AED + ∠CED = 180o

⇒ xo + 60o = 180o

⇒ xo = 180o – 60o = 120o

∴ x = 120 (iii)

∠EAF = ∠BAC [Vertically opposite angles]

⇒ ∠BAC = 60o

In ∆ABC, exterior ∠ACD is equal to the sum of two opposite interior angles. So, ∠ACD = ∠BAC + ∠ABC

⇒ 115o = 60o + xo

⇒ xo = 115o – 60o = 55o

∴ x = 55 (iv)

Since AB || CD and AD is a transversal. So, ∠BAD = ∠ADC

⇒ ∠ADC = 60o

In ∠ECD, we have,

∠E + ∠C + ∠D = 180o

⇒ xo + 45o + 60o = 180o

⇒ xo + 105o = 180o

⇒ xo = 180o – 105o = 75o

∴ x = 75 (v)

In ∆AEF,

Exterior ∠BED = ∠EAF + ∠EFA

⇒ 100o = 40o + ∠EFA

⇒ ∠EFA = 100o – 40o = 60o

Also, ∠CFD = ∠EFA [Vertically Opposite angles]

⇒ ∠CFD = 60o Now in ∆FCD,

Exterior ∠BCF = ∠CFD + ∠CDF

⇒ 90o = 60o + xo

⇒ xo = 90o – 60o = 30o

∴ x = 30 (vi)

In ∆ABE, we have,

∠A + ∠B + ∠E = 180o

⇒ 75o + 65o + ∠E = 180o

⇒ 140o + ∠E = 180o

⇒ ∠E = 180o – 140o = 40o

Now, ∠CED = ∠AEB [Vertically opposite angles]

⇒ ∠CED = 40o

Now, in ∆CED, we have,

∠C + ∠E + ∠D = 180o

⇒ 110o + 40o + xo = 180o

⇒ 150o + xo = 180o

⇒ xo = 180o – 150o = 30o

∴ x = 30

Question 18:

Produce CD to cut AB at E.

Now, in ∆BDE, we have,

Exterior ∠CDB = ∠CEB + ∠DBE

⇒ xo = ∠CEB + 45o In ∆AEC, we have,

…..(i)

Exterior ∠CEB = ∠CAB + ∠ACE

= 55o + 30o = 85o

Putting ∠CEB = 85o in (i), we get, xo = 85o + 45o = 130o

∴ x = 130

Question 19:

The angle ∠BAC is divided by AD in the ratio 1 : 3. Let ∠BAD and ∠DAC be y and 3y, respectively. As BAE is a straight line,

∠BAC + ∠CAE = 180o

[linear pair]

⇒ ∠BAD + ∠DAC + ∠CAE = 180o

⇒ y + 3y + 108o = 180o

⇒ 4y = 180o – 108o = 72o

⇒ y =

= 18o Now, in ∆ABC,

∠ABC + ∠BCA + ∠BAC = 180o

y + x + 4y = 180o

[Since, ∠ABC = ∠BAD (given AD = DB) and ∠BAC = y + 3y = 4y]

⇒ 5y + x = 180

⇒ 5 × 18 + x = 180

⇒ 90 + x = 180

∴ x = 180 – 90 = 90

Question 20:

Given : A ∆ABC in which BC, CA and AB are produced to D, E and F respectively. To prove : Exterior ∠DCA + Exterior ∠BAE + Exterior ∠FBD = 360o

Proof : Exterior ∠DCA = ∠A + ∠B ….(i) Exterior ∠FAE = ∠B + ∠C ….(ii) Exterior ∠FBD = ∠A + ∠C ….(iii) Adding (i), (ii) and (iii), we get,

Ext. ∠DCA + Ext. ∠FAE + Ext. ∠FBD

= ∠A + ∠B + ∠B + ∠C + ∠A + ∠C

= 2∠A + 2∠B + 2∠C

= 2 (∠A + ∠B + ∠C)

= 2 × 180o

[Since, in triangle the sum of all three angle is 180o]

= 360o

Hence, proved.

Question 21:

In ∆ACE, we have,

∠A + ∠C + ∠E = 180o ….(i) In ∆BDF, we have,

∠B + ∠D + ∠F = 180o ….(ii)

Adding both sides of (i) and (ii), we get,

∠A + ∠C +∠E + ∠B + ∠D + ∠F = 180o + 180o

⇒ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360o.

Question 22:

Given : In ∆ABC, bisectors of ∠B and ∠C meet at O and ∠A = 70o In ∆BOC, we have,

∠BOC + ∠OBC + ∠OCB = 180o

= 180o – 55o = 125o

∴ ∠BOC = 125o.

Question 23:

We have a ∆ABC whose sides AB and AC have been procued to D and E. A = 40o and bisectors of ∠CBD and ∠BCE meet at O.

In ∆ABC, we have, Exterior ∠CBD = C + 40o

And exterior ∠BCE = B + 40o

Now, in ∆BCO, we have,

= 50o + 20o

= 70o

Thus, ∠BOC = 70o

Question 24:

In the given ∆ABC, we have,

∠A : ∠B : ∠C = 3 : 2 : 1

Let ∠A = 3x, ∠B = 2x, ∠C = x. Then,

∠A + ∠B + ∠C = 180o

⇒ 3x + 2x + x = 180o

⇒ 6x = 180o

⇒ x = 30o

∠A = 3x = 3 30o = 90o

∠B = 2x = 2 30o = 60o

and, ∠C = x = 30o

Now, in ∆ABC, we have,

Ext ∠ACE = ∠A + ∠B = 90o + 60o = 150o

∠ACD + ∠ECD = 150o

⇒ ∠ECD = 150o – ∠ACD

⇒ ∠ECD = 150o – 90o

⇒ ∠ECD= 60o

[since , AD ⊥ CD, ∠ACD = 90o]

Question 25:

In ∆ABC, AN is the bisector of ∠A and AM ⊥ BC. Now in ∆ABC we have;

∠A = 180o – ∠B – ∠C

⇒ ∠A = 180o – 65o – 30o

= 180o – 95o

= 85o

Now, in ∆ANC we have;

Thus, ∠MAN =

Question 26:

(i) False (ii) True (iii) False (iv) False (v) True (vi) True.

CoordinateGeometry

Question 1:

Exercise 6A

Draw the perpendiculars from the AF, BG, CH, DI and EJ on the x-axis.

1. The distance of A from the y-axis = OF = -6 units The distance of A from the x-axis = AF = 5 units Hence, the coordinate of A are (-6, 5)
2. The distance of B from the y-axis = OG = 5 units The distance of B from the x-axis = BG = 4 units Hence, the coordinate of B are (5, 4)
3. The distance of C from the y-axis = OH = -3 units The distance of C from the x-axis = HC = 2 units Hence, the coordinate of C are (-3, 2)
4. The distance of D from the y-axis = OI = 2 units The distance of D from the x-axis = ID = -2 units Hence, the coordinate of D are (2, -2)
5. The distance of E from the y-axis = OJ = -1 unit The distance of E from the x-axis = JE = -4 units Hence, the coordinate of E are (-1, -4)

Thus, the coordinates of A, B, C, D and E are respectively, A(-6,5), B(5,4), C(-3,2), D(2,-2) and E(-1,-4)

Question 2:

Let X’OX and Y’OY be the coordinate axes.

Fix the side of the small squares as one units.

• 1. Starting from O, take +7 units on the x-axis and then +4 units on the y-axis to obtain the point P(7, 4)
  2. Starting from O, take -5 units on the x-axis and then +3 units on the y-axis to obtain the point Q(-5, 3)
  3. Starting from O, take -6 units on the x-axis and then -3 units on the y-axis to obtain the point R(-6, -3)
  4. Starting from O, take +3 units on the x-axis and then -7 units on the y-axis to obtain the point S(3, -7)
  5. Starting from O, take 6 units on the x-axis to obtain the point A(6, 0)
  6. Starting from O, take 9 units on the y-axis to obtain the point B(0,9)
  7. Mark the point O as O(0, 0)
  8. Starting from O, take -3 units on the x-axis and then -3 units on the y-axis to obtain the point C(-3, -3)

These points are shown in the following graph:

Question 3:

1. In (7, 0), we have the ordinate = 0.

Therefore, (7,0) lies on the x-axis

1. In (0, -5), we have the abscissa = 0.

Therefore, (0,-5) lies on the y-axis

1. In (0,1), we have the abscissa = 0.

Therefore, (0,1) lies on the y-axis

1. In (-4,0), we have the ordinate = 0.

Therefore, (-4,0) lies on the x-axis

Question 4:

1. Points of the type (-, +) lie in the second quadrant. Therefore, the point (-6,5) lies in the II quadrant.
2. Points of the type (-, -) lie in the third quadrant. Therefore, the point (-3,-2) lies in the III quadrant.
3. Points of the type (+, -) lie in the fourth quadrant. Therefore, the point (2,-9) lies in the IV quadrant.

Question 5:

The given equation is y = x + 1 Putting x = 1, we get y = 1 + 1 = 2

Putting x = 2, we get y = 2 + 1 = 3 Thus, we have the following table:

On a graph paper, draw the lines X’OX and YOY’ as the x-axis and y-axis respectively. Then, plot points P (1, 2) and Q (2, 3) on the graph paper. Join PQ and extend it to both sides.

Then, line PQ is the graph of the equation y = x + 1.

Question 6:

The give equation is y = 3x + 2 Putting x = 1, we get y = (3 1) + 2 = 5

Putting x = 2, we get y = (3 2) + 2 = 8 Thus, we have the following table:

On the graph paper, draw the lines X’OX and YOY’ as the x-axis and y-axis respectively. Now, plot points P(1,5) and Q(2,8) on the graph paper.

Join PQ and extend it to both sides.

Then, line PQ is the graph of the equation y = 3x + 2.

Question 7:

The given equation is y = 5x – 3 Putting x = 0, we get y = (5 × 0) – 3 = -3

Putting x = 1, we get y = (5 × 1) – 3 = 2 Thus, we have following table:

On a graph paper, draw the lines X’OX and YOY’ as the x-axis and y-axis respectively. Now plot the points P(0,-3) and Q(1,2).

Join PQ and extend it in both the directions.

Then, line PQ is the graph of the equation, y = 5x – 3.

Question 8:

The given equation is y = 3x Putting x = 1, we get y = (3 1) = 3

Putting x = 2, we get y = (3 2) = 6 Thus, we have the following table:

On a graph paper draw the lines X’OX and YOY’ as the x-axis and y-axis respectively. Now, plot points P(1,3) and Q(2,6).

Join PQ and extend it in both the directions. Then, line PQ is the graph of the equation y = 3x.

Question 9:

The given equation is y = -x

Putting x = 1, we get y = -1 Putting x = 2, we get y = -2

Thus, we have the following table:

On a graph paper, draw the lines X’OX and YOY’ as the x-axis and y-axis respectively. Now, plot the points P(1,-1) and Q(2,-2).

Join PQ and extend it in both the directions. Then, line PQ is the graph of the equation y = -x.

Introduction toEuclid’s Geometry

Question 1:

Exercise 3A

A theorem is a statement that requires a proof. Whereas, a basic fact which is taken for granted, without proof, is called an axiom.

Example of Theorem: Pythagoras Theorem

Example of axiom: A unique line can be drawn through any two points.

Question 2:

1. Line segment: The straight path between two points is called a line segment.
2. Ray: A line segment when extended indefinitely in one direction is called a ray.
3. Intersecting Lines: Two lines meeting at a common point are called intersecting lines, i.e., they have a common point.
4. Parallel Lines: Two lines in a plane are said to be parallel, if they have no common point, i.e., they do not meet at all.
5. Half-line: A ray without its initial point is called a half-line.
6. Concurrent lines: Three or more lines are said to be concurrent, if they intersect at the same point.
7. Collinear points: Three or more than three points are said to be collinear, if they lie on the same line.
8. Plane: A plane is a surface such that every point of the line joining any two points on it, lies on it.

Question 3:

1. Six points: A,B,C,D,E,F
2. Five line segments: , , , ,

1. Four rays: , , ,

1. Four lines: , , ,
2. Four collinear points: M,E,G,B

Question 4:

• 1. and their corresponding point of intersection is R. and their corresponding point of intersection is P.
  2. , , and their point of intersection is R.
  3. Three rays are: , ,
  4. Two line segments are: ,

Question 5:

1. An infinite number of lines can be drawn to pass through a given point.
2. One and only one line can pass through two given points.
3. Two given lines can at the most intersect at one and only one point.
4. , ,

Question 6:

1. False
2. False
3. False
4. True
5. False
6. True
7. True
8. True
9. True
10. False
11. False
12. True

Real Numbers

Question 1:

Exercise 1A

The numbers of the form , where p and q are integers and q ≠ 0 are known as rational numbers.

Ten examples of rational numbers are:

, , , , , , , , 1,

Question 2:

1. 5
2. -3

(v) 1.3

(vi) -2.4

(vii)

Question 3:

A rational number lying between and is

Therefore, we have < < < < Or we can say that, < < < < That is, < < < <

Therefore, three rational numbers between and are

, and

Question 5:

Let and

Then, x < y because <

Or we can say that, That is, < .

We know that, 8 < 9 < 10 < 11 < 12 < 13 < 14 < 15.

Therefore, we have, < < < < < < < Thus, 5 rational numbers between, < are:

, , , and

Question 6:

Let x = 3 and y = 4

Then, x < y, because 3 < 4 We can say that, < .

We know that, 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28.

Therefore, we have, < < < < < < < Therefore, 6 rational numbers between 3 and 4 are:

, , , and

Question 7:

Let x = 2.1 and y = 2.2

Then, x < y because 2.1 < 2.2 Or we can say that, < Or,

That is, we have, <

We know that, 2100 < 2105 < 2110 < 2115 < 2120 < 2125 < 2130 < 2135 < 2140 <

2145 < 2150 < 2155 < 2160 < 2165 < 2170 < 2175 < 2180 < 2185 < 2190 < 2195 <

2200

Therefore, we can have,

Therefore, 16 rational numbers between, 2.1 and 2.2 are:

So, 16 rational numbers between 2.1 and 2.2 are:

2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17,

2.175, 2.18

Exercise 1B

Question 1:

(i)

If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.

Since, 80 has prime factors 2 and 5, is a terminating decimal.

(ii)

If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.

Since, 24 has prime factors 2 and 3 and 3 is different from 2 and 5, is not a terminating decimal.

(iii)

If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.

Since 12 has prime factors 2 and 3 and 3 is different from 2 and 5, is not a terminating decimal.

(iv)

If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.

Since 35 has prime factors 5 and 7, and 7 is different from 2 and 5, is not a terminating decimal.

(v)

If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.

Since 125 has prime factor 5 only is a terminating decimal.

Question 2:

(i)

= 0.625

(ii)

= 0.5625

(iii)

= 0.28

(iv)

= 0.458

(v)

= 2.41

Question 3:

1. Let x =

i.e x = 0.333 …. (i)

⇒ 10x = 3.333 …. (ii) Subtracting (i) from (ii), we get 9x = 3

⇒ x = = Hence, 0. =

1. Let x = 1.

i.e x = 1.333 …. (i)

⇒10x = 13.333 …. (ii)

Subtracting (i) from (ii) we get; 9x = 12

⇒ x = =

Hence, 1. =

1. Let x = 0.

i.e x = 0.3434 …. (i)

⇒ 100x = 34.3434 …. (ii)

Subtracting (i) from (ii), we get 99x = 34

⇒ x =

Hence, 0. =

1. Let x = 3.

i.e x = 3.1414 …. (i)

⇒ 100x = 314.1414 …. (ii)

Subtracting (i) from (ii), we get 99x = 311

⇒ x =

Hence, 3. =

1. Let x = 0.

i.e. x = 0.324324 ….(i)

⇒ 1000x = 324.324324….(ii)

Subtracting (i) from (ii), we get 999x = 324

⇒ x = =

Hence, 0. =

1. Let x = 0.

i.e. x = 0.177 …. (i)

⇒ 10x = 1.777 …. (ii)

and 100x = 17.777…. (iii) Subtracting (ii) from (iii), we get 90x = 16

⇒ x = =

Hence, 0. =

1. Let x = 0.

i.e. x = 0.544 …. (i)

⇒ 10 x = 5.44 …. (ii) and 100x = 54.44 ….(iii)

Subtracting (ii) from (iii), we get 90x = 49

⇒ x =

Hence, 0. =

(vii) Let x = Let x = 0.1 i.e. x = 0.16363 …. (i)

⇒ 10x = 1.6363 …. (ii)

and 1000 x = 163.6363 …. (iii) Subtracting (ii) from (iii), we get

990x = 162

⇒ x = = Hence, 0.1 =

Question 4:

1. True. Since the collection of natural number is a sub collection of whole numbers, and every element of natural numbers is an element of whole numbers
2. False. Since 0 is whole number but it is not a natural number.
3. True. Every integer can be represented in a fraction form with denominator 1.
4. False. Since division of whole numbers is not closed under division, the value of , p and q are integers and q ≠ 0, may not be a whole number.
5. True. The prime factors of the denominator of the fraction form of terminating

decimal contains 2 and/or 5, which are integers and are not equal to zero.

1. True. The prime factors of the denominator of the fraction form of repeating decimal contains integers, which are not equal to zero.
2. True. 0 can considered as a fraction , which is a rational number.

Question 1:

Exercise 1C

Irrational number: A number which cannot be expressed either as a terminating decimal or a repeating decimal is known as irrational number. Rather irrational numbers cannot

be expressed in the fraction form, , p and q are integers and q ≠ 0

For example, 0.101001000100001 is neither a terminating nor a repeating decimal and so is an irrational number.

Also, etc are examples of irrational numbers.

Question 2:

(i)

We know that, if n is a perfect square, then is a rational number. Here, 4 is a perfect square and hence, = 2 is a rational number. So, is a rational number.

(ii)

We know that, if n is a perfect square, then is a rational number. Here, 196 is a perfect square and hence is a rational number. So, is rational.

(iii)

We know that, if n is a not a perfect square, then is an irrational number. Here, 21 is a not a perfect square number and hence, is an irrational number. So, is irrational.

(iv)

We know that, if n is a not a perfect square, then is an irrational number. Here, 43 is not a perfect square number and hence, is an irrational number. So, is irrational.

(v)

, is the sum of a rational number 3 and irrational number .

Theorem: The sum of a rational number and an irrational number is an irrational number.

So by the above theorem, the sum, , is an irrational number.

(vi)

= + (-2) is the sum of a rational number and an irrational number.

Theorem: The sum of a rational number and an irrational number is an irrational number.

So by the above theorem, the sum, + (-2) , is an irrational number.

So, is irrational.

(vii)

= × is the product of a rational number and an irrational number .

Theorem: The product of a non-zero rational number and an irrational number is an irrational number.

Thus, by the above theorem, × is an irrational number.

So, is an irrational number.

1. 0.

Every rational number can be expressed either in the terminating form or in the non- terminating, recurring decimal form.

Therefore, 0. = 0.6666

Question 3:

Let X’OX be a horizontal line, taken as the x-axis and let O be the origin. Let O represent 0.

Take OA = 1 unit and draw BA ⊥ OA such that AB = 1 unit, join OB. Then,

With O as centre and OB as radius, drawn an arc, meeting OX at P. Then, OP = OB = units

Thus the point P represents on the real line.

Now draw BC ⊥ OB such that BC = 1 units Join OC. Then,

With O as centre and OC as radius, draw an arc, meeting OX at Q. The, OQ = OC = units

Thus, the point Q represents on the real line. Now draw CD ⊥ OC such that CD = 1 units

Join OD. Then,

Now draw DE ⊥ OD such that DE = 1 units Join OE. Then,

With O as centre and OE as radius draw an arc, meeting OX at R. Then, OR = OE = units

Thus, the point R represents on the real line.

Question 4:

Draw horizontal line X’OX taken as the x-axis Take O as the origin to represent 0.

Let OA = 2 units and let AB ⊥ OA such that AB = 1 units

Join OB. Then,

With O as centre and OB as radius draw an arc meeting OX at P. Then, OP = OB =

Now draw BC ⊥ OB and set off BC = 1 unit Join OC. Then,

With O as centre and OC as radius, draw an arc, meeting OX at Q. Then, OQ = OC =

Thus, Q represents on the real line. Now, draw CD ⊥ OC as set off CD = 1 units Join OD. Then,

With O as centre and OD as radius, draw an arc, meeting OX at R. Then OR = OD =

Thus, R represents on the real line.

Question 5:

(i)

Since 4 is a rational number and is an irrational number.

So, is irrational because sum of a rational number and irrational number is always an irrational number.

(ii)

Since – 3 is a rational number and is irrational.

So, is irrational because sum of a rational number and irrational number is always an irrational number.

(iii)

Since 5 is a rational number and is an irrational number.

So, is irrational because product of a rational number and an irrational number is always irrational.

(iv)

Since -3 is a rational number and is an irrational number.

So, is irrational because product of a rational number and an irrational number is always irrational.

(v)

is irrational because it is the product of a rational number and the irrational number .

(vi)

is an irrational number because it is the product of rational number and irrational number .

Question 6:

1. True
2. False
3. True
4. False
5. True
6. False
7. False
8. True
9. True

Exercise 1D

Question 1:

(i)

We have:

(ii)

We have:

Question 2:

Question 3:

(i) by

(ii) by

(iii) by

Question 4:

Question 5:

Draw a line segment AB = 3.2 units and extend it to C such that BC = 1 units. Find the midpoint O of AC.

With O as centre and OA as radius, draw a semicircle. Now, draw BD AC, intersecting the semicircle at D. Then, BD = units.

With B as centre and BD as radius, draw an arc meeting AC produced at E. Then, BE = BD = units.

Question 6:

Draw a line segment AB = 7.28 units and extend it to C such that BC = 1 unit. Find the midpoint O of AC.

With O as centre and OA as radius, draw a semicircle.

Now, draw BD AC, intersecting the semicircle at D. Then, BD = units.

With D as centre and BD as radius, draw an arc, meeting AC produced at E. Then, BE = BD = units.

Question 7:

Closure Property: The sum of two real numbers is always a real number. Associative Law: (a + b) + c = a + (b + c) for al real numbers a, b, c.

Commutative Law: a + b = b + a, for all real numbers a and b.

Existence of identity: 0 is a real number such that 0 + a = a + 0, for every real number a. Existence of inverse of addition: For each real number a, there exists a real number (-a) such that

a + (-a) = (-a) + a= 0

a and (-a) are called the additive inverse of each other. Existence of inverse of multiplication:

For each non zero real number a, there exists a real number such that

a and are called the multiplicative inverse of each other.

Question 1:

Exercise 1E

On multiplying the numerator and denominator of the given number by , we get

Question 2:

On multiplying the numerator and denominator of the given number by , we get

Question 3:

Question 4:

Question 5:

Question 6:

Question 7:

Question 8:

Question 9:

Question 10:

Question 11:

Question 12:

Question 13:

Question 14:

Question 15:

Question 16:

Question 17:

Question 18:

Exercise 1F

Question 1:

(i)

(ii)

(iii)

Question 2:

(i)

(ii)

(iii)

Question 3:

(i)

(ii)

(iii)

Question 4:

(i)

(ii)

(iii)

Question 5:

(i)

(ii)

(iii)

Question 6:

(i)

(ii)

(iii)

Question 7:

(i)

(ii)

(iii)

Chapter 1 5 Probability

• Probability: Probability is a quantitative measure of certainty.
• Experiment: A job which produces some outcomes.
• Trial: Performing an experiment.
• Event: The group of outcomes, denoted by capital letter of English alphabets like A, B, E etc.
• The empirical (or experimental) probability P(E) of an event E is given by

• The probability of an event lies between 0 and 1 (0 and 1 are included)
• Impossible event: Event which never happen.
• Certain event: Event which definitely happen.
• The probability of sure event is 1.
• The probability of an impossible event is 0.
• The probability of an event E is a number P(E) such that 0 < P(E) < I.

1. Collection of Data
2. Presentation of Data
3. Graphical Representation of Data
4. Measures of Central Tendency

Chapter 13 Surface Areas and Volumes

1. Surface Area of a Cuboid and a Cube
2. Surface Area of a Right Circular Cylinder
3. Surface Area of a Right Circular Cone
4. Surface Area of a Sphere
5. Volume of a Cuboid
6. Volume of a Cylinder
7. Volume of a Right Circular Cone
8. Volume of a Sphere

Chapter 12 Heron’s Formula

1. Area of a Triangle – by Heron’s Formula
2. Application of Heron’s Formula in finding Areas of Quadrilaterals

• Triangle with base ‘b’ and altitude ‘h’ is