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Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 16 Preparation for Algebra Problem Set 56 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 16 Preparation for Algebra Problem Set 56

Question 1.
Use a letter for ‘any number’ and write the following properties in short.

(1) The sum of any number and zero is the number itself.
Answer:
a + 0 = a

(2) The product of any two numbers and the product obtained after changing the order of those numbers is the same.
Answer:
a x b = b x a

(3) The product of any number and zero is zero.
Answer:
a x 0 = 0

Question 2.
Write the following properties in words :

(1) m – 0 = m
Answer:
Subtracting zero from any number, gives the number itself.

(2) n ÷ 1 = n
Answer:
Dividing any number by 1, gives the number itself.

Preparation for Algebra Problem Set 56 Additional Important Questions and Answers

Use a letter for any number and write the following properties in short.

Question 1.
The product of any number and 1 is the number itself.
Answer:
a x 1 = a

Question 2.
The division of any two different numbers and the divisions obtained after changing the order of those numbers is not the same.
Answer:
a ÷ b ≠ b + a

Write the following properties in words:

Question 1.
p x 0 = 0
Answer:
The product of any number and zero is zero.

(4) a + b = b + a
Answer:
The sum of any two numbers and the sum obtained after changing the order of these numbers is the same.

Using brackets write three pairs of numbers whose

(1) Sum is 9
Answer:
5 + 4 = 9,
7 + 2 = 9,
8 + 1 = 9

(2) difference is 9
Answer:
12 – 3 = 9,
11 – 2 = 9,
10 – 1 = 9

(3) multiplication is 16 and
Answer:
4 x 4 = 16,
8 x 2 = 16,
16 x 1 = 16

(4) division is 16.
Answer:
32 ÷ 2 = 16,
48 ÷ 3 = 16,
64 ÷ 4 = 16,

Fill in the blanks.

(1) 4 + 2 = 7 – ……….
(2) 4 + 2 = 3 x ……….
(3) 4 + 2 = 12 ÷ ……….
Answer:
(1) 1
(2) 2
(3) 2

Match the columns:

(A)

A		B
(i)	8 + 6	(a)	6 x 2
(2)	9 + 3	(b)	6 + 2
(3)	5 + 1	(c)	16 – 2
(4)	10 – 2	(d)	12 + 2

Answer:
(1 – c),
(2 – a),
(3-d),
(4-b)

(B)

A	B
(1) a – b and b – a	(a) 0
(2) a x b and b x a	(b) 1
(3) a x 0	(c) =
(4) a + a	(d) ≠

Answer:
(1-d),
(2 – c),
(3 – a),
(4 – b)

Say whether right or wrong.

(1) (6 + 5) = (5 + 6)
(2) (8 + 5) > 10
(3) (8 + 5) < 10
(4) 108 > 108
(5) 108 = 108
(6) 108 < 108
(7) (6 x 3) = (20 – 2)
(8) 40 + 8 > 5
(9) (3 x 7) = (7 x 3)
(10) (5 + 0) = (5 x 1)
(11) (6 + 5) = 10
(12) (30 + 5) < (30 – 25)
Answer:
Right : (1), (2), (5), (7), (9), (10)
Wrong : (3), (4), (6), (8), (11), (12)

Fill in the blanks with the right symbol from <, > or =

(1) (24 ÷ 5) ……… (9 – 5)
(2) (4 + 2) ……… (5 x 1)
(3) (7 x 3) ……… (20 + 2)
(4) (8 x 2) (5 x 3)
(5) (5 x 6) ……… (25 + 5)
(6) (6 x 7) (9 x 5)
Answer:
(1) =
(2) >
(3) <
(4) >
(5) =
(6) <

Fill in the blanks in the expressions with the proper numbers.

(1) (4 x 4) = (………. x 2)
(2) (2 x 7) > (4 x ……….)
(3) (30 + 5) < ( x 3)
(4) (5 + 0)> (4 x ……….)
(5) (36 +3) = ( + )
(6) (9 – ……….) < (4 + 1)
(7) (8 + 9) < (3 x ……….)
(8) (0 + 3) > (4 x ……….)
(9) (28 ÷ 2) = (7 x ……….)
Answer:
(1) 8
(2) 3
(3) 9
(4) 1
(5) 7 + 5
(6) 5,
(7) 6
(8) 0
(9) 2

Use a letter for any number and write the following properties in short:

(1) Dividing zero by any non zero number is zero.
Answer:
0 + a = 0

(2) The difference of any two different numbers and the difference obtained after changing the order of those numbers is not same.
Answer:
a – b ≠ b – a

(3) Dividing non zero number by itself gives us 1.
Answer:
a ÷ a = 1

Write the followîng properties in words:

(1) a x 1 = a
Answer:
The product of any number and 1 is the number itself.

(2) a – a = 0
Answer:
Difference of the same two numbers is zero.

Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 7.4 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 7 Mensuration.

Practice Set 7.4 Geometry 10th Std Maths Part 2 Answers Chapter 7 Mensuration

Practice Set 7.4 Geometry Class 10 Question 1. In the adjoining figure, A is the centre of the circle. ∠ABC = 45° and AC = 7\(\sqrt { 2 }\) cm. Find the area of segment BXC, (π = 3.14)

Solution:
In ∆ABC,
AC = AB … [Radii of same circle]
∴ ∠ABC = ∠ACB …[Isosceles triangle theorem]
∴ ∠ABC = ∠ACB = 45°
In ∆ABC,
∠ABC + ∠ACB + ∠BAC = 180° … [Sum of the measures of angles of a triangle is 180° ]
∴ 45° + 45° + ∠BAC = 180°
∴ 90° + ∠BAC = 180°
∴ ∠BAC = 90°
Let ∠BAC = θ = 90°

∴ The area of segment BXC is 27.93 cm².

10th Class Geometry Practice Set 7.4 Question 2. In the adjoining figure, O is the centre of the circle.
m(arc PQR) = 60°, OP = 10 cm. Find the area of the shaded region.
(π = 3.14, \(\sqrt { 3 }\) = 1.73)

Given: m(arc PQR) = 60°, radius (r) = OP = 10 cm
To find: Area of shaded region.
Solution:
∠POR = m (arc PQR) …[Measure of central angle]
∴ ∠POR = θ = 60°


∴ The area of the shaded region is 9.08 cm².

7.4 Class 10 Question 3. In the adjoining figure, if A is the centre of the circle, ∠PAR = 30°, AP = 7.5, find the area of the segment PQR. (π = 3.14)

Given: Central angle (θ) = ∠PAR = 30°,
radius (r) = AP = 7.5
To find: Area of segment PQR.
Solution:
Let ∠PAR = θ = 30°

∴ The area of segment PQR is 0.65625 sq. units.

Chapter 7 Maths Class 10 Question 4. In the adjoining figure, if O is the centre of the circle, PQ is a chord, ∠POQ = 90°, area of shaded region is 114 cm2, find the radius of the circle, (π = 3.14)

Given: Central angle (θ) = ∠POQ= 90°,
A (segment PRQ) = 114 cm²
To find: Radius (r).
Solution:

…[Taking square root of both sides]
∴ r = 20 cm
∴ The radius of the circle is 20 cm.

Mensuration Questions for Class 10 Question 5. A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centre of the circle. Find the area of the minor as well as the major segment. (π = 3.14, \(\sqrt { 3 }\) = 1.73)
Given: Radius (r) =15 cm, central angle (θ) = 60°
To find: Areas of major and minor segments.
Solution:
Let chord PQ subtend ∠POQ = 60° at centre.
∴ θ = 60°

= 225 [0.0908]
= 20.43 cm²
∴ area of minor segment = 20.43 cm²
Area of circle = πr²
= 3.14 × 15 × 15
= 3.14 × 225
= 706.5 cm²
Area of major segment
= Area of circle – area of minor segment
= 706.5 – 20.43
= 686.07 cm²
Area of major segment 686.07 cm²
∴ The area of minor segment Is 20.43 cm² and the area of major segment is 686.07 cm².

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 15 Patterns Problem Set 53 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 15 Patterns Problem Set 53

Question 1.
Find the square numbers from the list given below.
5, 9, 12, 16, 50, 60, 64, 72, 80, 81
Answer:
9,16, 64, 81, 4, 25, 49 are square numbers.

Question 2.
Which are the triangular numbers in the given list?
3, 6, 8, 9, 12, 15, 16, 20, 21, 42
Answer:
3, 6, 15, 21, 28, 10, 45, 55 are triangular numbers.

Question 3.
Name a number which is square as well as triangular.
Answer:
36 is square as well as triangular number.

Question 4.
If 4 is the first square number, which is the tenth one?
Answer:
121 is the tenth square number.

Question 5.
If 3 is the first triangular number, which is the tenth one?
Answer:
66 is the tenth triangular number.

Think about it.

• How will you decide if a given number is a square number?
• How will you decide if a given number is a triangular number?
• How many square numbers do you think there are?
• How many triangular numbers do you think there are?

Activity

Make a collection of pictures in which you can see square or triangular numbers.

Patterns in floor tiles

The tiles in each picture below form a specific pattern. Observe that there is no gap or open ground between two tiles.

On a large piece of card sheet, draw several shapes like the one shown alongside. Colour half of them. Cut them all out and separate them.

One pattern made of these shapes is shown alongside. Make some other patterns of your own.

Cut out many pieces of each of the shapes shown alongside. Join them in a pattern like floor tiles.

Note the pattern and complete the design.

Make your own shapes and use them to make patterns for sari and shawl borders, etc.

Perimeter and Area Problem Set 49 Additional Important Questions and Answers

Solve the following :

Question 1.
If 4 is the first square number which is the eighth one?
Answer:
81 is the eighth square number.

Question 2.
If 3 is the first triangular number which is the eighth one?
Answer:
45 is the eighth triangular number.

Question 3.
Classify the following into square numbers and triangular numbers.
3, 4, 9,10,15,16; 45, 49, 64, 66, 81, 91
Answer:
Square Numbers : 4, 9,16, 49, 64, 81
Triangular Numbers : 3, 10, 15, 45, 66, 91

Question 4.
Find out the numbers which are neither square nor triangular numbers from the following.
4, 5, 6, 8, 9, 10, 14, 15, 16, 25, 26, 27, 28.
Answer:
5, 8 14, 26 and 27

Question 5.
(1) If 4 is the first square number, which is the fifth one?
(2) If 3 is the first triangular number, which is the sixth one?
(3) Write all the square numbers between 20 and 80.
(4) Write all the triangular numbers between 20 and 80.
(5) Write the greatest two-digit square numbers as well as triangular numbers.
(6) Write the next three square numbers, 36, 49, 64,…….,
(7) Write the next three triangular numbers 36, 45, 55,
Answer:
(1) 36
(2) 28
(3) 25, 36, 49, 64
(4) 21, 28, 36, 45, 55, 66, 78
(5) 81, 91
(6) 81, 100, 121
(7) 66, 78, 91

Question 6.
Match the columns

A	B
(1) Third square number	(a) 15
(2) Fourth triangular number	(b) 36
(3) Number neither square nor triangular	(c) 16
(4) Number is both square as well as triangular number	(d) 35

Answer:
(1 – c),
(2 – a),
(3 – d),
(4 – b).

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 14 Pictographs Problem Set 52 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 14 Pictographs Problem Set 52

Question 1.
Stocks of various types of grains stored in a warehouse are as given below. Make a pictograph based on the information given.

Grain	Sacks
Rice	40
Wheat	56
Bajra	8
Jowar	32

Answer:
Scale :1 picture = 8 sacks

Question 2.
Information about the various types of vehicles in Wadgaon is given below. Make a pictograph for this data.

Types of vehicles	Number
Bicycles	84
Automatic two-wheelers	60
Four-wheelers (cars/jeeps)	24
Heavy vehicles (truck, bus, etc.)	12
Tractors	24

Answer:

Question 3.
The numbers of the various books kept in a cupboard in the school library are given below. Make a pictograph showing the information given.

Type of book	Number
Science	28
Sports	14
Poetry	21
Literature	35
History	7

Answer:

Activity
Collect information based on the points given below and make a pictograph for each.

• Which crops are grown on the farms owned by students in your class? (Vegetables, grains, pulses, fruits, etc.)
• Which storybooks do your classmates like? (fairytales, stories about kings and queens, historical stories, stories about saints, picture stories, etc.)
• What do your classmates want to be when they grow up ? (doctor, teacher, farmer, engineer, officer, etc.)

Perimeter and Area Problem Set 49 Additional Important Questions and Answers

Solve the following

Question 1.
Information regarding the number of pages of novel book read in different days by Rosi are as follows. Make a pictograph showing the information given.

Days	1st day	2nd day	3rd day	4th day
Pages	60	40	30	20

Question 2.
Different types of currency notes had with Shamin are as follows. Make a pictograph showing the information given.

Types of Notes	₹ 500	₹ 100	₹ 50	₹ 10
Number of Notes	8	10	6	4

Question 3.
Different types of colour of scooters sold by a merchant are as follows. Make a pictograph showing the data given.

Colour	White	Red	Black	Yellow
No, of scooters sold	6	9	12	3

Question 4.
Ajhount of sales of goods in rupees for the first four days of a week are as follows. Make a pictograph from the information given below.

Days	Marks
Monday	₹ 150
Tuesday	₹ 200
Wednesday	₹ 250
Thursday	₹ 100

Answer:
(1) All the given numbers can be divided by 2, 5, and 10. 1 picture of 10 pages will be convenient scale so 6 pictures for 60 pages. 4 pictures for 40, 3 for 30 and 2 for 20.
(2) All the given numbers can divided by 2 only, so 1 picture for 2 notes will be the scale. So, 4 pictures for notes of ? 500, 5 for ? 100 notes, 3 for ? 50 and 2 for ? 10.
(3) All given numbers are divisible by 3, so 1 picture for 3 scooters will be the scale. So, 2 pictures for white, 3 for Red, 4 for Black and 1 pictures for Yellow.
(4) All the given numbers can be divided by 2, 5,10, 25 and 50. So, 1 picture for 50 rupees will be convenient scale. So, draw 3 pictures for Monday, 4 for Tuesday, 5 for Wednesday and 2 for Thursday.
(5) Number of books are multiples of 50. Therefore Take number of pictures = 5,4,2, 1, 3 respectively.

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50

Question 1.
The length of the side of each square is given below. Find its area.

(1) 12 metres
Solution:
Area of a square
= side x side
= 12 x 12

(2) 6 cm
Solution:
Area of a square
= side x side
= 6 x 6
= 36 sq.cm.

(3) 25 metres
Solution:
Area of a square
= side x side
= 25 x 25
= 625 sq.m.

(4) 18 cm
Solution:
Area of a square
= side x side
= 18 x 18
= 324 sq.cm.

Question 2.
If the cost of 1 sq m of a plot of land is 900 rupees, find the total cost of a plot of land that is 25 m long and 20 m broad.
Solution:
Area of the rectangular plot
= length x breadth
= 25 x 20
= 500 sq.m.

Cost of the plot of land
= Area of the plot x rate
= 500 x 900
= 4,50,000 rupees

Question 3.
The side of a square is 4 cm. The length of a rectangle is 8 cm and its width is 2 cm. Find the perimeter and area of both figures.
Solution:
Perimeter of a square = 4 x side
= 4 x 4
= 16 cm

Area of a square = side x side
= 4 x 4
= 16 sq.cm.

Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2 x 8 + 2 x 2
= 16 + 4
= 20 cm

Area of a rectangle
= length x breadth
= 8 x 2
= 16 sq.cm.

Question 4.
What will be the labour cost of laying the floor of an assembly hall that is 16 m long and 12 m wide if the cost of laying 1 sq m is 80 rupees?
Solution:
Area of rectangular floor
= length x breadth
= 16 x 12
= 192 sq.cm.

The cost of laying 1 sq.m, is 80 rupees.
Hence, the cost of laying 192 sq.m.
= 192 x 80
= 15,360 rupees.
∴ ₹ 15,360

Question 5.
The picture alongside shows some squares. Find out how many squares with the same measures will fit in the empty space in the figure.

Solution:
length of the empty space = 4 – 1 = 3 cm
breadth of the empty space = 3 – 1 = 2 cm
square in empty space
= length x breadth
= 3 x 2 = 6 sq.cm.

∴ 6 squares will fit in the empty space in the figure

Question 6.
Divide the figure given alongside into four parts in such a way that the area and shape of each part is the same. Colour the parts with different colours.

Fair and square

As shown in the figure alongside, a square plot of land owned by the government contains four houses and a well right in the centre. The government has to divide the houses and the land between four poor persons according to the following conditions.
(1) Each person must get only one house.
(2) The shape and area of the land must be the same.
(3) Each person must be able to use the well without trespassing on any one else’s land.

Show the appropriate divisions in four different colours.

Activity
Using a graph paper, find out the area of different rectangles and squares.

Perimeter and Area Problem Set 50 Additional Important Questions and Answers

Question 1.
15 cm
Solution:
Area of a square
= side x side
= 15 x 15
= 225 sq.cm.

Question 2.
21 cm
Solution:
Area of a square = side x side
= 21 x 21
= 441 sq.cm.

Solve the following:

Question 1.
The side of a square hall is of length 8 m. If it is tiled with a tile of length 4 m and breadth 2 m., how many tiles will be required?
Solution:
Area of the square hall
= side x side
= 8 x 8
= 64 sq.m.

Area of 1 rectangular tile
= length x breadth
= 4 x 2

∴ 8 sq.m.

For 8 sq.m., 1 tile is required,
but for 64 sq.m = \(\frac{64}{8}\) = 8 tiles required

∴ 8 tiles required

Question 2.
Perimeter of a square is 16 cm. What is the length of each side? What is the area of the square?
Solution:
Perimeter of square = 4 x side
16 = 4 x side

∴ side of a square = \(\frac{16}{4}\) = 4 cm
Area of the square
= side x side
= 4 x 4
= 16 sq.cm

∴ Side of square is 4 cm and area of the sqaure is 16 sq.cm

Question 3.
Write the perimeter of each figure in the box given below it.

Answer:
24 cm

Answer:
18 cm

Answer:
21 cm

Answer:
20 cm

Question 4.
Two squares of side 2 cm is cut out of two corners of a larger square with side 5 cm (see the figure). What will be the perimeter of the remaining shape?

Answer:
20 cm

Question 5.
Match the columns ‘A’ and ‘B’

Answer:
(1- d),
(2- a),
(3 – b),
(4 – c)

Question 6.
Solve the following word problems:
(1) What is the perimeter of a rectangle having length 9 cm and its breadth 6 cm?
(2) The sides of a rectangular field are having length 150 m and breadth 100 m. Find the perimeter of field.
(3) If each side of a square is 8 cm then what is the perimeter of the square?
(4) A rectangular garden of length 650 m and breadth 350 m. Mohan makes four rounds daily. How many kilometres does he walk everyday?
(5) One square field is having one side of it is of 225 m, Soham makes 6 rounds of the square field daily. How much distance is covered by him? Write it in km and m.
(6) A rectangular field whose length is 58 m and breadth is 32 m. Fencing the field by 4 rounds with a wire, what length of wire is required? If the cost of 1 m wire is ? 75, then what is the expenditure of fencing the field?
(7) A length of a rectangular classroom is 8 m and its breadth is 5 m. A wooden strip is to be fitted along the four walls to hang charts and pictures. What is the length of the wooden strip required?
(8) The side of a square table is 1.5 m. To fit a strip of tin sheet around the table, how many metres of strip is required?
(9) What is the perimeter of the triangle whose sides are 13.8 cm, 17.6 cm and 10.6 cm?
(10) The sides of some squares are given below. Find their areas.
(i) 11 cm (ii) 23 cm (iii) 9 cm
(11) Find the perimeter and area of the following:
(i) square of side 6 cm
(ii) Rectangle: length 12 cm and breadth 6 cm
(12) A rectangular land is having its length 25 m and breadth 16 m. If the cost of 1 sq.m, land is ? 1,500, then what will be cost of land?
(13) The side of a square plot is 10 metre. It is tiled at the rate of 50 rupees per sq.m. What will be cost of tiling the floor?
(14) A wall of length 25 m and breadth 12 m. It is painted at the rate of 60 rupees per sq.m. What will be cost of painting the wall?
Answer:
(1) 30 cm
(2) 500 m
(3) 32 cm
(4) 8 km
(5) 5 km 400 m
(6) ₹ 54,000
(7) 26 m
(8) 6 m
(9) 42 cm

(10) (i) 121 sq. cm
(ii) 529 sq.cm
(iii) 81sq.cm.

(11) (i) Perimeter = 24 cm, Area = 36 sq.cm,
(ii) Perimeter = 36 cm, Area = 72 sq. cm

(12) 6,00,000 rupees
(13) 5,000 rupees
(14) t 18,000

Question 7.
Look at the figures on the sheet of graph paper. Measure their sides with the help of the lines on the graph paper. Write the perimeter of each in the right box.

Perimeter of Rectangle
(1) XYZW = [ ] cm
(2) CDEF = [ ] cm
(3) JKLM = [ ] cm
(4) NOPQ = [ ] cm
Answer:
(1) 12 cm
(2) 10 cm
(3) 10 cm
(4) 8 cm

Question 8.
Find the area of the. following figures. (All small squares are having side 1 cm)

Answer:
5 sq.cm

Answer:
4 sq.cm

Answer:
9 sq.cm

Answer:
16 sq.cm.

Question 9.
The pictures below shows some squares. Find out how many squares with the same measures will fit in the empty space in the figures.

Answer:
(1) 9
(2) 9

Question 10.
Fill in the blanks:

Answer:
(1) Area = 24 sq.cm., Perimeter = 22 cm
(2) Breadth = 4 cm, Perimeter = 20 cm
(3) Length = 2 cm, Perimeter = 8 cm

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51

Question 1.
The first column shows a structure made of blocks. The other columns show different views of the structure in two dimensions. Say whether each view is from the front, from a side or from above.

Answer:

Question 2.
Draw three pictures of each of these three-dimensional objects – a table, a chair and a water bottle as viewed from the front, from a side and from above.
Answer:

Nets
Last year we saw that cutting some edges of a box and laying it out flat gives us the net from which it was made.
The two dimensional shape from which a three dimensional object can be made by folding is called the ‘net’ of that object.

1. By folding the cardboard shown below, along the lines shown in it, we get a three dimensional object (box). In this shape, all surfaces are square.
   An object of this shape is called a cube.
   
2. The net of another cardboard box is shown in the figure below. By folding along the lines in this net and joining the edges to each other, we can see that a three dimensional box is formed. The surfaces of this box are rectangular in shape.
   An object of this shape is called a cuboid.
   

Activity :
Draw the nets shown below on card sheet. Cut out the shapes and find out the shapes of the boxes they form.

A five-square net (Pentomino)

In the figure alongside, five squares of the same size are placed together with their sides joined.

Such an arrangement of five squares is called a ‘five-square net’ or a ‘pentomino’.

By folding along the edges of such a five-square net, an open box is formed.

Activity :
Some five-square nets are given below. Draw these nets on a card sheet. Make open boxes from these nets.

Try to find out other five-square nets that can be used to make open boxes.

A riddle

The net of a cube-shaped dice is given alongside. If a dice is made of this net, which of the following shapes will it definitely not resemble?

Chapter 12 Perimeter and Area Problem Set 49 Additional Important Questions and Answers

Question 1.
Draw the pictures of each of these three dimensional objects – Mobile, Oil tin as viewed from the front, from a side and from above.
Answer:

Question 2.
The three dimensional figure of block formation is shown in the figure along side. Draw as view from the front, from a side and from above (fig. drawn in answer part)
Answer:

Question 3.
Draw the nets shown below on card sheet. Cut out the shapes and find out the shapes of the boxes they form.
Answer:

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48

Question 1.
Write the perimeter of each figure in the box given below it.

(1)

Solution:
Perimeter [ ] DABCD
= 20 + 16 + 7 + 14
= 57 cm

∴ 57 cm

(2)

Solution:
Perimeter of the figure
= 12 + 18 + 8 + 8 + 18
= 64m

∴ 64m

(3)

Solution:
Perimeter of the figure
= 10 + 6 + 6 + 10 + 8 + 8
= 48 cm

∴ 48 cm

Question 2.
If a square of side 1 cm is cut out of the corner of a larger square with side 3 cm (see the figure), what will be the perimeter of the remaining shape?

Solution:

Perimeter
= 2 + 3 + 3 + 2 + 1 + 1
= 12 cm

∴ 12 cm

Formula for the perimeter of a rectangle

Perimeter = length + breadth + length + breadth Opposite sides of a rectangle are of the same length.
So, the perimeter of a rectangle
= twice the length + twice the breadth
= 2 × length + 2 × breadth

Perimeter of a rectangle = 2 × length + 2 × breadth

Example : The length of the rectangle below is 7 cm and its breadth, 3 cm. Let us find its perimeter.

Perimeter of rectangle PQRS = 2 × length + 2 × breadth
= 2 × 7 + 2 × 3
= 14 + 6
= 20
Therefore, the perimeter of the rectangle is 20 cm.

Formula for the perimeter of a square

The lengths of all the sides of a square are equal. Therefore, the perimeter of a square = four times the length of one of its sides.

Perimeter of a square = 4 × the length of one side

Example : The length of one side of a square is 6 cm. Find its perimeter. The perimeter of a square is four times the length of one side.

Perimeter of a square = 4 × length of one side
= 4 × 6
= 24

Therefore, the perimeter of the square is 24 cm.

Word problems

Example (1) The length of a rectangular park is 100 m, while its width is 80 m. What is its perimeter?

Perimeter of the rectangle = 2 × length + 2 × breadth
= 2 × 100 + 2 × 80
= 200 + 160
= 360

The perimeter of the rectangular park is 360 m.

Example (2) How much wire will be needed to put a triple fence around a square plot with side 30 m? What will be the total cost of the wire at ₹ 70 per metre ?

To put a single fence around the square plot, we need to find its perimeter.

Perimeter of a square = 4 × length of one side = 4 × 30 = 120

The perimeter of the square plot is 120 metres. Since the fence is to be a triple fence, we must triple the perimeter.

120 × 3 = 360 m of wire will be needed.

Now let us find out how much the wire will cost. One metre of wire costs ₹ 70.

Therefore, the cost of 360 m of wire will be 360 × 70 = 25, 200.

The total cost of wire for putting a triple fence around the plot will be ₹ 25, 200.

Perimeter and Area Problem Set 48 Additional Important Questions and Answers

Question 1.

Solution:
Perimeter of the figure
= 2 + 6 + 2 + 6
= 16 cm

∴ 16 cm

Question 2.

Solution:
Perimeter of the figure
= 3 + 3 + 3 + 3
= 12 cm

∴ 12 cm

Question 3.

Solution:
Perimeter of the figure
= 12 + 13 + 5
= 30 cm

∴ 30 cm

Question 4.
If four squares of side 1 cm ¡s cut out of all the corners of a larger square with side 4 cm (see the figure), what will be the perimeter of the remaining shape?

Solution:
Perimeter
= 2 + 1 + 1 + 2 + 1 + 1 + 2 + 1 + 1 + 2 + 1 + 1
= 16 cm

∴ 16 cm

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46

Question 1.
Add :

(1) ₹ 9, 50 paise + ₹ 14, 60 paise
Solution:

₹	Paise
1
9 + 14	5 0 6 0
2 4	1 0

50 paise + 60 paise
= 110 paise
= 1 ₹ 10 paise
∴ ₹ 24, 10 paise

(2) 6 cm 5 mm + 7 cm 9 mm
Solution:

cm	mm
1
6 + 7	5 9
1 4	4

5 mm + 9 mm
= 14 mm 14 mm
= 1 cm 4 mm
∴ 14 cm 4 mm

(3) 22 m 50 cm + 25 m 75 cm
Solution:

m	cm
1
2 2 + 2 5	5 0 7 5
4 8	2 5

50 cm + 75 cm
= 125 cm
= 1 m 25 cm
∴ 48 m 25 cm

(4) 15 km 740 m + 13 km 950 m
Solution:

km	m
1
1 5 + 13	7 4 0 9 5 0
2 9	6 9 0

740 m + 950 m
= 1690 m 1690 m
= 1km 690 m
∴ 29 km 690 m

(5) 25 kg 650 g + 29 kg 770 g
Solution:

kg	gm
1
2 5 + 29	6 5 0 7 7 0
5 5	4 2 0

650 gm + 770 gm
= 1420 gm
= 1 kg 420 gm
∴ 55 kg 420 gm

(6) 19l 840ml + 25l 250ml
Solution:

l	ml
1 1
1 9 + 2 5	8 4 0 2 5 0
4 5	0 9 0

840 ml + 250 ml
= 1090 ml
= 11 + 90 ml
∴ 45 l 90 ml

Question 2.
Subtract :

(1) ₹ 19, 50 paise – ₹ 12, 60 paise
Solution:

₹	Paise
1 8	1 5 0
1 9 – 1 2	5 0 6 0
6	9 0

We cannot subtract 60 paise from 50 paise. So convert 1 ₹ into 100 paise.
₹ 6, 90 paise

∴ ₹ 6, 90 paise

(2) 24 cm 2 mm – 3 cm 8 mm
Solution:

cm	mm
2 3	1 2
2 4 – 3	2 8
2 0	4

We cannot subtract 8 mm from 2 mm. So, convert 1 cm = 10 mm

∴ 20 cm 4 mm

(3) 20 m 30 cm – 17 m 60 cm
Solution:

m	cm
1 9	1 3 0
2 0 – 1 7	3 0 6 0
2	. 7 0

We cannot subtract 60 cm from 30 cm. So, convert 1 m = 100 cm

∴ 2 m 70 cm

(4) 40 km 255 m – 17 km 960 m
Solution:

km	m
3 9	12 2 5
4 0 -1 7	2 2 5 9 6 0
2 2	2 6 5

We cannot subtract 960 m from 225 m. So, convert 1 km = 1000 m

∴ 22 km 265 m

(5) 35 kg 150 g – 26 kg 470 g
Solution:

kg	gm
3 4	1 1 5 0
3 5 – 2 6	1 5 0 4 7 0
8	6 8 0

We cannot subtract 470 gm from 150 gm. So, convert I kg= 1000gm

∴ 8 kg 680 gm

(6) 46 l 200 ml – 38 l 750 ml
Solution:

l	ml
4 5	1 2 0 0
4 6 – 3 8	2 0 0 7 5 0
7	4 5 0

We cannot subtract 750 ml from 200 ml. So, convert 1 l = 1000 ml

∴ 7 l 450 ml

Word problems

Study the following examples.

Example (1) If a shopkeeper has 150 kg 500 g of rice and sells 75 kg 750 g, how much rice will be left?

74 kg 750 g of rice is left.

Example (2) A can of milk has 20 l 450 ml of milk. Another can has 18 l 800 ml. How much milk is there in the two cans altogether?

The total quantity of milk is 39l 250ml.

Example (3) At a speed of 90 km per hour, what distance will a train cover in two and a half hours?

The speed of the train is 90 kmph. That is, it travels 90 km in one hour. It travels 90 more km in the second hour.
In the next half an hour, 90 ÷ 2 = 45 km
The total distance travelled is 90 + 90 + 45 = 225 km.

Example (4) If one dress requires 3 m 25 cm of cloth, how much do 4 dresses need?

Manju’s method :
3 m 25 cm for the 1st dress
+ 3 m 25 cm for the 2nd dress
+ 3 m 25 cm for the 3rd dress
3 m 25 cm for the 4th dress
_________
12 m 100 cm
1 m is 100 cm, therefore 12 + 1 = 13 m

Example (5)
If a wire that is 9 m 50 cm long is cut into pieces of 5 cm each, how many pieces will be made?
9 m 50 cm = (900 + 50) cm
To find out how many pieces of 5 cm can be made from a wire 950 cm long, let us use division.
190 pieces will be made.

Example (6) A play started at 30 minutes past 6 in the evening and finished two and three quarter hours later. What time did the play get over?

The play got over at 15 minutes past 9 at night.

Note : The units for length, mass and capacity are written in decimal form. This makes it easy to carry out addition and subtraction of length, mass and capacity.

Units of measuring time are not in decimal form. It is a little more difficult to carry out additions and subtractions of those quantities.

Problems on Measurement Problem Set 46 Additional Important Questions and Answers

Add the following:

(1) 12 km 880 m + 7 km 620 m
Solution:

km	m
1
1 2 + 7	8 8 O 6 2 0
2 0	5 0 0

880m + 620 m = 1500 m
= 1km 500 m
∴ 20 km 500 m

(2) ₹ 62, 45 paise + ₹ 37, 55 paise
Solution:

₹	Paise
1
6 2 + 3 7	4 5 5 5
1 0 0	0 0

45 paise + 55 paise
100 paise = 1 ₹
∴ 100 rupees

Subtract the following:

(1) 15 m 15 cm – 4 m 65 cm
Solution:

kg	gm
1 4	1 1 5
1 5 – 4	1 5 6 5
1 0	5 0

We cannot subtract 65 cm from 15 cm. So, convert l m = 100 cm
∴ 10 m 50 cm

(2) 29 kg 880 gm – 8 kg 900 gm
Solution:

kg	gm
2 8	1 8 8 0
2 9 – 8	8 8 0 9 0 0
2 0	9 8 0

We cannot subtract 900 gin from 880 gm. So, convert 1 kg = 1000 gm
∴ 20 kg 980 gm

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 49 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 49

Question 1.
How much wire will be needed to make a rectangle 7 cm long and 4 cm wide?
Solution:
Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2 x 7 + 2 x 4
= 14 + 8
= 22 cm

∴ 22 cm wire will be needed to make a rectangle

Question 2.
If the length of a rectangle is 20 m and its width is 12m, what is its perimeter?
Solution:
Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2×20 + 2×12
= 40 + 24
= 64 m
∴ Perimeter is 64 m

Question 3.
Each side of a square is 9 m long. Find its perimeter.
Solution:
Perimeter of a square
= 4 x length of one side
= 4 x 9
= 36 m
∴ Perimeter is 36 m

Question 4.
If we take 4 rounds around a field that is 160 m long and 90 m wide, what is the distance we walk in kilometres?
Solution:
Perimeter of a rectangular field
= 2 x length + 2 x breadth
= 2 x 160 + 2 x 90
= 320 + 180
= 500 m

In one round distance walked is 500 m, hence, distance walked in 4 rounds
= 500 x 4
= 2000 m
= 2km
∴ The distance walked in 4 rounds is 2 km

Question 5.
Sanju completes 12 rounds around a square park every day. If one side of the park is 120 m, find out in kilometres and metres the distance that Sanju covers daily.
Solution:
Perimeter of a square
= 4 x length of one side
= 4 x 120
= 480 m

So, in one round the distance can be covered is 480 m, hence in 12 rounds the distance can be covered is
= 480 x 12
= 5760 m
= 5000 m + 760 m

∴ Sanju covers 5 km 760 m daily

Question 6.
The length of a rectangular plot of land is 50 m and its width is 30 m. A triple fence has to be put along its edges. If the wire costs 60 rupees permetre, what will be the total cost of the wire needed for the fence?
Solution:
Perimeter of a rectangular plot
= 2 x length + 2 x breadth
= 2 x 50 + 2 x 30
= 100 + 60 – 160 m
For a triple fence, wire needed
= 3 x 160 = 480 m

Cost of the wire needed
= wire needed x rate
= 480 x 60
= 28800 rupees
∴ The total cost of the wire needed for the fence is ₹ 28,800

Question 7.
A game requires its players to run around a square playground. Each side of the playground is 20 m long. One player took 5 rounds around the playground. How many metres did he run altogether?
Solution:
Perimeter of a square
= 4 x length of one side
= 4 x 20
= 80 m

In one round 80 m.
So in 5 round
= 80 x 5
= 400
= 400 m

∴ He runs altogether = 400 m

Question 8.
Four rounds of wire fence have to be put around a field. If the field is 60 m long and 40 m wide, how much wire will be needed?
Solution:
Perimeter of rectangular field
= 2 x length + 2 x breadth
= 2 x 60 + 2 x 40
= 120 + 80
= 200 m
Hence, wire required for 4 rounds
= 200 x 4
= 800 m

∴ Wire required for 4 rounds
= 800 m

Question 9.
The sides of a triangle are 24.7cm, 20.4 cm and 10.5 cm respectively. What is the perimeter of the triangle?
Solution:
Perimeter of triangle
= 24.7 + 20.4 + 10.5
= 55.6

∴ The perimeter of a triangle
= 55.6 cm

Question 10.
Look at the figures on the sheet of graph paper. Measure their sides with the help of the lines on the graph paper. Write the perimeter of each in the right box.

(1) Perimeter of
rectangle ABCD
= [ ] cm
(2) Perimeter of
rectangle EFGH
= [ ] cm
(3) Perimeter of
square PQRS
= [ ] cm
(4) Perimeter of
rectangle STUV
= [ ] cm
Solution:

(1) Perimeter of a rectangle ABCD
= 2 x length + 2 x breadth
= 2 x 3.5 + 2 x 2.5
= 7 + 5
= 12 cm

∴ 12 cm

(2) Perimeter of a rectangle EFGH
= 2 x length + 2 x breadth
= 2 x 3.8 + 2 x 1.3
= 7.6 + 2.6
= 10.2 cm

∴ 10.2 cm

(3) Perimeter of a rectangle PQRS
= 2 x length + 2 x breadth
= 2 x 2.4 + 2 x 2.4
= 4.8+ 4.8
= 9.6 cm

∴ 9.6 cm

(4) Perimeter of a rectangle STUV
= 2 x length + 2 x breadth
= 2 x 3 + 2 x 2
= 6 + 4
= 10 cm

∴ 10 cm

(5) Perimeter of a triangle LMN
= 1.5 + 2.5 + 2
= 6 cm

∴ 6 cm

Area : Revision

Of the figures given above, figure ABCD has six squares of 1 cm each inside it. It means that its area is 6 sq cm.

In the same way, count the squares in each figure and write its area.
(1) Area of MNRS = [ ] sq cm
(2) Area of EFGH = [ ] sq cm
(3) Area of PQRS = [ ] sq cm
(4) Area of IJKL = [ ] sq cm

Atul : Sir, why is the unit for area written as sq cm? We measure the sides in centimetres.

Teacher : Centimetre is a standard unit of length. In order to measure area, we need a standard unit of area. For this, a square with a side 1 cm is taken as the standard unit. The area of this square is 1 square centimetre. That is why this unit is written as sq cm, in short.

To measure large areas like fields, parks and playgrounds, a square with side 1 m, that is, an area of 1 sq m, is taken as the standard unit.

To measure the areas oftalukas or districts, a square with side 1km, or 1sq km is the standard unit used.

Formula for the area of a rectangle

(1) In the rectangle ABCD given alongside, 1 cm divisions were marked off on each side. The points on opposite sides were joined as shown in the figure. The length of the sides of each square thus created is 1cm. Therefore, the area of each square is 1 sq cm.

In ABCD, 3 rows with 5 squares each have been created.
The number of squares in rectangle ABCD is 3 × 5 = 15.
Therefore, the area of rectangle ABCD is 15 sq cm.
Here, the length of the figure is 5 cm and its breadth is 3 cm.
Note that the product of 3 and 5 is 15.

(2) In the rectangle with sides 4 cm and 2 cm, make squares of 1 sq cm each as shown above. Count the number of squares.

Note that here too, the number of squares formed are the same as the product of the length and width of the rectangle.

Therefore, The area of a rectangle = length × breadth

Formula for the area of a square

(1) Look at the square given alongside. The side of the square is 3 cm long. 9 squares of 1 cm each are formed within this square.

Therefore, the area of this square is 9 sq cm.

Here, there are 3 rows with 3 squares each, i.e., there are 3 × 3 = 9 squares.
The length of each side of the square is 3 cm.
The product of two sides of the square is 3 × 3 = 9.

(2) Measure the area of a square with side 5 cm, in the same way.
The answer will be 25 sq cm.
Note that 5 × 5 = 25

Therefore, The area of a square = length of a side × length of a side

It is not necessary to divide a square or rectangle into small squares every time you calculate their area. The advantage of a formula is that you can calculate the area simply by substituting the appropriate values.

Word problems
Example (1) What is the area of a rectangle of length 20 cm and width 15 cm?
Area of a rectangle = length × breadth
= 20 × 15 = 300.
Therefore, the area of the rectangle is 300 sq cm.

Example (2) A wall that is 4 m long and 3 m wide has to be painted. If the labour charges are ₹ 25 per sq m, what is the cost of labour for painting this wall?

First let us calculate the area of the wall to be painted.
Area of the wall = length of the wall × breadth of the wall = 4 × 3 = 12
Thus, the area of the wall is 12 sq m.
Labour cost of 1 sq m is 25 rupees.
So the labour cost for 12 sq m will be = 12 × 25 = 300
The cost of labour for painting the wall will be 300 rupees.

Example (3) What will be the area of a square with sides 15 cm?
Area of a square = length of side × length of side
= 15 × 15 = 225
The area of the square is 225 sq cm.

Example (4) One side of a square room is 4 m. If the cost of labour for laying 1 sq m of the floor is 35 rupees, what will be the total cost of labour?
First we must find the area of the square room.
Area of the square room = length of side × length of side = 4 × 4 = 16
Therefore, the area of the square room is 16 sq m.
The labour cost of laying 1 sq m of flooring is 35 rupees.
Therefore, the cost of laying 16 sq m of flooring is 16 × 35 = 560 rupees.

Perimeter and Area Problem Set 49 Additional Important Questions and Answers

Question 1.
Devendra walks five rounds of a square garden everyday. If the side of the garden is 150 m, how many kilometres does Devendra walk every morning?
Solution:
Perimeter of a square garden
= 4 x one side of the garden
= 4 x 150
= 600 m

In 5 rounds walking
= 5 x 600
= 3000 m
= 3 km
3 km

Question 2.
The length of a rectangular play ground is 75 m and its breadth is 50 m. Rupali walks four rounds. How many kilometres did she walk?
Solution:
Perimeter of rectangle
= 2 x length + 2 x breadth
= 2 x 75 + 2 x 50
= 150 + 100
= 250 m

In 4 rounds walking
= 4 x 250
= 1000 m
= 1 km

∴ 1 km

Question 3.
Length of the rectangle is 10 cm and its breadth is 8 cm and one square is side 9 cm. Whose perimetre is more? By how much?
Solution:
Perimeter of a rectangle
= 2 x length + 2 x breadth
= 2 x 10 + 2 x 8
= 20 + 16
= 36 cm ……………….. (i)

Perimeter of a square
= 4 x length of side
= 4 x 9
36 cm ……………….. (ii)
From (i) and (ii) perimeter of both is equal.

∴ perimeter of both is equal