Author Archives: Swati

Exercise 4.1

Question: 1

Fill in the blanks to make each of the following a true statement:

Solution:

(i) 359 + 476 = 476 + 359 (Commutativity)

(ii) 2008 + 1952 = 1952 + 2008 (Commutativity)

(iii) 90758 + 0 = 90758 (Additive identity)

(iv) 54321 + (489 + 699) = 489 + (54321 + 699) (Associativity)

Question: 2

Add each of the following and check by reversing the order of addends:

Solution:

(i) 5628 + 39784 = 45412

And,

39784 + 5628 = 45412

(ii) 923584 + 178 = 923762

And,

178 + 923584 = 923762

(iii) 15409 + 112 = 15521

And,

112 + 15409 = 15521

(iv) 2359 + 641 = 3000

And,

641 + 2359 = 3000

Question: 3

Determine the sum by suitable rearrangements:

Solution:

(i) 953 + 407 + 647

Therefore, 53 + 47 = 100

Therefore, (953 + 647) + 407 = 1600 + 407 = 2007

(ii) 15409 + 178 + 591 + 322

409 + 91 = 500

And,

78 + 22 = 100

Therefore, (15409 + 591) + (178 + 322) = (16000) + (500)

= 16500

(iii) 2359 + 10001 + 2641 + 9999

Therefore, 59 + 41 = 100

And, 99 + 01 = 100

Therefore, (2359 + 2641) + (10001 + 9999)

= (5000) + (20000)

= 25000

(iv) 1  + 2 + 3 + 4 + 1996 + 1997 + 1998 + 1999

Therefore, 99 + 1 = 100

98 + 2 = 100

97 + 3 = 100

And

96 + 4 = 100

Therefore, (1 + 1999) + (2 + 1998) + (3 + 1997) + (4 + 1996)

= 2000 + 2000 + 2000 + 2000

= 8000

(v) 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20

10 + 20 = 30

1 + 9 = 10

2 + 8 = 10

3 + 7 = 10

And,

4 + 6 = 10

Therefore, (10 + 20) + (11 + 19) + (12 + 18) + (13 + 17) + (14 + 16)

= 30 + 30 + 30 + 30 + 30 + 15

= 150 + 15

= 165

Question: 4

Which of the following statements are true and which are false?

(i) The sum of two odd numbers is an odd number.

(ii) The sum of two odd numbers is an even number.

(iii) The sum of two even numbers is an even number.

(iv)The sum of two even numbers is an odd number.

(v) The sum of an even number and an odd number is an odd number.

(vi)The sum of an odd number and an even number is an even number.

(vii) Every whole number is a natural number.

(viii) Every natural number is a whole number.

(ix) There is a whole number which when added to a whole number, gives that number

(x) There is a natural number which when added to a natural number, gives that number.

(xi) Commmutativity and associativity are properties of whole numbers.

(xii) Commmutativity and associativity are properties of addition of whole number.

Solution:

(i) FALSE (3 + 5 = 8; 8 is an even number)

(ii) TRUE (3 + 5 = 8; 8 is an even number)

(iii) TRUE (2 + 4 = 6; 6 is an even number)

(iv) FALSE (2 + 4 = 6; 6 is an even number)

(v) TRUE (2 + 3 = 5; 5 is an odd number)

(vi) FALSE (3 + 2 = 5; 5 is not an even number)

(vii) FALSE [The whole number set is {0, 1, 2, 3, 4 …), whereas the natural number set is {1, 2, 3, 4 …)]

(viii) TRUE [The whole number set is {0, 1, 2, 3, 4 …), whereas the natural number set is {1, 2, 3, 4 …)]

(ix) TRUE [That number is zero.]

(x) FALSE

(xi) FALSE

(xii) TRUE

Exercise 4.2

Question: 1

Perform the following subtractions and check your results by performing corresponding additions:

Solution:

(i) 57839 – 2983 = 54856

Verification: 54856 + 2983 = 57839

(ii) 92507 – 10879 = 81628

Verification: 81628 + 10879 = 92507

(iii) 400000 – 98798 = 301202

Verification: 301202 + 98798 = 400000

(iv) 5050501 – 969696 = 4080805

Verification: 4080805 + 969696 = 5050501

(v) 200000 – 97531 = 102469

Verification: 102469 + 97531 = 200000

(vi) 3030301 – 868686 = 2161615

Verification: 2161615 + 868686 = 3030301

Question: 2

Replace each * by the correct digit in each of the following:

Solution:

Here, we can we see that in the units digit, 6 – * = 7, which means that the value of * is 9, as 1 gets carried from 7 at tens place to 6 at unit place and 6 at unit digit becomes 16 then 16 – 9 = 7.

Now, when 7 gives 1 to 6, it becomes 6, so 6 – 3 = 3.

Also, it can be easily deduced that in (8 – * = 6 ), the value of * is 2.

(ii) Here, it is clear that in the units place, 9 – 4 = 5;

And in the tens place,

8 – 3 = 5.

We can now easily find out the other missing blanks by subtracting 3455 from 8989. Addend (difference) = 3455

Thus, the correct answer is:

(iii)

Here, in the units digit, 17 – 8 = 9; in the tens digit, 9 – 7 = 2;

in the hundreds place, 10 – 9 = 1;

and in the thousands place, 9 – 8 = 1.

Addend difference = 5061129.

So, in order to get the addend, we will subtract 5061129 from 6000107.

Thus, the correct answer is:

(iv)

In the units place, 10 -1 = 9;

Also, in the lakhs place, 9 -0 = 9;

Addend difference = 970429.

So, in order to get the addend, we will subtract 970429 from 1000000.

Thus, the correct answer is:

(v) 

Here, in the units digit, 13 – 7 = 6;

in the tens digit, 9 – 8 = 1;

in the hundreds place, 9 – 9 = 0;

and in the thousands place, 10 – 6 = 4.

Addend difference = 4844016.

So, in order to get the addend, we will subtract 4844016 from 5001003.

(vi)

It is clear from the units place that 11 – 9 = 2.

Addend difference = 54322.

To get the other addend, we will subtract 54322 from 111111.

Thus, the other addend is 56789.

The correct answer is:

Question: 3

What is the difference between the largest number of five digits and smallest number of six digits?

Solution:

The largest five – digit number is 99999.

The smallest six – digit number is 100000.

Therefore, difference between them = 100000 – 99999 = 1

Question: 4

Find the difference between the largest number of 4 digits and the smallest number of 6 digits.

Solution:

The largest four – digit number is 9999.

The smallest seven – digit number is 1000000.

Therefore, difference between them = 1000000 – 9999 = 990001

Question: 5

Rohit deposited Rs 125000 in his savings bank account. Later he withdrew Rs 35425 from it. How much money was left in his account?

Solution:

Money deposited by Rohit = Rs 125000

Money withdrawn by Rohit = Rs 35425

Therefore, money left in the account = Rs (125000 – 35425) = Rs 89575

Question: 6

The population of a town is 96209. If the number of men is 29642 and that of women is 29167, determine the number of children.

Solution:

Total population of the town = 96209

Number of men = 29642

Number of women = 29167

Sum of men and women = (29642 + 29167) = 58809

Therefore, Number of children in the town = ( Total population ) – ( Sum of men and women )

= 96209 – 58809 = 37400

Question: 7

The digits of 6 and 9 of the number 36490 are interchanged. Find the difference between the original number and the new number.

Solution:

Original number = 39460

New number = – 36490

Difference = 39460 – 36490 = 2970

Question: 8

The population of a town was 59000. In one year it was increased by 4563 due to new births. However, 9218 persons died or left the town during the year. What was the population at the end of the year?

Solution:

Population of the town = 59000

Increase in the population = 4536

Decrease in the population = 9218

New population = 59000 + 4536 – 9218 = 54318

Exercise 4.3

Question: 1

Fill in the blanks to make each of the following a true statement:

Solution:

(i) 785 × 0 = 0

(ii) 4567 × 1 = 4567 (Multiplicative identity)

(iii) 475 × 129 = 129 × 475 (Commutativity)

(iv) 1243 × 8975 = 8975 × 1243 (Commutativity)

(v) 10 × 100 × 10 = 10000

(vi) 27 × 18 = 27 × 9 + 27 × 4 + 27 × 5

(vii) 12 × 45 = 12 × 50 – 12 × 5

(viii) 78 × 89 = 78 × 100 – 78 × 16 + 78 × 5

(ix) 66 × 85 = 66 × 90 – 66 × 4 – 66

(x) 49 × 66 + 49 × 34 = 49 × (66 + 34)

Question: 2

Determine each of the following products by suitable rearrangements:

Solution:

(i) 2 × 1497 × 50

= (2 × 50) × 1497 = 100 × 1497 = 149700

(ii) 4 × 358 × 25

= (4 × 25) × 358 = 100 × 358 = 35800

(iii) 495 × 625 × 16

= (625 × 16) × 495 = 10000 × 495 = 4950000

(iv) 625 × 20 × 8 × 50

= (625 × 8) × (20 × 50) = 5000 × 1000 = 5000000

Question: 3

Using distributivity of multiplication over addition of whole numbers, find each of the following products:

Solution:

(i) 736 × 103 = 736 × (100 + 3)

{Using distributivity of multiplication over addition of whole numbers}

= (736 × 100) + (736 × 3)

= 73600 + 2208 = 75808

(ii) 258 × 1008 = 258 × (1000 + 8)

{Using distributivity of multiplication over addition of whole numbers}

= (258 × 1000) + (258 × 8)

= 258000 + 2064 = 260064

(iii) 258 × 1008 = 258 × (1000 + 8)

{Using distributivity of multiplication over addition of whole numbers}

= (258 × 1000) + (258 × 8)

= 258000 + 2064 = 260064

Question: 4

Find each of the following products:

Solution:

(i) 736 × 93

Since, 93 = (100 – 7)

Therefore, 736 × (100 – 7)

= (736 × 100) – (736 × 7)

(Using distributivity of multiplication over subtraction of whole numbers)

= 73600 – 5152 = 68448

(ii) 816 × 745

Since, 745 = (750 – 5)

Therefore, 816 × (750 – 5)

= (816 × 750) – (816 × 5)

(Using distributivity of multiplication over subtraction of whole numbers)

= 612000 – 4080 = 607920

(iii) 2032 × 613

Since, 613 = (600 +13)

Therefore, 2032 × (600 + 13)

= (2032 × 600) + (2032 × 13)

= 1219200 + 26416 = 1245616

Question: 5

Find the values of each of the following using properties:

Solution:

(i) 493 × 8 + 493 × 2

= 493 × (8 + 2)

(Using distributivity of multiplication over addition of whole numbers)

= 493 × 10 = 4930

(ii) 24579 × 93 + 7 × 24579

= 24579 × (93 + 7)

(Using distributivity of multiplication over addition of whole numbers)

= 24579 × 100 = 2457900

(iii) 1568 × 184 – 1568 × 84

= 1568 × (184 – 84)

(Using distributivity of multiplication over subtraction of whole numbers)

= 1568 × 100 = 156800

(iv) 15625 × 15625 – 15625 × 5625

= 15625 × (15625 – 5625)

(Using distributivity of multiplication over subtraction of whole numbers)

= 15625 × 10000 = 156250000

Question: 6

Determine the product of:

(i) the greatest number of four digits and the smallest number of three digits.

(ii) the greatest number of five digits and the greatest number of three digits.

Solution:

(i) The largest four-digit number = 9999

The smallest three – digit number = 100

Therefore, Product of the smallest three-digit number and the largest four-digit number = 9999 × 100 = 999900

(ii) The largest five – digit number = 9999

The largest number of three digits = 999

Therefore, Product of the largest three-digit number and the largest five-digit number

= 9999 × 999

= 9999 × (1000 — 1)

= (9999 × 1000) — (9999 × 1)

= 9999000 – 9999

= 9989001

Question: 7

In each of the following, fill in the blanks, so that the statement is true:

Solution:

(i) (500 + 7) (300 – 1)

= 507 × 299

= 299 × 507 (Commutativity)

(ii) 888 + 777 + 555

= 111 (8 + 7 + 5)

= 111 × 20 (Distributivity)

(iii) 75 × 425

= (70 + 5) × 425

= (70 + 5) (340 + 85)

(iv) 89 × (100 – 2)

= 89 × 98

= 98 × 89

= 98 × (100 – 11) (Commutativity)

(v) (15 + 5) (15 – 5)

= 20 × 10

= 200

= 225 – 25

(vi) 9 × (10000 + 974)

= 98766

Question: 8

A dealer purchased 125 color television sets. If the cost of each set is Rs 19820, determine the cost of all sets together.

Solution:

Cost of 1 color television set = Rs 19820

Therefore, Cost of 125 color television sets = Rs (19820 × 125)

= Rs 19820 × (100 + 25)

= Rs (19820 × 100) + (19820 × 25)

= Rs 1982000 + 495500

= Rs 2477500

Question: 9

The annual fee charged from a student of class 6^th in a school is Rs 8880. If there are, in all, 235 students in class 6^th, find the total collection.

Solution:

Fees charged from 1 student = Rs 8880

Therefore, Fees charged from 235 students = Rs 8880 × 235

= 2086800

Thus, the total collection from class VI students is Rs 2086800.

Question: 10

A group housing society constructed 350 flats. If the cost of construction for each flat is Rs 993570, what is the total cost of construction of all the flats.

Solution:

Cost of construction of 1 flat = Rs 993,570

Total number of flats constructed = 350

Total cost of construction of 350 flats = Rs (993,570 × 350)

= Rs 347,749,500

Question: 11

The product of two whole numbers is zero. What do you conclude?

Solution:

If the product of two whole numbers is zero, then it means that either one of them is zero or both of them are zero.

Question: 12

What are the whole numbers which when multiplied with itself gives the same number?

Solution:

There are two numbers which when multiplied with themselves give the same numbers.

(i) 0 × 0 = 0

(ii) 1 × 1 = 1

Question: 13

In a large housing complex, there are 15 small buildings and 22 large building. Each of the large buildings has 10 floors with 2 apartments on each floor. Each of the small buildings has 12 floors with 3 apartments on each floor. How many apartments are there in all.

Solution:

Number of large buildings = 22

Number of small buildings = 15

Number of floors in 1 large building = 10

Number of apartments on 1 floor = 2

Therefore, Total apartments in 1 large building = 10 × 2 = 20

Similarly,

Total apartments in 1 small building = 12 × 3 = 36

Therefore, Total apartments in the entire housing complex = (22 × 20) + (15 × 36)

= 440 + 540

= 980

Exercise 4.4

Question: 1

Does there exists a whole number ‘a’ such that a/a = a?

Solution:

Yes, there exists a whole number ‘a’ such that a/a = a.

The whole number is 1 such that,

1/1 = 1

Question: 2

Find the value of:

Solution:

(i) 23457 / 1 = 23457

(ii) 0 / 97 = 0

(iii) 476 + (840 / 84) = 476 + 10 = 486

(iv) 964 – (425 / 425) = 964 – 1 = 963

(v) (2758 / 2758) – (2758 + 2758) = 1 – 1 = 0

(vi) 72450 / (583 – 58) = 72450 + 525 = 138

Question: 3

Which of the following statements are true:

Solution:

(i) False

LHS: 10 / (5 × 2)

= 10 / 10

=1

RHS: (10 / 5) × (10 / 2)

= 2 × 5 = 10

(ii) True

LHS: (35 – 14) / 7

= 21 / 7

=3

RHS: 35 / 7 – 14 / 7

= 5 – 2 = 3

(iii) False

LHS: 35 – 14 / 7

= 35 – 2 = 33

RHS: 35 / 7 – 14 / 7

= 5 – 2

= 3

(iv) False

LHS: (20 – 5) / 5

= 15 / 5

= 3

RHS: 20 / 5 – 5

= 4 – 5 = -1

(v) False

LHS: 12 × (14 / 7)

= 12 × 2

= 24

RHS: (12 × 14) / (12 × 7)

= 168 / 84

=2

(vi) True

LHS: (20 / 5) / 2

= 4 / 2

=2

RHS : ( 20 / 2 ) / 5

= 10 / 5

= 2

Question: 4

Divide and check the quotient and remainder:

Solution:

(i) 7777 / 58 = 134

Verification: [Dividend = Divisor × Quotient + Remainder ]

7772 = 58 × 134 + 0

7772 = 7772

LHS = RHS

(ii) 6906/35 gives quotient = 197 and remainder = 11

Verification: [Dividend = Divisor × Quotient + Remainder ]

6906 = 35 × 197 + 11

6906 = 6895 + 11

6906 = 6906

LHS = RHS

(iii) 16135 / 875 gives quotient = 18 and remainder = 385.

Verification: [Dividend = Divisor × Quotient + Remainder ]

16135 = 875 × 18 + 385

16135 = 15750 + 385

16135 = 16135

LHS = RHS

(iv) 16025/1000 gives quotient and remainder = 25

Verification: [Dividend = Divisor × Quotient + Remainder ]

16025 = 1000 × 16 + 25

16025 = 16000 + 25

16025 = 16025

LHS = RHS

Question: 5

Find a number which when divided by 35 gives the quotient 20 and remainder 18.

Solution:

Dividend = Divisor × Quotient + Remainder

Dividend = 35 × 20 + 18

= 700 + 18

= 718

Question: 6

Find the number which when divided by 58 gives a quotient 40 and remainder 31.

Solution:

Dividend = Divisor × Quotient + Remainder

Dividend = 58 × 40 + 31

= 2320 + 31

= 2351

Question: 7

The product of two numbers is 504347. If one of the numbers is 1591, find the other.

Solution:

Product of two numbers = 504347

One of the two numbers = 1591

Let the number be A.

Therefore, A × 1591 = 504347

A = 5043471591 = 317

Question: 8

On dividing 59761 by a certain number, the quotient is 189 and the remainder is 37. Find the divisor.

Solution:

Dividend = 59761

Quotient = 189

Remainder = 37

Divisor = A

Now, Dividend = Divisor × Quotient + Remainder

59761 =A × 189 + 37

59761 – 37 = A × 189

59724 = A × 18

Therefore, A = 59724189

= 316

Question: 9

On dividing 55390 by 299, the remainder is 75. Find the quotient.

Solution:

Dividend = 55390

Divisor = 299

Remainder = 75

Quotient = A

Dividend = Divisor × Quotient + Remainder

55390 = 299 × A + 75

55390 – 75 = A × 299

55315 = A × 299

Therefore, A = 55315299 = 185

Exercise 4.5

Question: 1

Without drawing a diagram, find:

Solution:

(i) 10th square number:

A square number can easily be remembered by the following rule

N^th square number = n × n

10th square number = 10 × 10 = 100

(ii) 6th triangular number:

A triangular number can easily be remembered by the following rule

N^th triangular number = n × ( n + 1 )2

Therefore, 6^th triangular number = 6 × (6 + 1)2 = 21

Question: 2

(i) Can a rectangle number also be a square number?

(ii) Can a triangular number also be a square number?

Solution:

(i) Yes, a rectangular number can also be a square number; for example, 16 is a square number also a rectangular number.

(ii) Yes, there e×ists only one triangular number that is both a triangular number and a square number, and that number is 1.

Question: 3

Write the first four products of two numbers with difference 4 starting from in the following order:

1 , 2 , 3 , 4 , 5 , 6 , ………..

Identify the pattern in the products and write the next three products.

Solution:

1 × 5 = 5 (5 – 1 = 4)

2 × 6 = 12 (6 – 2 =4)

3 × 7 = 21 (7 – 3 = 4)

4 × 8 = 32 (8 – 4 = 4)

Question: 4

Observe the pattern in the following and fill in the blanks:

Solution:

9 × 9 + 7 =88

98 × 9 + 6 = 888

987 × 9 + 5 = 8888

9876 × 9 + 4 = 88888

98765 × 9 + 3 = 888888

987654 × 9 + 2 = 8888888

9876543 × 9 + 1 = 88888888

Question: 5

Observe the following pattern and extend it to three more steps:

Solution:

6 × 2 – 5 = 7

7 × 3 – 12 = 9

8 × 4 – 21 = 11

9 × 5 – 32 = 13

10 × 6 – 45 = 15

11 × 7 – 60 = 17

12 × 8 – 77 = 19

Question: 6

Study the following pattern:

1 + 3 = 2 × 2

1 + 3 + 5 = 3 × 3

1 + 3 + 5 + 7 = 4 × 4

1 + 3 + 5 + 7 + 9 = 5 × 5

By observing the above pattern, find:

Solution:

(i) 1 + 3 + 5 + 7 + 9 + 11

= 6 × 6

= 36

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

= 8 × 8

= 64

(iii) 21 + 23 + 25 + … + 51

= (21 + 23 + 25 +…+ 51) can also be written as

(1 + 3 + 5 + 7 +…+ 49 + 51) – (1 + 3 + 5 +…+ 17 + 19)

(1 + 3 + 5 + 7+…+ 49 + 51) = 26 × 26 = 676

and, (1 + 3 + 5 +…+ 17 + 19 ) = 10 × 10 = 100

Now,

(21 + 23 + 25 +…+ 51 ) =  676  – 100 = 576

Question: 7

Study the following pattern:

By observing the above pattern, write next two steps.

Solution:

The next two steps are as follows:

1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5

= 5 × 6 × 116

= 55

1 × 1+ 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6

= 6 × 7 × 136

= 91

Question: 8

Study the following pattern:

By observing the above pattern, find:

Solution:

(i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

= 10 × 112

= 55

(ii) 50 + 51 + 52 + …+ 100

This can also be written as

(1 + 2 + 3 + …+ 99 + 100) – (1 + 2 + 3 + 4 + …+ 47 + 49)

Now,

(1 + 2 + 3 + …+ 99 + 100 ) = 100 × 1012

and, (1 + 2 + 3 + 4 +…+ 47 + 49 ) = 49 × 502

So, (50 + 51 + 52 + …+ 100 ) = 100 × 1012 – 49 × 502

= 5050 – 1225

= 3825

(iii) 2 + 4 + 6 + 8 + 10 +…+ 100

This can also be written as 2 × (1 + 2 + 3 + 4 + …+ 49 + 50)

Now,

(1 + 2 + 3 + 4 + …+ 49 + 50 ) = 50 × 512

= 1275

Therefore, (2 + 4 + 6 + 8 + 10 + …+ 100) = 2 × 1275 = 2550

Exercise 4.6

Question: 1

Which one of the following is the smallest whole number?

(a) 1  (b)  2   (c) 0  (d) None of these

Solution:

The set of whole numbers is {0 , 1, 2, 3, 4, …}.

So, the smallest whole number is 0.

Hence, the correct option is (c).

Question: 2

Which one of the following is the smallest even whole number?

(a) 0 (b) 1 (c) 2 (d) None of these

Solution:

The natural numbers along with 0 form the collection of whole numbers.

So, the numbers 0, 1, 2, 3, 4, … form the collection of whole numbers.

The number which is divisible by 2 is an even number.

So, in the collection “0, 1, 2, 3, 4, …”, 2 is the smallest even number.

Hence, the correct option is (c).

Question: 3

Which one of the following is the smallest odd whole number?

(a) 0 (b) 1 (c) 3 (d) 5

Solution:

The natural numbers along with 0 form the collection of whole numbers.

So, the numbers 0, 1, 2, 3, 4, … form the collection of whole numbers.

A natural number which is not divisible by 2 is called an odd whole number.

So, in the collection “0, 1, 2, 3, 4, …”, 1 is the smallest odd whole number.

Hence, the correct option is (b).

Question: 4

How many whole numbers are between 437 and 487?

(a) 50 (b) 49 (c) 51 (d) None of these

Solution:

The whole numbers between 437 and 487 are 438, 439, 440, 441, … , 484, 485 and 486. To find the required number of whole numbers,

We need to subtract 437 from 487 and then subtract again 1 from the result.

Thus, there are (487 – 437) – 1 whole numbers between 437 and 487.

Now, (487 – 437) – 1 = 50 – 1 = 49

Hence, the correct option is (b).

Question: 5

The product of the successor 999 and predecessor of 1001 is:

(a) one lakh (b) one billion (c) one million (d) one crore

Solution:

Successor of 999 = 999 + 1 = 1000

Predecessor of 1001 = 1001 – 1 = 1000

Now,

Product = (Successor of 999) × (Predecessor of 1001)

= 1000 × 1000

= 1000000

= one million

Hence, the correct option is (c).

Question: 6

Which one of the following whole numbers does not have a predecessor?

(a) 1 (b) 0 (c) 2 (d) None of these

Solution:

The numbers 0, 1, 2, 3, 4, …. form the collection of whole numbers.

The smallest whole number is 0.

So, 0 does not have a predecessor.

Hence, the correct option is (b).

Question: 7

The number of whole numbers between the smallest whole number and the greatest 2 digit number is:

(a) 101 (b) 100  (c) 99 (d) 98

Solution:

Smallest whole number = 0

Greatest 2-digit whole number = 99

The whole numbers between 0 and 99 are 1, 2, 3, 4 …… 97, 98.

To find the number of whole numbers between 0 and 99,

Subtract 1 from the difference of 0 and 99.

Therefore, Number of whole numbers between 0 and 99 = (99 – 0) – 1

= 99 – 1

= 98

Hence, the correct option is (d).

Question: 8

If n is a whole number such that n + n = n, then n =?

(a) 1 (b) 2 (c) 3 (d) None of these

Solution:

Here, 0 + 0 = 0, 1 + 1 = 2 , 2 + 2 = 4 …..

So, the statement n + n = n is true only when n = 0.

Hence, the correct option is (d).

Question: 9

The predecessor of the smallest 3 digit number is:

(a) 999 (b) 99 (c) 100 (d) 101

Solution:

Smallest 3-digit number = 100

Predecessor of 3-digit number = 100 — 1 = 99

Hence, the correct option is (b).

Question: 10

The least number of 4 digits which is exactly divisible by 9 is:

(a)1008  (b)1009  (c)1026  (d)1018

Solution:

Least 4-digit number = 1000

The least 4-digit number exactly divisible by 9 is 1000 + (9 – 1) = 1008.

Hence, the correct option is (a).

Question: 11

The number which when divided by 53 gives 8 as quotient and 5 as remainder is:

(a) 424 (b) 419 (c) 429 (d) None of these

Solution:

Here, Divisor = 53, Quotient = 8 and Remainder = 5.

Now, using the relation Dividend = Divisor x Quotient + Remainder

We get

Dividend = 53 x 8 + 5

= 424 + 5

= 429

Thus, the required number is 429.

Hence, the correct option is (c).

Question: 12

The whole number n satisfying n + 35 = 101 is:

(a) 65 (b) 67 (c) 64 (d) 66

Solution:

Here, n+ 35 = 101.

Adding – 35 on both sides, we get

n + 35 + (- 35) = 101 + (- 35)

n + 0 = 66

n = 66

Hence, the correct option is (d).

Question: 13

The value of 4 x 378 x 25 is:

(a) 37800 (b) 3780 (c) 9450 (d) 30078

Solution:

By regrouping, we get

4 × 378 × 25 = 4 × 25 × 378

= 100 × 378

= 37800

Hence, the correct option is (a).

Question: 14

The value of 1735 x 1232 – 1735 x 232 is:

(a) 17350 (b) 173500 (c) 1735000 (d) 173505

Solution:

Using distributive law of multiplication over subtraction, we get

1735 × 1232 – 1735 × 232 = 1735 (1232 – 232)

= 1735 × 1000

= 1735000

Hence, the correct option is (c).

Question: 15

The value of 47 × 99 is:

(a) 4635 (b) 4653 (c) 4563 (d) 6453

Solution:

Since, 99 = 100 — 1

Therefore, 47 × 99 = 47 × (100 — 1)

= 47 × 100 — 47

= 4700 — 47

= 4653

Thus, the value of 47 × 99 is 4653.

Hence, the correct option is (b).

Exercise 3.1

Whole Number – Exercise – 3.1 – Q.1

Ans. 

Smallest natural number 1.

Whole Number – Exercise – 3.1 – Q.2

Ans. 

Smallest whole number 0.

Whole Number – Exercise – 3.1 – Q.3

Ans. 

Does not exist

Whole Number – Exercise – 3.1 – Q.4

Ans. 

Does not exist.

Whole Number – Exercise – 3.1 – Q.5

Ans. 

Yes all natural numbers also whole numbers.

Whole Number – Exercise – 3.1 – Q.6

Ans. 

No, all whole numbers are not natural numbers.

Whole Number – Exercise – 3.1 – Q.7

Ans.

(i) Successor of 1000909 is 1000910.

(ii) Successor of 2340900 is 2340901.

(iii) Successor of 7039999 is 7039100.

Whole Number – Exercise – 3.1 – Q.8

Ans.

(i) Predecessor of 10000 is 9999.

(ii) 807000 → 806999

(iii) Predecessor of 7005000 is 7004999.

Whole Number – Exercise – 3.1 – Q.9

Ans.

Numbers are as shown in the above line

Whole Number – Exercise – 3.1 – Q.10

Ans.

39 whole number are there between 21 and 61.

Whole Number – Exercise – 3.1 – Q.11

Ans.

(i) 25 < 205

(ii) 170 > 107

(iii) 415 > 514

(iv) 10001 < 100001

(v) 2300014 < 2300041.

Whole Number – Exercise – 3.1 – Q.12

Ans.

1100, 925, 886, 786, 325, 270, 141, 0.

Whole Number – Exercise – 3.1 – Q.13

Ans.

999999, 1000000, 1000000 is 1 airier by 1.

Whole Number – Exercise – 3.1 – Q.14

Ans.

8510000, 8509999, 8509998.

Whole Number – Exercise – 3.1 – Q.15

Ans.

4009999, 4010000, 4010001.

Whole Number – Exercise – 3.1 – Q.16

Ans.

Every natural number has its successes.

Whole Number – Exercise – 3.1 – Q.17

Ans.

(i) true

(ii) false

(iii) false

(iv) false

(v) true

(vi) false

(vii) true

(viii) true

(ix) true

(x) false

(xi) false

(xii) true

(xiii) true

(xiv) true.

Exercise 1.1

Question: 1

Write each of the following in numeral form :

i) Eight thousand twelve

ii) Seventy thousand fifty-three

iii) Five lakh seven thousand four hundred six

iv) Six lakh tow thousand nine

v) Thirty lakh eleven thousand one

vi) Eight crore four lakh twenty-five.

vii) Three crore three thousand three hundred three

viii) Seventeen crores sixty lakh thirty thousand fifty-seven.

Solution:

i) 8,012

ii) 70,053

iii) 5,07,406

iv) 6,02,009

v) 30,11,001

vi) 8,04,00,025

vii) 3,03,03,303

viii) 17,60,30,057

Question: 2

Write the following numbers in words in the Indian system of numeration.

i) 42,007

ii) 4,05,045

iii) 35,42,012

iv) 7,06,04,014

v) 25,05,05,500

vi) 5,50,50,050

vii) 5,03,04,012

Solution:

i) Forty two thousand seven.

ii) Four lakh five thousand forty five.

iii) Thirty five lakh forty two thousand twelve.

iv) Seven crore six lakh four thousand fourteen.

v) Twenty five crore five lakh five thousand five hundred.

vi) Five crore fifty lakh fifty thousand fifty.

vii) Five crore three lakh four thousand twelve.

Question: 3

Insert commas in the correct positions to separate periods and write the following numbers in words:

i) 4375

ii) 24798

iii) 857367

iv) 9050784

v) 10105607

vi) 10000007

vii) 910107104

Solution:

i) 4,357

ii) 24,798

iii) 8,57,367

iv) 90,50,784

v) 1,01,05,607

vi) 1,00,00,007

vii) 91,01,07,104

Question: 4

Write each of the following in expanded form:

i) 3057

ii) 12345

iii) 10205

iv) 235060

Solution:

i) 3000 + 50 + 7

ii) 10000 + 2000 + 300 + 40 + 5

iii) 10000 + 200 + 5

iv) 200000 + 30000 + 5000 + 60

Question: 5

Write the corresponding numeral for each of the following :

i) 7 x 10000 + 2 x 1000 + 5 x 100 + 9 x 10 + 6 x 1

ii) 4 x 100000 + 5 x 1000 + 1 x 100 + 7 x 1

iii) 8 x 1000000 + 3 x 1000 + 6 x 1

iv) 5 x 10000000 + 7 x 1000000 + 8 x 1000 + 9 x 10 + 4

Solution:

i) 70000 + 2000 + 500 + 90 + 6 = 72,596

ii) 400000 + 5000 + 100 + 7 = 4,05,107

iii) 8000000 + 3000 + 6 = 80,03,006

iv) 50000000 + 7000000 + 8000 + 90 + 4 = 5,70,08,094

Question: 6

Find the place value of the digit 4 in each of the following:

i) 74983160

ii) 8745836

Solution:

i)  Place value of 4 = 4 × 10,00,00 = 40,00,00

ii) Place value of 4 = 4 × 10,000 = 40,000

Question: 7

Determine the product of the place values of two fives in 450758.

Solution:

Place value of first 5 = 5 × 10 = 50

Place value of second 5 = 5 × 10,000 = 50,000

Required product = 50 × 50,000 = 25, 00,000

Question: 8

Determine the difference of the place values of 7’s in 257839705.

Solution:

Place value of first 7 = 7 × 10 = 700

Place value of second 7 = 7 × 10,000 = 70, 00,000

Required difference = 70, 00,000 – 700 = 69, 99,300

Question: 9

Determine the difference between the place value and the face value of 5 in 78654321.

Solution:

The number = 7, 86, 54, 321

The place value of 5 = 5 ten thousands = 50,000

The face value of 5 = 5

Therefore, the difference = 50,000 – 5 = 49,995

Question: 10

Which digits have the same face value and place value in 92078634?

Solution:

The place value of a digit depends on the place where it occurs, while the face value is the value of the digit itself.

In a number, the digits that have same face value and place value are the ones digit and all the zeroes of the number.

Therefore, in 9, 20, 78,634, 4 ( the ones digit ) and 0 ( the lakhs digit ) have the same face value and place value

Question: 11

How many different 3- digit numbers can be formed by using the digits 0, 2, 5 without repeating any digit in the number?

Solution:

The three-digit numbers formed using the digits 0, 2 and 5 ( without repeating any digit in the number ) are 250 , 205 , 502 and 520.

Therefore, four such numbers can be formed.

Question: 12

Write all possible 3- digit numbers using 6, 0, 4 when

i) Repetition of digits is not allowed

ii) Repetition of digits is allowed

Solution:

i) 604, 640, 460, 406

ii) 666, 664, 646, 660, 606, 600, 644, 640, 604, 444, 466, 440, 446,464, 400, 404, 406, 460

Question: 13

Fill in the blanks:

i)1 Iakh = —— ten thousand

ii) 1 Iakh = —— thousand

iii) 1 Iakh = ——- hundred

iv) 1 Iakh = ——- ten

v) 1 crore = —– ten Iakh

vi) 1 crore = —– Iakh

vii)1 crore = —— ten thousand

viii) 1 crore = ——- thousand

ix) 1 crore = ——- hundred

x) 1 crore = —— ten

Solution:

i) 1 Iakh = 10 ten thousand

ii) 1 Iakh = 100 thousand

iii) 1 Iakh = 1000 hundred

iv) 1 Iakh = 10000 ten

v) 1 crore = 10 ten Iakh

vi) 1 crore = 100 Iakh

vii) 1 crore = 1000 ten thousand

viii) 1 crore = 10000 thousand

ix) 1 crore = 100000 hundred

x) 1 crore = 1000000 ten

Exercise 1.2

Question: 1

Write each of the following numbers in digits by using international place value chart. Also, write them in expanded form.

i) Seven million three hundred three thousand two hundred six

ii) Fifty five million twenty nine thousand seven

iii) Six billion one hundred ten million three thousand seven

Solution:

i) 7,303,206

Expanded form = 7 × 1000000 + 3 × 100000 + 0 × 10000 + 3 × 1000 + 2 × 100 + 0 × 10 + 6 × 1

ii) 55,029,007

Expanded form = 5 × 10000000 + 5 × 1000000 + 0 × 100000 + 2 × 10000 + 9 × 1000 + 0 × 100 + 0 × 10 + 7 × 1

iii) 6,110,003,007

Expanded form = 6 × 1000000000 + 1 × 100000000 + 1 × 10000000 + 0 × 1000000 + 0 × 100000 + 0 × 10000 + 3 × 1000 + 0 × 100 + 0 × 10 + 7 × 1

Question: 2

Rewrite each of the following numerals with proper commas in the international system of numeration

i) 513625

ii) 4035672

iii) 65954923

iv) 70902005

Solution:

i) 513,625 or Five hundred thirteen thousand six hundred twenty five.

ii) 4,035,672 or Four million thirty five thousand six hundred seventy two.

iii) 65,954,923 or Sixty five million nine hundred fifty four thousand nine hundred twenty three

(iv) 70,902,005 or Seventy million nine hundred two thousand five

Question: 3

Write each of the following numbers in the international system of numeration :

i) Forty three lakh four thousand eighty four.

ii) Six crore thirty four lakh four thousand forty four.

iii) Seven lakh thirty five thousand eight hundred ninety nine only.

Solution:

i) 4,304,084 or Four million three hundred four thousand eighty four.

ii) 63,404,044 or Sixty three million four hundred four thousand forty four.

iii) 735,899 or Seven hundred thirty five thousand eight hundred ninety nine.

Question: 4

Write the following numbers in the Indian system of numeration :

i)Six million five hundred forty three thousand two hundred ten.

ii)Seventy six million eighty five thousand nine hundred eighty seven

iii) Three hundred twenty five million four hundred seventy nine thousand eight hundred thirty eight.

Solution:

i) 65, 43,210 or Sixty five lakh forty three thousand two hundred ten.

ii) 7, 60,85,987 or Seven crore sixty lakh eighty five thousand nine hundred eighty seven.

iii) 32, 54,79,838 or Thirty two crore fifty four lakh seventy nine thousand eight hundred thirty eight.

Question: 5

A certain nine digit number has only ones in ones period, only twos in the thousands period and only threes in millions period. Write this number in words in the Indian system.

Solution:

The number is 333,222,111

In Indian system , the number is written as 33,32,22,111 thirty – three crore thirty – two lakh twenty thousand one hundred and eleven.

Question: 6

How many thousands make a million?

Solution:

1,000 thousands makes a million

Question: 7

How many millions make a billion?

Solution:

1,000 millions make a billion

Question: 8

i) How many lakhs make a million?

ii) How many lakhs make billion?

Solution:

i) Ten lakhs make a million

ii) Ten thousand lakhs make a billion

Question: 9

Write each of the following in numerical form:

i) Nighty-Eight million seven hundred eight thousand four.

ii) Six hundred seven million twelve thousand eighty four.

iii) Four billion twenty five million forty five thousand.

Solution:

i) 98,708,004

ii) 607,012,084

iii) 4,025,045,000

Question: 10

Write the number names of each of the following in international system of numeration :

i) 435,002

ii) 1,047,509

iii) 59,064,523

iv) 25,201,905

Solution:

i) Four hundred thirty-five thousand and two

ii) One million, forty-seven thousand, five hundred and nine

iii) Fifty-nine million, sixty-four thousand, five hundred and twenty-three

iv) Twenty-five million, two hundred one thousand, nine hundred and five

Exercise 1.3

Question: 1

How many four – digit numbers are there in all?

Solution:

There are 10 digits i.e., 0, 1, 2, 3, 4, 5, 6 ,7, 8, 9.

We cannot use ‘0’ at thousand’s place.

So, we can use only 9 digits at thousand’s place.

Also, we can use 10 digits at hundred’s, 10 digits at ten’s and 10 digits at unit’s place.

So, total numbers of four-digit numbers = 9 × 10 × 10 × 10 = 9000

Question: 2

Write the smallest and the largest six digit numbers. How many numbers are between these two.

Solution:

The smallest digit is 0. But we cannot use 0 at the place having the highest place value in six digit numbers. So, we will use the second smallest digit i.e., 1. All other places are filled by 9.

Hence, the required number = 100000

Smallest six digit number will be 100000.

The largest digit is 9.

We can use 9 at any place. In fact , we can use 9 in all places in six digit numbers.

Hence, the required number = 999999

Largest six digit number will be 999999

Required difference = 999999 – 100000 = 899999

So, the total numbers between 999999 and 100000 will be 899998.

Question: 3

How many 8 – digit numbers are there in all ?

Solution:

There are 10 digits i.e., 0, 1, 2, 3, 4, 5, 6 ,7, 8, 9.

We cannot use ‘0’ at the place having the highest place value in 8 digit numbers.

So, we can use only 9 digits at the place having the highest place value in 8 digit numbers.

Also, we can use 10 digits at the remaining places in 8 digit numbers So, total numbers of 8-digit numbers = 9 × 10 × 10 × 10 × 10 × 10 × 10 × 10 = 90000000

Question: 4

Write 10075302 in words and rearrange the digits to get the smallest and the largest numbers.

Solution:

One crore seventy-five thousand three hundred two.

In order to write the smallest 8-digit number using digits 0, 1, 2, 3, 5 and 7, we put the smallest digit 1 ( Except 0 ) at the place having the highest place value. The largest digit 7 is put at the rightmost place i.e. at unit’s place, the digit 5 is put at the ten’s place, the digit 3 is put at the hundred’s place and the digit 2 is put at the thousand’s place. All other places are filled by 0. Hence, the required largest number is 10002357.

In order to write the largest 8-digit number using digits 0, 1, 2, 3, 5 and 7, we put the largest digit 7 at the place having the highest place value. The smallest digit 5 is put at the place after the highest place value. We put the next smallest digit ( i.e., 3 ) after the previous one. After it we place the next smallest digit ( i.e., 2 ) and after that we put the digit 1. All other places are filled by 0. Hence, the required largest number is 75321000.

Question: 5

What is smallest 3-digit number with unique digits?

Solution:

The smallest three-digit number with unique digits is 102.

Question: 6

What is the largest 5- digits number with unique digits?

Solution:

The largest five – digit number with unique digits 98,765.

Question: 7

Write is smallest 3-digit number which does not change if the digits are written in reverse order.

Solution:

The smallest three – digit number that does not change if the digits are written in reverse order is 101.

Question: 8

Find the difference between the number 279 and that obtained on reversing its digits.

Solution:

The number obtained on reversing 279 = 972

Difference = 972 – 279 = 693

Thus, the difference between 279 and that obtained on reversing its digits is 693.

Question: 9

Form the largest and smallest 4- digit numbers using each of digits 7,1,0,5 only once.

Solution:

The largest and smallest four- digit numbers formed using 7,1,0 and and 5are 7,510 and 1,057.

Exercise 1.4

Question: 1

Put the appropriate symbol ( < > ) in each of the following boxes :

i) 102394 ___ 99887

ii) 2507324 ___ 2517324

iii) 3572014 ____ 10253104

iv) 47983505 ____ 47894012

Solution:

i) 102394 > 99887

ii) 2507324 < 2517324

iii) 3572014 < 10253104

iv) 47983505 > 47894012

Question: 2

Arrange the following numbers in ascending order :

i) 6,35,47,201, 10,23,45,694 , 65,39,542 , 83,54,208 , 1,23,45,678

ii) 18,08,088, 1,81,888, 1,90,909, 18,08,090, 1,60,60,666

Solution:

i) 65,39,542, 83,54,208, 1,23,45,678, 6,35,47,201, 10,23,45,694

ii) 1,81,888, 1,90,909, 18,08,088, 18,08,090, 1,60,60,666

Question: 3

Arrange the following numbers in descending order :

i) 05,69,44,000, 5,69,43,201 , , 56,95,440, 5,69,43,300, 56,94,437

ii) 10,20,216, 10,20,308 , 10,21,430, 8,93,425, 8,93,245

Solution:

i) 5,69,44,000, 5,69,43,300, 5,69,43,201, 56,95,440, 56,94,437

ii) 10,21,430, 10,20,308, 10,20,216, 8,93,425, 8,93,245

Exercise 1.5

Question: 1

How many milligrams make one kilogram?

Solution:

Ten lakh or one million (10, 00, 000) milligrams make one kilogram.

Question: 2

A box of medicine tablets contains 2, 00,000 tablets each weighing 20mg. what is the total weight of all the tablets in the box in grams? in kilograms ?

Solution:

Given data: Each tablet weighs = 20 mg

Therefore, The weight of 2, 00,000 tablets = 2, 00,000 × 20 = 40, 00,000 mg

Therefore, The total weight of all the tablets in the box = 40, 00,000 mg

We know 1 g = 1,000 mg

Weight of the box having all tablets = 40,00,000 ÷ 1,000=4000g

And, as 1 kg = 1,000 g

Therefore, Weight of the box having all tablets = 4,000 ÷ 1,000 = 4000g = 4 kg

Question: 3

Population of sundarnagar was 2, 35,471 in the year 1991. In the year 2001 it was found to have increased by 72,958. What was the population of the city in 2001?

Solution:

The population of Sundar Nagar in 2001 = Sum of the population of city in 1991 + Increase in population over the given time period

As given in the question, The population of Sundar Nagar in 1991 = 2, 35,471

As given in the question,

Increase in population over the given time period = 72.958

Therefore, The population of Sundar Nagar in 2001

= 2, 35,471 + 72,958 = 3, 08,429

Question: 4

A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final days were respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.

Solution:

Total number of tickets sold on all four days is the sum of the tickets sold on the first, second, third and final days.

Therefore, total number of tickets sold on all four days is given by:

= 1094 + 1812 + 2050 + 2751 = 7707

Question: 5

The town newspaper is published every day. One copy has 12 pages. Everyday 11,980 copies are printed. How many pages are in all printed every day? Every month?

Solution:

As given in the question,

Number of pages in 1 copy of newspaper = 12

Therefore, Number of pages in 11,980 copies of newspaper

= 11,980 × 12 = 1, 43,760

Thus, 1, 43,760 pages are printed every day.

Now, number of pages printed every day = 1, 43,760

Therefore, Number of pages printed in a month = 1, 43,760 × 30 = 43, 12,800

Thus, 43, 12,800 pages are printed in a month.

Question: 6

A machine, on an average, manufactures 2825 screws a day. How many screws did it produce in the month of January 2006?

Solution:

As given in the question,

Number of screws produced by a machine in a day = 2,825

Therefore, Number of screws produced by the same machine in the month of January 2006 = 2, 825 × 31 = 87,575

Thus, machine-produced 87,575 screws in the month of January 2006.

Question: 7

A famous cricket player has so far scored 6978 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?

Solution:

Runs scored by cricket player in test matches = 6,978

Therefore, Remaining runs required to complete 10,000 runs

= 10,000 – 6,978 = 3,022

Thus, the player needs to score 3,022 more runs to complete 10,000 runs.

Question: 8

Ravish has Rs. 78,592 with him. He placed an order for purchasing 39 radio sets at Rs. 1234 each. How much money will remain with him after the purchase?

Solution:

Ravish’s initial money = Rs.78, 592

He purchased 39 radio sets at Rs.1, 234 each.

Therefore, Money spent by him on purchasing 39 radio sets

= 1,234 × 39 = Rs. 48,126

Therefore, Remaining money with Ravish after the purchase = Initial money – Money spent on purchasing 39 radio sets = Rs. 78,592 – Rs. 48,126 = Rs. 30,466

Thus, 230,466 are left with him after the purchase.

Question: 9

In an election, the successful candidate registered 5, 77,570 votes and his nearest rival secured 3, 48,685 votes. By what margin did the successful candidate win the election?

Solution:

Margin of victory in the election for the successful candidate = Number of votes registered by the winner – Number of votes secured by nearest rival candidate

Votes registered by the winner = 5, 77,570

Votes secured by the rival = 3, 48,685

Therefore, Margin of victory for the successful candidate

= 5, 77, 570 – 3, 48, 685 = 2, 28, 885

Question: 10

To stitch a shirt 2m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?

Solution:

As given in the question, Total length of available cloth = 40 m = 4,000 cm (1 m = 100 cm)

As given in the question, Length of cloth required to stitch a shirt

= 215 cm = 200 + 15 = 215 cm

Therefore, The number of shirts that can be stitched from the 40-metre cloth

= 4,000 / 215 = 18.60

As the number of shirts has to be a whole number, we consider the whole part only. That is, 18 such shirts can be stitched.

Therefore, Cloth required for stitching 18 shirts = 215 x 18 = 3870 cm. Therefore, Remaining cloth = 4,000 — 3870 = 130 cm = 1.3 m

Question: 11

A vessel has 4 litre and 650 ml of curd. In how many glasses, each of 25 ml capacity, can it be distributed?

Solution:

The number of glasses in which curd can be distributed = Total amount of curd/Capacity of each glass.

Total amount of curd in the vessel = 4,650 ml = 4,000 + 650 = 4,650 ml

( 1 L = 1,000 ml )

Capacity of each glass = 25 ml

Therefore, Number of glasses in which curd can be distributed = 4,650/25 = 186

Question: 12

Medicine in packed in boxes, each such boxes weighing 4kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 Kg?

Solution:

Sol :

As given in the question,

Total capacity of a van carrying boxes of medicines = 800 kg = 8, 00,000 g (1 kg = 1,000 g)

As given in the question, Weight of each packed box

= 4,500 g = 4,000 + 500 = 4,500 g

Therefore, Total number of boxes that can be loaded in the van

= 8, 00,000 / 4,500 = 177.77

The obtained number of boxes is not a whole number.

Therefore, Weight of 177 boxes = 177 × 4,500 = 7, 96,500 g (under permissible limit)

Therefore, Weight of 178 boxes = 178 × 4,500 = 8, 01,000 g (beyond permissible limit)

Therefore, we can’t load 178 boxes; hence, we can say that 177 boxes can be loaded in the van.

Question: 13

The Distance between the school and the house of a student is 1 Km 875 m. Every day she walks both ways between her school and home. Find the total distance covered by her in a week?

Solution:

Therefore, Distance between the school and the house of a student

= 1,875 m = 1,000 + 875 = 1,875 m (1 km = 1,000 m)

As given in the question, Distance covered by a student in a day

= 2 × 1,875 = 3,750 m

Total distance covered by her in a week = 7 × 3,750 = 26,250 m = 26.25 km

Exercise 1.6

Question: 1

Round off each of the following numbers to nearest tens :

i) 84

ii) 98

iii) 984

iv) 808

v) 998

vi) 12,096

vii) 10,908

viii) 28,925

Solution:

i) 80

ii) 100

iii) 980

iv) 810

v) 1,000

vi) 12,100

vii) 10,910

viii) 28,930

Question: 2

Round off each of the following numbers to nearest hundreds :

i) 3,985

ii) 7289

iii) 8074

iv) 14,627

v) 28,826

vi) 4,20,387

vii) 43,68,973

viii) 7,42,898

Solution:

i) 4,000

ii) 7,300

iii) 8,100

iv) 14,600

v) 28,800

vi) 4,20,400

vii) 43,69,000

viii) 7,42,900

Question: 3

Round off each of the numbers to nearest thousands :

i) 2401

ii) 9600

iii) 4278

iv) 7832

v) 9567

vi) 26,019

vii) 20,963

viii) 4,36,952

Solution:

i) 2000

ii) 10000

iii) 4000

iv) 8000

v) 10000

vi) 26000

vii) 21000

viii) 4,37,000

Question: 4

Round off each of the following numbers to nearest tens, hundreds and thousands.

i) 964

ii) 1049

iii) 45,634

iv) 79,085

Solution:

Tens :

i) 970

ii) 1050

iii) 45,630

iv) 79,090

Hundreds :

i)1000

ii) 1000

iii) 45,600

iv) 79,100

Thousands :

i) 1000

ii) 1000

iii) 46000

iv) 79000

Question: 5

Round off the following measures to the nearest hundreds :

i) Rs 666

ii) Rs 850

iii) Rs 3,428

iv) Rs 9,080

v) 1265 km

vi) 417 m

vii) 550 cm

viii) 2486 m

ix) 360 gm

x) 940 kg

xi) 273 l

xii) 820 mg

Solution:

i) Rs. 700

ii) Rs. 900

iii) Rs. 3,500

iv) Rs.9100

v) 1300 km

vi) 400 m

vii) 600 cm

viii) 2500 m

ix) 400 gm

x) 900 kg

xi) 300 l

xii) 800 mg

Question: 6

List all numbers which are rounded off to the nearest ten as 370.

Solution:

365 , 366 , 367 , 368 , 369 , 370 , 371 , 372 , 373 , 374

Question: 7

Find the smallest and the greatest numbers which are rounded off to the nearest hundreds as 900.

Solution:

Smallest number: 850

Greatest number: 949

Question: 8

Find the smallest and the greatest numbers which are rounded off to the nearest thousands as 9000.

Solution:

Smallest number: 8,500

Greatest number: 9,499

Exercise 1.7

Question: 1

Estimate the following by rounding off each factor to nearest hundreds:

i) 730 + 998

ii) 796 – 314

iii) 875 – 384

Solution:

i) 700 + 1000 = 1700

ii) 800 – 300 = 500

iii) 900 – 400 = 500

Question: 2

Estimate the following by rounding off each factor to nearest thousands:

i) 12904 + 2888

ii) 28292 – 21496

Solution:

i) 13000 + 3000 = 16000

ii) 28000 – 21000 = 7000

Question: 3

Estimate the following by rounding off each number to its greatest place:

i) 439 + 334 + 4317

ii) 8325 – 491

iii) 108734 – 47599

iv) 898 × 785

v) 9 × 795

vi) 87 × 317

Solution:

i) 400 + 300 + 4000 = 4700

ii)8000 – 500 = 7500

iii)100000 -500000 = 50000

iv) 900 × 800 = 720000

v) 10 × 800 = 8000

vi) 90 x 300 = 27000

Question: 4

Find the estimated quotient for each of the following by rounding off each number to its greatest place :

i) 878 ÷ 28

ii) 745 ÷ 24

iii) 4489 ÷ 394

Solution:

i) 900 ÷ 30 = 30

ii) 700 ÷ 20 = 35

iii) 4000 ÷ 400 = 10

Question: 5

Write the expression for each of the following statements using brackets:

i) Four multiplied by the sum of 13 and 7

ii) Eight multiplied by the difference of four from nine.

iii) Divide the difference of twenty eight and seven by 3.

The sum of 3 and 7 in multiplied by the difference of twelve and eight.

Solution:

i) 4 × (13 + 7)

ii) 8 × (9 – 4)

iii) 28 – 73

iv) (3 + 7) x (12 – 8)

Question: 6

Simplify each of the following:

i) 124 – (12 – 2) × 9

ii) (13 + 7) × (9 – 4) – 18

iii) 210 – (14 – 4) x (18 + 2) – 10

Solution:

i) 34

ii) 82

iii) 0

Question: 7

Simplify each of the following:

i) 7 × 109

ii) 6 × 112

iii) 9 × 105

iv) 17 × 109

v) 16 × 108

vi) 12 × 105

vii) 102 × 103

viii) 101 × 105

ix) 109 × 107

Solution:

i) 763

ii) 672

iii) 945

iv) 1853

v) 1728

vi) 1260

vii) 10506

viii)10605

ix)11663

Question: 8

Write the roman – numerals for each of the following:

i) 33

ii) 48

iii) 76

iv) 95

Solution:

i) XXXIII

ii) XLVIII

iii) LXXVI

iv) XCV

Question: 9

Write the following in roman numerals:

i) 154

ii) 173

iii) 248

iv) 319

Solution:

i) CLIV

ii) CLXXIII

iii) CCXLVIII

iv) CCCXIX

Question: 10

Write the following in roman numerals:

i) 1008

ii) 2718

iii) 3906

iv) 3794

Solution:

i) KVIII

ii) KKDCCXVIII

iii) KKKCKVI

iv) KKKDCCXCIV

Question: 11

Write the following in roman numerals:

i) 4201

ii) 10009

iii) 44000

iv) 25819

Solution:

Question: 12

Write the following in Hindu – Arabic numerical:

i) XXVI

ii) XXIX

iii) LXXII

iv) XCI

Solution:

i) 26

ii) 29

iii) 72

iv) 91

Question: 13

Write the corresponding Hindu – Arabic numerical for each of the following:

i) CIX

ii) CLXXII

iii) CCLIV

iv) CCCXXIX

Solution:

i) 109

ii) 172

iii) 254

iv) 329

Question: 14

Write the corresponding Hindu – Arabic numerical for each of the following:

i) KXIX

ii) KDLXV

iii) KKCXXIII

iv) KKKDCXL

Solution:

i) 1019

ii) 1565

iii) 2123

iv) 3640

Question: 15

Write the following in Hindu – Arabic numerical:

Solution:

i) 4444

ii) 6949

iii) 9391

iv) 70009

Question: 16

Which of the following is meaningless?

ii) KKKCCXI

iii) XD

iv) VC

Solution:

(i) and (iii) are meaningless.

Exercise 25.1

Question: 1

A coin is tossed 1000 times with the following frequencies:

Head: 445, Tail: 555

When a coin is tossed at random, what is the probability of getting?

(i). a head?

(ii). a tail?

Solution:

Total number of times a coin is tossed = 1000

Number of times a head comes up = 445

Number of times a tail comes up = 555

Question: 2

A box contains two pair of socks of two colours (black and white). I have picked out a white sock. I pick out one more with my eyes closed. What is the probability that I will make a pair?

Solution:

No. of socks in the box = 4

Let B and W denote black and white socks respectively. Then we have:

S = {B,B,W,W}

If a white sock is picked out, then the total no. of socks left in the box = 3

No. of white socks left = 2 – 1 = 1

Question: 3

Two coins are tossed simultaneously 500 times and the outcomes are noted as given below:

If same pair of coins is tossed at random, find the probability of getting:

(i) Two heads (ii) One head (iii) No head.

Solution:

Number of trials = 500

Number of outcomes of two heads (HH) = 105

Number of outcomes of one head (HT or TH) = 275

Number of outcomes of no head (TT) = 120

Exercise 23.1

Question: 1

Ashish studies for 4 hours, 5 hours and 3 hours on three consecutive days. How many hours does he study daily on an average?

Solution:

Average number of study hours = (4 + 5 + 3) ÷ 3

= 12 ÷ 3

= 4 hours

Thus, Ashish studies for 4 hours on an average.

Question: 2

A cricketer scores the following runs in 8 innings: 58, 76, 40, 35, 48, 45, 0, 100.

Find the mean score.

Solution:

We have:

The mean score = (58 + 76 + 40 + 35 + 48 + 45 + 0 + 100) ÷ 8

= 402 ÷ 8

= 50.25 runs.

Question: 3

The marks (out of 100) obtained by a group of students in science test are 85, 76, 90, 84, 39, 48, 56, 95, 81 and 75. Find the

(i) Highest and the lowest marks obtained by the students.

(ii) Range of marks obtained.

(iii) Mean marks obtained by the group.

Solution:

In order to find the highest and lowest marks, let us arrange the marks in ascending order as follows:

39, 48, 56, 75, 76, 81, 84, 85, 90, 95

(i) Clearly, the highest mark is 95 and the lowest is 39.

(ii) The range of the marks obtained is: (95 – 39) = 56.

(iii) We have:

Mean marks = Sum of the marks ÷ Total number of students

→ Mean marks = (39 + 48 + 56 + 75 + 76 + 81 + 84 + 85 + 90 + 95) ÷ 10

= 729 ÷ 10

= 72.9.

Hence, the mean mark of the students is 72.9.

Question: 4

The enrolment of a school during six consecutive years was as follows:

1555, 1670, 1750, 2019, 2540, 2820

Find the mean enrollment of the school for this period.

Solution:

The mean enrolment = Sum of the enrolments in each year ÷ Total number of years

The mean enrolment = (1555 + 1670 + 1750 + 2019 + 2540 + 2820) ÷ 6

= 12354 ÷ 6

= 2059.

Thus, the mean enrolment of the school for the given period is 2059.

Question: 5

The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

Day	Mon	Tue	Wed	Thu	Fri	Sat	Sun
Rainfall (in mm)	0.0	12.2	2.1	0.0	20.5	5.3	1.0

(i) Find the range of the rainfall from the above data.

(ii) Find the mean rainfall for the week.

(iii) On how many days was the rainfall less than the mean rainfall.

Solution:

(i) The range of the rainfall = Maximum rainfall – Minimum rainfall

= 20.5 – 0.0

= 20.5 mm.

(ii) The mean rainfall = (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.3 + 1.0) ÷ 7

= 41.1 ÷ 7

= 5.87 mm.

(iii) Clearly, there are 5 days (Mon, Wed, Thu, Sat and Sun), when the rainfall was less than the mean, i.e., 5.87 mm.

Question: 6

If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively, find the mean height.

Solution:

The mean height = Sum of the heights ÷ Total number of persons

= (140 + 150 + 152 + 158 + 161) ÷ 5

= 761 ÷ 5

= 152.2 cm.

Question: 7

Find the mean of 994, 996, 998, 1002 and 1000.

Solution:

Mean = Sum of the observations ÷ Total number of observations

Mean = (994 + 996 + 998 + 1002 + 1000) ÷ 5

= 4990 ÷ 5

= 998.

Question: 8

Find the mean of first five natural numbers.

Solution:

The first five natural numbers are 1, 2, 3, 4 and 5.

Question: 9

Find the mean of all factors of 10.

Solution:

Question: 10

Find the mean of first 10 even natural numbers.

Solution:

Question: 11

Find the mean of x, x + 2, x + 4, x + 6, x + 8

Solution:

Mean = Sum of observations ÷ Number of observations

→ Mean = (x + x + 2 + x + 4 + x + 6 + x + 8) ÷ 5

→ Mean = (5x + 20) ÷ 5

→ Mean = 5(x + 4)5

→ Mean = x + 4

Question: 12

Find the mean of first five multiples of 3.

Solution:

The first five multiples of 3 are 3, 6, 9, 12 and 15.

Question: 13

Following are the weights (in kg) of 10 new born babies in a hospital on a particular day: 3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6 Find the mean 

Solution:

We Have

Question: 14

The percentage of marks obtained by students of a class in mathematics are:

64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1 Find their mean.

Solution:

We have.

Question: 15

The numbers of children in 10 families of a locality are:

2, 4, 3, 4, 2, 3, 5, 1, 1, 5 Find the mean number of children per family.

Solution:

= 3.

Thus, on an average there are 3 children per family in the locality.

Question: 16

The mean of marks scored by 100 students was found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct mean.

Solution:

We have:

n = The number of observations = 100, Mean = 40

→ Sum of the observations = 40 x 100

Thus, the incorrect sum of the observations = 40 x 100 = 4000.

Now,

The correct sum of the observations = Incorrect sum of the observations – Incorrect observation + Correct observation

→ The correct sum of the observations = 4000 – 83 + 53

→ The correct sum of the observations = 4000 – 30 = 3970

Question: 17

The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.

Solution:

We have:

So, sum of the five numbers = 5 x 27 = 135.

Now,

So, sum of the four numbers = 4 x 25 = 100.

Therefore, the excluded number = Sum of the five number – Sum of the four numbers

→ The excluded number = 135 – 100 = 35.

Question: 18

The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.

Solution:

We have:

Let the weight of the seventh student be x kg.

Thus, the weight of the seventh student is 61 kg.

Question: 19

The mean weight of 8 numbers is 15 kg. If each number is multiplied by 2, what will be the new mean?

Solution:

Let x₁, x₂, x₃…x₈ be the eight numbers whose mean is 15 kg. Then,

x₁ + x₂ + x₃ + …+ x₈ = 15 × 8

→x₁ + x₂ + x₃ +…+ x₈ = 120.

Let the new numbers be 2x₁, 2x₂, 2x₃ …2x₈. Let M be the arithmetic mean of the new numbers.

Then,

Question: 20

The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.

Solution:

Let x₁, x₂, x₃, x₄ and x₅ be five numbers whose mean is 18. Then,

18 = Sum of five numbers ÷ 5

∴ Sum of five numbers = 18 × 5 = 90

Now, if one number is excluded, then their mean is 16.

So,

16 = Sum of four numbers ÷ 4

∴ Sum of four numbers = 16 × 4 = 64.

The excluded number = Sum of five observations – Sum of four observations

∴ The excluded number = 90 – 64

∴ The excluded number = 26.

Question: 21

The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.

Solution:

n = Number of observations = 200

→ Sum of the observations = 50 x 200 = 10,000.

Thus, the incorrect sum of the observations = 50 x 200

Now,

The correct sum of the observations = Incorrect sum of the observations – Incorrect observations + Correct observations

→ Correct sum of the observations = 10,000 – (92 + 8) + (192 + 88)

→ Correct sum of the observations = 10,000 – 100 + 280

→ Correct sum of the observations = 9900 + 280

→ Correct sum of the observations = 10,180.

Question: 22

The mean of 5 numbers is 27. If one more number is included, then the mean is 25. Find the included number.

Solution:

We have:

Mean = Sum of five numbers ÷ 5

→ Sum of the five numbers = 27 × 5 = 135.

Now, New mean = 25

25 = Sum of six numbers ÷ 6

→ Sum of the six numbers = 25 × 6 = 150.

The included number = Sum of the six numbers – Sum of the five numbers

→ The included number = 150 – 135

→ The included number = 15.

Question: 23

The mean of 75 numbers is 35. If each number is multiplied by 4, find the new mean.

Solution:

Let x₁, x₂, x₃…x₇₅ be 75 numbers with their mean equal to 35. Then,

x₁ + x₂ + x₃ +…+x₇₅ = 35 × 75

→ x₁ + x₂ + x₃ +…+ x₇₅ = 2625

The new numbers are 4 x 1, 4 x 2, 4 x 3…4 x 75

Let M be the arithmetic mean of the new numbers. Then,

→ M = 140.

Exercise 23.2

Question: 1

A die was thrown 20 times and the following scores were recorded:

5, 2, 1, 3, 4, 4, 5, 6, 2, 2, 4, 5, 5, 6, 2, 2, 4, 5, 5, 1

Prepare the frequency table of the scores on the upper face of the die and find the mean score.

Solution:

Question: 2

The daily wages (in Rs) of 15 workers in a factory are given below:

200, 180, 150, 150, 130, 180, 180, 200, 150, 130, 180, 180, 200, 150, 180

Prepare the frequency table and find the mean wage.

Solution:

Exercise 23.3

Question: 1

Data – 83, 37, 70, 29, 45, 63, 41, 70, 34, 54

Solution:

Question: 2

Data – 133, 73, 89, 108, 94,104, 94, 85, 100, 120

Solution:

Question: 3

Data – 31, 38, 27, 28, 36, 25, 35, 40

Solution:

Question: 4

Data – 15, 6, 16, 8, 22, 21, 9, 18, 25

Solution:

Question: 5

Data – 41, 43,127, 99, 71, 92, 71, 58, 57

Solution:

Question: 6

Data – 25, 34, 31, 23, 22, 26, 35, 29, 20, 32

Solution:

Question: 7

Data –  12, 17, 3, 14, 5, 8, 7, 15

Solution:

Question: 8

Data – 92, 35, 67, 85, 72, 81, 56, 51, 42, 69

Solution:

Question: 9

Numbers 50, 42, 35, 2x +10, 2x – 8, 12, 11, 8, 6 are written in descending order and their median is 25, find x.

Solution:

Question: 10

Find the median of the following observations: 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?

Solution:

Question: 11

Find the median of the following data: 41, 43, 127, 99, 61, 92, 71, 58, 57, If 58 is replaced by 85, what will be the new median?

Solution:

Question: 12

The weights (in kg) of 15 students are: 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median.

Solution:

Question: 13

The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95

Solution:

Exercise 23.4

Question: 1

Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14

By using the empirical relation also find the mean.

Solution:

Arranging the data in ascending order such that same numbers are put together, we get:

12, 12, 13, 13, 14, 14, 14, 16, 19

Here, n = 9.

Here, 14 occurs the maximum number of times, i.e., three times. Therefore, 14 is the mode of the data.

Now,

Mode = 3 Median – 2 Mean

→ 14 = 3 x 14 – 2 Mean

→ 2 Mean = 42 – 14 = 28

→ Mean = 28 ÷ 2 = 14.

Question: 2

Find the median and mode of the data: 35, 32, 35, 42, 38, 32, 34

Solution:

Arranging the data in ascending order such that same numbers are put together, we get:

32, 32, 34, 35, 35, 38, 42

Here, n = 7

Here, 32 and 35, both occur twice. Therefore, 32 and 35 are the two modes.

Question: 3

Find the mode of the data: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4

Solution:

Arranging the data in ascending order such that same values are put together, we get:

0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6

Here, 2, 3 and 4 occur three times each. Therefore, 2, 3 and 4 are the three modes.

Question: 4

The runs scored in a cricket match by 11 players are as follows:

6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 10

Find the mean, mode and median of this data.

Solution:

Arranging the data in ascending order such that same values are put together, we get:

6, 8, 10, 10, 15, 15, 50, 80, 100, 120

Here, n = 11

Here, 10 occur three times. Therefore, 10 is the mode of the given data.

Now,

Mode = 3 Median – 2 Mean

→ 10 = 3 x 15 – 2 Mean

→ 2 Mean = 45 – 10 = 35

→ Mean = 35 ÷ 2 = 17.5

Question: 5

Find the mode of the following data:

12, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14

Solution:

Arranging the data in ascending order such that same values are put together, we get:

10, 12, 12, 14, 14, 14, 14, 14, 14, 16, 18

Here, clearly, 14 occurs the most number of times.

Therefore, 14 is the mode of the given data.

Exercise 21.1

Question: 1

Find the circumference of a circle whose radius is

(i) 14 cm                      (ii) 10 m                      (iii) 4 km

Solution:

Question: 2

Find the circumference of a circle whose diameter is

(i) 7 cm                        (ii) 4.2 cm                    (iii) 11.2 km

Solution:

Question: 3

Find the radius of a circle whose circumference is

(i) 52.8 cm                   (ii) 42 cm                     (iii) 6.6 km

Solution:

Question: 4

Find the diameter of a circle whose circumference is

(i) 12.56 cm                 (ii) 88 m                      (iii) 11.0 km

Solution:

Question: 5

The ratio of the radii of two circles is 3 : 2. What is the ratio of their circumferences?

Solution:

We have, the ratio of the radii = 3 : 2

So, let the radii of the two circles be 3r and 2r respectively.

Question: 6

A wire in the form of a rectangle 18.7 cm long and 14.3 cm wide is reshaped and bent into the form of a circle. Find the radius of the circle so formed.

Solution:

Question: 7

A piece of wire is bent in the shape of an equilateral triangle of each side 6.6 cm. It is re-bent to form a circular ring. What is the diameter of the ring?

Solution:

Question: 8

The diameter of a wheel of a car is 63 cm. Find the distance travelled by the car during the period, the wheel makes 1000 revolutions.

Solution:

It may be noted that in one revolution, the cycle covers a distance equal to the circumference of the wheel.

Now, the diameter of the wheel = 63 cm

∴ Circumference of the wheel = πd = 227 x 63 = 198 cm.

Thus, the cycle covers 198 cm in one revolution.

∴ The distance covered by the cycle in 1000 revolutions = (198 x 1000) = 198000 cm = 1980 m.

Question: 9

The diameter of a wheel of a car is 98 cm. How many revolutions will it make to travel 6160 metres.

Solution:

Question: 10

The moon is about 384400 km from the earth and its path around the earth is nearly circular. Find the circumference of the path described by the moon in lunar month.

Solution:

Question: 11

How long will John take to make a round of a circular field of radius 21 m cycling at the speed of 8 km/hr?

Solution:

Question: 12

The hour and minute hands of a clock are 4 cm and 6 cm long respectively. Find the sum of the distances travelled by their tips in 2 days.

Solution:

Question: 13

A rhombus has the same perimeter as the circumference of a circle. If the side of the rhombus is 2.2 m. Find the radius of the circle.

Solution:

Question: 14

A wire is looped in the form of a circle of radius 28 cm. It is re-bent into a square form. Determine the length of the side of the square.

Solution:

Question: 15

A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.

Solution:

Question: 16

A boy is cycling such that the wheels of the cycle are making 140 revolutions per minute. If the diameter of the wheel is 60 cm, calculate the speed per hour with which the boy is cycling.

Solution:

Question: 17

The diameter of the driving wheel of a bus is 140 cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 km per hour?

Solution:

Question: 18

A water sprinkler in a lawn sprays water as far as 7 m in all directions. Find the length of the outer edge of wet grass.

Solution:

Question: 19

A well of diameter 150 cm has a stone parapet around it. If the length of the outer edge of the parapet is 660 cm. then find the width of the parapet.

Solution:

Question: 20

An ox in a kolhu (an oil processing apparatus) is tethered to a rope 3 m long. How much distance does it cover in 14 rounds?

Solution:

Exercise 21.2

Question: 1

Find the area of a circle whose radius is

(i) 7 cm

(ii) 2.1 m

(iii) 7 km

Solution:

Question: 2

Find the area of a circle whose diameter is

(i) 8.4 cm

(ii) 5.6 m

(iii) 7 km

Solution:

Question: 3

The area of a circle is 154 cm2. Find the radius of the circle.

Solution:

Question: 4

Find the radius of a circle, if its area is

(i) 4 it cm²

(ii) 55.44 m²

(iii) 1.54 km²

Solution:

Question: 5

The circumference of a circle is 3.14 m, find its area.

Solution:

Question: 6

If the area of a circle is 50.24 m², find its circumference.

Solution:

Question: 7

A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (Take π = 22 / 7).

Solution:

Question: 8

A steel wire when bent in the form of a square encloses an area of 121 cm². If the same wire is bent in the form of a circle, find the area of the circle.

Solution:

Question: 9

A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of of road.

Solution:

Question: 10

Prove that the area of a circular path of uniform width h surrounding a circular region of radius r is πh(2r + h).

Solution:

Question: 11

The perimeter of a circle is 4πr cm. What is the area of the circle?

Solution:

Question: 12

A wire of 5024 m length is in the form of a square. It is cut and made a circle. Find the ratio of the area of the square to that of the circle.

Solution:

Question: 13

The radius of a circle is 14 cm. Find the radius of the circle whose area is double of the area of the circle.

Solution:

Question: 14

The radius of one circular field is 20 m and that of another is 48 m. find the radius of the third circular field whose area is equal to the sum of the areas of two fields.

Solution:

Question: 15

The radius of one circular field is 5 m and that of the other is 13 m. Find the radius of the circular field whose area is the difference of the areas of first and second field.

Solution:

Question: 16

Two circles are drawn inside a big circle with diameters 2/3rd and 1/3rd of the diameter of the  big circle as shown in Figure. Find the area of the shaded portion, if the length of the diameter of the circle is 18 cm.

Solution:

Question: 17

In Figure, the radius of quarter circular plot taken is 2 m and radius of the flower bed is 2 m. Find the area of the remaining field.

Solution:

Question: 18

Four equal circles, each of radius 5 cm, touch each other as shown in Figure. Find the area included between them. (Take π = 3.14).

Solution:

Question: 19

The area of circle is 100 times the area of another circle. What is the ratio of their circumferences?

Solution:

Exercise 20.1

Question: 1

Find the area, in square meters, of a rectangle whose

(i) Length = 5.5 m, breadth = 2.4 m

(ii) Length = 180 cm, breadth = 150 cm

Solution:

We have,

(i) Length = 5.5 m, Breadth = 2.4 m Therefore, Area of rectangle = Length x Breadth = 5.5 m x 2.4 m = 13.2 m²

(ii) Length = 180 cm = 1.8 m, Breadth = 150 cm = 1.5 m [ Since 100 cm = 1 m] Therefore, Area of rectangle = Length x Breadth = 1.8 m x 1.5 m = 2.7 m²

Question: 2

Find the area, in square centimeters, of a square whose side is

(i) 2.6 cm

(ii) 1.2 dm

Solution:

We have,

(i) Side of the square = 2.6 cm

Therefore, area of the square = (Side)² = (2.6 cm)2= 6.76 cm²

(ii) Side of the square = 1.2 dm = 1.2 x 10 cm = 12 cm

Therefore, area of the square = (Side)² = (12 cm)²= 144 cm² [ Since 1 dm = 10 cm]

Question: 3

Find in square metres, the area of a square of side 16.5 dam.

Solution:

We have,

Side of the square = 16.5

dam = 16.5 x 10 m = 165 m

Area of the square = (Side)² = (165 m)² = 27225 m²

[Since 1 dam/dm (decameter) = 10 m]

Question: 4

Find the area of a rectangular field in acres whose sides are:

(1) 200 m and 125 m

(ii) 75 m 5 dm and 120 m

Solution:

We have,

(i) Length of the rectangular field = 200 m

Breadth of the rectangular field = 125 m

Therefore, Area of the rectangular field = Length x Breadth = 200 m x 125 m

= 25000 m² = 250 acres [Since 100 m² = 1 are]

(ii) Length of the rectangular field =75 m 5 dm = (75 + 0.5) m

= 75.5 m [Since 1 dm = 10 cm = OA m]

Breadth of the rectangular field = 120 m

Therefore, Area of the rectangular field = Length x Breadth

= 75.5 m x 120 m = 9060 m² = 90.6 acres [Since 100 m² = 1 are]

Question: 5

Find the area of a rectangular field in hectares whose sides are:

(i) 125 m and 400 m

(ii) 75 m 5 dm and 120 m

Solution:

We have,

(i) Length of the rectangular field = 125 m

Breadth of the rectangular field = 400 m

Therefore, Area of the rectangular field = Length x Breadth

= 125 m x 400 m = 50000 m² = 5 hectares [Since 10000 m² = 1 hectare]

(ii) Length of the rectangular field =75 m 5 dm = (75 + 0.5) m

= 75.5 m [Since 1 dm = 10 cm = 0.1 m]

Breadth of the rectangular field = 120 m

Therefore, Area of the rectangular field = Length x Breadth

= 75.5 m x 120 m = 9060 m² = 0.906 hectares [Since 10000 m² = 1 hectare]

Question: 6

A door of dimensions 3 m x 2m is on the wall of dimension 10 m x 10 m. Find the cost of painting the wall if rate of painting is Rs 2.50 per sq. m.

Solution:

We have,

Length of the door = 3 m

Breadth of the door = 2 m

Side of the wall = 10 m

Area of the wall = Side x Side = 10 m x 10 m

= 100 m²

Area of the door = Length x Breadth = 3 m x 2 m = 6 m

Thus, required area of the wall for painting = Area of the wall – Area of the door

= (100 – 6) m²= 94 m²

Rate of painting per square metre = Rs. 2.50

Hence, the cost of painting the wall = Rs. (94 x 2.50) = Rs. 235

Question: 7

A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is bent in the shape of a square, what will be the measure of each side? Also, find which side encloses more area?

Solution:

We have,

Perimeter of the rectangle = 2(Length + Breadth)

= 2(40 cm + 22 cm) = 124 cm

It is given that the wire which was in the shape of a rectangle is now bent into a square.

Therefore, the perimeter of the square = Perimeter of the rectangle

→ Perimeter of the square = 124 cm

4 x side = 124 cm

Side = 124/4 = 31 cm

Now, Area of the rectangle = 40 cm x 22 cm = 880 cm²

Area of the square = (Side)² = (31 cm)² = 961 cm².

Therefore, the square-shaped wire encloses more area.

Question: 8

How many square metres of glass will be required for a window, which has 12 panes, each pane measuring 25 cm by 16 cm?

Solution:

We have,

Length of the glass pane = 25 cm

Breadth of the glass pane = 16 cm

Area of one glass pane = 25 cm x 16 cm

= 400 cm² = 0.04 m²

[Since 1 m² = 10000 cm² ]

Thus, Area of 12 such panes = 12 x 0.04 = 0.48 m²

Question: 9

A marble tile measures 10 cm x 12 cm. How many tiles will be required to cover a wall of size 3 m x 4 m? Also, find the total cost of the tiles at the rate of Rs 2 per tile.

Solution:

We have,

Area of the wall = 3 m x 4 m = 12 m²

Area of one marble tile = 10 cm x 12 cm

= 120 c m² = 0.012 m² [Since 1 m² = 10000 c m² ]

Thus, Number of tiles = Area of wall

Area of one tile=12 m² = 0.012 m²=1000

Cost of one tile = Rs. 2

Total cost = Number of tiles x Cost of one tile

= Rs. (1000 x 2)

= Rs. 2000

Question: 10

A table top is 9 dm 5 cm long 6 dm 5 cm broad. What will be the cost to polish it at the rate of 20 paise per square centimetre?

Solution:

We have,

Length of the table top = 9 dm 5 cm = (9 x 10 + 5) cm = 95 cm [ Since 1 dm = 10 cm]

Breadth of the table top = 6 dm 5 cm = (6 x 10 + 5) cm = 65 cm

Area of the table top = Length x Breadth = (95 cm x 65 cm) = 6175 c m²

Rate of polishing per square centimetre = 20 paise = Rs. 0.20

Total cost = Rs. (6175 x 0.20) = Rs. 1235

Question: 11

A room is 9.68 m long and 6.2 m wide. Its floor is to be covered with rectangular tiles of size 22 cm by 10 cm. Find the total cost of the tiles at the rate of Rs 2.50 per tile.

Solution:

We have,

Length of the floor of the room = 9.68 m

Breadth of the floor of the room = 6.2 m

Area of the floor = 9.68 m x 6.2 m = 60.016 m²

Length of the tile = 22 cm

Breadth of the tile = 10 cm

Area of one tile = 22 cm x 10 cm = 220 c m² = 0.022 m² [Since 1 m² = 10000 c m²]

Thus, Number of tiles = 60.016 m²/0.022 m²=2728

Cost of one tile = Rs. 2.50

Total cost = Number of tiles x Cost of one tile = Rs. (2728 x 2.50) = Rs. 6820

Question: 12

One side of a square field is 179 m. Find the cost of raising a lawn on the field at the rate of Rs 1.50 per square metre.

Solution:

We have,

Side of the square field = 179 m

Area of the field = (Side) ² = (179 m) ² = 32041 m²

Rate of raising a lawn on the field per square metre = Rs. 1.50 Thus,

Total cost of raising a lawn on the field = Rs. (32041 x 1.50) = Rs. 48061.50

Question: 13

A rectangular field is measured 290 m by 210 m. How long will it take for a girl to go two times round the field, if she walks at the rate of 1.5 m/sec?

Solution:

We have,

Length of the rectangular field = 290 m

Breadth of the rectangular field = 210 m

Perimeter of the rectangular field = 2(Length + Breadth) = 2(290 + 210) = 1000 m

Distance covered by the girl = 2 x Perimeter of the rectangular field = 2 x 1000 = 2000 m

The girl walks at the rate of 1.5 m/sec. Or, Rate = 1.5 x 60 m/min = 90 m/min

Thus, required time to cover a distance of 2000 m = 2000 m/90 m/min

= 2229min

Hence, the girl will take 2229 min to go two times around the field.

Question: 14

A corridor of a school is 8 m long and 6 m wide. It is to be covered with canvas sheets. If the available canvas sheets have the size 2 m x 1 m, find the cost of canvas sheets required to cover the corridor at the rate of Rs 8 per sheet.

Solution:

We have,

Length of the corridor = 8 m

Breadth of the corridor = 6 m

Area of the corridor of a school = Length x Breadth = (8 m x 6 m) = 48 m²

Length of the canvas sheet = 2 m

Breadth of the canvas sheet = 1 m

Area of one canvas sheet = Length x Breadth = (2 m x 1 m) = 2 m²

Thus, Number of canvas sheets = 48 m² /2m²=24

Cost of one canvas sheet = Rs. 8

Total cost of the canvas sheets = Rs. (24 x 8) = Rs. 192

Question: 15

The length and breadth of a playground are 62 m 60 cm and 25 m 40 cm respectively. Find the cost of turfing it at Rs 2.50 per square metre. How long will a man take to go three times round the field, if he walks at the rate of 2 metres per second?

Solution:

We have,

Length of a playground = 62 m 60 cm = 62.6 m [ Since 10 cm = 0.1 m]

Breadth of a playground = 25 m 40 cm = 25.4 m

Area of a playground = Length x Breadth= 62.6 m x 25.4 m = 1590.04 m²

Rate of turfing = Rs. 2.50/ m² Total cost of turfing = Rs. (1590.04 x 2.50) = Rs. 3975.10

Again, Perimeter of a rectangular field = 2(Length + Breadth) = 2(62.6 + 25.4) = 176 m

Distance covered by the man in 3 rounds of a field = 3 x Perimeter of a rectangular field

= 3 x 176 m = 528 m

The man walks at the rate of 2 m/sec. Or, Rate = 2 x 60 m/min = 120 m/min

Thus, required time to cover a distance of 528 m = 528 m120 m/min=4.4 min

= 4 minutes 24 seconds [Since 0.1 minutes = 6 seconds]

Question: 16

A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cm. Find the cost of bricks that are required, at the rate of Rs 750 per thousand.

Solution:

We have,

Length of the lane = 180 m

Breadth of the lane = 5 m

Area of a lane = Length x Breadth = 180 m x 5 m = 900 m²

Length of the brick = 20 cm

Breadth of the brick = 15 cm

Area of a brick = Length x Breadth = 20 cm x 15 cm

= 300 cm² = 0.03 m² [Since 1 m² = 10000 cm²]

Required number of bricks = 900 m²/0.03 m² = 30000

Cost of 1000 bricks = Rs. 750

Total cost of 30,000 bricks = Rs. 750×30,000/1000 = Rs. 22,500

Question: 17

How many envelopes can be made out of a sheet of paper 125 cm by 85 cm; supposing one envelope requires a piece of paper of size 17 cm by 5 cm?

Solution:

We have,

Length of the sheet of paper = 125 cm

Breadth of the sheet of paper = 85 cm

Area of a sheet of paper = Length x Breadth = 125 cm x 85 cm = 10,625 cm²

Length of sheet required for an envelope = 17 cm

Breadth of sheet required for an envelope = 5 cm

Area of the sheet required for one envelope = Length x Breadth

= 17 cm x 5 cm = 85 c m²

Thus, required number of envelopes = 10,625 cm²/85 c m² = 125

Question: 18

The width of a cloth is 170 cm. Calculate the length of the cloth required to make 25 diapers, if each diaper requires a piece of cloth of size 50 cm by 17 cm.

Solution:

We have,

Length of the diaper = 50 cm

Breadth of the diaper = 17 cm

Area of cloth to make 1 diaper = Length x Breadth = 50 cm x 17 cm = 850 cm²

Thus, Area of 25 such diapers = (25 x 850) c m² = 21,250 cm²

Area of total cloth = Area of 25 diapers = 21,250 cm²

It is given that width of a cloth = 170 cm

Length of the cloth = Area of cloth

Width of a cloth = 21,250 cm 2170 cm = 125 cm

Hence, length of the cloth will be 125 cm.

Question: 19

The carpet for a room 6.6 m by 5.6 m costs Rs 3960 and it was made from a roll 70 cm wide. Find the cost of the carpet per metre.

Solution:

We have,

Length of a room = 6.6 m

Breadth of a room = 5.6 m

Area of a room = Length x Breadth = 6.6 m x 5.6 m = 36.96 m²

Width of a carpet = 70 cm = 0.7 m [Since 1 m = 100 cm]

Length of a carpet = Area of a room

Width of a carpet = 36.96 m 20.7 m = 52.8 m

Cost of 52.8 m long roll of carpet = Rs. 3960

Therefore, Cost of 1 m long roll of carpet = Rs. 396052.8 = Rs. 75

Question: 20

A room is 9 m long, 8 m broad and 6.5 m high. It has one door of dimensions 2 m x 1.5 m and three windows each of dimensions 1.5 m x 1 m. Find the cost of white T4L washing the walls at Rs 3.80 per square metre.

Solution:

We have,

Length of a room = 9 m

Breadth of a room = 8 m

Height of a room = 6.5 m

Area of 4 walls = 2(1 + b)h = 2(9 m + 8 m) x 6.5 m = 2 x 17 m x 6.5 m = 221 m²

Length of a door = 2 m

Breadth of a door = 1.5 m

Area of a door = Length x Breadth = 2 m x 1.5 m = 3 m²

Length of a window = 1.5 m

Breadth of a window = 1 m

Since, area of one window = Length x Breadth = 1.5 m x 1 m = 1.5 m²

Thus, Area of 3 such windows = 3 x 1.5 m² = 4.5 m²

Area to be white-washed = Area of 4 walls – (Area of one door + Area of 3 windows)

Area to be white-washed = [221 – (3 + 4.5)] m² = (221 – 7.5) m² = 213.5 m²

Cost of white-washing for 1 m² area = Rs. 3.80

Cost of white-washing for 213.5 m² area = Rs. (213.5 x 3.80) = Rs. 811.30

Question: 21

A hall 36 m long and 24 m broad allowing 80 m2 for doors and windows, the cost of papering the walls at Rs 8.40 per m2 is Rs 9408. Find the height of the hall.

Solution:

We have,

Length of the hall = 36 m

Breadth of the hall = 24 m

Let h be the height of the hall.

Now, in papering the wall, we need to paper the four walls excluding the floor and roof of the hall. So, the area of the wall which is to be papered = Area of 4 walls

= 2h(I + b)

= 2h (36 + 24)

= 120h m²

Now, area left for the door and the windows = 80 m²

So, the area which is actually papered = (120h – 80) m²

Again, The cost of papering the walls at Rs 8.40 per m² = Rs. 9408

→ (120h – 80) m² x Rs. 8.40 per m²= Rs. 9408

→ (120h – 80) m² = Rs. 9408/Rs. 8.40

→ (120h – 80) m² = 1120 m²

→ 120h m² = (1120 + 80) m²

→ 120h m²= 1200 m²

h = 1200 m² 120 m = 10 m

Hence, the height of the wall would be 10 m.

Exercise 20.2

Question: 1

A rectangular grassy lawn measuring 40 m by 25 m is to be surrounded externally by a path which is 2 m wide. Calculate the cost of leveling the path at the rate of Rs 8.25 per square metre.

Solution:

We have,

Length AB = 40 m and breadth BC = 25 m

Area of lawn ABCD = 40 m x 25 m = 1000 m²

Length PQ = (40 + 2 + 2 ) m = 44 m

Breadth QR = ( 25 + 2 + 2 ) m = 29 m

Area of PQRS = 44 m x 29 m = 1276 m²

Now, Area of the path = Area of PQRS – Area of the lawn ABCD

= 1276 m² – 1000 m² = 276 m²

Rate of leveling the path = Rs. 8.25 per m²

Cost of leveling the path = Rs.( 8.25 x 276) = Rs. 2277

Question: 2

One metre wide path is built inside a square park of side 30 m along its sides. The remaining part of the park is covered by grass. If the total cost of covering by grass is Rs 1176, find the rate per square metre at which the park is covered by the grass.

Solution:

We have,

Side of square garden (a) = 30 m

Area of the square garden including the path = a²= (30)² = 900 m²

From the figure, it can be observed that the side of the square garden, when the path is not included, is 28 m.

Area of the square garden not including the path = (28)2 = 784 m²

Total cost of covering the park with grass = Area of the park covering with green grass x Rate per square metre

1176 = 784 x Rate per square metre

Rate per square metre at which the park is covered with grass = Rs. (1176 ÷ 784 ) = Rs. 1.50

Question: 3

Through a rectangular field of sides 90 m x 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the field. if the width of the roads is 3 m, find the total area covered by the two roads.

Solution:

Length of the rectangular sheet = 90 m

Breadth of the rectangular sheet = 60 cm

Area of the rectangular field = 90 m x 60 m = 5400 m²

Area of the road PQRS = 90 m x 3 m = 270 m²

Area of the road ABCD = 60 m x 3 m = 180 m²

Clearly, area of KLMN is common to the two roads.

Thus, area of KLMN = 3 m x 3 m = 9 m²

Hence, Area of the roads = Area (PQRS) + Area (ABCD) – Area (KLMN)

= (270 + 180) m² – 9 m² = 441 m²

Question: 4

from a rectangular sheet of tin, of size 100 cm by 80 cm, are cut four squares of side 10 cm from each corner. Find the area of the remaining sheet.

Solution:

Length of the rectangular sheet = 100 cm

Breadth of the rectangular sheet = 80 cm

Area of the rectangular sheet of tin = 100 cm x 80 cm

= 8000 c m²

Side of the square at the corner of the sheet = 10 cm

Area of one square at the corner of the sheet = (10 cm)² = 100 cm²

Area of 4 squares at the corner of the sheet = 4 x 100 cm² = 400 cm²

Hence, Area of the remaining sheet of tin = Area of the rectangular sheet – Area of the 4 squares

Area of the remaining sheet of tin = (8000 – 400) cm² = 7600 cm²

Question: 5

A painting 8 cm long and 5 cm wide is painted on a cardboard such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.

Solution:

We have, Length of the cardboard = 8 cm and breadth of the cardboard = 5 cm

Area of the cardboard including the margin = 8 cm x 5 cm = 40 c m²

From the figure, it can be observed that,

New length of the painting when the margin is not included = 8 cm – (1.5 cm + 1.5 cm)

= (8 – 3) cm = 5 cm

New breadth of the painting when the margin is not included = 5 cm – (1.5 cm + 1.5 cm)

= (5 – 3) cm = 2 cm

Area of the painting not including the margin = 5 cm x 2 cm = 10 cm²

Hence, Area of the margin = Area of the cardboard including the margin – Area of the painting

= (40 – 10) cm² = 30 cm²

Question: 6

Rakesh has a rectangular field of length 80 m and breadth 60 m. In it, he wants to make a garden 10 m long and 4 m broad at one of the corners and at another corner, he wants to grow flowers in two floor-beds each of size 4 m by 1.5 m. In the remaining part of the field, he wants to apply manures. Find the cost of applying the manures at the rate of Rs 300 per area.

Solution:

Length of the rectangular field = 80 m

Breadth of the rectangular field = 60 m

Area of the rectangular field = 80 m x 60 = 4800 m²

Again, Area of the garden = 10 m x 4 m = 40 m²

Area of one flower bed = 4 m x 1.5 m = 6 m²

Thus, Area of two flower beds = 2 x 6 m² = 12 m²

Remaining area of the field for applying manure = Area of the rectangular field – (Area of the garden + Area of the two flower beds)

Remaining area of the field for applying manure = 4800 m² – (40 + 12) m²

= (4800 – 52 ) m² = 4748 m²

Since 100 m² = 1 acre → 4748 m² = 47.48 acres

So, cost of applying manure at the rate of Rs. 300 per are will be Rs. (300 x 47.48) = Rs. 14244

Question: 7

Each side of a square flower bed is 2 m 80 cm long. It is extended by digging a strip 30 cm wide all around it. Find the area of the enlarged flower bed and also the increase in the area of the flower bed.

Solution:

We have ,

Side of the flower bed = 2 m 80 cm = 2.80 m [since 100 cm = 1 m ]

Area of the square flower bed = (Side)² = (2.80 m )² = 7.84 m²

Side of the flower bed with the digging strip = 2.80 m + 30 cm + 30 cm

= (2.80 + 0.3 + 0.3) m

= 3.4 m

Area of the enlarged flower bed with the digging strip = (Side) 2 = (3.4)² = 11.56 m²

Thus, Increase in the area of the flower bed = 11.56 m² – 7.84 m² = 3.72 m²

Question: 8

A room 5 m long and 4 m wide is surrounded by a verandah. If the verandah occupies an area of 22 m² , find the width of the varandah.

Solution:

Let the width of the verandah be x m.

Length of the room AB = 5 m and BC = 4 m

Area of the room = 5 m x 4 m = 20 m²

Length of the verandah PQ = (5 + x + x) = (5 + 2x) m

Breadth of the verandah QR = ( 4 + x + x) = (4 + 2x) m

Area of verandah PQRS = (5 + 2x) x (4 + 2x) = (4×2 + 18x + 20 ) m²

Area of verandah = Area of PQRS – Area of ABCD

→ 22 = 4x² + 18x + 20 – 20

22 = 4x² + 18x

11 = 2x² + 9x

2x² + 9x – 11 = 0

2x² + 11x – 2x – 11 =0

x(2x+11)-1(2x+11)=0

(x- 1)(2x+11)= 0

When x – 1 = 0, x = 1

When 2x + 11 = 0, x = -11/2

The width cannot be a negative value. So, width of the verandah = x = 1 m.

Question: 9

A square lawn has a 2 m wide path surrounding it. If the area of the path is 136 m² , find the area of the lawn.

Solution:

Let ABCD be the square lawn and PQRS be the outer boundary of the square path.

Let side of the lawn AB be x m.

Area of the square lawn = x²

Length PQ = (x m + 2 m + 2 m) = (x + 4) m

Area of PQRS = (x + 4)² = (x² + 8x + 16) m²

Now, Area of the path = Area of PQRS – Area of the square lawn

136 = x² + 8x + 16 – x²

136 = 8x + 16

136 – 16 = 8x

120 = 8x

x = 120/ 8 = 15

Side of the lawn = 15 m Hence, Area of the lawn = (Side)² = (15 m)² = 225 m²

Question: 10

A poster of size 10 cm by 8 cm is pasted on a sheet of cardboard such that there is a margin of width 1.75 cm along each side of the poster. Find (i) the total area of the margin (ii) the cost of the cardboard used at the rate of Re 0.60 per c m² .

Solution:

We have,

Length of poster = 10 cm and breadth of poster = 8 cm

Area of the poster = Length x Breadth = 10 cm x 8 cm = 80 cm²

From the figure, it can be observed that,

Length of the cardboard when the margin is included = 10 cm + 1.75 cm + 1.75 cm = 13.5 cm

Breadth of the cardboard when the margin is included = 8 cm + 1.75 cm + 1.75 cm = 11.5 cm

Area of the cardboard = Length x Breadth = 13.5 cm x 11.5 cm = 155.25 c m²

Hence,

(i) Area of the margin = Area of cardboard including the margin – Area of the poster

= 155.25 c m² – 80 c m²

= 75.25 c m²

(ii) Cost of the cardboard = Area of cardboard x Rate of the cardboard Rs 0.60 per cm²

= Rs. (155.25 x 0.60)

= Rs. 93.15

Question: 11

A rectangular field is 50 m by 40 m. It has two roads through its centre, running parallel to its sides. The widths of the longer and shorter roads are 1.8 m and 2.5 m respectively. Find the area of the roads and the area of the remaining portion of the field.

Solution:

Let ABCD be the rectangular field and KLMN and PQRS the two rectangular roads with width   1.8 m and 2.5 m, respectively.

Length of the rectangular field CD = 50 cm and breadth of the rectangular field BC = 40 m

Area of the rectangular field ABCD = 50 m x 40 m = 2000 m²

Area of the road KLMN = 40 m x 2.5 m = 100 m²

Area of the road PQRS = 50 m x 1.8 m = 90 m²

Clearly area of EFGH is common to the two roads.

Thus, Area of EFGH = 2.5 m x 1.8 m = 4.5 m²

Hence, Area of the roads = Area (KLMN) + Area (PQRS) – Area (EFGH)

= (100 m² + 90 m²) – 4.5 m²

= 185.5 m²

Area of the remaining portion of the field = Area of the rectangular field ABCD – Area of the roads

= (2000 – 185.5) m²

= 1814.5 m²
 

Question: 12

There is a rectangular field of size 94 m x 32 m. Three roads each of 2 m width pass through the field such that two roads are parallel to the breadth of the field and the third is parallel to the length. Calculate: (i) area of the field covered by the three roads (ii) area of the field not covered by the roads.

Solution:

Let ABCD be the rectangular field.

Here, Two roads which are parallel to the breadth of the field KLMN and EFGH with width 2 m each. One road which is parallel to the length of the field PQRS with width 2 m.

Length of the rectangular field AB = 94 m and breadth of the rectangular field BC = 32 m

Area of the rectangular field = Length x Breadth = 94 m x 32 m = 3008 m²

Area of the road KLMN = 32 m x 2 m = 64 m²

Area of the road EFGH = 32 m x 2 m = 64 m²

Area of the road PQRS = 94 m x 2 m = 188 m²

Clearly area of TUVI and WXYZ is common to these three roads.

Thus, Area of TUV1 = 2 m x 2 m = 4 m²

Area of WXYZ = 2 m x 2 m = 4 m²

Hence,

(i) Area of the field covered by the three roads: = Area (KLMN) + Area (EFGH) + Area (PQRS) – {Area (TUVI) + Area (WXYZ)}

= [ 64+ 64 + 188 – (4 + 4 )] m²

= 316 m² – 8 m²

= 308 m²

(ii) Area of the field not covered by the roads: = Area of the rectangular field ABCD – Area of the field covered by the three roads

= 3008 m² – 308 m²

= 2700 m²

Question: 13

A school has a hall which is 22 m long and 15.5 m broad. A carpet is laid inside the hall leaving all around a margin of 75 cm from the walls. Find the area of the carpet and the area of the strip left uncovered. If the width of the carpet is 82 cm, find the cost at the rate of Rs 18 per metre.

Solution:

We have,

Length of hall PQ = 22 m and breadth of hall QR = 15.5 m

Area of the school hall PQRS = 22 m x 15.5 m = 341 m²

Length of the carpet AB = 22 m – ( 0.75 m + 0.75 m) = 20.5 m [ Since 100 cm = 1 m]

Breadth of the carpet BC = 15.5 m – ( 0.75 m + 0.75 m) = 14 m

Area of the carpet ABCD = 20.5 m x 14 m = 287 m²

Area of the strip = Area of the school hall PQRS – Area of the carpet ABCD

= 341 m² – 287 m² = 54 m²

Again, Area of the 1 m length of carpet = 1 m x 0.82 m = 0.82 m²

Thus, Length of the carpet whose area is 287 m² = 287 m² + 0.82 m² = 350 m

Cost of the 350 m long carpet = Rs. 18 x 350 = Rs. 6300

Question: 14

Two cross roads, each of width 5 m, run at right angles through the centre of a rectangular park of length 70 m and breadth 45 m parallel to its sides. Find the area of the roads. Also, find the cost of constructing the roads at the rate of Rs 105 per m² .

Solution:

Let ABCD be the rectangular park then EFGH and IJKL the two rectangular roads with width 5 m.

Length of the rectangular park AD = 70 cm

Breadth of the rectangular park CD = 45 m

Area of the rectangular park = Length x Breadth = 70 m x 45 m = 3150 m²

Area of the road EFGH = 70 m x 5 m = 350 m²

Area of the road JKIL = 45 m x 5 m = 225 m²

Clearly area of MNOP is common to the two roads.

Thus, Area of MNOP = 5 m x 5 m = 25 m²

Hence,

Area of the roads = Area (EFGH) + Area (JKIL) – Area (MNOP)

= (350 + 225) m²– 25 m² = 550 m²

Again, it is given that the cost of constructing the roads = Rs. 105 per m²

Therefore,

Cost of constructing 550 m² area of the roads = Rs. (105 x 550)

= Rs. 57750.

Question: 15

The length and breadth of a rectangular park are in the ratio 5: 2. A 2.5 m wide path running all around the outside the park has an area 305 m² . Find the dimensions of the park.

Solution:

We have,

Area of path = 305 m²

Let the length of the park be 5x m and the breadth of the park be 2x m

Thus,

Area of the rectangular park = (5x) x (2x) = 10x² m²

Width of the path = 2.5 m

Outer length PQ = 5x m + 2.5 m + 2.5 m = (5x + 5) m

Outer breadth QR = 2x + 2.5 m + 2.5 m = (2x + 5) m

Area of PQRS = (5x + 5) m x (2x + 5) m = (10x² + 25x + 10x + 25) m²= (10x² + 35x + 25) m²

Area of the path = [(10x² + 35x + 25) – 10x²] m²

→ 305 = 35x + 25

→ 305 – 25 = 35x

→ 280 = 35x

→ x = 280 + 35 = 8

Therefore,

Length of the park = 5x = 5 x 8 = 40 m

Breadth of the park = 2x = 2 x 8 = 16 m

Question: 16

A square lawn is surrounded by a path 2.5 m wide. If the area of the path is 165 m² , find the area of the lawn.

Solution:

Let the side of the lawn be x m.

Given that width of the path = 2.5 m

Side of the lawn including the path = (x + 2.5 + 2.5) m = (x + 5 ) m

So, area of lawn = (Area of the lawn including the path) – (Area of the path)

We know that the area of a square = (Side)²

Area of lawn (x²) = (x + 5)² – 165

→ x² = (X² + 10X + 25) – 165

→ 165 = 10x + 25

→ 165 – 25 =10x

→ 140 = 10x

Therefore x = 140 / 10 = 14

Thus the side of the lawn = 14 m

Hence,

The area of the lawn = (14 m) ² = 196 m²

Exercise 20.3

Question: 1

Find the area of a parallelogram with base 8 cm and altitude 4.5 cm.

Solution:

We have,

Base = 8 cm and altitude = 4.5 cm

Thus, Area of the parallelogram = Base x Altitude

= 8 cm x 4.5 cm

= 36 cm²

Question: 2

Find the area in square metres of the parallelogram whose base and altitudes are as under

(i) Base =15 dm, altitude = 6.4 dm

(ii) Base =1 m 40 cm, altitude = 60 cm

Solution:

We have,

(i) Base = 15 dm = (15 x 10) cm = 150 cm = 1.5 m

Altitude = 6.4 dm = (6.4 x 10) cm = 64 cm = 0.64 m

Thus, Area of the parallelogram = Base x Altitude

= 1.5 m x 0.64 m

= 0.96 m²

(ii) Base = 1 m 40 cm = 1.4 m [Since 100 cm = 1 m]

Altitude = 60 cm = 0.6 m

Thus, Area of the parallelogram = Base x Altitude

= 1.4 m x 0.6 m

= 0.84 m²     [Since 100 cm = 1 m]

Question: 3

Find the altitude of a parallelogram whose area is 54 d m² and base is 12 dm.

Solution:

We have,

Area of the given parallelogram = 54 d m²

Base of the given parallelogram = 12 dm

Altitude of the given parallelogram = Area/Base = 54/12 dm = 4.5 dm

Question: 4

The area of a rhombus is 28 m². If its perimeter be 28 m, find its altitude.

Solution:

We have,

Perimeter of a rhombus = 28 m 4(Side) = 28 m [Since perimeter = 4(Side)]

Side = 28 m4=7 m

Now, Area of the rhombus = 28 m²

(Side x Altitude) = 28 m² (7 m x Altitude) = 28 m²

Altitude = 28 m 27m = 4 m

Question: 5

In Fig., ABCD is a parallelogram, DL ⊥ AB and DM ⊥ BC. If AB = 18 cm, BC =12 cm and DM= 9.3 cm, find DL.

Solution:

We have,

Taking BC as the base, BC = 12 cm and altitude DM = 9.3 cm

Area of parallelogram ABCD = Base x Altitude = (12 cm x 9.3 cm) = 111.6 c m²  —–(i)

Now, Taking AB as the base,

We have, Area of the parallelogram ABCD = Base x Altitude

= (18 cm x DL)                                          —–(ii)

From (i) and (ii), we have 18 cm x DL = 111.6 c m²

DL = 111.6 cm 218 cm = 6.2 cm

Question: 6

The longer side of a parallelogram is 54 cm and the corresponding altitude is 16 cm. If the altitude corresponding to the shorter side is 24 cm, find the length of the shorter side.

Solution:

We have,

ABCD is a parallelogram with the longer side AB = 54 cm and corresponding altitude AE = 16 cm. The shorter side is BC and the corresponding altitude is CF = 24 cm.

Area of a parallelogram = base x height.

We have two altitudes and two corresponding bases.

So,

½ x BC x CF = ½ x AB x AE

= BC x CF = AB x AE

= BC x 24 = 54 x 16

= BC = (54×16)/24 = 36 cm

Hence, the length of the shorter side BC = AD = 36 cm.

Question: 7

In Fig. 21, ABCD is a parallelogram, DL ⊥ AB. If AB = 20 cm, AD = 13 cm and area of the parallelogram is 100 c m², find AL.

Solution:

We have,

ABCD is a parallelogram with base AB = 20 cm and corresponding altitude DL.

It is given that the area of the parallelogram ABCD = 100 c m²

Now, Area of a parallelogram = Base x Height

100 c m² = AB x DL

100 c m² = 20 cm x DL

DL = 100 c m²= 5 cm

Again by Pythagoras theorem, we have,

(AD)² = (AL)² + (DL)²

= (13)² = (AL)²+ (5)²

(AL)2= (13)² – (5)²

= 169 – 25 = 144

(AL)² = (12)²

AL = 12 cm

Hence, length of AL is 12 cm.

Question: 8

In Fig. 21, if AB = 35 cm, AD= 20 cm and area of the parallelogram is 560 cm², find LB.

Solution:

We have,

ABCD is a parallelogram with base AB = 35 cm and corresponding altitude DL.

The adjacent side of the parallelogram AD = 20 cm.

It is given that the area of the parallelogram ABCD = 560 c m²

Now, Area of the parallelogram = Base x Height

560 cm² = AB x DL 560 c m² = 35 cm x DL

DL = 560 cm/235 cm= 16 cm

Again by Pythagoras theorem, we have, (AD)² = (AL)² + (DL)²

(20)²= (AL)² + (16)²

(AL)² = (20)² – (16)²

= 400 – 256

= 144

(AL)² = (12)2

= AL = 12 cm

From the figure, AB = AL + LB 35 cm

= 12 cm + LB

LB = 35 cm – 12 cm = 23 cm

Hence, length of LB is 23 cm.

Question: 9

The adjacent sides of a parallelogram are 10 m and 8 m. If the distance between the longer sides is 4 m, find the distance between the shorter sides.

Solution:

We have,

ABCD is a parallelogram with side AB = 10 m and corresponding altitude AE = 4 m.

The adjacent side AD = 8 m and the corresponding altitude is CF.

Area of a parallelogram = Base x Height

We have two altitudes and two corresponding bases.

So, AD x CF = AB x AE = 8 m x CF = 10 m x 4 m

= CF = (10 x 4)8 = 5 m

Hence, the distance between the shorter sides is 5 m.

Question: 10

The base of a parallelogram is twice its height. If the area of the parallelogram is 512 cm², find the base and height.

Solution:

Let the height of the parallelogram be x cm.

Then the base of the parallelogram is 2x cm.

It is given that the area of the parallelogram = 512 cm²

So, Area of a parallelogram = Base x Height

512 c m² = (2x) (x)

512 c m² = 2x²

X² = 512 cm²/2 = 256 c m²

X² = (16 cm)²

X = 16 cm

Hence, base = 2x = 2 x 16 = 32 cm and height = x = 16 cm.

Question: 11

Find the area of a rhombus having each side equal to 15 cm and one of whose diagonals is 24 cm.

Solution:

Let ABCD be the rhombus where diagonals intersect at 0.

Then AB = 15 cm and AC = 24 cm.

The diagonals of a rhombus bisect each other at right angles.

Therefore, triangle A0B is a right-angled triangle, right angled at O such that

OA = ½(AC) = 12 cm and AB = 15 cm.

By Pythagoras theorem, we have,

(AB)² = (OA)² + (OB)²

(15)² = (12)² + (OB)²

(OB)²= (15)²– (12)²

(OB)² = 225 – 144 = 81

(OB)² = (9)²

OB = 9 cm

BD = 2 x OB = 2 x 9 cm = 18 cm

Hence, Area of the rhombus ABCD = (½ x AC x BD)

=( 1/2 x 24 x 18 )

= 216 cm²

Question: 12

Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.

Solution:

Let ABCD be the rhombus whose diagonals intersect at 0.

Then AB = 20 cm and AC = 24 cm.

The diagonals of a rhombus bisect each other at right angles.

Therefore Triangle AOB is a right-angled triangle, right angled at O

Such that;

OA = ½ AC =12 cm and AB =20 cm

By Pythagoras theorem, we have,

(AB)² = (OA)²+ (OB) ²

(20)² = (12)² + (OB) ²

(OB) ²= (20) ²– (12) ²

(OB)² = 400 – 144

= 256

(OB) ² = (16) ²

= OB = 16 cm

BD =2 x OB = 2 x 16 cm = 32 cm

Hence, Area of the rhombus ABCD = ½ x AC x BD

= 1/2 x 24 x 32

= 384 c m²

Question: 13

The length of a side of a square field is 4 m. What will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonals is 2 m?

Solution:

We have,

Area of the rhombus = Area of the square of side 4 m

= 1/2 x AC x 130 = 4 m²

= 1/2 x AC x 2 m =16 m²

= AC = 16 m

We know that the diagonals of a rhombus are perpendicular bisectors of each other.

= AO = 1/2 (AC) = 8 m and BO = 1/2 (BD) = 1 m

By Pythagoras theorem, we have:

AO² + BO² = AB²

AB² = (8 m) ² + (1 m) ² = 64 m² + 1 m² = 65 m²

Side of a rhombus = AB = √65 m.

Let DX be the altitude.

Area of the rhombus = AB x DX 16 m²

= √65m × DX

DX = 16/(√65) m

Hence, the altitude of the rhombus will be 16/√65 m.

Question: 14

Two sides of a parallelogram are 20 cm and 25 cm. If the altitude corresponding to the sides of length 25 cm is 10 cm, find the altitude corresponding to the other pair of sides.

Solution:

We have,

ABCD is a parallelogram with longer side AB = 25 cm and altitude AE = 10 cm.

As ABCD is a parallelogram. Hence AB = CD (opposite sides of parallelogram are equal).

The shorter side is AD = 20 cm and the corresponding altitude is CF.

Area of a parallelogram = Base x Height

We have two altitudes and two corresponding bases.

So, = AD x CF = CD x AE

= 20 x CF = 25 x 10

CF = 12.5 cm

Hence, the altitude corresponding to the other pair of the side AD is 12.5 cm.

Question: 15

The base and corresponding altitude of a parallelogram are 10 cm and 12 cm respectively. If the other altitude is 8 cm, find the length of the other pair of parallel sides.

Solution:

We have,

ABCD is a parallelogram with side AB = CD = 10 cm (Opposite sides of parallelogram are equal) and corresponding altitude AM = 12 cm. The other side is AD and the corresponding altitude is CN = 8 cm.

Area of a parallelogram = Base x Height

We have two altitudes and two corresponding bases.

So,

= AD x CN = CD x AM

= AD x 8 = 10 x 12

= AD = (10×12)/8=15 cm

Hence, the length of the other pair of the parallel side = 15 cm.

Question: 16

A floral design on the floor of a building consists of 280 tiles. Each tile is in the shape of a parallelogram of altitude 3 cm and base 5 cm. Find the cost of polishing the design at the rate of 50 paise per cm².

Solution:

We have,

Attitude of a tile = 3 cm

Base of a tile = 5 cm

Area of one tile = Attitude x Base = 5 cm x 3 cm = 15 c m²

Area of 280 tiles = 280 x 15 c m² = 4200 c m²

Rate of polishing the tiles at 50 paise per c m² = Rs. 0.5 per c m²

Thus, Total cost of polishing the design = Rs. (4200 x 0.5) = Rs. 2100

Exercise 20.4

Question: 1

Find the area in square centimeters of a triangle whose base and altitude are as under :

(i) base =18 cm, altitude = 3.5 cm

(ii) base = 8 dm, altitude =15 cm

Solution:

We know that the area of a triangle = 1/2 (Base x Height)

(i) Here, base = 18 cm and height = 3.5 cm

Area of the triangle = 1/2 x 18 x 3.5

= 31.5 cm²

(ii) Here, base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm] and height = 3.5 cm

Area of the triangle = 1/2 x 80 x 15

= 600 c m²

Question: 2

Find the altitude of a triangle whose area is 42 cm² and base is 12 cm.

Solution:

We have,

Attitude of a triangle = (2 x Area)/Base

Here, base = 12 cm and area = 42 cm²

Attitude = (2 x 42)/12 = 7 cm

Question: 3

The area of a triangle is 50 cm². If the altitude is 8 cm, what is its base?

Solution:

We have,

Base of a triangle = (2 x Area)/ Altitude

Here, altitude = 8 cm and area = 50 cm²

Altitude = (2 x 50)/ 8 = 12.5 cm

Question: 4

Find the area of a right angled triangle whose sides containing the right angle are of lengths 20.8 m and 14.7 m.

Solution:

In a right-angled triangle,

The sides containing the right angles are of lengths 20.8 m and 14.7 m.

Let the base be 20.8 m and the height be 141 m.

Then,

Area of a triangle = 1/2 (Base x Height)

= 1/2 (20.8 × 14.7)

= 152.88 m²

Question: 5

The area of a triangle, whose base and the corresponding altitude are 15 cm and 7 cm, is equal to area of a right triangle whose one of the sides containing the right angle is 10.5 cm. Find the other side of this triangle.

Solution:

For the first triangle, we have,

Base = 15 cm and altitude = 7 cm

Thus, area of a triangle = 1/2 (Base x Altitude)

= 1/2 (15 x 7)

= 52.5 cm²

It is given that the area of the first triangle and the second triangle are equal.

Area of the second triangle = 52.5 c m²

One side of the second triangle = 10.5 cm

Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle

= (2x 52.5)/10.5

=10 cm

Hence, the other side of the second triangle will be 10 cm.

Question: 6

A rectangular field is 48 m long and 20 m wide. How many right triangular flower beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this field?

Solution:

We have,

Length of the rectangular field = 48 m

Breadth of the rectangular field = 20 m

Area of the rectangular field = Length x Breadth = 48 m x 20 m = 960 m²

Area of one right triangular flower bed = ½ (12 m x 5m) = 30 m²

Therefore,

Required number of right triangular flower beds = 960 m²/30 m²= 32

Question: 7

In Figure, ABCD is a quadrilateral in which diagonal AC = 84 cm; DL ⊥AC, BM ⊥ AC, DL = 16.5 cm and BM = 12 cm. Find the area of quadrilateral ABCD.

Solution:

We have,

AC = 84 cm, DL = 16.5 cm and BM = 12 cm

Area of triangle ADC = 1/2 (AC x DL) = 1/2 (84 cm x 16.5 cm) = 693 cm²

Area of triangle ABC = 1/2 (AC x BM) = 1/2 (84 cm x 12 cm) = 504 cm²

Hence, Area of quadrilateral ABCD = Area of ADC + Area of ABC = (693 + 504) cm² = 1197 cm²

Question: 8

Find the area of the quadrilateral ABCD given in Figure. The diagonals AC and BD measure 48 m and 32 m respectively and are perpendicular to each other.

Solution:

We have,

Diagonal AC = 48 cm and diagonal BD = 32 m

Area of a quadrilateral = 1/2 (Product of diagonals)

= 1/2(AC x BD) = 1/2 (48 x 32) m²

= (24 x 32) m² = 768 m²

Question: 9

In Fig below, ABCD is a rectangle with dimensions 32 m by 18 m. ADE is a triangle such that EF⊥ AD and EF= 14 cm. Calculate the area of the shaded region.

Solution:

We have,

Area of the rectangle = AB x BC = 32 m x 18 m = 576 m²

Area of the triangle = 1/2 (AD x FE)

= 1/2 (BC x FE) [Since AD = BC]

= 1/2 (18 m x 14 m)

= 9 m x 14 m

= 126 m²

Area of the shaded region = Area of the rectangle – Area of the triangle

= (576 – 126) m²

= 450 m²

Question: 10

In Fig. below, ABCD is a rectangle of length AB = 40 cm and breadth BC = 25 cm. If P, Q, R, S be the mid-points of the sides AB, BC, CD and DA respectively, find the area of the shaded region.

Solution:

Join points PR and SQ. These two lines bisect each other at point 0.

Here, AB = DC = SQ = 40 cm and AD = BC = RP = 25 cm

Also OP = OR = RP/2 = 25/2 = 12.5 cm

From the figure we observed that,

Area of Triangle SPQ = Area of Triangle SRQ

Hence, area of the shaded region = 2 x (Area of SPQ)

= 2 x (1/2 (SQ x OP))

= 2 x (1/2 (40 cm x 12.5 cm))

= 500 cm²

Question: 11

Calculate the area of the quadrilateral ABCD as shown in Figure, given that BD = 42 cm, AC = 28 cm, OD = 12 cm and AC ⊥ BO.

Solution:

We have,

BD = 42 cm, AC = 28 cm, OD= 12 cm

Area of Triangle ABC = 1/2 (AC x OB)

= 1/2 (AC x (BD – OD))

= 1/2 (28 cm x (42 cm – 12 cm))

= 1/2 (28 cm x 30 cm)

= 14 cm x 30 cm

= 420 cm²

Area of Triangle ADC = 1/2 (AC x OD) = 1/2 (28 cm x 12 cm)

= 14 cm x 12 cm

= 168 cm²

Hence, Area of the quadrilateral ABCD = Area of ABC + Area of ADC

= (420 + 168) cm²

= 588 cm²

Question: 12

Find the area of a figure formed by a square of side 8 cm and an isosceles triangle with base as one side of the square and perimeter as 18 cm.

Solution:

Let x cm be one of the equal sides of an isosceles triangle.

Given that the perimeter of the isosceles triangle = 18 cm

Then, x + x + 8 =18

2x = (18 – 8) cm = 10 cm

x = 5 cm

Area of the figure formed = Area of the square + Area of the isosceles triangle

Question: 13

Find the area of Figure, in the following ways: (i) Sum of the areas of three triangles (ii) Area of a rectangle — sum of the areas of five triangles

Solution:

We have,

(i) P is the midpoint of AD.

Thus AP = PD = 25 cm and AB = CD = 20 cm

From the figure, we observed that,

Area of Triangle APB = Area of Triangle PDC

Area of Triangle APB = 1/2 (AB x AP ) = 1/2 (20 cm x 25 cm) = 250 cm²

Area of Triangle PDC = Area of Triangle APB = 250 c m²

Area of Triangle RPQ = 1/2 (Base x Height) = 1/2 (25 cm x 10 cm) = 125 cm²

Hence, Sum of the three triangles = (250 + 250 + 125) cm² = 625 cm²

(ii) Area of the rectangle ABCD = 50 cm x 20 cm = 1000 cm²

Thus, Area of the rectangle – Sum of the areas of three triangles

= (1000 – 625 ) cm² = 375 cm²

Question: 14

Calculate the area of quadrilateral field ABCD as shown in Figure, by dividing it into a rectangle and a triangle.

Solution:

We have,

Join CE , which intersect AD at point E.

Here, AE = ED = BC = 25 m and EC = AB = 30 m

Area of the rectangle ABCE = AB x BC = 30 m x 25 m = 750 m²

Area of Triangle CED = 1/2 (EC x ED) = 1/2 ( 30 m x 25 m) = 375 m²

Hence, Area of the quadrilateral ABCD = (750 + 375) m² = 1125 m²

Question: 15

Calculate the area of the pentagon ABCDE, where AB = AE and with dimensions as shown in Figure.

Solution:

Join BE.

Area of the rectangle BCDE = CD x DE

= 10 cm x 12 cm

= 120 c m²

Area of Triangle ABE = 1/2 (BE x height of the triangle)

= 1/2 (10 cm x (20 – 12) cm)

= 1/2 (10 cm x 8 cm)

= 40 cm²

Hence, Area of the pentagon ABCDE = (120 + 40) cm² = 160 cm²

Question: 16

The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs 24.60 per hectare is Rs 332.10, find its base and height.

Solution:

Let altitude of the triangular field be h m

Then base of the triangular field is 3h m.

Area of the triangular field = 1/2 (h x 3h )=3h²/2 m² —–(i)

The rate of cultivating the field is Rs 24.60 per hectare.

Therefore,

Area of the triangular field = 332.10 /24.60

= 13.5 hectare = 135000 m² [Since 1 hectare = 10000 m²] —–(ii)

From equation (i) and (ii) we have,

3h²/2 = 135000 m²

3h² = 135000 x 2 = 270000 m²

h² = 270000/3 m²= 90000 m² = (300 m)²

h = 300 m

Hence, Height of the triangular field = 300 m and base of the triangular field = 3 x 300 m = 900 m

Question: 17

A wall is 4.5 m long and 3 m high. It has two equal windows, each having form and dimensions as shown in Figure. below. Find the cost of painting the wall (leaving windows) at the rate of Rs 15 per m².

Solution:

We have,

Length of a wall = 4.5 m

Breadth of the wall = 3 m

Area of the wall = Length x Breadth

= 4.5 m x 3 m = 13.5 m²

From the figure we observed that,

Area of the window = Area of the rectangle + Area of the triangle

= (0.8 m x 0.5 m) + (12 x 0.8 m x 0.2 m) [Since 1 m = 100 cm]

= 0.4 m² + 0.08 m²

= 0.48 m²

Area of two windows = 2 x 0.48 = 0.96 m²

Area of the remaining wall (leaving windows) = (13.5 – 0.96) m²

= 12.54 m²

Cost of painting the wall per m² = Rs. 15

Hence, the cost of painting on the wall = Rs. (15 x 12.54)

= Rs. 188.1

(In the book, the answer is given for one window, but we have 2 windows.)

Exercise 18.1

Question: 1

State the number of lines of symmetry for the following figures:

(i) An equilateral triangle

(ii) An isosceles triangle

(iii) A scalene triangle

(iv) A rectangle

(v) A rhombus

(vi) A square

(vii) A parallelogram

(viii) A quadrilateral

(ix) A regular pentagon

(x) A regular hexagon

(xi) A circle

(xii) A semi-circle

Solution:

(i) An equilateral triangle has 3 lines of symmetry.

(ii) An isosceles triangle has 1 line of symmetry.

(iii) A scalene triangle has no line of symmetry.

(iv) A rectangle has 2 lines of symmetry.

(v) A rhombus has 2 lines of symmetry.

(vi) A square has 4 lines of symmetry.

(vii) A parallelogram has no line of symmetry.

(viii) A quadrilateral has no line of symmetry.

(ix) A regular pentagon has 5 lines of symmetry.

(x) A regular hexagon has 6 lines of symmetry.

(xi) A circle has an infinite number of lines of symmetry all along the diameters.

(xii) A semicircle has only one line of symmetry.

Question: 2

What other name can you give to the line of symmetry of

(i) An isosceles triangle?

(ii) A circle?

Solution:

(i) An isosceles triangle has only 1 line of symmetry.

This line of symmetry is also known as the altitude of an isosceles triangle.

(ii) A circle has infinite lines of symmetry all along its diameters.

Question: 3

Identify three examples of shapes with no line of symmetry.

Solution:

A scalene triangle, a parallelogram and a trapezium do not have any line of symmetry.

Question: 4

Identify multiple lines of symmetry, if any, in each of the following figures:

Solution:

(A) The given figure has 3 lines of symmetry. Therefore it has multiple lines of symmetry.

(B) The given figure has 2 lines of symmetry. Therefore it has multiple lines of symmetry.

(C) The given figure has 3 lines of symmetry. Therefore it has multiple lines of symmetry.

(D) The given figure has 2 lines of symmetry. Therefore it has multiple lines of symmetry.

(E) The given figure has 4 lines of symmetry. Therefore it has multiple lines of symmetry.

(F) The given figure has only 1 line of symmetry.

(G) The given figure has 4 lines of symmetry. Therefore it has multiple lines of symmetry.

(H) The given figure has 6 lines of symmetry. Therefore it has multiple lines of symmetry.

Exercise 18.2

Question: 1

In the following figures, the mirror line (i.e. the line of symmetry) is given as dotted line. Complete each figure performing reflection in the dotted (mirror) line. Also, try to recall name of the complete figure.

Solution:

(a)it will be a rectangle.

(b) It will be a triangle.

(c) It will be a rhombus.

(d) It will be a circle.

(e) It will be a pentagon.

(f) It will be an octagon.

Question: 2

Each of the following figures shows paper cuttings with punched holes. Copy these figures on a plane sheet and mark the axis of symmetry so that if the paper is folded along it, then the wholes on one side of it coincide with the holes on the other side.

Solution:

The lines of symmetry in the given figures are as follows:

Question: 3

In the following figures if the dotted lines represent the lines of symmetry, find the other hole (s).

Solution:

The other holes in the figure are as follows:

Exercise 18.3

Question: 1

Give the order of rotational symmetry for each of the following figures when rotated about the marked point (x):

Solution:

(i) The given figure has its rotational symmetry as 4.

(ii) The given figure has its rotational symmetry as 3.

(iii) The given figure has its rotational symmetry as 3.

(iv) The given figure has its rotational symmetry as 4.

(v) The given figure has its rotational symmetry as 2.

(vi) The given figure has its rotational symmetry as 4.

(vii) The given figure has its rotational symmetry as 5.

(viii) The given figure has its rotational symmetry as 6.

(ix) The given figure has its rotational symmetry as 3.

Question: 2

Name any two figures that have both line symmetry and rotational symmetry.

Solution:

An equilateral triangle and a square have both lines of symmetry and rotational symmetry.

Question: 3

Give an example of a figure that has a line of symmetry but does not have rotational symmetry.

Solution:

A semicircle and an isosceles triangle have a line of symmetry but do not have rotational symmetry.

Question: 4

Give an example of a geometrical figure which has neither a line of symmetry nor a rotational symmetry.

Solution:

A scalene triangle has neither a line of symmetry nor a rotational symmetry.

Question: 5

Give an example of a letter of the English alphabet which has

(i) No line of symmetry

(ii) Rotational symmetry of order 2.

Solution:

(i) The letter of the English alphabet which has no line of symmetry is Z.

(ii) The letter of the English alphabet which has rotational symmetry of order 2 is N.

Question: 6

What is the line of symmetry of a semi-circle? Does it have rotational symmetry?

Solution:

A semicircle (half of a circle) has only one line of symmetry. In the figure, there is one line of symmetry. The figure is symmetric along the perpendicular bisector I of the diameter XY.

A semi-circle does not have any rotational symmetry.

Question: 7

Draw, whenever possible, a rough sketch of

(i) a triangle with both line and rotational symmetries.

(ii) a triangle with only line symmetry and no rotational symmetry.

(iii) a quadrilateral with a rotational symmetry but not a line of symmetry.

(iv) a quadrilateral with line symmetry but not a rotational symmetry.

Solution:

(i) An equilateral triangle has 3 lines of symmetry and a rotational symmetry of order 3.

(ii) An isosceles triangle has only 1 line of symmetry and no rotational symmetry.

(iii) A parallelogram is a quadrilateral which has no line of symmetry but a rotational symmetry of order 2.

(iv) A kite is a quadrilateral which has only one line of symmetry and no rotational symmetry.

Exercise 17.1

Question: 1

Draw an ∠BAC of measure 50° such that AB = 5 cm and AC = 7 cm. Through C draw a line parallel to AB and through B draw a line parallel to AC, intersecting each other at D. Measure BD and CD

Solution:

Steps of construction:

Draw angle BAC = 50° such that AB = 5 cm and AC = 7 cm.

Cut an arc through C at an angle of 50°

Draw a straight line passing through C and the arc. This line will be parallel to AB since ∠CAB =∠RCA=50°

Alternate angles are equal; therefore the line is parallel to AB.

Again through B, cut an arc at an angle of 50° and draw a line passing through B and this arc and say this intersects the line drawn parallel to AB at D.

∠SBA =∠BAC = 50°, since they are alternate angles. Therefore BD parallel to AC

Also we can measure BD = 7 cm and CD = 5 cm.

Question: 2

Draw a line PQ.  Draw another line parallel to PQ at a distance of 3 cm from it.

Solution:

Steps of construction:

Draw a line PQ.

Take any two points A and B on the line.

Construct ∠PBF = 90° and ∠QAE = 90°

With A as centre and radius 3 cm cut AE at C.

With B as centre and radius 3 cm cut BF at D.

Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.

Question: 3

Take any three non-collinear points A, B, C and draw ∠ABC. Through each vertex of the triangle, draw a line parallel to the opposite side.

Solution:

Steps of construction:

Mark three non collinear points A, B and C such that none of them lie on the same line.

Join AB, BC and CA to form triangle ABC.

Parallel line to AC

With A as centre, draw an arc cutting AC and AB at T and U, respectively.

With centre B and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at X.

With centre X and radius equal to TU, draw an arc cutting the arc drawn in the previous step at Y.

Join BY and produce in both directions to obtain the line parallel to AC.

Parallel line to AB

With B as centre, draw an arc cutting BC and BA at W and V, respectively.

With centre C and the same radius as in the previous step, draw an arc on the opposite side of BC to cut BC at P.

With centre P and radius equal to WV, draw an arc cutting the arc drawn in the previous step at Q.

Join CQ and produce in both directions to obtain the line parallel to AB.

Parallel line to BC

With B as centre, draw an arc cutting BC and BA at W and V, respectively (already drawn).

With centre A and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at R.

With centre R and radius equal to WV, draw an arc cutting the arc drawn in the previous step at S.

Join AS and produce in both directions to obtain the line parallel to BC.

Question: 4

Draw two parallel lines at a distance of 5kms apart.

Solution:

Steps of construction:

Draw a line PQ.

Take any two points A and B on the line.

Construct ∠PBF = 90° and ∠QAE = 90°

With A as centre and radius 5 cm cut AE at C.

With B as centre and radius 5 cm cut BF at D.

Join CD and produce it on either side to get the required line parallel to AB and at a distance of 5 cm from it.

Exercise 17.2

Question: 1

Draw △ABC in which AB = 5.5 cm. BC = 6 cm and CA = 7 cm. Also, draw perpendicular bisector of side BC.

Solution:

Steps of construction:

• Draw a line segment AB of length 5.5 cm.
• From B, cut an arc of radius 6 cm.
• With centre A, draw an arc of radius 7 cm intersecting the previously drawn arc at C.
• Join AC and BC to obtain the desired triangle.
• With centre B and radius more than half of BC, draw two arcs on both sides of BC.
• With centre C and the same radius as in the previous step, draw two arcs intersecting the arcs drawn in the previous step at X and Y.
• Join XY to get the perpendicular bisector of BC.
   

Question: 2

Draw ∆PQR in which PQ = 3 cm, QR. 4 cm and RP = 5 cm. Also, draw the bisector of ∠Q

Solution:

Steps of construction:

• Draw a line segment PQ of length 3 cm.
• With Q as centre and radius 4 cm, draw an arc.
• With P as centre and radius 5 cm, draw an arc intersecting the previously drawn arc at R.
• Join PR and OR to obtain the required triangle.
• From Q, cut arcs of equal radius intersecting PQ and QR at M and N, respectively.
• From M and N, cut arcs of equal radius intersecting at point S.
• Join QS and extend to produce the angle bisector of angle PQR.
• Verify that angle PQS and angle SQR are equal to 45° each.
   

Question: 3

Draw an equilateral triangle one of whose sides is of length 7 cm.

Solution:

Steps of construction:

• Draw a line segment AB of length 7 cm.
• With centre A, draw an arc of radius 7 cm.
• With centre B, draw an arc of radius 7 cm intersecting the previously drawn arc at C.
• Join AC and BC to get the required triangle.
   

Question: 4

Draw a triangle whose sides are of lengths 4 cm, 5 cm and 7 cm. Draw the perpendicular bisector of the largest side.

Solution:

Steps of construction:

Draw a line segment PR of length 7 cm.

• With centre P, draw an arc of radius 5 cm.
• With centre R, draw an arc of radius 4 cm intersecting the previously drawn arc at Q.
• Join PQ and QR to obtain the required triangle.
• From P, draw arcs with radius more than half of PR on either sides.
• With the same radius as in the previous step, draw arcs from R on either sides of PR intersecting the arcs drawn in the previous step at M and N.
• MN is the required perpendicular bisector of the largest side.
   

Question: 5

Draw a triangle ABC with AB = 6 cm, BC = 7 cm and CA = 8 cm. Using ruler and compass alone, draw (i) the bisector AD of ∠A and (ii) perpendicular AL from A on BC. Measure LAD.

Solution:

Steps of construction:

Draw a line segment BC of length 7 cm.

With centre B, draw an arc of radius 6 cm.

With centre C, draw an arc of radius 8 cm intersecting the previously drawn arc at A.

Join AC and BC to get the required triangle.

Angle bisector steps:

• From A, cut arcs of equal radius intersecting AB and AC at E and F, respectively.
• From E and F, cut arcs of equal radius intersecting at point H.
• Join AH and extend to produce the angle bisector of angle A, meeting line BC at D.

Perpendicular from Point A to line BC steps:

• From A, cut arcs of equal radius intersecting BC at P and Q, respectively (Extend BC to draw these arcs).
• From P and Q, cut arcs of equal radius intersecting at M.
• Join AM cutting BC at L.
• AL is the perpendicular to the line BC.
• Angle LAD is 15°.
   

Question: 6

Draw △DEF such that DE= DF= 4 cm and EF = 6 cm. Measure ∠E  and ∠F.

Solution:

Steps of construction:

• Draw a line segment EF of length 6 cm.
• With E as centre, draw an arc of radius 4 cm.
• With F as centre, draw an arc of radius 4 cm intersecting the previous arc at D.
• Join DE and DF to get the desired triangle DEF.
• By measuring we get, ∠E= ∠F= 40°..
   

Question: 7

Draw any triangle ABC. Bisect side AB at D. Through D, draw a line parallel to BC, meeting AC in E. Measure AE and EC.

Solution:

Steps of construction:

We first draw a triangle ABC with each side = 6 cm.

Steps to bisect line AB:

• Draw an arc from A on either side of line AB.
• With the same radius as in the previous step, draw an arc from B on either side of AB intersecting the arcs drawn in the previous step at P and Q.
• Join PQ cutting AB at D. PQ is the perpendicular bisector of AB.

Parallel line to BC:

• With B as centre, draw an arc cutting BC and BA at M and N, respectively.
• With centre D and the same radius as in the previous step, draw an arc on the opposite side of AB to cut AB at Y.
• With centre Y and radius equal to MN, draw an arc cutting the arc drawn in the previous step at X.
• Join XD and extend it to intersect AC at E.
• DE is the required parallel line.

Exercise 17.3

Question: 1

Draw △ABC in which AB = 3 cm, BC = 5 cm and ∠Q = 70°.

Solution:

Steps of construction:

• Draw a line segment AB of length 3 cm.
• Draw ∠XBA=70°.
• Cut an arc on BX at a distance of 5 cm at C.
• Join AC to get the required triangle.
   

Question: 2

Draw △ABC  in which ∠A=70°., AB = 4 cm and AC= 6 cm. Measure BC.

Solution:

Steps of construction:

• Draw a line segment AC of length 6 cm.
• Draw ∠XAC=70°.
• Cut an arc on AX at a distance of 4 cm at B.
• Join BC to get the desired triangle.
• We see that BC = 6 cm.
   

Question: 3

Draw an isosceles triangle in which each of the equal sides is of length 3 cm and the angle between them is 45°.

Solution:

Steps of construction:

Draw a line segment PQ of length 3 cm.

Draw ∠QPX=45°.

Cut an arc on PX at a distance of 3 cm at R.

Join QR to get the required triangle.
 

Question: 4

Draw △ABC  in which ∠A = 120°, AB = AC = 3 cm. Measure ∠B and ∠C.

Solution:

Steps of construction:

• Draw a line segment AC of length 3 cm.
• Draw ∠XAC = 120°.
• Cut an arc on AX at a distance of 3 cm at B.
• Join BC to get the required triangle.

By measuring, we get ∠B = ∠C = 30°.
 

Question: 5

Draw △ABC  in which ∠C = 90° and AC = BC = 4 cm.

Solution:

Steps of construction:

• Draw a line segment BC of length 4 cm.
• At C, draw ∠BCY=90°.
• Cut an arc on CY at a distance of 4 cm at A.
• Join AB. ABC is the required triangle.
   

Question: 6

Draw a triangle ABC in which BC = 4 cm, AB = 3 cm and ∠B = 45°. Also, draw a perpendicular from A on BC.

Solution:

Steps of construction:

• Draw a line segment AB of length 3 cm.
• Draw an angle of 45° and cut an arc at this angle at a radius of 4 cm at C.
• Join AC to get the required triangle.
• With A as centre, draw intersecting arcs at M and N.
• With centre M and radius more than half of MN, cut an arc on the opposite side of A.
• With N as centre and radius the same as in the previous step, cut an arc intersecting the previous arc at E.
• Join AE, it meets BC at D, then AE is the required perpendicular.
   

Question: 7

Draw a triangle ABC with AB = 3 cm, BC = 4 cm and ∠B = 60°. Also, draw the bisector of angles C and A of the triangle, meeting in a point O. Measure ∠COA.

Solution:

Steps of construction:

Draw a line segment BC = 4 cm.

Draw ∠CBX = 60°.

Draw an arc on BX at a radius of 3 cm cutting BX at A.

Join AC to get the required triangle.

Angle bisector for angle A:

• With A as centre, cut arcs of the same radius cutting AB and AC at P and Q, respectively.
• From P and Q cut arcs of same radius intersecting at R.
• Join AR to get the angle bisector of angle A.

Angle bisector for angle C:

• With A as centre, cut arcs of the same radius cutting CB and CA at M and N, respectively.
• From M and N, cut arcs of the same radius intersecting at T
• Join CT to get the angle bisector of angle C.

Mark the point of intersection of CT and AR as 0.

Angle ∠COA = 120°.

Exercise 17.4

Question: 1

Construct ∆ABC in which BC = 4 cm, ∠B = 50° and ∠C = 70°.

Solution:

Steps of construction:

• Draw a line segment BC of length 4 cm.
• Draw ∠CBX such that ∠CBX=50°.
• Draw ∠BCY with Y on the same side of BC as X such that ∠BCY=70°.
• Let CY and BX intersect at A.
• ABC is the required triangle.
   

Question: 2

Draw ∆ABC in which BC = 8 cm, ∠B = 50° and ∠A = 50°.

∠ABC + ∠BCA + ∠CAB = 180°

∠BCA = 180° − ∠CAB − ∠ABC

∠BCA = 180°− 100° = 80°

Solution:

Steps of construction:

• Draw a line segment BC of length 8 cm.
• Draw ∠CBX such that ∠CBX = 50°.
• Draw ∠BCY with Y on the same side of BC as X such that ∠BCY = 80°.
• Let CY and BX intersect at A.
   

Question: 3

Draw ∆ABC in which ∠Q = 80°, ∠R = 55° and QR = 4.5 cm. Draw the perpendicular bisector of side QR.

Solution:

Steps of construction:

• Draw a line segment QR = 4.5 cm.
• Draw ∠RQX = 80° and ∠QRY = 55°.
• Let QX and RY intersect at P so that PQR is the required triangle.
• With Q as centre and radius more that 2.25 cm, draw arcs on either sides of QR.
• With R as centre and radius more than 2.25 cm, draw arcs intersecting the previous arcs at M and N.
• Join MN

MN is the required perpendicular bisector of QR.

Question: 4

Construct ∆ABC in which AB = 6.4 cm, ∠A = 45° and ∠B = 60°

Solution:

Steps of construction:

Draw a line segment AB = 6.4 cm.

Draw ∠BAX = 45°.

Draw ∠ABY with Y on the same side of AB as X such that ∠ABY = 60°.

Let AX and BY intersect at C.

ABC is the required triangle.

Question: 5

Draw ∆ABC in which AC = 6 cm, ∠A = 90° and ∠B = 60°.

∠A + ∠B + ∠C = 180°

Therefore ∠C = 180°− 60°− 90°= 30°

Solution:

Steps of construction:

• Draw a line segment AC = 6 cm.
• Draw ∠ACX = 30°.
• Draw ∠CAY with Y on the same side of AC as X such that ∠CAY = 90°.
• Join CX and AY. Let these intersect at B.

ABC is the required triangle where angle ∠ABC = 60°.

Exercise 17.5

Question: 1

Draw a right triangle with hypotenuse of length 5 cm and one side of length 4 cm.

Solution:

Steps of construction:

• Draw a line segment QR = 4 cm.
• Draw ∠QRX of measure 90°.
• With centre Q and radius PQ = 5 cm, draw an arc of the circle to intersect ray RX at P.
• Join PQ to obtain the desired triangle PQR.

PQR is the required triangle.

Question: 2

Draw a right triangle whose hypotenuse is of length 4 cm and one side is of length 2.5 cm.

Solution:

Steps of construction:

• Draw a line segment QR = 2.5 cm.
• Draw ∠QRX of measure 90°.
• With centre Q and radius PQ = 4 cm, draw an arc of the circle to intersect ray RX at P.
• Join PQ to obtain the desired triangle PQR.

PQR is the required triangle.

Question: 3

Draw a right triangle having hypotenuse of length 5.4 cm, and one of the acute angles of measure 30° 

Solution:

Let ABC be the right triangle at A such that hypotenuse BC = 5.4 cm. Let cC = 30°.

Therefore ∠A + ∠B + ∠C = 180°∠B = 180°− 30°− 90° = 60°

Steps of construction:

• Draw a line segment BC = 5.4 cm.
• Draw angle CBY = 60°
• Draw angle BCX of measure 30° with X on the same side of BC as Y.
• Let BY and CX intersect at A.

Then ABC is the required triangle.

Question: 4

Construct a right triangle ABC in which AB = 5.8 cm, BC = 4.5 cm and ∠C = 90°.

Solution:

Steps of construction:

• Draw a line segment BC = 4.5 cm.
• Draw ∠BCX of measure 90°..
• With centre B and radius AB = 5.8 cm, draw an arc of the circle to intersect ray BX at A.
• Join AB to obtain the desired triangle ABC.

ABC is the required triangle.
 

Question: 5

Construct a right triangle, right angled at C in which AB = 5.2 cm and BC= 4.6 cm.

Solution:

Steps of construction:

• Draw a line segment BC = 4.6 cm.
• Draw ∠BCX of measure 90°
• With centre B and radius AB = 5.2 cm, draw an arc of the circle to intersect ray CX at A.
• Join AB to obtain the desired triangle ABC.

ABC is the required triangle.

Exercise 16.1

Question: 1

Explain the concept of congruence of figures with the help of certain examples.

Solution:

Congruent objects or figures are exact copies of each other or we can say mirror images of each other. The relation of two objects being congruent is called congruence.

Consider Ball A and Ball B. These two balls are congruent.

Now consider the two stars below. Star A and Star B are exactly the same in size, colour and shape. These are congruent stars.

Question: 2

Fill in the blanks:

(i) Two line segments are congruent if ___

(ii) Two angles are congruent if ___

(iii) Two square are congruent if ___

(iv) Two rectangles are congruent if ___

(v) Two circles are congruent if ___

Solution:

(i) They have the same length, since they can superpose on each other.

(ii) Their measures are the same. On superposition, we can see that the angles are equal.

(iii) Their sides are equal. All the sides of a square are equal and if two squares have equal sides, then all their sides are of the same length. Also angles of a square are 90° which is also the same for both the squares.

(iv) Their lengths are equal and their breadths are also equal. The opposite sides of a rectangle are equal. So if two rectangles have lengths of the same size and breadths of the same size, then they are congruent to each other.

(v) Their radii are of the same length. Then the circles will have the same diameter and thus will be congruent to each other.

Question: 3

In Figure, ∠POQ ≅ ∠ROS, can we say that ∠POR ≅ ∠QOS

Solution:

We have,

∠POQ ≅ ∠ROS (1) Also, ∠ROQ ≅ ∠ROQ Therefore adding ∠ROQ to both sides of (1), Weget, ∠POQ + ∠ROQ ≅ ∠ROQ + ∠ROS Therefore, ∠PQR = ∠QOS

Question: 4

In figure, a = b = c, name the angle which is congruent to ∠AOC

Solution:

We have,

∠ AOB = ∠ BOC = ∠ COD

Therefore, ∠ AOB = ∠ COD

Also, ∠ AOB + ∠ BOC = ∠ BOC + ∠ COD

∠ AOC = ∠ BOD

Hence, ∠ BOD is congruent to ∠ AOC

Question: 5

Is it correct to say that any two right angles are congruent? Give reasons to justify your answer.

Solution:

Two right angles are congruent to each other because they both measure 90 degrees.

We know that two angles are congruent if they have the same measure.

Question: 6

In figure, ∠AOC ≅ ∠PYR and ∠BOC ≅ ∠QYR. Name the angle which is congruent to ∠AOB.

Solution:

∠AOC ≅ ∠PYR…. (i) Also, ∠BOC ≅ ∠QYR Now, ∠AOC = ∠AOB + ∠BOC ∠PYR = ∠PYQ +∠QYR By putting the value of ∠AOC and ∠PYR in equation (i), we get, ∠AOB + ∠BOC ≅ ∠PYQ + ∠QYR ∠AOB ≅ ∠PYQ (∠BOC ≅ ∠QYR) Hence, ∠AOB ≅ ∠PYQ

 

Question: 7

Which of the following statements are true and which are false;

(i) All squares are congruent.

(ii) If two squares have equal areas, they are congruent.

(iii) If two rectangles have equal areas, they are congruent.

(iv) If two triangles have equal areas, they are congruent.

Solution:

(i) False.

All the sides of a square are of equal length.

However, different squares can have sides of different lengths. Hence all squares are not congruent.

(ii) True.

Area of a square = side x side

Therefore, two squares that have the same area will have sides of the same lengths. Hence they will be congruent.

(iii) False Area of a rectangle = length x breadth

Two rectangles can have the same area. However, the lengths of their sides can vary and hence they are not congruent.

Example: Suppose rectangle 1 has sides 8 m and 8 m and area 64 meter square. Rectangle 2 has sides 16 m and 4 m and area 64 meter square. Then rectangle 1 and 2 are not congruent.

(iv) False

Area of a triangle = 12 x base x height

Two triangles can have the same area but the lengths of their sides can vary and hence they cannot be congruent.

Exercise 16.2

Question: 1

In the following pairs of triangle (Figures), the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic.

Solution:

1) In Δ ABC and Δ DEF

AB = DE = 4.5 cm (Side)

BC = EF = 6 cm (Side) and

AC = DF = 4 cm (Side)

Therefore, by SSS criterion of congruence, ΔABC ≅ ΔDEF

2)

In Δ ACB and Δ ADB

AC = AD (Side)

BC = BD (Side) and

AB = AB (Side)

Therefore, by SSS criterion of congruence, ΔACB ≅ ΔADB

3) In Δ ABD and Δ FEC,

AB = FE (Side)

AD = FC (Side)

BD = CE (Side)

Therefore, by SSS criterion of congruence, ΔABD ≅ ΔFEC

4) In Δ ABO and Δ DOC,

AB = DC (Side)

AO = OC (Side)

BO = OD (Side)

Therefore, by SSS criterion of congruence, ΔABO ≅ ΔODC

Question: 2

In figure, AD = DC and AB = BC

(i) Is ΔABD ≅ ΔCBD?

(ii) State the three parts of matching pairs you have used to answer (i).

Solution:

Yes ΔABD = ΔCBD by the SSS criterion. We have used the three conditions in the SSS criterion as follows:

AD = DC

AB = BC and

DB = BD

Question: 3

In Figure, AB = DC and BC = AD.

(i) Is ΔABC ≅ ΔCDA?

(ii) What congruence condition have you used?

(iii) You have used some fact, not given in the question, what is that?

Solution:

We have AB = DC

BC = AD

and AC = AC

Therefore by SSS ΔABC ≅ ΔCDA

We have used Side congruence condition with one side common in both the triangles.

Yes, have used the fact that AC = CA.
 

Question: 4

In ΔPQR ≅ ΔEFD,

(i) Which side of ΔPQR equals ED?

(ii) Which angle of ΔPQR equals angle E?

Solution:

ΔPQR ≅ ΔEFD

(i) Therefore PR = ED since the corresponding sides of congruent triangles are equal.

(ii) ∠QPR = ∠FED since the corresponding angles of congruent triangles are equal.

Question: 5

Triangles ABC and PQR are both isosceles with AB = AC and PO = PR respectively. If also, AB = PQ and BC = QR, are the two triangles congruent? Which condition do you use?

It ∠B = 50°, what is the measure of ∠R?

Solution:

We have AB = AC in isosceles ΔABC

And PQ = PR in isosceles ΔPQR.

Also, we are given that AB = PQ and QR = BC.

Therefore, AC = PR (AB = AC, PQ = PR and AB = PQ)

Hence, ΔABC ≅ ΔPQR

Now

∠ABC = ∠PQR (Since triangles are congruent)However, ΔPQR is isosceles.

Therefore, ∠PRQ = ∠PQR = ∠ABC = 50°

Question: 6

ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If ∠BAC = 40° and ∠BDC = 100°, then find ∠ADB.

Solution:

∠BAD = ∠CAD (c.p.c.t)

∠BAD + ∠CAD = 40°/ 2 ∠BAD = 40°

∠BAD = 40°/2 =20°

∠ABC + ∠BCA + ∠BAC = 180° (Angle sum property)

Since ΔABC is an isosceles triangle,

∠ABC = ∠BCA ∠ABC +∠ABC + 40°= 180°

2 ∠ABC = 180° – 40° = 140° ∠ABC = 140°/2 = 70°

∠DBC + ∠ BCD + ∠ BDC = 180° (Angle sum property)

Since ΔABC is an isosceles triangle, ∠ DBC = ∠BCD ∠DBC + ∠DBC + 100^o = 180°

2 ∠DBC = 180° – 100^o = 80°

∠DBC = 80°/2 = 40°

In Δ BAD,

∠ABD + ∠BAD + ∠ADB = 180°(Angle sum property)

30° + 20° + ∠ADB = 180° (∠ADB = ∠ABC – ∠DBC), ∠ADB = 180°- 20° – 30°

∠ADB = 130°

∠ADB =130°

Question: 7

Δ ABC and ΔABD are on a common base AB, and AC = BD and BC = AD as shown in Figure. Which of the following statements is true?

(i) ΔABC ≅ ΔABD

(ii) ΔABC ≅ ΔADB

(iii) ΔABC ≅ ΔBAD

Solution:

In ΔABC and ΔBAD we have,

AC = BD (given)

BC = AD (given)

and AB = BA (common)

Therefore by SSS criterion of congruency, ΔABC ≅ ΔBAD

There option (iii) is true.

Question: 8

In Figure, ΔABC is isosceles with AB = AC, D is the mid-point of base BC.

(i) Is ΔADB ≅ ΔADC?

(ii) State the three pairs of matching parts you use to arrive at your answer.

Solution:

We have AB = AC.

Also since D is the midpoint of BC, BD = DC

Also, AD = DA

Therefore by SSS condition,

ΔADB ≅ ΔADC

We have used AB, AC : BD, DC AND AD, DA
 

Question: 9

In figure, ΔABC is isosceles with AB = AC. State if ΔABC ≅ ΔACB. If yes, state three relations that you use to arrive at your answer.

Solution:

Yes, ΔABC ≅ ΔACB by SSS condition.

Since, ABC is an isosceles triangle, AB = BC, BC = CB and AC = AB

Question: 10

Triangles ABC and DBC have side BC common, AB = BD and AC = CD. Are the two triangles congruent? State in symbolic form, which congruence do you use? Does ∠ABD equal ∠ACD? Why or why not?

Solution:

Yes,

Given,

Δ ABC and Δ DBC have side BC common, AB = BD and AC = CD

By SSS criterion of congruency, ΔABC ≅ ΔDBC

No, ∠ABD and ∠ACD are not equal because AB ≠ AC

Exercise 16.3

Question: 1

By applying SAS congruence condition, state which of the following pairs of triangle are congruent. State the result in symbolic form

Solution:

(i)

We have OA = OC and OB = OD and

∠AOB = ∠COD which are vertically opposite angles. Therefore by SAS condition, ΔAOC ≅ ΔBOD

(ii)

We have BD = DC

∠ADB = ∠ADC = 90° and

Therefore, by SAS condition, ΔADB ≅ ΔADC.

(iii)

We have AB = DC

∠ABD = ∠CDB and

Therefore, by SAS condition, ΔABD ≅ ΔCBD

(iv)

We have BC = QR

ABC = PQR = 90°

And AB = PQ

Therefore, by SAS condition, ΔABC≅ ΔPQR.

Question: 2

State the condition by which the following pairs of triangles are congruent.

Solution:

(i)

AB = AD

BC = CD and AC = CA

Therefore by SSS condition, ΔABC≅ ΔADC

(ii)

AC = BD

AD = BC and AB = BA

Therefore, by SSS condition, ΔABD ≅ ΔADC

(iii)

AB = AD

∠BAC = ∠DAC and

Therefore by SAS condition, ΔBAC ≅ ΔBAC

(iv)

AD = BC

∠DAC = ∠BCA and

Therefore, by SAS condition, ΔABC ≅ ΔADC

Question: 3

In figure, line segments AB and CD bisect each other at O. Which of the following statements is true?

(i) ΔAOC ≅ ΔDOB

(ii) ΔAOC ≅ ΔBOD

(iii) ΔAOC ≅ ΔODB

State the three pairs of matching parts, you have used to arrive at the answer.

Solution:

We have,

And, CO = OD

Also, AOC = BOD

Therefore, by SAS condition, ΔAOC ≅ ΔBOD

Question: 4

Line-segments AB and CD bisect each other at O. AC and BD are joined forming triangles AOC and BOD. State the three equality relations between the parts of the two triangles that are given or otherwise known. Are the two triangles congruent? State in symbolic form, which congruence condition do you use?

Solution:

We have AO = OB and CO = OD since AB and CD bisect each other at 0.

Also ∠AOC = ∠BOD since they are opposite angles on the same vertex.

Therefore by SAS congruence condition, ΔAOC ≅ ΔBOD

Question: 5

ΔABC is isosceles with AB = AC. Line segment AD bisects ∠A and meets the base BC in D.

(i) Is ΔADB ≅ ΔADC?

(ii) State the three pairs of matching parts used to answer (i).

(iii) Is it true to say that BD = DC?

Solution:

(i) We have AB = AC (Given)

∠BAD = ∠CAD (AD bisects ∠BAC)

Therefore by SAS condition of congruence, ΔABD ≅ ΔACD

(ii) We have used AB, AC; ∠BAD = ∠CAD; AD, DA.

(iii) Now, ΔABD≅ΔACD therefore by c.p.c.t BD = DC.

Question: 6

In Figure, AB = AD and ∠BAC = ∠DAC.

(i) State in symbolic form the congruence of two triangles ABC and ADC that is true.

(ii) Complete each of the following, so as to make it true:

(a) ∠ABC =

(b) ∠ACD =

(c) Line segment AC bisects ___ and ___

Solution:

i) AB = AD (given)

∠BAC = ∠DAC (given)

AC = CA (common)

Therefore by SAS condition of congruency, ΔABC ≅ ΔADC

ii) ∠ABC = ∠ADC (c.p.c.t)

∠ACD = ∠ACB (c.p.c.t)

Question: 7

In figure, AB || DC and AB = DC.

(i) Is ΔACD ≅ ΔCAB?

(ii) State the three pairs of matching parts used to answer (i).

(iii) Which angle is equal to ∠CAD ?

(iv) Does it follow from (iii) that AD || BC?

Solution:

(i) Yes by SAS condition of congruency, ΔDCA ≅ ΔBAC

(ii) We have used AB = DC, AC = CA and ∠DCA = ∠BAC.

(iii) ∠CAD = ∠ACB since the two triangles are congruent.

(iv) Yes this follows from AD // BC as alternate angles are equal. lf alternate angles are equal the lines are parallel

Exercise 16.4

Question: 1

Which of the following pairs of triangle are congruent by ASA condition?

Solution:

i)

We have,

Since ∠ABO = ∠CDO = 45° and both are alternate angles, AB // DC, ∠BAO = ∠DCO (alternate angle, AB // CD and AC is a transversal line)

∠ABO = ∠CDO = 45° (given in the figure) Also, AB = DC (Given in the figure)

Therefore, by ASA ΔAOB ≅ ΔDOC

ii)

In ABC,

Now AB =AC (Given)

∠ABD = ∠ACD = 40° (Angles opposite to equal sides)

∠ABD + ∠ACD + ∠BAC = 180° (Angle sum property)

40° + 40° + ∠BAC=180°

∠BAC =180°- 80° =100°

∠BAD + ∠DAC = ∠BAC ∠BAD = ∠BAC – ∠DAC = 100° – 50° = 50°

∠BAD = ∠CAD = 50°

Therefore, by ASA, ΔABD ≅ ΔADC

iii)

In Δ ABC,

∠A + ∠B + ∠C = 180°(Angle sum property)

∠C = 180°- ∠A – ∠B ∠C = 180° – 30° – 90° = 60°

In PQR,

∠P + ∠Q + ∠R = 180°(Angle sum property)

∠P = 180° – ∠Q – ∠R ∠P = 180°- 60°- 90° = 30°

∠BAC = ∠QPR = 30°

∠BCA = ∠PRQ = 60° and AC = PR (Given)

Therefore, by ASA, ΔABC ≅ ΔPQR

iv)

We have only

BC = QR but none of the angles of ΔABC and ΔPQR are equal.

Therefore, ΔABC and Cong ΔPRQ

Question: 2

In figure, AD bisects A and AD and AD ⊥ BC.

(i) Is ΔADB ≅ ΔADC?

(ii) State the three pairs of matching parts you have used in (i)

(iii) Is it true to say that BD = DC?

Solution:

(i) Yes, ΔADB≅ΔADC, by ASA criterion of congruency.

(ii) We have used ∠BAD = ∠CAD ∠ADB = ∠ADC = 90°

Since, AD ⊥ BC and AD = DA

(iii) Yes, BD = DC since, ΔADB ≅ ΔADC

Question: 3

Draw any triangle ABC. Use ASA condition to construct other triangle congruent to it.

Solution:

We have drawn

Δ ABC with ∠ABC = 65° and ∠ACB = 70°

We now construct ΔPQR ≅ ΔABC has ∠PQR = 65° and ∠PRQ = 70°

Also we construct ΔPQR such that BC = QR

Therefore by ASA the two triangles are congruent

Question: 4

In Δ ABC, it is known that ∠B = C. Imagine you have another copy of Δ ABC

(i) Is ΔABC ≅ ΔACB

(ii) State the three pairs of matching parts you have used to answer (i).

(iii) Is it true to say that AB = AC?

Solution:

(i) Yes ΔABC ≅ ΔACB

(ii) We have used ∠ABC = ∠ACB and ∠ACB = ∠ABC again.

Also BC = CB

(iii) Yes it is true to say that AB = AC since ∠ABC = ∠ACB.

Question: 5

In Figure, AX bisects ∠BAC as well as ∠BDC. State the three facts needed to ensure that ΔACD ≅ ΔABD

Solution:

As per the given conditions, ∠CAD = ∠BAD and ∠CDA = ∠BDA (because AX bisects ∠BAC)

AD = DA (common)

Therefore, by ASA, ΔACD ≅ ΔABD

Question: 6

In Figure, AO = OB and ∠A = ∠B.

(i) Is ΔAOC ≅ ΔBOD

(ii) State the matching pair you have used, which is not given in the question.

(iii) Is it true to say that ∠ACO = ∠BDO?

Solution:

We have

∠OAC = ∠OBD,

AO = OB

Also, ∠AOC = ∠BOD (Opposite angles on same vertex)

Therefore, by ASA ΔAOC ≅ ΔBOD

Exercise 16.5

Question: 1

In each of the following pairs of right triangles, the measures of some part are indicated alongside. State by the application of RHS congruence conditions which are congruent, and also state each result in symbolic form.

Solution:

i)

∠ADC = ∠BCA = 90°

AD = BC and hyp AB = hyp AB

Therefore, by RHS ΔADB ≅ ΔACB

ii)

AD = AD (Common)

hyp AC = hyp AB (Given)

∠ADB + ∠ADC = 180° (Linear pair)

∠ADB + 90° = 180°

∠ADB = 180° – 90° = 90°

∠ADB = ∠ADC = 90°

Therefore, by RHS Δ ADB = Δ ADC

iii)

hyp AO = hyp DOBO = CO ∠B = ∠C = 90°

Therefore, by RHS, ΔAOB≅ΔDOC

iv)

Hyp A = Hyp CABC = DC ∠ABC = ∠ADC = 90°

Therefore, by RHS, ΔABC ≅ ΔADC

v)

BD = DB Hyp AB = Hyp BC, as per the given figure,

∠BDA + ∠BDC = 180°

∠BDA + 90° = 180°

∠BDA= 180°- 90° = 90°

∠BDA = ∠BDC = 90°

Therefore, by RHS, ΔABD ≅ ΔCBD

Question: 2

Δ ABC is isosceles with AB = AC. AD is the altitude from A on BC.

i) Is ΔABD ≅ ΔACD?

(ii) State the pairs of matching parts you have used to answer (i).

(iii) Is it true to say that BD = DC?

Solution:

(i) Yes, ΔABD ≅ ΔACD by RHS congruence condition.

(ii) We have used Hyp AB = Hyp AC

AD = DA

∠ADB = ∠ADC = 90° (AD ⊥ BC at point D)

(iii)Yes, it is true to say that BD = DC (c.p.c.t) since we have already proved that the two triangles are congruent.

Question: 3

ΔABC is isosceles with AB = AC. Also. AD ⊥ BC meeting BC in D. Are the two triangles ABD and ACD congruent? State in symbolic form. Which congruence condition do you use? Which side of ADC equals BD? Which angle of Δ ADC equals ∠B?

Solution:

We have AB = AC            …… (i)

AD = DA (common)           ……(ii)

And, ∠ADC = ∠ADB (AD ⊥ BC at point D)                ……(iii)

Therefore, from (i), (ii) and (iii), by RHS congruence condition, ΔABD ≅ ΔACD, the triangles are congruent.

Therefore, BD = CD.

And ∠ABD = ∠ACD (c.p.c.t)

Question: 4

Draw a right triangle ABC. Use RHS condition to construct another triangle congruent to it.

Solution:

Consider

Δ ABC with ∠B as right angle.

We now construct another triangle on base BC, such that ∠C is a right angle and AB = DC

Also, BC = CB

Therefore, BC = CB

Therefore by RHS, ΔABC ≅ ΔDCB

Question: 5

In figure, BD and CE are altitudes of Δ ABC and BD = CE.

(i) Is ΔBCD ≅ ΔCBE?

(ii) State the three pairs or matching parts you have used to answer (i)

Solution:

(i) Yes, ΔBCD ≅ ΔCBE by RHS congruence condition.

(ii) We have used hyp BC = hyp CB

BD = CE (Given in question)

And ∠BDC = ∠CBE = 90^o