“Fields and Fences” is Chapter 13 from Merry Math IV for students of Class 4th of JKBOSE. This post is about Fields and Fences Class 4 Math JKBOSE Solutions. You read about How Heavy How Light Class 4 Math JKBOSE Solutions in a previous post. Let’s get started:
Fields and Fences Class 4 Math JKBOSE Solutions
Page No. 183-184 Solutions
Gulam Nabi is a farmer. He needs a fence around his field. So, he bought a roll of 70 m wire for the fence. The boundary of his field is 54 metres long.
How much wire did Gulam give to Satish Ji?
Sol. Total wire = 70 metres
The wire used by Gulam Nabi = 54 metres
Wire given to Satish Ji = 70 m – 54 m = 16 m
Page No. 185 Solutions
Satish Ji thanked Gulam Nabi and started fencing his own field. But he needed to get more wire.
How long is the boundary of Satish Ji’s field?
Sol. The boundary of Satish Ji’s field = 18 m + 9 m + 15 m + 15 m + 9 m = 66 metres.
How much more wire will Satish Ji need for his field?
Sol. The total boundary of Satish Ji’s field = 66 metres
Wire given by Gulam Nabi = 16 metres
Therefore, the length of wire needed by Satish Ji = (66 – 16) m = 50 metres
Page No. 186 Solutions
Let’s Us Try These
- Here are pictures of some more fields. Find out which one has the longest boundary.
Sol. Boundary= 24 m + 15 m + 15 m + 6 m = 60 metres
Sol. Boundary = 12 m + 6 m + 6m+ 3 m + 6 m + 9 m = 42 metres
Sol. Boundary = 15 m + 12 m + 9 m = 36 metres
Sol. Boundary 9 m + 15 m + 15 m + 9 m + 15 m + 15 m = 78 metres
Ans. D field has the longest boundary.
Page No. 187 Solutions
- Nandu’s father is called the young old man in his village. At 70 years of age, he is fully fit. Do you know his secret?
He goes for a walk around the field every morning. Every day he takes four rounds of Nandu’s field.
What is the total distance he covers?
4 × ______ = ______m = ______km
Sol. Total boundary of Nandu’s field = 100 m + 150 m + 100 m + 150 m = 500 m
The total distance covered by Nandu’s father
4 × 500 m = 2000 m = 2 km (1000 m = 1 km)
Satish Ji’s wife works in a tailor’s shop. She has to fix Lace, around a tablecloth. She bought a 100-metre roll of lace.
Look at the picture of the tablecloth and tell how much lace is used for one table cloth.
Sol. Boundary of the table cloth = 1 m 50 cm + 1 m 50 cm + 50 cm + 50 cm = 4.00 m Ans.
So, 4 metre lace is used for one table cloth.
How much lace will be used in 3 such tablecloths?
Sol. Lace used for one tablecloth = 4 m
Lace used for 3 table cloths = (4 × 3) m = 12 m Ans.
How much lace will be left in the roll?
Sol. Total lace = 100 m
Lace used for 3 table cloths = 12 m
Lace left in the roll = Total lace – Used lace = 100 m – 12 m = 88 m Ans.
Page No. 188 Solutions
Activity
- Find out the length of the boundary of these shapes.
(Hint: You can use a thread.)
Now count the square to find out:
How many squares are there in each shape?
Ans. There are 4 squares in shape ‘A’, 6 squares in shape ‘B’, 5 squares in shape ‘C’ and 4 squares in shape ‘D’.
Which shape covers the least number of squares?
Ans. Shapes ‘A’ and ‘D’ cover the least number of squares.
Which shape covers the most number of squares?
Ans. Shape ‘B’ covers the most number of squares.
- Take a 20-centimetre-long thread. Make different shapes by joining the ends.
Find out
How many squares are there in each shape?
Ans. There are 24 squares in shape (i), 25 squares in shapes (ii) 16 squares in shape (iii) and 21 squares in shape (iv)
Which is the biggest shape?
Ans. Shape (ii) is the biggest shape.
Which is the smallest shape?
Ans. Shape (iii) is the smallest shape.
How long is the boundary of each shape?
Ans. The boundary of each shape is 20 centimetres lòng.
Page No. 189 Solutions
- How many different shapes can you make by joining two squares? Draw them on the squared sheet given below. How long is the boundary of each shape?
Sol.
Try this activity with three squares also.
Sol.
Page No. 190 – 191 Solutions
- A square has a boundary of 12 cm.
(a) From the corner of this square, a small square of side 1 cm is cut off. Will the boundary of B be less or more? Find its length.
Sol. After cutting 1 cm square,
The length of the boundary of B = 3 cm + 2 cm + 1 cm + 1 cm + 2 cm + 3 cm = 12 cm
Since the boundary of ‘A’ is also 12 cm.
So, the boundary of ‘B’ is neither less nor more than that of ‘A’.
(b) If you cut a 1 cm square to get shape C, what will be the length of the boundary of C?
Sol. The length of the boundary of ‘C’ = 3 cm + 3 cm + 3 m + 1 cm + 1 cm + 1 m + 1 cm + 1 cm = 14 cm.
- (a) Find the length of the boundary of square D.
Sol. The length of the boundary of ‘D’ = 5 cm + 5 cm + 5 cm + 5 cm = 20 cm
(b) 8 squares of side 1 cm is cut out of square D. Now it looks like shape E. What is the length of the boundary of shape E?
Sol. The length of the boundary of ‘E’ = 36 cm
There are 36 sides of 1 cm
So, 36 × 1 cm= 36 cm
(c) The boundary of this square of 1 cm is 1 cm + 1 cm + 1 cm + 1 cm = 4 cm. Can we also say that the boundary is 4 x 1 cm?
Ans. Yes, we can also say that the boundary is 4 x 1 cm.
- A hockey field is 91 metres 40 cm long and 55 metres wide. How long is the boundary of the field?
Sol. Length of a hockey field = 91 m 40 cm
Width of a hockey field = 55 m
The length of the boundary of the hockey field = (91 m 40 cm + 55 m + 91 m 40 cm + 55 m) = 292 m 80 cm Ans.
- Nisha and Salma are running a race. Nisha is running on the inner circle. Salma is running on the outer circle.
Salma runs faster than Nisha. But still, she loses the race. Can you guess why?
Ans. Because. the inner boundary of the circle is smaller than the outer boundary. Salma is running on the outer circle as such she has to run for more distance. Therefore, she loses the race.
- Have you seen any race where runners start from different places – like in this picture? Guess why?
Ans. Yes, I have seen the race, which is organized on a circular track. It is made in such a manner to make the distances equal.
Page No. 192 Solutions
School Garden
The students of Class III and IV thought of making a vegetable garden. They chose a place which looked like this.
Students of both classes thought that Garden I was bigger so both wanted to take Garden II. Suddenly Geetika said, “I think both gardens are equally big.”
How will Geetika find out if the two gardens are equally big?
Ans. Geetika will find the length and breadth of both the gardens i.e. Garden I and Garden II. Then she will find the area of both gardens by using the formula A = L × B. By comparing areas, she can know which garden is bigger and which is smaller.
Page No. 193 Solutions
Activity
- Look at the table in your classroom. Guess how many Merry-Math books you can place on it.
(Remember — The books should not overlap. Do not leave gaps between the books.)
Write your guess here. __________
Ans. 8 Merry-Math books can be placed on the table.
Now check if your guess was right. How many books could you place? __________
Ans. 8 Merry-Math books can be placed on the table.
What is the difference between your guess and the actual number of books? __________
Ans. The difference between your guess and the actual number of books is 0.
- Now look for another table.
a) Is this table bigger than the last table? Yes/No
Ans. Yes. This table is bigger than the last table.
b) Make a guess on how many Merry-Math books can be kept on this table. ____________
Ans. 12 Merry-Math books can be kept on this table.
c) Check if your guess was correct. How many Merry-Math books could you keep? __________
Ans. 10 Merry-Math books could be kept on the table.
d) The difference between the sizes of the two tables is _________ books.
Ans. The difference between the sizes of the two tables is 2 books.
- a) How many Merry-Math books can be covered with one sheet of newspaper?
Ans. 4 Merry-Math books can be covered with one sheet of newspaper.
b) Try covering your Merry-Math book with half a sheet of newspaper.
Ans. Try it yourself.
c) Can you cover your book with a smaller sheet?
Ans. Yes, we can cover the book with a smaller sheet.
d) Find the smallest sheet which can cover your book. Check if your friend used a smaller sheet than you did.
Ans. Try it yourself
Page No. 194 Solutions
- a) Collect some leaves from the garden. Place each of them here on this squared sheet. Trace out their edges and check how many squares there are in each leaf.
(b) Which is the biggest leaf in this picture?
Ans. ‘b’ is the biggest leaf in this piece and should have the same number picture.
(c) Which is the smallest leaf?
Ans. ‘c’ is the smallest leaf in this picture.
Page No. 195 Solutions
- (a) How many small squares of size 1 cm are there in this big green square?
Sol. There are 36 small squares of size 1 cm in this big green square
In each line number of squares = 6
In 6 lines the number of squares = 6 × 6 = 36
(b) Can you think of a faster way to know the total number of small squares without counting each?
Ans. Just find 6 x 6 = 36.
- Guess how many squares of one centimetre can fill this blue rectangle.
Ans. There are 32 squares of one centimetre in this blue rectangle.
- Look at the picture. Can you divide it into 4 equal pieces? Each piece should have the same number of squares.
Ans. Yes, we can divide it into 4 equal pieces as shown:
Page No. 196 Solutions
Puzzle: A House and the Well
Rehman has a piece of land.
There are 4 houses on his land and in the middle, there is a well. He wants to divide this land equally among his four children. Each should get one house and be able to use the well without entering the other’s land.
Can you help him to divide the land? Give different colours to each one’s share.
Ans.
Perimeter of a Figure
Page No. 198 Solutions
Now Let’s Try These
- Find the perimeter of each of the following figures:
Sol. (A) Perimeter of figure = 6 cm + 6 cm + 6 cm + 6 cm = 24 cm
Ans. (B) Perimeter of figure = 12 cm + 8 cm + 12 cm + 8 cm = 40 cm Ans.
- Find the perimeter of each following figure. Express the perimeter so obtained in metres and centimetres.
Sol. (A) Perimeter of the figure 70 cm + 40 cm + 70 cm + 40 cm = 220 cm = 220/100 m (100 cm = 1 m) = 2 m 20 cm Ans.
(B) Perimeter of the figure= 40 cm + 40 cm + 40 cm + 40 cm = 160 cm = 160/100 m= 1 m 60 cm Ans.
(C) Perimeter of the figure = 50 cm + 60 cm + 50 cm = 160 cm = 160/100 m = 1 m 60 cm Ans.
(D) Perimeter of the figure 30 cm + 50 cm + 30 cm + 50 cm = 160 cm = 160/100 m = 1 m 60 cm Ans.
Page No. 199 Solutions
- Find the perimeters of the following figures. Express the perimeter so obtained in metres and decimetres.
Sol. (A) Perimeter = 6 dm + 6 dm + 5 dm = 17 dm = 17/10 m = 1 m 7 dm. Ans
(B) Perimeter = 8 dm + 8 dm + 8 dm + 8 dm = 32 dm = 32/10 m = 3 m 2 dm. Ans.
(C) Perimeter = 18 dm + 5 dm + 18 dm + 5 dm = 46 dm = 46/10 m = 4 m 6 dm. Ans.
(D) Perimeter = 6 dm + 5 dm + 4 dm + 10 dm = 25 dm = 25/10 m = 2m 5 dm Ans.
- Find the perimeter of the following figures.
Express the perimeter so obtained in kilometres and metres.
Sol. (A) Perimeter = 500 m + 400 m + 1200 m + 400 m = 1500 m = 1500/1000 km = 1 km 500 m Ans.
(B) Perimeter = 700 m + 400 m + 700 m + 400 m = 2200 m = 2200/1000 km = 2 km 200 m Ans.
(C) Perimeter of figure = 600 m + 600 m + 600 m = 1800 m = 1800/1000 km = 1 km 800 m Ans.
(D) Perimeter of figure = 200 m + 200 m + 200 m + 200 m + 200 m + 200 m = 1200 m = 1200/1000 km = 1 km 200 m Ans.
Page No. 200 Solutions
- Measure the sides of these shapes. Work out the perimeter of each shape.
Sol. (A) The perimeter of the shape = 5 cm + 3 cm + 5 cm + 3 cm = 16 cm Ans.
(B) Perimeter of the shape = 4.5 cm + 3 cm + 4.5 cm + 3 cm = 15 cm Ans.
(C) Perimeter of the shape = 7.5 cm + 3.5 cm + 4.5 cm + 3.5 cm = 19 cm Ans.
That’s it about Fields and Fences Class 4 Math JKBOSE Solutions. Hope you found it useful. Do share your views about this post in the comment section below.
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