Yearly Archives: 2022

Chapter 17 – Heron’s Formula Exercise Ex. 17.1

Question 1

Find the area of the triangle whose sides are respectively 150 cm, 120 cm and 200 cm.Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5The perimeter of triangular field is 540 m and its sides are in the ratio 25:17:12. Find the area of the triangle.Solution 5The sides of the triangular field are in the ratio 25:17:12.
Let the sides of triangle be 25x, 17x, and 12x.
  Perimeter of this triangle = 540 m
             25x + 17x + 12x = 540 m
                         54x = 540 m
                            x = 10 m
  Sides of triangle will be 250 m, 170 m, and 120 m. Semi-perimeter (s) =  By Heron’s formula:

    So, area of the triangle is 9000 m².Question 6

The perimeter of right triangle is 300m. If its sides are in the ratio 3 : 5 : 7. Find the area of the triangle.Solution 6

Question 7

The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.Solution 7

Question 8

A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.Solution 8

Question 9

The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side.Solution 9

Question 10

The perimeter of an isosceles triangle is 42 cm and its base is (3/2) times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.Solution 10

Question 11

Find the area of the shaded region in fig.12.12

Solution 11

Chapter 17 – Heron’s Formula Exercise Ex. 17.2

Question 1Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.Solution 1

For ABC
AC² = AB² + BC²
(5)² = (3)² + (4)²
So, ABC is a right angle triangle, right angled at point B.
Area of ABCFor ADC
Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm
            s = 7 cm
By Heron’s formula
Area of triangle 

Area of ABCD = Area of ABC + Area of ACD
    = (6 + 9.166) cm² = 15.166 cm² = 15.2 cm² (approximately)

Question 2

Solution 2

Question 3

Solution 3

Question 4A park, in the shape of a quadrilateral ABCD, has  = 90^o, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?Solution 4Let us join BD.
In BCD applying Pythagoras theorem
BD² = BC² + CD²
       = (12)² + (5)²
       = 144 + 25
BD² = 169
  BD = 13 m

Area of BCD

                  For ABD

                    By Heron’s formula Area of triangle  

                               Area of park = Area of ABD + Area of BCD
                         = 35.496 + 30 m²                         = 65.496 m²                         = 65. 5 m² (approximately)

Question 5

Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.Solution 5

Question 6

A rhombus sheet, whose perimeter is 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of Rs 5 per m². Find the cost of painting.Solution 6

Question 7

Solution 7

Question 8

Find the area of a quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.Solution 8

Question 09

The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.Solution 09

Question 10

Find the area of the blades of the magnetic compass shown in fig.

(Take √11 = 3.32)

Solution 10

Question 11

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown in fig., The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each type of paper needed to make the hand fan.

Solution 14

Chapter 16 – Constructions Exercise Ex. 16.1

Question 1

Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.Solution 1

Question 2

Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this segment.Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Draw a line segment AB and bisect it. Bisect one of the equal parts of obtain a line segment of length  (AB)Solution 6

Question 7

Solution 7

Chapter 16 – Constructions Exercise Ex. 16.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Using ruler and compasses only, draw a right angle.Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10(i)

Solution 10(i)

Question 10(ii)

Solution 10(ii)

Question 11(i)

Solution 11(i)

Question 11(ii)

Solution 11(ii)

Question 11(iii)

Solution 11(iii)

Question 11(iv)

Solution 11(iv)

Question 11(v)

Construct an angle of 15^o.Solution 11(v)

Question 11(vi)

Solution 11(vi)

Chapter 16 – Constructions Exercise Ex. 16.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Construct a right angled triangle whose perimeter is equal to 10 cm and one actute angle equal to 60^o.Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Page No 113:

Exercise 7A

Question 1:

Write each of the following in figures:
(i) Fifty-eight point six three
(ii) One hundred twenty four point four two five
(iii) Seven point seven six
(iv) Nineteen point eight
(v) Four hundred four point zero four four
(vi) Point one seven three
(vii) Point zero one five

ANSWER:

(i) 58.63
(ii) 124.425
(iii) 7.76
(iv) 19.8
(v) 404.044 
(vi) 0.173
(vii) 0.015

Page No 113:

Question 2:

Write the place value of each digit in each of the following decimals:
(i) 14.83
(ii) 275.269
(iii) 46.075
(iv) 302.459
(v) 5370.34
(vi) 186.209

ANSWER:

(i) In 14.83, we have:
Place value of 1 = 1 tens = 10
Place value of 4 = 4 ones = 4
Place value of 8 = 8 tenths = 810810
Place value of 3 = 3 hundredths = 31003100

(ii) In 275.269, we have:
Place value of 2 = 2 hundreds = 200
Place value of 7 = 7 tens = 70
Place value of 5 = 5 ones = 5
Place value of 2 = 2 tenths = 210210
Place value of 6 = 6 hundredths = 61006100
Place value of 9 = 9 thousandths = 9100091000

(iii) In 46.075, we have:
Place value of 4 = 4 tens = 40
Place value of 6 = 6 ones = 6
Place value of 0 = 0 tenths = 010010= 0
Place value of 7 = 7 hundredths = 71007100
Place value of 5 = 5 thousandths = 5100051000

(iv) In 302.459, we have:
Place value of 3 = 3 hundreds = 300
Place value of 0 = 0 tens = 0
Place value of 2 = 2 ones = 2
Place value of 4 = 4 tenths = 410410
Place value of 5 = 5 hundredths = 51005100
Place value of 9 = 9 thousandths = 9100091000

(v) In 5370.34, we have:
Place value of 5 = 5 thousands = 5000
Place value of 3 = 3 hundreds = 300
Place value of 7 = 7 tens = 70
Place value of 0 = 0 ones = 0
Place value of 3 = 3 tenths = 310310
Place value of 4 = 4 hundredths = 41004100

(vi) In 186.209, we have:
Place value of 1 = 1 hundreds = 100
Place value of 8 = 8 tens = 80
Place value of 6 = 6 ones = 6
Place value of 2 = 2 tenths = 210210
Place value of 0 = 0 hundredths = 0
Place value of 9 = 9 thousandths = 9100091000

Page No 113:

Question 3:

Write each of the following decimals in expanded form:
(i) 67.83
(ii) 283.61
(iii) 24.675
(iv) 0.294
(v) 8.006
(vi) 4615.72

ANSWER:

(i) 67.83
 = 6 tens + 7 ones + 8 tenths + 3 hundredths
 = 60 + 7 + 810 + 310060 + 7 + 810 + 3100

(ii) 283.61
 = 2 hundreds  + 8 tens + 3 ones + 6 tenths + 1 hundredths
 = 200 + 80 + 3 + 610  + 1100200 + 80 + 3 + 610  + 1100

(iii) 24.675
 = 2 tens + 4 ones + 6 tenths + 7 hundredths + 5 thousandths
 = 20 + 4 + 610 + 7100 + 5100020 + 4 + 610 + 7100 + 51000

(iv) 0.294
 = 2 tenths + 9 hundredths + 4 thousandths
 = 210  + 9100 + 41000210  + 9100 + 41000

(v) 8.006
 = 8 ones + 0 tenths + 0 hundredths  + 6 thousandths
 = 8 + 010 + 0100 + 610008 + 010 + 0100 + 61000

(vi) 4615.72
 = 4 thousands + 6 hundreds + 1 tens + 5 ones + 7 tenths + 2 hundredths
 = 4000 + 600 + 10 + 5 + 710 + 21004000 + 600 + 10 + 5 + 710 + 2100

Page No 113:

Question 4:

Write each of the following in decimals form:
(i) 40 + 6 + 710 + 910040 + 6 + 710 + 9100
(ii) 500 + 70 + 8 + 310 + 1100 + 61000500 + 70 + 8 + 310 + 1100 + 61000
(iii) 700 + 30 + 1 + 810 + 4100700 + 30 + 1 + 810 + 4100
(iv) 600 + 5 + 7100 + 91000600 + 5 + 7100 + 91000
(v) 800 + 5 + 810 + 61000800 + 5 + 810 + 61000
(vi) 30 + 9 + 4100 + 8100030 + 9 + 4100 + 81000

ANSWER:

(i) 40 + 6 + 710 + 910040 + 6 + 710 + 9100 = 46 + 0.7 + .09 = 46.79

(ii) 500 + 70 + 8 + 310 + 1100 + 61000500 + 70 + 8 + 310 + 1100 + 61000 = 578 + 0.3 + 0.01 + 0.006 = 578.316

(iii) 700 + 30 + 1 + 810 + 4100700 + 30 + 1 + 810 + 4100 = 731 + 0.8 + 0.04 = 731.84

(iv) 600 + 5 + 7100 + 91000600 + 5 + 7100 + 91000 = 605 + 0.07 + 0.009 = 605.079

(v) 800 + 5 + 810 + 61000800 + 5 + 810 + 61000 = 805 + 0.8 + 0.006 = 805.806

(vi) 30 + 9 + 4100 + 8100030 + 9 + 4100 + 81000 = 39 + 0.04 + 0.008 = 39.048

Page No 114:

Question 5:

Convert each of the following into like decimals:
(i) 7.5, 64.23, 0.074
(ii) 0.6, 5.937, 2.36, 4.2
(iii) 1.6, 0.07, 3.58, 2.9
(iv) 2.5, 0.63, 14.08, 1.637

ANSWER:

(i) Each of the numbers has maximum 3 decimal places. So, we convert them into numbers having three decimal places by annexing suitable number of zeroes to the extreme right of the decimal part.
  7.5 = 7.500
64.23 = 64.230
0.074 = 0.074

(ii) Each of the numbers has maximum 3 decimal places. So, we convert them into numbers having three decimal places by annexing suitable number of zeroes to the extreme right of the decimal part.
  0.6 = 0.600
5.937 = 5.937
2.36 = 2.360
4.2 = 4.200

(iii) Each of the numbers has maximum 2 decimal places. So, we convert them into numbers having three decimal places by annexing suitable number of zeroes to the extreme right of the decimal part.
  1.6 = 1.60
0.07 = 0.07
3.58 = 3.58
2.9 = 2.90

(iv) Each of the numbers has maximum 3 decimal places. So, we convert them into numbers having three decimal places by annexing suitable number of zeroes to the extreme right of the decimal part.
  2.5 = 2.500
0.63 = 0.630
14.08 = 14.080
1.637 = 1.637

Page No 114:

Question 6:

Fill in each of the place holders with the correct symbol > or <:
(i) 84.23      76.3584.23      76.35
(ii) 7.608      7.687.608      7.68
(iii) 8.34      8.438.34      8.43
(iv) 12.06      12.00612.06      12.006
(v) 3.85      3.8053.85      3.805
(vi) 0.97      1.070.97      1.07

ANSWER:

(i) 84.23 > 76.35
Since 84 is greater than 76, 84.23 is greater than 76.35. (Comparing the whole number parts)

(ii) 7.608 < 7.680
Since 8 is greater than 0 at the hundredths place, 7.608 is smaller than 7.680.

(iii) 8.34 < 8.43
Since 4 is greater than 3 at the tenths place, 8.34 is smaller than 8.43.

(iv) 12.06 > 12.006
Since 6 is greater than 0 at the hundredths place, 12.06 is greater than 12.006.

(v) 3.850 > 3.805
Since 5 is greater than 0 at the hundredths place, 3.850 is greater than 3.805.

(vi) 0.97 < 1.07
Since 1 is greater than 0, 0.97 is smaller than 1.07. (Comparing the whole number parts)

Page No 114:

Question 7:

Arrange the following decimals in ascending order:
(i) 5.8, 7.2, 5.69, 7.14, 5.06
(ii) 0.6, 6.6, 6.06, 66.6, 0.06
(iii) 6.54, 6.45, 6.4, 6.5, 6.05
(iv) 3.3, 3.303, 3.033, 0.33, 3.003

ANSWER:

(i) 5.8, 7.2, 5.69, 7.14, 5.06
 Converting the given decimals into like decimals:
    5.80, 7.20, 5.69, 7.14, 5.06
  Clearly, 5.06 < 5.69 < 5.80 < 7.14 < 7.20
  Hence, the given decimals can be arranged in the ascending order as follows:
          5.06, 5.69, 5.80, 7.14 and 7.2

(ii) 0.6, 6.6, 6.06, 66.6, 0.06
  Converting the given decimals into like decimals:
   0.60, 6.60, 6.06, 66.60, 0.06
  Clearly, 0.06 < 0.60 < 6.06 < 6.60 < 66.60
  Hence, the given decimals can be arranged in the ascending order as follows:
           0.06, 0.60, 6.06, 6.60 and 66.60

(iii) 6.54, 6.45, 6.4, 6.5, 6.05
Converting the given decimals into like decimals:
  6.54, 6.45, 6.40, 6.50, 6.05
  Clearly, 6.05 < 6.40 < 6.45 < 6.50 < 6.54
  Hence, the given decimals can be arranged in the ascending order as follows:
         6.05, 6.40, 6.45, 6.50 and 6.54

(iv) 3.3, 3.303, 3.033, 0.33, 3.003
Converting the given decimals into like decimals:
  3.300, 3.303, 3.033, 0.330, 3.003
  Clearly, 0.330 < 3.003 < 3.033 < 3.300 < 3.303
  Hence, the given decimals can be arranged in the ascending order as follows:
         0.33, 3.003, 3.033, 3.300 and 3.303

Page No 114:

Question 8:

Arrange the following decimals in descending order:
(i) 7.3, 8.73, 73.03, 7.33, 8.073
(ii) 3.3, 3.03, 30.3, 30.03, 3.003
(iii) 2.7, 7.2, 2.27, 2.72, 2.02, 2.007
(iv) 8.88, 8.088, 88.8, 88.08, 8.008

ANSWER:

(i) 7.3, 8.73, 73.03, 7.33, 8.073
 Converting each decimal into like decimals:
  7.300, 8.730, 73.030, 7.330, 8.073
 Clearly, 73.030 > 8.730 > 8.073 > 7.330 > 7.300
 Hence, the given decimals can be arranged in the descending order as follows:
   73.03, 8.73, 8.073, 7.33 and 7.3

(ii) 3.3, 3.03, 30.3, 30.03, 3.003
 Converting each decimal into like decimals:
   3.300, 3.030, 30.300, 30.030, 3.003
 Clearly, 30.300 > 30.030 > 3.300 > 3.030 > 3.003
 Hence, the given decimals can be arranged in the descending order as follows:
   30.3, 30.03, 3.3, 3.03 and 3.003

(iii) 2.7, 7.2, 2.27, 2.72, 2.02, 2.007
 Converting each decimal into like decimals:
   2.700, 7.200, 2.270, 2.720, 2.020, 2.007
 Clearly, 7.200 > 2.720 > 2.700 > 2.270 > 2.020 > 2.007
 Hence, the given decimals can be arranged in the descending order as follows:
      7.2, 2.72, 2.7, 2.27, 2.02 and 2.007

(iv) 8.88, 8.088, 88.8, 88.08, 8.008
 Converting each decimal into like decimals:
        8.880, 8.088, 88.800, 88.080, 8.008
 Clearly, 88.800 > 88.080 > 8.880 > 8.088 > 8.008
 Hence, the given decimals can be arranged in the descending order as follows:
        88.8, 88.08, 8.88, 8.088 and 8.008

Page No 118:

Exercise 7B

Question 1:

Convert each of the following into a fraction in its simplest form:
.9

ANSWER:

We have:
 .9 = 910910

Page No 118:

Question 2:

Convert each of the following into a fraction in its simplest form:
0.6

ANSWER:

We have:
 0.6 = 610 = 35610 = 35

Page No 118:

Question 3:

Convert each of the following into a fraction in its simplest form:
.08

ANSWER:

We have:
 0.08 = 8100 = 450 = 2258100 = 450 = 225
    

Page No 118:

Question 4:

Convert each of the following into a fraction in its simplest form:
0.15

ANSWER:

We have:
 0.15 = 15100 = 32015100 = 320

Page No 118:

Question 5:

Convert each of the following into a fraction in its simplest form:
0.48

ANSWER:

We have:
 0.48 = 48100 = 122548100 = 1225

Page No 118:

Question 6:

Convert each of the following into a fraction in its simplest form:
.053

ANSWER:

We have:
 0.053 = 531000 531000 

Page No 118:

Question 7:

Convert each of the following into a fraction in its simplest form:
0.125

ANSWER:

We have:
 0.125 = 1251000 = 25200 =540= 181251000 = 25200 =540= 18

Page No 118:

Question 8:

Convert each of the following into a fraction in its simplest form:
.224

ANSWER:

We have:
0.224 = 2241000 = 56250 = 281252241000 = 56250 = 28125

Page No 118:

Question 9:

Convert each of the following as a mixed-fraction:
6.4

ANSWER:

We have:
 6.4 = 6410 = 325 = 6256410 = 325 = 625

Page No 118:

Question 10:

Convert each of the following as a mixed-fraction:
16.5

ANSWER:

We have:
 16.5 = 16510 = 332 =161216510 = 332 =1612

Page No 118:

Question 11:

Convert each of the following as a mixed-fraction:
8.36

ANSWER:

We have:
 8.36 = 836100 = 20925 = 8925836100 = 20925 = 8925

Page No 118:

Question 12:

Convert each of the following as a mixed-fraction:
4.275

ANSWER:

We have:
 4.275 = 42751000 = 17140 = 4114042751000 = 17140 = 41140

Page No 118:

Question 13:

Convert each of the following as a mixed-fraction:
25.06

ANSWER:

We have:
 25.06 = 2506100 = 125350 =253502506100 = 125350 =25350

Page No 118:

Question 14:

Convert each of the following as a mixed-fraction:
7.004

ANSWER:

We have:
7.004 = 70041000 = 1751250 = 7125070041000 = 1751250 = 71250

Page No 118:

Question 15:

Convert each of the following as a mixed-fraction:
2.052

ANSWER:

We have:
2.052 = 20521000 = 513250 = 21325020521000 = 513250 = 213250

Page No 118:

Question 16:

Convert each of the following as a mixed-fraction:
3.108

ANSWER:

We have:
3.108 = 31081000 = 777250 = 32725031081000 = 777250 = 327250

Page No 118:

Question 17:

Convert each of the following into a decimal:
23102310

ANSWER:

We have:
2310 = 2310 = 2 + 0.3 = 2.32310 = 2310 = 2 + 0.3 = 2.3

Page No 118:

Question 18:

Convert each of the following into a decimal:
167100167100

ANSWER:

We have:
167100 = 167100 = 1 + 0.67 = 1.67167100 = 167100 = 1 + 0.67 = 1.67

Page No 118:

Question 19:

Convert each of the following into a decimal:
15891001589100

ANSWER:

We have:
1589100 = 1589100 = 15+ 0.89 = 15.891589100 = 1589100 = 15+ 0.89 = 15.89

Page No 118:

Question 20:

Convert each of the following into a decimal:
5413100054131000

ANSWER:

We have:
54131000 = 54131000 = 5 + 0.413 = 5.41354131000 = 54131000 = 5 + 0.413 = 5.413

Page No 118:

Question 21:

Convert each of the following into a decimal:
214151000214151000

ANSWER:

We have:
214151000 = 214151000 = 21 + 0.415 = 21.415214151000 = 214151000 = 21 + 0.415 = 21.415

Page No 118:

Question 22:

Convert each of the following into a decimal:
254254

ANSWER:

We have:
254 = 614 = 6 + 0.25 = 6.25254 = 614 = 6 + 0.25 = 6.25

Page No 118:

Question 23:

Convert each of the following into a decimal:
335335

ANSWER:

We have:
335= 3 + 0.6 = 3.6335= 3 + 0.6 = 3.6

Page No 118:

Question 24:

Convert each of the following into a decimal:
14251425

ANSWER:

We have:
 1425 = 1 + 0.16 = 1.16 1425 = 1 + 0.16 = 1.16

Page No 118:

Question 25:

Convert each of the following into a decimal:
5175051750

ANSWER:

We have:
51750 = 5 + 0.34 = 5.3451750 = 5 + 0.34 = 5.34

Page No 118:

Question 26:

Convert each of the following into a decimal:
12381238

ANSWER:

We have:
1238 = 12 + 0.375 = 12.3751238 = 12 + 0.375 = 12.375

Page No 118:

Question 27:

Convert each of the following into a decimal:
2194021940

ANSWER:

We have:
21940 = 2 + 0.475 = 2.47521940 = 2 + 0.475 = 2.475

Page No 118:

Question 28:

Convert each of the following into a decimal:
19201920

ANSWER:

We have:
1920 = 0.951920 = 0.95

Page No 118:

Question 29:

Convert each of the following into a decimal:
37503750

ANSWER:

We have:
3750 =  0.743750 =  0.74

Page No 118:

Question 30:

Convert each of the following into a decimal:
107250107250

ANSWER:

We have:
107250 = 0.428107250 = 0.428

Page No 118:

Question 31:

Convert each of the following into a decimal:
340340

ANSWER:

We have:
340 = 0.075340 = 0.075

Page No 118:

Question 32:

Convert each of the following into a decimal:
7878

ANSWER:

We have:
78 = 0.87578 = 0.875

Page No 118:

Question 33:

Using decimals, express
(i) 8 kg 640 g in kilograms
(ii) 9 kg 37 g in kilograms
(iii) 6 kg 8 g in kilograms

ANSWER:

(i) 8 kg 640 g in kilograms:
     8 kg + 640 gm = 8 kg + 64010006401000 kg
     8 kg + 0.640 kg = 8.640 kg

(ii) 9 kg 37 g in kilograms:
     9 kg + 37 gm = 9 kg + 371000371000 kg
     9 kg + 0.037 kg = 9.037 kg

(iii) 6 kg 8 g in kilograms:
      6 kg + 8 gm = 6 kg + 8100081000 kg
      6 kg + 0.008 kg = 6.008 kg

Page No 118:

Question 34:

Using decimals, express
(i) 4 km 365 m in kilometres
(ii) 5 km 87 m in kilometres
(iii) 3 km 6 m in kilometres
(iv) 270 m in kilometres
(v) 35 m in kilometres
(vi) 6 m in kilometres

ANSWER:

(i) 4 km 365 m in kilometres:
    4 km + 365 m = 4 km + 36510003651000 km     [Since 1 km = 1000 m]
    4 km + 0.365 km = 4.365 km

(ii) 5 km 87 m in kilometres:
    5 km + 87 m = 5 km + 871000871000 km     [Since 1 km = 1000 m]
    5 km + 0.087 km = 5.087 km

(iii) 3 km 6 m in kilometres:
     3 km + 6 m = 3 km + 6100061000 km      [Since 1 km = 1000 m]
     3 km + 0.006 km = 3.006 km

(iv) 270 m in kilometres:
      27010002701000 km = 0.270 km                     [Since 1 km = 1000 m]

(v) 35 m in kilometres:
      351000351000 km = 0.035 km                      [Since 1 km = 1000 m]

(vi) 6 m in kilometres:
       6100061000 km = 0.006 km                    [Since 1 km = 1000 m]

Page No 118:

Question 35:

Using decimals, express
(i) 15 kg 850 g in kilograms
(ii) 8 kg 96 g in kilograms
(iii) 540 g in kilograms
(iv) 8 g in kilograms

ANSWER:

(i) 15 kg 850 g in kilograms:
    15 kg + 850 gm = 15 kg + 85010008501000 kg        [Since 1 kg = 1000 gm]
    15 kg + 0.850 kg = 15.850 kg

(ii) 8 kg 96 g in kilograms:
     8 kg + 96 gm = 8 kg + 961000 961000 kg         [Since 1 kg = 1000 gm]
     8 kg + 0.096 kg = 8.096 kg

(iii) 540 g in kilograms:
       540 gm = 54010005401000 kg = 0.540 kg        [Since 1 kg = 1000 gm]

(iv) 8 g in kilograms:
       8 gm = 8100081000 kg = 0.008 kg           [Since 1 kg = 1000 gm]

Page No 118:

Question 36:

Using decimals, express
(i) Rs 18 and 25 paise in rupees
(ii) Rs 9 and 8 paise in rupees
(iii) 32 paise in rupees
(iv) 5 paise in rupees

ANSWER:

(i) Rs 18 and 25 paise in rupees:
    Rs 18 + 25 paise = Rs 18 + Rs 2510025100         [Since Re 1 = 100 paise]
    Rs 18 + Rs 0.25 = Rs 18.25

(ii) Rs 9 and 8 paise in rupees:
    Rs 9 + 8 paise = Rs 9 + Rs 81008100         [Since Re 1 = 100 paise]
    Rs 9 + Rs 0.08 = Rs 9.08

(iii) 32 paise in rupees:
      32 paise = Rs 3210032100 = Rs 0.32          [Since Re 1 = 100 paise]

(iv) 5 paise in rupees:
      5 paise = Rs 51005100 = Rs 0.05          [Since Re 1 = 100 paise]

Page No 120:

Exercise 7C

Question 1:

Add the following decimals:
9.6, 14.8, 37 and 5.9

ANSWER:

9.6, 14.8, 37 and 5.9
Converting the decimals into like decimals:
9.6, 14.8, 37.0 and 5.9
Let us write the given numbers in the column form.
Now, adding:
     9.6
   14.8
   37.0
     5.9
   67.3
Hence, the sum of the given numbers is 67.3.

Page No 120:

Question 2:

Add the following decimals:
23.7, 106.94, 68.9 and 29.5

ANSWER:

23.7, 106.94, 68.9 and 29.5
Converting the decimals into like decimals:
23.70, 106.94, 68.90 and 29.50
Let us write the given numbers in the column form.
Now, adding:
   23.70
 106.94
  68.90
  29.50
 229.04
Hence, the sum of the given numbers is 229.04.

Page No 120:

Question 3:

Add the following decimals:
72.8, 7.68, 16.23 and 0.7

ANSWER:

72.8, 7.68, 16.23 and 0.7
Converting the decimals into like decimals:
72.80, 7.68, 16.23 and 0.70
Let us write the given numbers in the column form.
Now, adding:
   72.80
     7.68
   16.23
     0.70
   97.41
Hence, the sum of the given numbers is 97.41.

Page No 120:

Question 4:

Add the following decimals:
18.6, 84.75, 8.345 and 9.7

ANSWER:

18.6, 84.75, 8.345 and 9.7
Converting the decimals into like decimals:
18.600, 84.750, 8.345 and 9.700
Let us write the given numbers in the column form.
Now, adding:
   18.600
   84.750
     8.345
     9.700
  121.395
Hence, the sum of the given numbers is 121.395.

Page No 120:

Question 5:

Add the following decimals:
8.236, 16.064, 63.8 and 27.53

ANSWER:

8.236, 16.064, 63.8 and 27.53
Converting the decimals into like decimals:
8.236, 16.064, 63.800 and 27.530
Let us write the given numbers in the column form.
Now, adding:
     8.236
   16.064
   63.800
   27.530  
  115.630
Hence, the sum of the given numbers is 115.630.

Page No 120:

Question 6:

Add the following decimals:
28.9, 19.64, 123.697 and 0.354

ANSWER:

28.9, 19.64, 123.697 and 0.354
Converting the decimals into like decimals:
28.900, 19.640, 123.697 and 0.354
Let us write the given numbers in the column form.
Now, adding:
    28.900
    19.640
  123.697
     0.354 
  172.591
Hence, the sum of the given numbers is 172.591.

Page No 120:

Question 7:

Add the following decimals:
4.37, 9.638, 17.007 and 6.8

ANSWER:

4.37, 9.638, 17.007 and 6.8
Converting the decimals into like decimals:
4.370, 9.638, 17.007 and 6.800
Let us write the given numbers in the column form.
Now, adding:
    4.370
    9.683
  17.007
    6.800 
  37.815  
Hence, the sum of the given numbers is 37.815.

Page No 120:

Question 8:

Add the following decimals:
14.5, 0.038, 118.573 and 6.84

ANSWER:

14.5, 0.038, 118.573 and 6.84
Converting the decimals into like decimals:
14.500, 0.038, 118.573 and 6.840
Let us write the given numbers in the column form.
Now, adding:
   14.500
     0.038
  118.573
     6.840 
  139.951 
Hence, the sum of the given numbers is 139.951.

Page No 120:

Question 9:

During three days of a week, a rickshaw puller earns Rs 32.60, Rs 56.80 and Rs 72 respectively. What is his total earning during these days?

ANSWER:

Earning on the 1st day of the week =  Rs 32.60
Earning on the 2nd day of the week = Rs 56.80
Earning on the 3rd day of the week = Rs 72.00
Total earning =                                 Rs 161.40

Page No 120:

Question 10:

A man purchases an almirah for Rs 11025, gives Rs 172.50 as its cartage and spends Rs 64.80 on its repair. How much does the almirah cost him?

ANSWER:

Cost of the almirah =          Rs 11025.00
Money spent on cartage =       Rs 172.50
Money spent on repair =           Rs 64.800
Total cost of the almirah =  Rs 11262.3    

Page No 120:

Question 11:

Ramesh covers 36 km 235 m by taxi, 4 km 85 m by rickshaw and 1 km 80 m on foot What is the total distance covered by him?

ANSWER:

Distance covered by the taxi =        36 km 235 m
Distance covered by the rickshaw =  4 km 085 m
Distance covered on foot =               1 km 080 m
Total distance covered =                41 km  400 m

Page No 120:

Question 12:

A bag contains 45 kg 80 g of sugar and the mass of the empty bag is 950 g. What is the mass of the bag containing this much of sugar?

ANSWER:

Weight of sugar in the bag =   45 kg 080 g
Weight of the empty bag =       0 kg 950 g
Total weight of the bag =         46 kg 030 g

Page No 120:

Question 13:

Ramu bought 2 m 70 cm cloth for his shirt and 2 m 60 cm cloth for his pyjamas. Find the total length of cloth bought by him.

ANSWER:

Length of cloth for his shirt =         2 m 70 cm
Length of cloth for his pyjamas =    2 m 60 cm
Total length of cloth bought =         5 m 30 cm

Page No 120:

Question 14:

Radhika bought 2 m 5 cm cloth for her salwar and 3 m 35 cm cloth for her shirt. Find the total length of cloth bought by her.

ANSWER:

Length of cloth for her salwar =        2 m 05 cm
Length of cloth for her shirt =           3 m 35 cm
Total length of cloth bought =           5 m 40 cm

Page No 122:

Question 1:

Exercise 7D

Subtract:
27.86 from 53.74

ANSWER:

Let us write the numbers in the column form with the larger one at the top.
Now, subtracting:
   53.74
 – 27.86
   25.88
∴∴ 53.74 – 27.86 = 25.88

Page No 122:

Question 2:

Subtract:
64.98 from 103.87

ANSWER:

Let us write the numbers in the column form with the larger one at the top.
Now, subtracting:
 103.87
– 64.98
  38.89
∴∴ 103.87 – 64.98 = 38.89

Page No 122:

Question 3:

Subtract:
59.63 from 92.4

ANSWER:

Converting the given numbers into like decimals:
59.63 and 92.40
Let us write them in the column form with the larger number at the top.
Now, subtracting:
   92.40
 – 59.63
   32.77
∴∴ 53.74 – 27.86 = 32.77

Page No 122:

Question 4:

Subtract:
56.8 from 204

ANSWER:

Converting the given numbers into like decimals:
56.80 and 204.00
Let us write them in the column form with the larger number at the top.
Now, subtracting:
  204.00
 – 56.80
   147.2
∴∴ 204.00 – 56.80 = 147.2

Page No 122:

Question 5:

Subtract:
127.38 from 216.2

ANSWER:

Converting the given numbers into like decimals:
127.38 and 216.20
Let us write them in the column form with the larger number at the top.
Now, subtracting:
  216.20
– 127.38
    88.82
∴∴ 216.20 – 127.38 = 88.82

Page No 122:

Question 6:

Subtract:
39.875 from 70.68

ANSWER:

Converting the given numbers into like decimals:
39.875 and 70.680
Let us write them in the column form with the larger number at the top.
Now, subtracting:
  70.680
– 39.875
  30.805
∴∴ 70.680 – 39.875 = 30.805

Page No 122:

Question 7:

Subtract:
348.237 from 523.12

ANSWER:

Converting the given numbers into like decimals:
348.237 and 523.120
Let us write them in the column form with the larger number at the top.
Now, subtracting:
  523.120
– 348.237
  174.883
∴∴ 523.120 – 348.237 = 174.883

Page No 122:

Question 8:

Subtract:
458.573 from 600

ANSWER:

Converting the given numbers into like decimals:
458.573 and 600.000
Let us write them in the column form with the larger number at the top.
Now, subtracting:
  600.000
– 458.573
  141.427
∴∴ 600.000 – 458.573 =141.427

Page No 122:

Question 9:

Subtract:
149.456 from 206.321

ANSWER:

Let us write the numbers in the column form with the larger one at the top.
Now, subtracting:
  206.321
– 149.456
    56.865
∴∴ 206.321 – 149.456 = 56.865

Page No 122:

Question 10:

Subtract:
0.612 from 3.4

ANSWER:

Converting the given numbers into like decimals:
3.400 and 0.612
Let us write them in the column form with the larger number at the top.
Now, subtracting:
  3.400
– 0.612
  2.788
∴∴ 3.400 – 0.612 = 2.788

Page No 122:

Question 11:

Simplify:
37.6 + 72.85 − 58.678 − 6.09

ANSWER:

Converting the given decimals into like decimals, then adding and, finally, subtracting:
                         
        37.60                        58.678
     + 72.85                      +  6.090                
       110.45                        64.768 
                         
                                 
                                      110.450
                                   −-  64.768
                                        45.682    

(37.60 + 72.85) −- (58.678 + 6.090)
= 110.450 −- 64.768
= 45.682                     

Page No 122:

Question 12:

Simplify:
75.3 − 104.645 + 178.96 − 47.9

ANSWER:

Converting the given decimals into like decimals, then adding and, finally, subtracting:
                         
      178.96                        104.645
     + 75.30                      + 47.900              
      254.26                        152.545

(75.30 + 178.96) −- (104.645 + 47.900)
254.260 −- 152.545
101.715                           254.260
                                   −- 152.545
                                       101.715
                                 
                              

Page No 122:

Question 13:

Simplify:
213.4 − 56.84 − 11.87 − 16.087

ANSWER:

Converting the given decimals into like decimals, then adding and, finally, subtracting:
                         
       56.840
       11.870                     
     +16.087                                  
       84.797                 

(213.400) −- (56.840 + 11.870 + 16.087)
213.400 −- 84.797
128.603                          213.400
                                   −-  84.797
                                      128.603
                                 
                              

Page No 122:

Question 14:

Simplify:
76.3 − 7.666 − 6.77

ANSWER:

Converting the given decimals into like decimals, then adding and, finally, subtracting:
                         
         7.666                       
     +  6.770                            
       14.436                   

(76.300) −- (7.666 + 6.770)
= 76.300 −- 14.436
= 61.864                          

  76.300
-14.436
 61.864
                                 
                              

Page No 122:

Question 15:

What is to be added to 74.5 to get 91?

ANSWER:

In order to get the number that must be added to 74.5 to get 91, we must subtract 74.5 from 91.0.

         91.0
    −-  74.5
         16.5

Thus, 16.5 is the required number.

Page No 122:

Question 16:

What is to be subtracted from 7.3 to get 0.862?

ANSWER:

In order to get the number that must be subtracted from 7.300 to get .0862, we have to subtract 0.862 from 7.300.

         7.300
    −-  0.862
        6.438 

Thus, 6.438 is the required number.

Page No 122:

Question 17:

By how much should 23.754 be increased to get 50?

ANSWER:

In order to get the number by which 23.754 must be increased to get 50, we have to subtract 23.754 from 50.000.
 
      50.000
   −-23.754
      26.246

Page No 122:

Question 18:

By how much should 84.5 be decreased to get 27.84?

ANSWER:

In order to get the number by which 84.50 must be decreased to get 27.84, we have to subtract 27.84 from 84.50.
 
      84.50
   −-27.84
      56.66

Page No 122:

Question 19:

If the school bags of Neelam and Garima weigh 6 kg 80 g and 5 kg 265 g respectively, whose bag is heavier and by how much?

ANSWER:

Weight of Neelam’s school bag =       6080 g  {Converting into grams: 6 kg + 80 g = (6000 + 80) g =  6080 g}
Weight of Garima’s school bag =     −-5265 g  {Converting into grams: 5 kg + 265 g = (5000 + 265)g = 5265 g}
Difference of the weights of bags =      815 g
Thus, the weight of Neelam’s school bag is more than that of Garima’s school bag by 815 grams, i.e. by 0.815 kg.

Page No 122:

Question 20:

Kunal purchased a notebook for Rs 19.75, a pencil for Rs 3.85 and a pen for Rs 8.35 from a book shop. He gave a 50-rupee note to the shopkeeper. What amount did he get back?

ANSWER:

Cost of the notebook =  Rs 19.75
Cost of the pencil =        Rs 3.85
Cost of the pen =        + Rs 8.35
Total cost payable =      Rs 31.95

Total money paid =      Rs 50.00
Total money spent = −-Rs 31.95
Balance =                   Rs 18.05 

Thus, Kunal got back Rs 18.05 from the shopkeeper.

Page No 122:

Question 21:

Sunita purchased 5 kg 75 g of fruits and 3 kg 465 g of vegetables, and put them in a bag. If this bag with these contents weighs 9 kg, find the weight of the empty bag.

ANSWER:

Weight of the fruits =                                  5 kg 075 g
Weight of the vegetables =                       + 3 kg 465 g
Total weight of the contents of the bag =      8 kg 540 g

Total weight of the bag with its contents =    9 kg 000 g
Total weight of the contents of the bag =  −- 8 kg 540 g
Weight of the empty bag =                           0 kg 460 g
Thus, the weight of the empty bag is 460 grams.

Page No 123:

Question 22:

The distance between Reeta’s house and her office is 14 km. She covers 10 km 65 m by scooter, 3 km 75 m by bus and the rest on foot. How much distance does she cover by walking?

ANSWER:

Converting into metres:
10 km 65 m = (10 + 0.065) m = 10.065 m
3 km 75 m = (3 + 0.075) m = 3.075 m

Distance covered by the scooter =                             10.065 km
Distance covered by the bus =                                 + 3.075 km
Total distance covered by the bus and the scooter =   13.140 km

Total distance between the house and the office =       14.000 km
Total distance covered by the bus and the scooter = −-13.140 km
Distance covered on foot =                                          0.860 km

∴∴ Distance covered by walking = 0.860 km = 860 metres

Page No 123:

Exercise 7E

Question 1:

Mark (✓) against the correct answer in each of the following:
710=?710=?
(a) 7.1
(b) 1.7
(c) 0.7
(d) 0.07

ANSWER:

(c) 0.7

710710 = 7 tenths = 0.7

Page No 123:

Question 2:

Mark (✓) against the correct answer in each of the following:
5100=?5100=?
(a) 5.1
(b) 5.01
(c) 0.5
(d) 0.05

ANSWER:

(d) 0.05
 51005100 = 5 hundredths = 0.05

Page No 123:

Question 3:

Mark (✓) against the correct answer in each of the following:
91000=?91000=?
(a) 0.0009
(b) 0.009
(c) 9.001
(d) none of these

ANSWER:

(b) 0.009
9100091000 = 9 thousandths = 0.009

Page No 123:

Question 4:

Mark (✓) against the correct answer in each of the following:
161000=?161000=?
(a) 0.016
(b) 0.16
(c) 0.0016
(d) 1.006

ANSWER:

(a) 0.016
161000161000 = 16 thousandths = 0.016

Page No 123:

Question 5:

Mark (✓) against the correct answer in each of the following:
1341000=?1341000=?
(a) 13.4
(b) 1.34
(c) 0.134
(d) 0.0134

ANSWER:

(c) 0.134
 13410001341000 = 134 thousandths = 0.134

Page No 123:

Question 6:

Mark (✓) against the correct answer in each of the following:
217100=?217100=?
(a) 2.17
(b) 2.017
(c) 0.217
(d) 21.7

ANSWER:

(a) 2.17
 21710017100 = 2 + 1710017100 = 2 + 0.17 = 2.17

Page No 123:

Question 7:

Mark (✓) against the correct answer in each of the following:
43100=?43100=?
(a) 4.3
(b) 4.03
(c) 4.003
(d) 43.10

ANSWER:

(b) 4.03
 431003100 = 4 + 31003100 = 4 + 0.03 = 4.03

Page No 123:

Question 8:

Mark (✓) against the correct answer in each of the following:
6.25 = ?
(a) 612612
(b) 614614
(c) 62126212
(d) none of these

ANSWER:

b) 614614614
6.25 = 6 + 0.25 = 6 + 2510025100 = 6 + 1414 = 614614

Page No 123:

Question 9:

Mark (✓) against the correct answer in each of the following:
625=?625=?
(a) 2.4
(b) 0.24
(c) 0.024
(d) none of these

ANSWER:

(b) 0.24
     625625 =  0.24
                                  
            25) 60 (0.24
               −-50
                  100
              −- 100
                   0  

Page No 123:

Question 10:

Mark (✓) against the correct answer in each of the following:
478=?478=?
(a) 4.78
(b) 4.87
(c) 4.875
(d) none of these

ANSWER:

(c) 4.875
478478 =  4 + 7878 = 4 + 0.875 = 4.875

Page No 123:

Question 11:

Mark (✓) against the correct answer in each of the following:
24.8 = ?
(a) 24452445
(b) 24252425
(c) 24152415
(d) none of these

ANSWER:

(a)  24452445
24.8 = 24 + 0.8 = 24 + 810810 = 24 + 4545 = 24452445

Page No 124:

Question 12:

Mark (✓) against the correct answer in each of the following:
2125=?2125=?
(a) 2.4
(b) 2.04
(c) 2.004
(d) none of these

ANSWER:

(b) 2.04
21252125 = 2 + 125125 = 2 + 0.04 = 2.04

Page No 124:

Question 13:

Mark (✓) against the correct answer in each of the following:
2+310+41002+310+4100
(a) 2.304
(b) 2.403
(c) 2.34
(d) none of these

ANSWER:

(c) 2.34
2 + 310 + 4100310 + 4100 = 2 + 0.3 + 0.04 = 2.34

Page No 124:

Question 14:

Mark (✓) against the correct answer in each of the following:
2+6100=?2+6100=?
(a) 2.006
(b) 2.06
(c) 2.6
(d) none of these

ANSWER:

(b) 2.06
2 + 61006100 = 2 + 0.06 = 2.06

Page No 124:

Question 15:

Mark (✓) against the correct answer in each of the following:
4100+710000=?4100+710000=?
(a) 0.47
(b) 0.407
(c) 0.0407
(d) none of these

ANSWER:

(c) 0.0407
4100 + 7100004100 + 710000 = 0.04 + 0.0007 = 0.0407

Page No 124:

Question 16:

The correct expanded from of 2.06 is
(a) (2 × 10) + (6 × 110)2 × 10 + 6 × 110
(b) (2 × 1) + (6 ×110)2 × 1 + 6 ×110
(c) (2 × 1) + (6 × 1100)2 × 1 + 6 × 1100
(d) none of these

ANSWER:

(c) (2 × 1) + (6 × 1100)(2 × 1) + (6 × 1100)

2.06 =  2 + 61006100 = (2 × 1) + (6 × 1100)(2 × 1) + (6 × 1100)2 × 1) + (6 × 1(2 × 1) + (6 (2 × 1) + (6 

Page No 124:

Question 17:

Amoung 2.6, 2.006, 2.66 and 2.08, the largest number is
(a) 2.006
(b) 2.08
(c) 2.6
(d) 2.66

ANSWER:

(d) 2.66
Converting the given decimals into like decimals:
2.600, 2.006, 2.660 and 2.080
Among the given decimals, 2.660 is the largest.

Page No 124:

Question 18:

Which of the following is the correct order?
(a) 2.2 < 2.02 < 2.002 < 2.222
(b) 2.002 < 2.02 < 2.2 < 2.222
(c) 2.02 < 2.22 < 2.002 < 2.222
(d) none of these

ANSWER:

(b) 2.002 < 2.02 < 2.2 < 2.222

Converting the given decimals into like decimals:
2.002, 2.020, 2.200, 2.222

∴∴ 2.002 < 2.02 < 2.2 < 2.222

Page No 124:

Question 19:

Which is larger: 2.1 or 2.055?
(a) 2.1
(b) 2.055
(c) cannot be compared

ANSWER:

(a) 2.1
If we convert the given decimals into like decimals, we get 2.100 and 2.055.
At tenths place, 1 is greater than 0. Thus, 2.100 is greater than 2.055.

Page No 124:

Question 20:

Mark (✓) against the correct answer in each of the following:
1 cm = ?
(a) 0.1 m
(b) 0.01 m
(c) 0.001 m
(d) none of these

ANSWER:

(b) 0.01 m
1 m = 100 cm
∴∴ 1 cm = 11001100m = 0.01 m   

Page No 124:

Question 21:

Mark (✓) against the correct answer in each of the following:
2 m 5 cm = ?
(a) 2.5 m
(b) 2.05 m
(c) 2.005 m
(d) 0.25 m

ANSWER:

(b) 2.05 m
2 m 5 cm = (2 + 51005100) m = (2 + 0.05) m = 2.05 m

Page No 124:

Question 22:

Mark (✓) against the correct answer in each of the following:
2 kg 8 g = ?
(a) 2.8 kg
(b) 2.08 kg
(c) 2.008 kg
(d) none of these

ANSWER:

(c) 2.008 kg
1 kg = 1000 g
∴∴ 2 kg 8 g = 2 kg + 8100081000 kg = (2 + 0.008) kg = 2.008 kg

Page No 124:

Question 23:

Mark (✓) against the correct answer in each of the following:
2 kg 56 g = ?
(a) 2.56 kg
(b) 2.056 kg
(c) 2.560 kg
(d) none of these

ANSWER:

(b) 2.056 kg
2 kg + 56 g = (2 + 561000561000) kg = (2 + 0.056) kg = 2.056 kg

Page No 124:

Question 24:

Mark (✓) against the correct answer in each of the following:
2 km 35 m = ?
(a) 2.35 km
(b) 2.350 km
(c) 2.035 km
(d) none of these

ANSWER:

(c) 2.035 km
1 km = 1000 m
∴∴ 2 km 35 m  = (2 + 351000351000) km = (2 + 0.035) km = 2.035 km

Page No 124:

Question 25:

Mark (✓) against the correct answer in each of the following:
0.4 + 0.004 + 4.4 = ?
(a) 4.444
(b) 5.2
(c) 4.804
(d) 5.404

ANSWER:

(c) 4.804
0.4 + 0.004 + 4.4
Converting into like decimals and then adding:
   4.400
   0.004
+ 0.400
   4.804

Page No 124:

Question 26:

Mark (✓) against the correct answer in each of the following:
3.5 + 4.05 − 6.005 = ?
(a) 1.545
(b) 1.095
(c) 1.6
(d) none of these

ANSWER:

(a) 1.545
Converting into like decimals:
3.500 + 4.050 − 6.005
   3.500
+ 4.050
   7.550

   7.550
− 6.005
   1.545

Page No 124:

Question 27:

Mark (✓) against the correct answer in each of the following:
6.3 − 2.8 = ?
(a) 0.35
(b) 3.5
(c) 3.035
(d) none of these

ANSWER:

(b) 3.5

   6.3
− 2.8
   3.5

Page No 124:

Question 28:

Mark (✓) against the correct answer in each of the following:
5.01 − 3.6 = ?
(a) 4.65
(b) 1.95
(c) 1.41
(d) none of these

ANSWER:

(c) 1.41
Converting into like decimals and then subtracting:
   5.01
− 3.60
   1.41

Page No 125:

Question 29:

Mark (✓) against the correct answer in each of the following:
2 − 0.7 = ?
(a) 1.3
(b) 1.5
(c) 2.03
(d) none of these

ANSWER:

(a) 1.3
Converting into like decimals and then subtracting:
   2.0
− 0.7
   1.3

Page No 125:

Question 30:

Mark (✓) against the correct answer in each of the following:
1.1 − 0.3 = ?
(a) 0.8
(b) 0.08
(c) 8
(d) none of these

ANSWER:

(a) 0.8
Converting into like decimals and then subtracting:
   1.1
− 0.3
   0.8

Page No 126:

Exercise 7F

Question 1:

Convert 458458 into a decimal fraction.

ANSWER:

458458 = 4 + 5858 =  4 + 0.625 = 4.625

Page No 126:

Question 2:

Express 105 cm into metres using decimals.

ANSWER:

1 cm = 11001100 m
∴∴ 105 cm = 105100105100 m = (1 + 51005100) m = (1 + 0.05) m = 1.05 m

Page No 126:

Question 3:

Express 6 km 5 m as km using decimals.

ANSWER:

1 m = 1100011000 km
  ∴∴ 5 m = 5100051000 km = 0.005 km
6 km 5 m = (6 + 0.005) km = 6.005 km

Page No 126:

Question 4:

Express 8 m as kilometre using decimals.

ANSWER:

1 m = 1100011000 km
∴∴8 m = 8100081000 km = 0.008 km

Page No 126:

Question 5:

Add 26.4, 163.05, 8.75 and 5.6.

ANSWER:

Converting into like decimals and then adding:
    26.40
  163.05
     8.75
+   5.60
  203.80

Page No 126:

Question 6:

Subtract 0.528 from 3.2.

ANSWER:

Converting into like decimals and then subtracting:
    3.200
−- 0.528
    2.672

Page No 126:

Question 7:

What is to be added to 63.5 to get 71?

ANSWER:

We subtract 63.5 from 71 in order to find the number that is to be added to 63.5 to make it 71.
    Converting into like decimals and then subtracting:  
     71.0
 −- 63.5
      7.5

Page No 126:

Question 8:

What is to be subtracted from 13 to get 5.4?

ANSWER:

We subtract 5.4 from 13 in order to find the number that is to be subtracted from 13 to get 5.4.
Converting into like decimals and then subtracting:
      13.0
   −-  5.4
       7.6 
Thus, we need to subtract 7.6 from 13 to get 5.4.

Page No 126:

Question 9:

Arrange the following decimals in descending order:
6.5, 6.05, 6.54, 6.4 and 6.45

ANSWER:

6.5, 6.05, 6.54, 6.4 and 6.45
Converting into like decimals:
6.50, 6.05, 6.54, 6.40 and 6.45
Clearly, 6.54 > 6.50 > 6.45 > 6.40 > 6.05
Hence, the given decimals can be arranged in the descending order in the following manner:
 6.54, 6.50, 6.45, 6.40 and 6.05

Page No 126:

Question 10:

Convert each of the following into a fraction in simplest form:
(i) .4
(ii) .35
(c) 0.08
(iv) 0.075

ANSWER:

(i) 0.4 = 410 = 25410 = 25

(ii) 0.35 = 35100 = 72035100 = 720

(iii) 0.08 = 8100 = 2258100 = 225

(iv) 0.075 = 751000 = 340751000 = 340

Page No 126:

Question 11:

Mark (✓) against the correct answer in each of the following:
325=?325=?
(a) 1.2
(b) 0.12
(c) 0.012
(d) none of these

ANSWER:

(b) 0.12
 325325 = 0.12
                  
      25)  30 (0.12
         −- 25
               50
           −- 50  
                 0  

Page No 126:

Question 12:

Mark (✓) against the correct answer in each of the following:
61000=?61000=?
(a) 6.001
(b) 0.0006
(c) 0.006
(d) 0.06

ANSWER:

(c) 0.006
 6100061000 = 6 thousandths = 0.006

Page No 126:

Question 13:

Mark (✓) against the correct answer in each of the following:
23100=?23100=?
(a) 2.003
(b) 2.03
(c) 2.3
(d) none of these

ANSWER:

(b) 2.03
2310023100 = 2 + 31003100 = 2 + 0.03 = 2.03

Page No 126:

Question 14:

Mark (✓) against the correct answer in each of the following:
The place value of 3 in 16.534 is
(a) 310310
(b) 31003100
(c) 3100031000
(d) 3

ANSWER:

(b) 31003100
 In 16.534, place value of 3 = 3 hundredths = 0.03 = 31003100

Page No 126:

Question 15:

Mark (✓) against the correct answer in each of the following:
478=?478=?
(a) 4.78
(b) 4.87
(c) 4.875
(d) none of these

ANSWER:

(c) 4.875
478478 = 4 + 7878 = 4 + 0.875 = 4.875

Page No 126:

Question 16:

Mark (✓) against the correct answer in each of the following:
5.01 − 3.6 = ?
(a) 4.65
(b) 1.95
(c) 1.41
(d) none of these

ANSWER:

(c) 1.41
Converting into like decimals:
     5.01
 −  3.60
     1.41

Page No 126:

Question 17:

Mark (✓) against the correct answer in each of the following:
3.5 + 4.05 − 6.005 = ?
(a) 1.545
(b) 1.095
(c) 1.6
(d) none of these

ANSWER:

(a) 1.545

(3.50 + 4.05) −- 6.005
 7.550 −- 6.005
Converting into like decimals and then adding:
      3.50
   + 4.05
      7.55
Converting into like decimals and then subtracting:
     7.550
−-  6.005
     1.545

Page No 126:

Question 18:

Mark (✓) against the correct answer in each of the following:
4100+710000=?4100+710000=?
(a) 0.47
(b) 0.407
(c) 0.0407
(d) none of these

ANSWER:

(c) 0.0407

  4100 + 7100004100 + 710000 = 0.04  + 0.0007
Converting into like decimals:
0.0400 + 0.0007 = 0.0407

Page No 126:

Question 19:

Mark (✓) against the correct answer in each of the following:
Among 2.6, 2.006, 2.66 and 2.08, the largest number is
(a) 2.006
(b) 2.08
(c) 2.6
(d) 2.66

ANSWER:

(d) 2.66
Converting all the given decimals into like decimals:
2.600, 2.006, 2.660 and 2.080
Among these given like decimals, 2.660 is the largest.

Page No 126:

Question 20:

Fill in the blanks.
(i) 1 m = …… km
(ii) 10 ml = …… 1
(iii) 16 kg 5 g = …… kg
(iv) 2 m 8 cm = …… m
(v) 3.02, 4.75, 1.63 are examples of …… decimals.

ANSWER:

(i) 1 m = 1100011000 km = 0.001 km

(ii) 10 mL = 11001100 1 = 0.01 L

(iii) 16 kg 5 g = (16 + 0.005) kg = 16.005 kg
(iv) 2 m 8 cm = (2 + 0.08) m = 2.08 m
(v) 3.02, 4.75, 1.63 are examples of like decimals.

Page No 127:

Question 21:

Write ‘T’ for true and ‘F’ for false for each of the statements given below:
(i) 3.02 < 3.2.
(ii) 3 g = 0.003 kg.
(iii) 3411000=3.410.3411000=3.410.
(iv) 6.2 and 6.200 are equivalent decimals.
(v) 2.3, 3.41, 4.53, 5.61 are examples of like decimals.

ANSWER:

(i) True
This is because after converting them into like decimals, we get 3.20 > 3.02, which is true.

(ii) True
1 g = 1100011000 kg
∴∴ 3 g = 3100031000 = 0.003 kg

(iii) False
34110003411000 = 0.341
This is because it is equal to 341 thousandths.

(iv) True

(v) False
In this case, every decimal should have 2 decimals in order to be like decimals. But, 2.3 has only 1 decimal.

Chapter 15 – Circles Exercise Ex. 15.1

Question 1

Fill in the blanks:

(i) All points lying inside/outside a circle are called …… points/ … points.

(ii) Circles having the same centre and different radii are called … circles.

(iii) A point whose distance from the centre of a circle is greater than its radius lies in … of the circle.

(iv) A continuous piece of a circle is … of the circle.

(v) The longest chord of a circle is a … of the circle.

(vi) An arc is a … when its ends are the ends of a diameter.

(vii) Segment of a circle is the region between an arc and … of the circle.

(viii) A circle divides the plane, on which it lies, in …. parts.Solution 1

(i) interior/exterior

(ii) concentric

(iii) the exterior

(iv) arc

(v) diameter

(vi) semi-circle

(vii) centre

(viii) threeQuestion 2

Write the truth value (T/F) of the following with suitable reasons:

(i) A circle is a plane figure.

(ii) Line segment joining the centre to any point on the circle is a radius of the circle.

(iii) If a circle is divided into three equal arcs each is a major arc.

(iv) A circle has only finite number of equal chords.

(v) A chord of a circle, which is twice as long is its radius is a diameter of the circle.

(vi) Sector is the region between the chord and its corresponding arc.

(vii) The degree measure of an arc is the complement of the central angle containing the arc.

(viii) The degree measure of a semi-circle is 180^o.Solution 2

(i) T

(ii) T

(iii) T

(iv) F

(v) T

(vi) T

(vii) F

(viii) T

Chapter 15 – Circles Exercise Ex. 15.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Give a method to find the centre of a given circle.Solution 4

Steps of construction:

(1) Take three point A, B and C on the given circle.

(2) Join AB and BC.

(3) Draw the perpendicular bisectors of chord AB and BC which interesect each other at O.

(4) Point O will be the required circle because we know that the perpendicular bisector of a chord always passes through the centre.Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord form the centre?Solution 11

                                              Distance of smaller chord AB from centre of circle = 4 cm.
OM = 4 cm

In OMB

In ONDOD=OB=5cm             (radii of same circle)

 So, distance of bigger chord from centre is 3 cm.Question 12

Solution 12

Question 13

Solution 13

Question 14

Prove that two different circles cannot intersect each other at more than two points.Solution 14

Suppose two different circles can intersect each other at three points then they will pass through the three common points but we know that there is one and only one circle with passes through three non-collinear points, which contradicts our supposition.

Hence, two different circles cannot intersect each other at more than two points.Question 15Two chords AB and CD of lengths 5 cm and 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.Solution 15Draw OM  AB and ON  CD. Join OB and OD 

                     (Perpendicular from centre bisects the chord)

Let ON be x, so OM will be 6 – x
In MOB

In NOD

 We have OB = OD             (radii of same circle)
So, from equation (1) and (2) 

From equation (2) 

So, radius of circle is found to be  cm.

Chapter 15 – Circles Exercise Ex. 15.3

Question 1

Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha, Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha?Solution 1

Question 2

A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.Solution 2

Chapter 15 – Circles Exercise Ex. 15.4

Question 1

In fig., O is the centre of the circle. If ∠APB = 50°, find ∠AOB and ∠OAB.

Solution 1

Question 2

In fig., O is the centre of the circle. Find ∠BAC.

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 3(vii)

If O is the centre of the circle. Find the value of x in the following figure:

Solution 3(vii)

Question 3(viii)

Solution 3(viii)

Question 3(ix)

If O is the centre of the circle. Find the value of x in the following figure:

Solution 3(ix)

Question 3(x)

If O is the centre of the circle. Find the value of x in the following figure:

Solution 3(x)

Question 3(xi)

If O is the centre of the circle. Find the value of x in the following figure:

Solution 3(xi)

Question 3(xii)

Solution 3(xii)

Question 4

Solution 4

Question 5

In fig., O is the centre of the circle, BO is  the bisector of ∠ABC. Show that AB = AC.

Solution 5

Question 6

In fig., O and O’ are centres of two circles intersecting at B and C. ABD is straight line, find x.

Solution 6

Question 7

In fig., if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.

Solution 7

Question 8

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.Solution 8

Question 9

In fig., it is given given that O is the centre of the circle and ∠AOC = 150°. Find ∠ABC.

Solution 9

Question 10

In fig., O is the centre of the circle, prove that ∠x = ∠y + ∠z.

Solution 10

Question 11

in fig., O is the centre of a circle and PQ is a diameter. If ∠ROS = 40°, find. ∠RTS.

Solution 11

Chapter 15 – Circles Exercise Ex. 15.5

Question 1

In fig., ΔABC is an equilateral triangle. Find m∠BEC.

Solution 1

Question 2

In fig., ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°. find m∠QSR and m∠QTR.

Solution 2

Question 3

In fig., O is the centre of the circle. If ∠BOD = 160°, find the values of x and y.

Solution 3

Question 4

In fig., ABCD is a cyclic qudrilateral. If ∠BCD = 100° and ABD = 70°, find ∠ADB.

Solution 4

Question 5

If ABCD is a cyclic quadrilateral in which AD ∥ BC. Prove that ∠B = ∠C.

Solution 5

Question 6

In fig., O is the centre of the circle. find ∠CBD.

Solution 6

Question 7

In fig., AB and CD are diameters of a circle with centre O. If ∠OBD = 50°, find ∠AOC.

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

In fig., O is the centre of the circle and DAB = 50. calculate the values of x and y.

Solution 11

Question 12

In fig., if ∠BAC = 60°, and ∠BCA = 20°, find ∠ADC.

Solution 12

Question 13

In fig., if ABC is an equilateral triangle. Find ∠BDC and ∠BEC.

Solution 13

Question 14

In fig., O is the centre of the circle. If ∠CEA = 30°, find the values of x, y and z.

Solution 14

Question 15

In fig., ∠BAD = 78°, ∠DCF = x° and DEF = y° find the values of x and y.

Solution 15

Question 16

Solution 16

Question 17

In fig., ABCD is cyclic qudrilateral. Find the value of x.

Solution 17

Question 18(i)

Solution 18(i)

Question 18(ii)

Solution 18(ii)

Question 18(iii)

Solution 18(iii)

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

In fig., ABCD is cyclic quadrilaterial in which AC an BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.

Solution 23

Question 24

Solution 24

Question 25

Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.Solution 25

Let O be the centre of the circle circumscribing the cyclic rectangle ABCD. Since ABC = 90^o and AC is a chord of the circle, so, AC is a diameter of the circle. Similarly, BD is a diameter.

Hence, point of intersection of AC and BD is the centre of the circle.Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half of the hypotenuse.Solution 29

Chapter 14 – Areas of Parallelograms and Triangles Exercise Ex. 14.1

Question 1

Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:

Solution 1

(i) ΔAPB and trapezium ABCD are on the same base AB and between the same parallels AB and CD.

(ii) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.

(iii) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC but they are not on the same base.

(iv) ΔQRT and parallelogram PQRS are on the same base QR and between the same parallels QR and PS.

(v) Parallelogram PQRS and trapezium SMNR are on the same base SR but they are not between the same parallels.

(vi) Parallelograms PQRS, AQRD, BQRC are between the same parallels. Also, parallelograms PQRS, BPSC and APSD are between the same parallels.

Chapter 14 – Areas of Parallelograms and Triangles Exercise Ex. 14.2

Question 1In the given figure, ABCD is parallelogram, AE  DC and CF  AD. If AB = 16 cm. AE = 8 cm and CF = 10 cm, find AD.  

Solution 1In parallelogram ABCD, CD = AB = 16 cm     [Opposite sides of a parallelogram are equal]
We know that,
Area of parallelogram = Base x corresponding attitude
Area of parallelogram ABCD = CD x AE = AD x CF  
16 cm x 8 cm = AD x 10 cmAD =  cm = 12.8 cm.Thus, the length of AD is 12.8 cm.Question 2

In Q. No. 1, if AD = 6 cm, CF = 10 cm, and AE = 8, find AB.Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 14 – Areas of Parallelograms and Triangles Exercise Ex. 14.3

Question 1

In fig., compute the area of quadrilateral ABCD.

Solution 1

Question 2

In the fig., PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of ΔOTS if PQ = 8 cm.

Solution 2

Question 3

Compute the area of trapezium PQRS in fig.

Solution 3

Question 4

In fig., ∠AOB = 90, AC = BC, OA = 12 cm and OC = 6.5 cm. find the area of ΔAOB

Solution 4

Question 5

In fig., ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

Solution 5

Question 6

Solution 6

Question 7

In fig., ABCD is a trapezium in which AB ∥ DC. PRove that ar (ΔAOD) = ar (ΔBOC)

Solution 7

Question 8

Solution 8

Question 9

In fig., ABC and ABD are two triangles on the base Ab. If the line segment CD is bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD).

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

In fig., D and E are two points on BC such that BD = DE = EC. Show that ar (ΔABD) = ar (ΔADE) = ar(ΔAEC).

Solution 15

Draw a line l through A parallel to BC.

Given that, BD = DE = EC.

We observe that the triangles ABD, ADE and AEC are on the equal bases and between the same parallels l and BC. Therefore, their areas are equal.

Hence, ar (ΔABD) = ar (ΔADE) = ar(ΔAEC).Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

(i) 
(ii) 

(iii)

Question 20

Solution 20

Question 21

In fig., CD ∥ AE and CY ∥ BA.

(i) Name a triangle equal in area of ΔCBX 

(ii) Prove that ar (ΔZDE) = ar (ΔCZA)

(iii) Prove that ar (BCZY) = ar (ΔEDZ)

Solution 21

Question 22

In fig., PSDA is a parallelogram in which PQ = QR = RS and AP ∥ BQ ∥CR. Prove that ar(Δ PQE) = ar (Δ CFD).

Solution 22

Question 23

In fig., ABCD is a trapezium in which AB DC and DC = 40 cm and AB = 60 cm. If X and Y are, respectively, the mid – points of AD and BC, prove that:

(i) XY = 50 cm

(ii) DCYX is a trapezium

(iii) ar (trap. DCYX) = (9/11)ar (trap.(XYBA)

Solution 23

Question 24

Solution 24

Question 25

In fig., X and Y are the mid-points of AC and AB respectively, QP ∥ BC and CYQ and BXP are straight lines. Prove that ar(Δ ABP) = ar (Δ ACQ)

Solution 25

Question 26

In fig., ABCD and AEFD are two parallelograms. Prove that

(i) PE = FQ

(ii) ar(ΔAPE) : ar(ΔPFA) = ar Δ(QFD) : ar (ΔPFD)

(iii) ar(ΔPEA) = ar (ΔQFD)

Solution 26

Question 27

In fig. ABCD is a ∥^gm. O is any point on AC. PQ ∥ AB and LM ∥ AD. Prove that ar(∥^gm DLOP) = ar (∥^gm BMOQ).

Solution 27

Question 28

Solution 28

Question 29

In fig., ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F. Prove that

(i) ar (BDE) = ar (ABC)

(ii) ar(BDE) = ar(BAE)

(iii) ar (BFE) = ar(AFD)

(iv) ar(ABC) = 2 ar(BEC)

(v) ar (FED) = ar(AFC)

(vi) ar(BFE) = 2 ar (EFD)

Solution 29

Given, ABC and BDE are two equilateral triangles.

Let AB = BC = CA = x. Then, BD = = DE = BE

(i) We have,

ar(ABC) = x²

ar (BDE) = 

ar(BDE) = ar (ABC)

(ii) It is given that triangles ABC and BED are equilateral triangles.

ACB = DBE = 60^o

BE||AC(Since, alternate angles are equal)

Triangles BAE and BEC are on the same base BE and between the same parallels BE and AC.

ar (BAE) = ar(BEC)

ar (BAE) =2 ar (BDE)

[ ED is a median of EBC ar(BEC) = 2ar(BDE)]

ar (BDE) = ar(BAE)

(iii) Since ABC and BDE are equilateral triangles.

ABC = 60^o and BDE = 60^o

ABC = BDE

AB||DE(Since, alternate angles are equal)

Triangles BED and AED are on the same base ED and between the same parallels AB and DE.

ar (BED) = ar(AED)

ar (BED) ar(EFD) = ar(AED) ar(EFD)

ar(BEF) = ar(AFD)

(iv) Since ED is a median of BEC

ar (BEC) = 2 ar (BDE)

ar (BEC) = ar (ABC)[From (i), ar (BDE) = ar (ABC)]

ar(BEC) = ar (ABC)

ar (ABC) = 2 ar (BEC)

(v) Let h be the height of vertex E, corresponding to the side BD in triangle BDE.

Let H be the height of vertex A, corresponding to the side BC in triangle ABC.

From part (i),

ar(BDE) = ar (ABC)

From part (iii),

ar (BFE) = ar (AFD)

(vi) ar (AFC) = ar (AFD) + ar (ADC)

= ar (BFE) + ar (ABC)

(Using part (iii); and AD is the median of ABC)

= ar (BFE) + 4 ar (BDE)(Using part (i))

= ar (BFE) + 2 ar (BDE) (2)

Now, from part (v),

ar (BFE) = 2ar (FED) (3)

ar (BDE) = ar (BFE) + ar (FED)

= 2 ar (FED) + ar (FED)

= 3 ar (FED) (4)

From (2), (3) and (4), we get,

ar (AFC) = 2ar (FED) + 2 3 ar (FED) = 8 ar (FED)

Hence, ar (FED) = ar(AFC)Now, fromQuestion 30

If fig., ABC is a right triangle right angled at A, BCED, ACFG and ABMN are square on the sides BC, CA and AB respectively. Line segment AX DE meets BC at Y. Show that

(i) MBC ABD

(ii) ar (BYXD) = 2ar(MBC)

(iii) ar(BYXD) = ar(ABMN)

(iv) FCB ACE

(v) ar(CYXE) = 2ar (FCB)

(vi) ar(CYXE) = ar (ACFG)

(vii) ar(BCED) = ar (ABMN) + ar (ACFG)

Solution 30

(i) In MBC and ABD, we have

MB = AB

BC = BD

And MBC = ABD

[MBC and ABD are obtained by adding ABC to a right angle]

So, by SAS congruence criterion, we have

MBC ABD

ar (MBC) = ar(ABD) (1)

(ii) Clearly, triangle ABD and rectangle BYXD are on the same base BD and between the same parallels AX and BD.

ar(ABD) = ar (rect. BYXD)

ar (rect. BYXD) = 2 ar(ABD)

ar (rect. BYXD) = 2 ar (MBC)…(2)

[ ar (ABD) = ar (MBC), from (1)]

(iii) Since triangle MBC and square MBAN are on the same base MB and between the same parallels MB and NC.

2 ar (MBC) = ar (MBAN) (3)

From (2) and (3), we have

ar (sq. MBAN) = ar(rec. BYXD)

(iv) In triangles FCB and ACE, we have

FC = AC

CB = CE

And, FCB = ACE

[FCB and ACE are obtained by adding ACB to a right angle]

So, by SAS congruence criterion, we have

FCB ACE

(v) We have,

FCB ACE

ar (FCB) = ar (ACE)

Clearly, ACE and rectangle CYXE are on the same base CE ad between the same parallels CE and AX.

2 ar (ACE) = ar (CYXE)

2 ar (FCB) = ar (CYXE) (4)

(vi) Clearly, FCB and rectangle FCAG are on the same base FC and between the same parallels FC and BG.

2ar (FCB) = ar(FCAG) (5)

From (4) and (5), we get

ar(CYXE) = ar (ACFG)

(vii) Applying Pythagoras theorem in ACB, we have

BC² = AB² + AC²

Chapter 13 – Quadrilaterals Exercise Ex. 13.1

Question 1

Solution 1

Question 2

Solution 2

Question 3The angles of quadrilateral are in the ratio 3: 5: 9: 13, Find all the angles of the quadrilateral.Solution 3Let the common ratio between the angles is x. So, the angles will be 3x, 5x, 9x and 13x respectively.
Since the sum of all interior angles of a quadrilateral is 360^o.
 3x + 5x + 9x + 13x = 360^o
30x = 360^o
    x = 12^o
Hence, the angles are
3x = 3  12 = 36^o
5x = 5  12 = 60^o
9x = 9  12 = 108^o
13x = 13  12 = 156^o

Question 4

Solution 4

Chapter 13 – Quadrilaterals Exercise Ex. 13.2

Question 1

Two opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of each of the parallelogram.Solution 1

Question 2

If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.Solution 2

Question 3

Find the measure of all the angles of a parallelogram, if one angle is 24^o less than twice the smallest angle.Solution 3

Question 4

The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side?Solution 4

Question 5

In a parallelogram ABCD, ∠D = 135°, determine the measures of ∠A and ∠B.Solution 5

Question 6

ABCD is a parallelogram in which ∠A = 70. Compute ∠B, ∠C and ∠D.Solution 6

Question 7

In fig., ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute ∠CDB and ∠ADB.

Solution 7

Question 8

Solution 8

i. F

ii. T

iii. F

iv. F

v. T

vi. F

vii. F

viii. TQuestion 9

In fig., ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠b meet at P, prove that AD = DP, PC = BC and DC = 2AD.

Solution 9

Question 10

In fig., ABCD is a parallelogram and E is the mid-point of side BC. IF DE and AB when produced meet at F, prove that AF = 2AB.

Solution 10

Chapter 13 – Quadrilaterals Exercise Ex. 13.3

Question 1

In a parallelogram ABCD, determine sum of angles ∠C and ∠D.Solution 1

C and D are cosecutive interior angles on the same side of the transversal CD. Therefore, 

C + D = 180^oQuestion 2

In a parallelogram ABCD, if ∠B = 135°, determine the measures of its other angles.Solution 2

Question 3

ABCD is a square. AC and BD intersect at O. State the measure of AOB.Solution 3

Since, diagonals of a square bisect each other at right angle. Therefore, AOB = 90^oQuestion 4

Solution 4

Question 5

The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.Solution 5

Question 6

P and Q are the points of trisection of the diagonal BD of a parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ.Solution 6

Question 7

ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is square.Solution 7

Question 8

ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.Solution 8

Question 9

ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC.Solution 9

Chapter 13 – Quadrilaterals Exercise Ex. 13.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

In fig., triangle ABC is right-angled at B. Give that AB = 9 cm, AC = 15 cm and D, E are the mid-points of the sides AB and AC respectively, calculate

(i) The length of BC

(ii) The area of ADE.

Solution 7

Question 8

In fig., M, N, and P are the mid-points of AB, AC and BC respectively. If MN = 3 cm, Np = 3.5 cm and MP = 2.5 cm, Calculate BC, AB and AC.

Solution 8

Question 9

In fig., AB = AC and CP ∥ BA and AP is the bisector of exterior ∠CAD of ΔABC. Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram.

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.Solution 12

                                          Let ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively.
Join PQ, QR, RS, SP and BD.
In ABD, S and P are mid points of AD and AB respectively.
So, By using mid-point theorem, we can say that
SP || BD and SP =   BD             … (1)  
Similarly in BCD
QR || BD and QR =  BD               … (2)
From equations (1) and (2), we have
SP || QR and SP = QR  
As in quadrilateral SPQR one pair of opposite sides are equal and parallel to
each other.
So, SPQR is a parallelogram.Since, diagonals of a parallelogram bisect each other.
Hence, PR and QS bisect each other.

Question 13

Fill in the blanks to make the following statements correct:

(i) The triangle formed by joining the mid-points of the sides of an isosceles traingle is ______.

(ii) The triangle formed by joining the mid-points of the sides of a right triangle is ______ .

(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is ______ .Solution 13

(i) isosceles

(ii) right triangle

(iii) parallelogramQuestion 14

Solution 14

Question 15

In fig., BE ⊥ AC. AD is any line from A to BC interesting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°.

Solution 15

Question 16

Solution 16

Question 17

In fig., ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = (1/4)AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.

Solution 17

Question 18

In fig., ABCD and PQRC are rectangle and Q is the mid-point of AC. Prove that 

i. DP = PC ii. PR = (1/2) AC

Solution 18

Question 19

ABCD is a parallelogram, E and F are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC and G, P and H respectively. Prove that GP = PH.Solution 19

Question 20

BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If L is the mid-point of BC, prove that LM = LN.Solution 20

Chapter 12 – Congruent Triangles Exercise Ex. 12.1

Question 1

In fig., the sides BA and CA have been produced such that BA = AD and CA = AE.

Prove that segment DE || BC

Solution 1

Question 2

Solution 2

Question 3

Prove that the medians of an equilateral triangle are equal.Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

The vertical angle of an isosceles triangle is 100^o. Find its base angles.Solution 6

Question 7

 In fig., AB = Ac and ∠ACD = 105°, find ∠BAC. 

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

In fig., AB =AC and DB = DC, find the ratio ∠ABD = ∠ACD. 

Solution 10

Question 11

Determine the measure of each of the equal angles of a right-angled isosceles triangle.

                                                                                   OR

ABC is a right-angled triangle in which A = 90^o and AB = AC. Find B and C.Solution 11

Question 12

Solution 12

Question 13

AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See fig.). Show that the line PQ is perpendicular bisector of AB.

Solution 13

Chapter 12 – Congruent Triangles Exercise Ex. 12.2

Question 1

Solution 1

Question 2

In fig., it is given RT = TS, ∠1 = 2∠2 and ∠4 = 2∠3 prove that ΔRBT ≅ ΔSAT.

Solution 2

Question 3

Solution 3

Chapter 12 – Congruent Triangles Exercise Ex. 12.3

Question 1

In two right triangles one side and acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.Solution 1

Let ABC and DEF be two right triangles.

Question 2

Solution 2

Question 3

Solution 3

Question 4Show that the angles of an equilateral triangle are 60^o each.Solution 4

 Let us consider that ABC is an equilateral triangle.
So, AB = BC = AC
Now, AB = AC⇒ ∠C = ∠B         (angles opposite to equal sides of a triangle are equal)

We also have
AC = BC    
⇒ ∠B = ∠A             (angles opposite to equal sides of a triangle are equal)

So, we have
∠A = ∠B = ∠C
    Now, in ΔABC
∠A + ∠B + ∠C = 180^o
⇒ ∠A + ∠A + ∠A = 180^o
⇒ 3∠A = 180^o
⇒ ∠A = 60^o
⇒ ∠A = ∠B = ∠C = 60^o
Hence, in an equilateral triangle all interior angles are of 60^o.

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 12 – Congruent Triangles Exercise Ex. 12.4

Question 1

In fig., it is given that Ab = CD and AD = BC. prove that ΔADC ≅ ΔCBA

Solution 1

Question 2

Solution 2

Chapter 12 – Congruent Triangles Exercise Ex. 12.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

In fig., AD ⊥ CD and CB ⊥ CD. If AQ = BP an DP = CQ, prove that ∠DAQ = ∠CBP.

Solution 4

Question 5

Which of the following statements are True (T) and which are False (f):

(i) Sides opposite to equal angles of a triangle may be unequal.

(ii) Angles opposite to equal sides of a triangle are equal.

(iii) The measure of each angle of an equilaterial triangle is 60^o.

(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isoscles.

(v) The bisectors of two equal angles of a traingle are equal.

(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.

(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.

(viii) If any two sides of a right triangle are respectively equal to two sides of other right triagnle, then the two triangles are congruent.

(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.Solution 5

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) False

(vii) False

(viii) False

(ix) TrueQuestion 6

Solution 6

(i) equal

(ii) equal

(iii) equal

(iv) BC

(v) AC

(vi) equal to

(vii) EFDQuestion 7

Solution 7

Chapter 12 – Congruent Triangles Exercise Ex. 12.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Is it possible to draw a triangle with sides of length 2cm, 3cm and 7 cm?Solution 4

Here, 2 + 3 < 7

Hence, it is not possible because triangle can be drawn only if the sum of any two sides is greater than third side.Question 5

Solution 5

Question 6

Solution 6

Question 7

In fig., prove that:

i. CD + DA + AB + BC > 2AC

ii. CD + DA + AB > BC

Solution 7

Question 8

Which of the following statements are true (T) and which are false (F)?

(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.

(ii) Sum of any two sides of a triangle is greater than twice the median drawn to the third side.

(iii) Sum of any two sides of a triangle is greater than the third side.

(iv) Difference of any two sides of a triangle is equal to the third side.

(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it.

(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.Solution 8

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) TrueQuestion 9

Solution 9

(i) largest

(ii) less

(iii) greater

(iv) smaller

(v) less

(vi) greaterQuestion 10

Solution 10

Question 11

Solution 11

Chapter 11 Triangle and its Angles Exercise Ex. 11.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Two angles of a triangle are equal and the third angle is greater than each of those angles by 30^o. Determine all the angles of the triangle.Solution 4

Question 5

Solution 5

Question 6

Can a triangle have:

(i) Two right angles?

(ii) Two obtuse angles?

(iii) Two acute angles?

(iv) All angles more than 60^o?

(v) All angles less than 60^o?

(vi) All angles equal to 60^o?

Justify your answer in each case.Solution 6

(i) No

As two right angles would sum up to 180^o, and we know that the sum of all three angles of a triangle is 180^o, so the third angle will become zero. This is not possible, so a triangle cannot have two right angles.

(ii) No

A triangle cannot have 2 obtuse angles, since then the sum of those two angles will be greater than 180^o which is not possible as the sum of all three angles of a triangle is 180^o.

(iii) Yes

A triangle can have 2 acute angles.

(iv) No

The sum of all the internal angles of a triangle is 180^o. Having all angles more than 60^o will make that sum more than 180^o, which is impossible.

(v) No

The sum of all the internal angles of a triangle is 180^o. Having all angles less than 60^o will make that sum less than 180^o, which is impossible.

(vi) Yes

The sum of all the internal angles of a triangle is 180^o.  So, a triangle can have all angles as 60^o. Such triangles are called equilateral triangles.Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 11 Triangle and its Angles Exercise Ex. 11.2

Question 1

The exterior angles, obtained on producing both the base of a triangle both ways are 104^o and 136^o. Find all the angles of the triangle.

Solution 1

Question 2

In fig., the sides BC, CA and AB of a ΔABC have been produced to D, E, and F respectively. If ∠ACD = 105° and ∠EAF = 45°, find all the angles of the ΔABC.

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4

In fig., AC ⊥ CE and ∠A : ∠B : ∠C = 3 : 2 : 1, find the value of ∠ECD.

Solution 4

Question 5

In fig. AB ∥ DE. Find ∠ACD.

Solution 5

Question 6

Which of the following statements are true (T) and which are false (F):

Solution 6

Question 7

Fill in the blanks to make the following statements true:

(i) Sum of the angle of triangle is ______ .

(ii) An exterior angle of a triangle is equal to the two ______ opposite angles.

(iii) An exterior angle of a traingle is always _______ than either of the interior oppsite angles.

(iv) A traingle cannot have more than ______ right angles.

(v) A triangles cannot have more than ______ obtuse angles.Solution 7

(i) 180^o

(ii) interior

(iii) greater

(iv) one

(v) oneQuestion 8

Solution 8

Question 9

Solution 9

Question 10

In fig., AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Determine the value of x.

Solution 10

Question 11

Solution 11

Question 12

In fig., AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

In fig. AE bisects ∠CAD and ∠B = ∠C. Prove that AE ∥ BC.

Solution 15

Chapter 10 – Lines and Angles Exercise Ex. 10.1

Question 1

Write the complement of each of the following angles:

(i) 20^o

(ii) 35^o

(iii) 90^o

(iv) 77^o

(v) 30^oSolution 1

Question 2

Write the supplement of each of the following angles:

(i) 54^o

(ii) 132^o

(iii) 138^oSolution 2

(i) 54°

Since, the sum of an angle and its supplement is 180°

∴Its supplement will be 180° – 54° = 126°.

(ii) 132°

Since, the sum of an angle and its supplement is 180°

∴Its supplement will be 180° – 132° = 48°.

(iii) 138°

Since, the sum of an angle and its supplement is 180°

∴Its supplement will be 180° – 138° = 42°.Question 3

If an angle is 28^o less than its complement, find its measure.Solution 3

Question 4

If an angle is 30^o more than one half of its complement, find the measure of the angle.Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

If the complement of an angle is equal to the supplement of the thrice of it. Find the measure of angle.Solution 13

Let the measure of the angle be x^o.

Its complement will be (90^o – x^o) and its supplement will be (180^o – x^o).

Supplement of thrice of the angle = (180^o – 3x^o)

According to the given information:

(90^o – x^o) = (180^o – 3x^o)

3x – x = 180 – 90

2x = 90

x = 45

Thus, the measure of the angle is 45^o.

The measure of the angle is 45^oQuestion 14

Solution 14

Chapter 10 – Lines and Angles Exercise Ex. 10.2

Question 1

In fig., OA and OB are opposite rays:

(i) If x = 25°, what is the value of y?

(ii) if y = 35°, what is the value of x?

Solution 1

Question 2

In fig., write all pairs of adjacent angles and all the linear pairs.

Solution 2

Question 3

In fig., find x. further find ∠BOC, ∠COD and ∠AOD

Solution 3

Question 4

In fig., rays OA, OB, OC, OD and OE have the common end point O. Show that  ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360ºSolution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

How many pairs of adjacent angles, in all, can you name in fig.

Solution 7

Question 8

In fig., determine the value of x.

Solution 8

Question 9

In fig., AOC is a line, find x.

Solution 9

Question 10

In Fig., POS is a line, find x.

Solution 10

Question 11

In fig., ACB is a line such that ∠DCA = 5x and ∠DCB = 4x. Find the values of ∠DCA and∠DCB

Solution 11

Question 12

Give POR = 3x and QOR = 2x + 10, find the value of x for which POQ will be aline.

Solution 12

Question 13

What value of y would make AOB a line in fig., if ∠AOC = 4y and ∠BOC = (6y + 30)?

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

In Fig., Lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5:7, find all the angles.

Solution 16

Question 17

In Fig. If a greater than b by one third of a right-angle. find the values of a and b.

Solution 17

Question 18

Solution 18

Question 19In the given figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

                                    Solution 19Given that OR ⊥ PQ   ∴ POR = 90^o        ⇒ ∠POS  + ∠SOR = 90^o ∠ROS = 90^o  – ∠POS                … (1)     ∠QOR = 90^o                     (As OR ⊥ PQ)     ∠QOS – ∠ROS = 90^o     ∠ROS = ∠QOS – 90^o             … (2)     On adding equations (1) and (2), we have     2 ∠ROS = ∠QOS – ∠POS

Chapter 10 – Lines and Angles Exercise Ex. 10.3

Question 1

In fig., lines l₁ aans l₂ intersect at O, forming angles as shown in the figure. If x = 45, find the values of y, z and u.

Solution 1

Question 2

In fig., three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.

Solution 2

Question 3

In fig. , find the values of x, y and z.

Solution 3

Question 4

In Fig., find the value of X.

Solution 4

Question 5

Solution 5

Question 6

In fig., rays AB and CD intersect at O.

(i) Determine y when x = 60^o

(ii) Determine x when y = 40

Solution 6

Question 7

In fig., lines AB, CD and EF intersect at O. Find the measures of ∠AOC, ∠COF, ∠DOE and ∠BOF.

Solution 7

Question 8

Solution 8

Question 9

In fig., lines AB, and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution 9

Question 10

Which of the following statements are true (T) and which are false (F)?

(i) Angles forming a linear pair are supplementary.

(ii) If two adjacent angles are equal, then each angle measures 90^o.

(iii) Angles forming a linear pair can both be acute angles.

(iv) If angles forming a linear pair are equal, then each of these angles is of measure 90^o.Solution 10

(i) True

(ii) False

(iii) False

(iv) TrueQuestion 11

Fill in the blanks so as to make the following statements true:

(i) If one angle of a linear pair is acute, then its other angle will be ________.

(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is _______.

(iii) If the sum of two adjacent angles is 180^o, then the ______ arms of the two angles are opposite rays.Solution 11

(i) obtuse.

(ii) 180^o

(iii) uncommonQuestion 12

Solution 12

Question 13

Solution 13

Chapter 10 – Lines and Angles Exercise Ex. 10.4

Question 1

In fig., AB ∥ CD and ∠1 and ∠2 are in the ratio 3:2. determine all angles from 1 to 8.

Solution 1

Question 2

In fig., l, m and n are parallel lines intersected by transversal p at x, y and z respectively. find ∠1, ∠2, ∠3.

Solution 2

Question 3

In fig., if AB ∥ CD and CD ∥ EF, find ∠ACE.

Solution 3

Question 4

In fig., state which lines are parallel and why.

Solution 4

Question 5

In fig. if l ∥ m, n ∥ p and ∠1 = 85°, find ∠2.

Solution 5

Question 6

If two straight lines are perpendicular to the same line, prove that they are parallel to each other.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Let AB and CD be perpendicuar to line MN.

Question 9

In fig., ∠1 = 60° and ∠2 = (2/3)^rd of a right angle. prove that l ∥ m

Solution 9

Question 10

In fig., if l ∥ m ∥ n and ∠1 = 60°, find ∠2.

Solution 10

Question 11

Solution 11

Let AB and CD be perpendicuar to line MN.

Question 12

Solution 12

Question 13

Solution 13

Question 14

In fig., p is transversal to lines m and n, ∠2 = 120° and ∠5 = 60°. Prove that m ∥ n. 

Solution 14

Question 15

In fig., transceral l intersects two lines m and n, ∠4 = 110° and ∠7 = 65°. is m ∥ n ?

Solution 15

Question 16

Which pair of lines in Fig., are parallel? give reasons.

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

In fig., AB ∥ CD ∥ EF and GH ∥ KL. Find ∠HKL

Solution 20

Question 21

In fig., show that AB ∥ EF.

Solution 21

Question 22

In Fig., PQ ∥ AB and PR BC. IF ∠QPR = 102º, determine ∠ABC. Give reasons.

Solution 22

Question 23

Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.Solution 23

Consider the angles AOB and ACB.

Question 24

In fig.,  lines AB and CD are parallel and p is any point as shown in the figure. Show that ∠ABP + ∠CDP = ∠DPB.

Solution 24

Question 25

In fig., AB ∥ CD and P is any point shown in the figure. Prove that:

∠ABP + ∠BPD + ∠CDP = 360°

Solution 25

Question 26

In fig., arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF. Prove that ∠ABC = ∠DEF

 Solution 26

Question 27

 Solution 27

Chapter 9 Introduction to Euclid’s Geometry Exercise Ex. 9.1

Question 1

Solution 1

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 4

Write the truth value (T/F) of each of the following statements:

(i) Two lines intersect in a point.

(ii) Two lines may intersect in two points.

(iii) A segment has no length.

(iv) Two distinct points always determine a line.

(v) Every ray has a finite length.

(vi) A ray has one end-point only.

(vii) A segment has one end-point only.

(viii) The ray AB is same as ray BA.

(ix) Only a single line may pass through a given point.

(x) Two lines are coincident if they have only one point in common.Solution 4

(i) False

(ii) False

(iii) False

(iv) True

(v) False

(vi) True

(vii) False

(viii) False

(ix) False

(x) FalseQuestion 5

In fig., name the following:

(i) Five line segments.

(ii) Five rays.

(iii) Four collinear points.

(iv) Two pairs of non-intersecting line segments.Solution 5

Question 6

Fill in the blanks so as to make the following statements true:

(i) Two distinct points in a plane determine a ______ line.

(ii) Two distinct ______ in a plane cannot have more than one point in common.

(iii) Given a line and a point, not on the line, there is one and only ______ line which passes through the given point and is ______ to the given line.

(iv) A line separates a plane into ______ parts namely the ______ and the _______ itself.Solution 6

(i) unique

(ii) lines

(iii) perpendicular, perpendicular

(iv) three, two half planes, line.