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

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.
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 1
Answer:
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 10
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 18

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

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:
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 12
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 13

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

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.
    Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 2
  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.
    Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 3

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

Activity :
Draw the nets shown below on card sheet. Cut out the shapes and find out the shapes of the boxes they form.
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 4

A five-square net (Pentomino)

In the figure alongside, five squares of the same size are placed together with their sides joined.
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 5
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.

Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 6

Activity :
Some five-square nets are given below. Draw these nets on a card sheet. Make open boxes from these nets.
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 7
Try to find out other five-square nets that can be used to make open boxes.

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

A riddle
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 8
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?
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 9

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:
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 16

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

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:
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 17

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

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:
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 14
Maharashtra Board Class 5 Maths Solutions Chapter 13 Three Dimensional Objects and Nets Problem Set 51 15

Maharashtra Board Class 5 Marathi Solutions Chapter 15 नदीचे गाणे

Balbharti Maharashtra State Board Class 5 Marathi Solutions Sulabhbharati Chapter 15 नदीचे गाणे Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Marathi Sulabhbharati Solutions Chapter 15 नदीचे गाणे

5th Standard Marathi Digest Chapter 15 नदीचे गाणे Textbook Questions and Answers

1. एका वाक्यात उत्तरे लिहा.

प्रश्न 1.
मंजुळ गाणे कोण गाते?
उत्तर:
मंजूळ गाणे नदी गाते.

Maharashtra Board Class 5 Marathi Solutions Chapter 15 नदीचे गाणे

प्रश्न 2.
गावे कोठे वसली आहेत?
उत्तर:
गावे नदीच्या तीरावर वसली आहेत.

प्रश्न 3.
नदीवर शीतल छाया कोण धरते?
उत्तर:
नदीवर शीतल छाया आंब्याची झाडे धरतात.

प्रश्न 4.
नदी जेथे जाईल तेथे काय करेल?
उत्तर:
नदी जेथे जाईल तेथे मनोहर आनंदाची बाग फुलवेल.

2. कवितेच्या ओळी पूर्ण करा.

प्रश्न 1.
फुलवेली मज …………………………. देती,
कुठे …………………… खेळत बसती,
कुठे ……………………… माझ्यावरती
……………………. अपुली छाया धरती.
उत्तरः
फुलवेली मज सुमने देती,
कुठे लव्हाळी खेळत बसती,
कुठे आम्रतरू माझ्यावरती,
शीतल अपुली छाया धरती.

Maharashtra Board Class 5 Marathi Solutions Chapter 15 नदीचे गाणे

3. जोड्या जुळवा.

प्रश्न 1.
जोड्या जुळवा.

अ ‘गट’ब ‘गट’
1. झुळझुळ(अ) गाणे
2. मंजूळ(आ) छाया
3. शीतल(इ) पाणी

उत्तरः

अ ‘गट’ब ‘गट’
1. झुळझुळ(इ) पाणी
2. मंजूळ(अ) गाणे
3. शीतल(आ) छाया

4. हे शब्द असेच लिहा.

प्रश्न 1.
हे शब्द असेच लिहा.
उत्तर:

  1. मंजूळ – मधुर
  2. शीतल – थंड
  3. लव्हाळी – लव्हाळं
  4. लतावृक्ष – आंब्याचे झाड.
  5. लव्हाळी – पहिल्या पावसानंतर लतावृक्ष बहरून गेले.
  6. आम्रतरू – पाण्यात किंवा पाण्याजवळ वाढणारी एक वनस्पती

5. कवितेच्या चित्राचे निरीक्षण करा व पाच वाक्ये लिहा.

प्रश्न 1.
कवितेच्या चित्राचे निरीक्षण करा व पाच वाक्ये लिहा.
उत्तर:

  1. नदीच्या तीरावरती गावे वसली आहेत.
  2. नदीकाठी झाडे, वेली दिसत आहेत.
  3. गुरे-वासरे नदीचे पाणी पित आहेत.
  4. मुले लाटांवरती खेळ खेळत आहेत.
  5. बायका नदीचे पाणी मडक्यात भरून नेत आहे.

Marathi Sulabhbharati Class 5 Solutions Chapter 15 नदीचे गाणे Additional Important Questions and Answers

प्रश्न 1.
नदीला सुमने कोण देतात?
उत्तर:
नदीला सुमने फुलवेली देतात.

Maharashtra Board Class 5 Marathi Solutions Chapter 15 नदीचे गाणे

प्रश्न 2.
नदीत कोण खेळत आहेत?
उत्तर:
नदीत लव्हाळी खेळत आहेत.

प्रश्न 3.
घटात काय भरतात?
उत्तर:
घटात पाणी भरतात.

प्रश्न 4.
नदीच्या काठावर पाणी पिण्यासाठी कोण येतात?
उत्तर:
नदीच्या काठावर पाणी पिण्यासाठी गुरे-वासरे येतात.

प्रश्न 5.
मुले कुठे खेळतात?
उत्तर:
मुले लाटांवर खेळतात.

प्रश्न 6.
‘नदीचे गाणे’ कवितेचे कवी कोण आहेत?
उत्तर:
‘नदीचे गाणे’ कवितेचे कवी ‘वि. म. कुलकर्णी आहेत.

प्रश्न 7.
नदी कोणाची आहे?
उत्तर:
नदी सर्वांची आहे.

Maharashtra Board Class 5 Marathi Solutions Chapter 15 नदीचे गाणे

कवितेच्या ओळी पूर्ण करा.

प्रश्न 1.
पाणी पिऊनी ………………………….. जाती,
……………….. भरुनी कोणी ………………….. नेती,
…………………….. जवळी येती,
मुले खेळती ……………………….
उत्तरः
पाणी पिउनी पक्षी जाती,
घट भरुनी कोणी जल नेती,
गुरे-वासरे जवळी येती,
मुले खेळती लाटांवरती.

थोडक्यात उत्तरे लिहा.

प्रश्न 1.
कवितेत नदीचे वर्णन कसे केले आहे?
उत्तरः
नदी ही दरी, वनातून वाहते. ती झुळझुळ वाहते. तिच्या तीरावर अनेक गावे वसली आहेत. अनेक वृक्षवेली नदीच्या काठावर आहेत. आंब्याची झाडे नदीवर सावली धरतात. अनेक पक्षी आपली तहान भागवतात. कुणी नदीवर पाणी भरण्यासाठी येतात. गुरे-वासरे नदीवर येतात. मुले तिच्या लाटांवर खेळतात. नदी ही सर्वांची आहे. नदी जिथे जाईल तेथे मनोहर आनंदाची बाग फुलवते. अशाप्रकारे कवितेते नदीचे वर्णन केले आहे.

व्याकरण व भाषाभ्यास:

प्रश्न 1.
समानार्थी शब्द लिहा.

  1. गाव
  2. तरू
  3. छाया
  4. धती
  5. लाट
  6. आनंद
  7. वन
  8. लता
  9. आम्र
  10. बाग
  11. गुरे
  12. नदी
  13. मनोहर
  14. पक्षी
  15. जल
  16. घट
  17. सुमन

उत्तर:

  1. ग्राम
  2. झाड
  3. सावली
  4. धरणी
  5. तरंग
  6. हर्ष
  7. रान
  8. वेली
  9. आंबा
  10. उदयान
  11. जनावरे
  12. सरिता
  13. सुंदर
  14. खग
  15. पाणी
  16. माठ, मडके
  17. फूल

Maharashtra Board Class 5 Marathi Solutions Chapter 15 नदीचे गाणे

प्रश्न 2.
विरुद्धार्थी शब्द लिहा.

  1. मंजूळ
  2. पुढे
  3. शीतल
  4. जाणे
  5. जवळ
  6. मला
  7. बसणे
  8. छाया
  9. जाईन
  10. आनंद

उत्तरः

  1. कर्कश
  2. मागे
  3. उष्ण
  4. येणे
  5. दूर
  6. तुला
  7. उठणे
  8. ऊन/सूर्यप्रकाश
  9. येईन
  10. दु:ख

प्रश्न 3.
वचन बदला.

  1. दरी
  2. वन
  3. गाणे
  4. गाव
  5. काठ
  6. फुले
  7. लाटा
  8. बाग
  9. सुमने
  10. वासरू
  11. मुले
  12. फुलवेली

उत्तर:

  1. दऱ्या
  2. वने
  3. गाणी
  4. गावे
  5. काठ
  6. फूल
  7. लाट
  8. बागा
  9. सुमन
  10. वासरे
  11. मूल
  12. फुलवेल

प्रश्न 4.
शब्दाचे अर्थ लिहा.
उत्तर:
घट – मातीचा घडा (माठ)

Maharashtra Board Class 5 Marathi Solutions Chapter 15 नदीचे गाणे

प्रश्न 5.
पुढील शब्दांचा उपयोग करून अर्थपूर्ण वाक्य बनवा.
जसे – झुळझुळ. नदी झुळझुळ वाहते.
उत्तर:
1. शीतल – झाडे शीतल छाया देतात.
2. मनोहर – निसर्गाच्या मनोहर दृश्याने सारेच मंत्रमुग्ध झाले.
3. मंजूळ – रमाने सर्वांसमोर मंजुळ गाणे म्हटले.

नदीचे गाणे Summary in Marathi

पदयपरिचय:

या कवितेत कवी वि. म. कुलकर्णी यांनी नदीचे मनोगत व नदीकाठच्या जीवनाचे सुंदर वर्णन केले आहे.

शब्दार्थ:

  1. दरी – दोन टेकड्यांमधील खोलगट भाग – (a valley)
  2. वन – जंगल, अरण्य (forest)
  3. झुळझुळ – मंदपणे, हळुवारपणे (softly)
  4. मंजुळ – मधुर, सुरेल (a sweet, melodious)
  5. वसणे – राहणे (to stay)
  6. तीर – काठ (shore)
  7. लता – वेल (creeper)
  8. वृक्ष – झाड (a tree)
  9. भूमी – जमिन (land)
  10. सुमने – चांगले, पवित्र मन (clean mind)
  11. लव्हाळी – पाण्याजवळ वाढणारी एक वनस्पती (rush like grass)
  12. आम्रतरू – आंब्याचे झाड (a mango tree)
  13. शीतल – गार (cool)
  14. छाया – सावली (shadow)
  15. घट – घडा, घागर (a vessel for holding water)
  16. गुरे – गाय, बैल इ. जनावरे (cattle)
  17. वासरू – गाईचे पारडू (a calf)
  18. लाटा – लहरी (waves)
  19. मनोहर – आकर्षक (attractive)

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

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 41 0

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

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

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

cmmm
1
6
+ 7
5
9
1 44

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:

mcm
1
2 2
+ 2 5
5 0
7 5
4 82 5

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

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

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

kmm
1
1 5
+ 13
7 4 0
9 5 0
2 96 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:

kggm
1
2 5
+ 29
6 5 0
7 7 0
5 54 2 0

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

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

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

lml
1 1
1 9
+ 2 5
8 4 0
2 5 0
4 50 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 81 5 0
1 9
– 1 2
5 0
6 0
69 0

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

∴ ₹ 6, 90 paise

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

(2) 24 cm 2 mm – 3 cm 8 mm
Solution:

cmmm
2 31 2
2 4
– 3
2
8
2 04

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:

mcm
1 91 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

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

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

kmm
3 912 2 5
4 0
-1 7
2 2 5
9 6 0
2 22 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:

kggm
3 41 1 5 0
3 5
– 2 6
1 5 0
4 7 0
86 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:

lml
4 51 2 0 0
4 6
– 3 8
2 0 0
7 5 0
74 5 0

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

∴ 7 l 450 ml

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

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?
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46 1

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?
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46 2

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

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

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46 3

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.
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 46 4

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

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.

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

Problems on Measurement Problem Set 46 Additional Important Questions and Answers

Add the following:

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

kmm
1
1 2
+ 7
8 8 O
6 2 0
2 05 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 00 0

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

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

Subtract the following:

(1) 15 m 15 cm – 4 m 65 cm
Solution:

kggm
1 41 1 5
1 5
– 4
1 5
6 5
1 05 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:

kggm
2 81 8 8 0
2 9
– 8
8 8 0
9 0 0
2 09 8 0

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

Maharashtra Board Class 5 Marathi Solutions Chapter 10 बैलपोळा

Balbharti Maharashtra State Board Class 5 Marathi Solutions Sulabhbharati Chapter 10 बैलपोळा Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Marathi Sulabhbharati Solutions Chapter 10 बैलपोळा

5th Standard Marathi Digest Chapter 10 बैलपोळा Textbook Questions and Answers

1. एका वाक्यात उत्तरे लिहा.

प्रश्न (अ)
बैलांच्या सणाला काय म्हणतात?
उत्तर:
बैलांच्या सणाला पोळा म्हणतात.

Maharashtra Board Class 5 Marathi Solutions Chapter 10 बैलपोळा

प्रश्न (आ)
बैलांना कसे सजवले आहे?
उत्तर:
बैलांना पाठीवर मखमली झुली घातल्या आहेत. त्यांची शिंगे रंगवली आहेत व त्यांच्या कपाळी रेशमी बाशिंगे बांधली आहेत.

प्रश्न (इ)
बैलांची नावे कोणती आहेत?
उत्तर:
बैलांची नावे ढवळ्या-पवळ्या आहेत.

प्रश्न (ई)
बैलपोळ्याच्या दिवशी बैलांना काय खाऊ घालतात?
उत्तर:
बैलपोळ्याच्या दिवशी बैलांना पुरणपोळी खाऊ घालतात.

2. शेवट समान असणारे कवितेतील शब्द लिहा.

प्रश्न 1.
शेवट समान असणारे कवितेतील शब्द लिहा.
मखमली
उत्तर:
मखमली – पाऊस पडल्यावर धरणीने हिरवागार मखमली शालू घातल्यासारखे भासते.

Maharashtra Board Class 5 Marathi Solutions Chapter 10 बैलपोळा

3. बैल गोठ्यात राहतात, तसे खालील प्राणी कोठे राहतात ते लिहा.

प्रश्न 1.
बैल गोठ्यात राहतात, तसे खालील प्राणी कोठे राहतात ते लिहा.
(अ) घोडा
(आ) वाघ
(इ) माकडे
(ई) हत्ती
(उ) मासा
(ऊ) कासव
उत्तर:
(अ) तबेला
(आ) गुहा
(इ) झाड
(ई) जंगल
(उ) पाणी
(ऊ) जमीन/पाणी

4. खालील शब्दांपासून अर्थपूर्ण वाक्य बनवा.

प्रश्न 1.
खालील शब्दांपासून अनेक शब्द बनवा.
उदा., कपाळी – कळी, पाळी, पाक, पाकळी.
(अ) पुरणपोळी
(आ) गावभर
उत्तर:
(अ) रण, पुळी, पोर, पोळी
(आ) रव, गार, वर, भर

5. ‘सजलेधजले’ अशा शब्दांना जोडशब्द म्हणतात. कवितेत आलेले खालील जोडशब्द वाचा.

प्रश्न 1.
‘सजलेधजले’ अशा शब्दांना जोडशब्द म्हणतात. कवितेत आलेले खालील जोडशब्द वाचा. जसेच्या तसे पाहून लिहा. असे आणखी काही जोडशब्द लिहा.
(अ) कामधाम
(आ) पुरणपोळी
उत्तर:
(अ) आराम
(आ) पुरणपोळी – आई छान पुरणपोळ्या बनवते.

Maharashtra Board Class 5 Marathi Solutions Chapter 10 बैलपोळा

6. ‘गावभर मिरवणे’ म्हणजे संपूर्ण गावातून फिरणे, खालील शब्दांना ‘भर’ हा शब्द जोडा आणि वाक्यात उपयोग करा.

प्रश्न 1.
‘गावभर मिरवणे’ म्हणजे संपूर्ण गावातून फिरणे. खालील शब्दांना ‘भर’ हा शब्द जोडा आणि वाक्यांत उपयोग करा. सांगा.
उदा., बाबांनी पोतंभर धान्य आणलं.
Maharashtra Board Class 5 Marathi Solutions Chapter 10 बैलपोळा 1
उत्तर:

1. तोंडभर – बाबांनी माझी तोंडभर स्तुती केली.
2. पिशवीभर – आईने पिशवीभर भाजी आणली.
3. हातभर – मी हातभर मेंदी काढून घेतली.
4. घरभर – हसण्याचे आवाज घरभर येत होते.

7. तुमच्या आवडत्या प्राण्याची माहिती पाच वाक्यांत लिहा.

प्रश्न 1.
तुमच्या आवडत्या प्राण्याची माहिती पाच वाक्यांत लिहा.
उत्तर:
माझा आवडता प्राणी सिंह आहे. त्याला जंगलाचा राजा म्हणजेच वनराज म्हणतात. त्याच्या मानेभोवती आयाळ असते. त्याच्या ओरडण्याला ‘गर्जना’ म्हणतात. सिंह शूर असतो.

8. खालील चौकटीत काही अक्षरे दिलेली आहेत त्या अक्षरांपासून काही फुलांची नावे तयार होतात ती शोधा व लिहा.

प्रश्न 1
खालील चौकटीत काही अक्षरे दिलेली आहेत त्या अक्षरांपासून काही फुलांची नावे तयार होतात ती शोधा व लिहा.
Maharashtra Board Class 5 Marathi Solutions Chapter 10 बैलपोळा 2
उत्तरः
फुलांची नावे:

  1. पारिजातक
  2. मोगरा
  3. जास्वंद
  4. शेवंती
  5. कमळ
  6. अबोली
  7. गुलछडी
  8. चाफा
  9. झेंडू
  10. सदाफुली
  11. गुलाब
  12. कमळ
  13. जुई
  14. जाई
  15. झेंडू
  16. चाफा

Marathi Sulabhbharati Class 5 Solutions Chapter 10 बैलपोळा Additional Important Questions and Answers

1. खालील प्रश्नांची उत्तरे एका वाक्यात लिहा.

प्रश्न 1.
गोठ्यातील बैलांना काय करायचे आहे?
उत्तर:
गोठ्यातील बैलांना न्हाऊ घालायचे आहे.

Maharashtra Board Class 5 Marathi Solutions Chapter 10 बैलपोळा

प्रश्न 2.
बैल कोठे मिरवू लागले?
उत्तर:
बैल गावभर मिरवू लागले.

प्रश्न 3.
पोळ्याच्या दिवशी बैलांची कोणती विशेष काळजी घेतली जाते?
उत्तर:
पोळ्याच्या दिवशी बैलांकडून काहीही काम करून घेत नाही. त्यांना आराम दिला जातो.

प्रश्न 4.
‘बैलपोळा’ या कवितेचे कवी कोण आहेत?
उत्तर:
‘बैलपोळा’ या कवितेचे कवी धोंडीरामसिंह राजपूत’ आहेत.

2. पुढील प्रश्नांची उत्तरे थोडक्यात लिहा.

प्रश्न 1.
पोळ्याच्या दिवशी बैलांना कसे सजवतात?
उत्तर:
पोळ्याच्या दिवशी सर्वप्रथम बैलांना स्नान घातले जाते. त्यांच्या अंगावर मखमली झुली घातल्या जातात. शिंगे रंगवून कपाळावर रेशमी बाशिंगे बांधली जातात. अशा प्रकारे पोळ्याच्या दिवशी बैलांना सजवतात.

Maharashtra Board Class 5 Marathi Solutions Chapter 10 बैलपोळा

प्रश्न 2.
पोळा सण का साजरा करतात?
उत्तर:
आपल्या संस्कृतीमध्ये प्राणी व पक्ष्यांविषयी कृतज्ञता व्यक्त करण्याची पद्धत आहे. पोळा सण हा त्याचाच एक भाग आहे. वर्षभर शेतात राबणाऱ्या या बैलाला एक दिवस स्वत:चा मिळावा व त्याच्याप्रती कृतज्ञता व्यक्त करता यावी, यासाठी श्रावण अमावस्येला पोळा सण साजरा करतात.

3. शेवट समान असणारे कवितेतील शब्द लिहा.

प्रश्न 1.
(अ) ढवळ्या
(ब) झुली
(क) सजलेधजले
उत्तर:
(अ) पवळ्या
(ब) बांधली
(क) लागले

प्रश्न 2.
‘सजलेधजले’ यासारखे जोडशब्द लिहा.
उत्तर:
(अ) कामधाम
(आ) पुरणपोळी
(इ) मित्रमैत्रिणी
(ई) घरदार
(उ) रामलक्ष्मण
(ऊ) धनुष्यबाण
(ए) राघूमैना
(ऐ) नोकरचाकर

व्याकरण व भाषाभ्यास:

प्रश्न 1.
समानार्थी शब्द लिहा.

  1. सण
  2. बैल
  3. कपाळ
  4. गाव
  5. दिन
  6. काम
  7. पोळा
  8. आराम

उत्तर:

  1. उत्सव
  2. वृषभ
  3. भाल
  4. खेडे
  5. दिवस
  6. कार्य
  7. बेंदूर
  8. विश्रांती

Maharashtra Board Class 5 Marathi Solutions Chapter 10 बैलपोळा

प्रश्न 2.
विरुद्धार्थी शब्द लिहा.

  1. आज
  2. रंगीत
  3. वर
  4. बघणे
  5. नाही
  6. काम

उत्तर:

  1. उदया
  2. रंगहीन
  3. खाली
  4. दाखवणे
  5. होय
  6. आराम

प्रश्न 3.
वचन बदला.

  1. सण
  2. गोठा
  3. झुली
  4. शिंगे
  5. गाव
  6. पुरणपोळी
  7. बाशिंगे

उत्तर:

  1. सण
  2. गोठे
  3. झूल
  4. शिंग
  5. गावे
  6. पुरणपोळ्या
  7. बाशिंग

प्रश्न 4.
लिंग बदला.

  1. बैल
  2. घोडा
  3. वाघ
  4. माकड
  5. हत्ती
  6. शेतकरी
  7. गाववाली
  8. पुरुष

उत्तरः

  1. गाय
  2. घोडी
  3. वाघीण
  4. माकडीण
  5. हत्तीण
  6. शेतकरीण
  7. गाववाला
  8. बाई, स्त्री

Maharashtra Board Class 5 Marathi Solutions Chapter 10 बैलपोळा

प्रश्न 5.
पुढील शब्दांपासून अर्थपूर्ण वाक्य बनवा.
1. रंगीत
2. सण
उत्तरः
1. रंगीत – रंगीत फुग्यांनी वाढदिवसाला घर सजवले गेले.
2. सण – वेगवेगळे सण जीवनात आनंद. आणतात.

बैलपोळा Summary in Marathi

पदयपरिचय:

बैलपोळा म्हणजे वर्षभर शेतकऱ्याबरोबर राबणाऱ्या बैलाचा सण. शेतकरी हा सण कशा प्रकारे साजरा करतात त्याचे वर्णन प्रस्तुत कवितेत कवींनी केले आहे.

Maharashtra Board Class 5 Marathi Solutions Chapter 10 बैलपोळा

शब्दार्थ:

  1. सण – उत्सव (festival)
  2. बैल – वृषभ (bullock)
  3. गोठा – गाय, बैल राहण्याची राहण्याची जागा (shed)
  4. मखमली – एक प्रकारचे अतिशय तलम मऊ वस्त्र (of velvet)
  5. झुली – घोडा, हत्ती, बैलांच्या पाठीवर घालायचे आच्छादन (a cloth of rich brocade to cover the body of animals such as horse, elephant, bullock)
  6. शिंगे – जनावरांच्या डोक्यावरील अवयव (horns)
  7. रेशमी – रेशमाचे (silken)
  8. बाशिंग – विवाह प्रसंगी नवरा, नवरीच्या कपाळावर बांधायचे एक आभूषण (an ornamental headgear usually of paper)
  9. धजणे – धाडस करणे (to have courage)
  10. गाव – ग्राम (a village)
  11. दिन – दिवस (a day)
  12. आराम – विश्रांती (rest)
  13. कपाळ – भाळ (forehead)

Maharashtra Board Class 5 Marathi Solutions Chapter 8 कोणापासून काय घ्यावे?

Balbharti Maharashtra State Board Class 5 Marathi Solutions Sulabhbharati Chapter 8 कोणापासून काय घ्यावे Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Marathi Sulabhbharati Solutions Chapter 8 कोणापासून काय घ्यावे?

5th Standard Marathi Digest Chapter 8 कोणापासून काय घ्यावे Textbook Questions and Answers

1. रिकाम्या जाग भरा.

प्रश्न 1.
रिकाम्या जाग भरा.
(अ) …………………………… ” घेऊ सावली.
………………………………. पासुन जगणे.
(आ) ……………………………… एक शिकूया.
…………………………………… जीव जगूया.
उत्तर:
(अ) झाडापासून, मातीपासून
(आ) प्रभातकाळी, जगवित

Maharashtra Board Class 5 Marathi Solutions Chapter 8 कोणापासून काय घ्यावे

2. कवितेतील कोणापासून काय घेता येईल ते लिहा.

प्रश्न 1.
कवितेतील कोणापासून काय घेता येईल ते लिहा.
(अ) सूर्य
(आ) चंद्र
(इ) तारा
(ई) फूल
(उ) कोकिळ
(ऊ) झरा
(ए) झाड
(ऐ) माती
उत्तर:
(अ) रंग
(आ) शांती
(इ) दिव्य कांती
(ई) गंध
(उ) गाणे
(ऊ) नवे तराणे
(ए) सावली
(ऐ) जगणे

3. कवितेत ‘चमचम’ शब्द आलेला आहे. यासारखे आणखी काही शब्द माहीत करून घ्या व लिहा.

प्रश्न 1.
कवितेत ‘चमचम’ शब्द आलेला आहे. यासारखे आणखी काही शब्द माहीत करून घ्या व लिहा.
उत्तर:

  1. घमघम
  2. कटकट
  3. लटपट
  4. छमछम
  5. सरसर
  6. फडफड
  7. कटकट
  8. भरभर
  9. लटपट
  10. झरझर.

Marathi Sulabhbharati Class 5 Solutions Chapter 8 कोणापासून काय घ्यावे Additional Important Questions and Answers

1. खालील प्रश्नांची एका वाक्यात उत्तरे लिहा.

प्रश्न 1.
निसर्गाचे विविध घटक कोणते?
उत्तरः
सूर्य, चंद्र, तारे, आकाश, वृक्ष, पक्षी, माती, पाऊस हे सर्व निसर्गाचे घटक आहेत.

Maharashtra Board Class 5 Marathi Solutions Chapter 8 कोणापासून काय घ्यावे

प्रश्न 2.
तुम्ही कधी निसर्गातील घटकांचा अनुभव घेतला आहे का? कशाप्रमाणे?
उत्तर:
हो. पौर्णिमेच्या रात्री चांदण्या बघण्याचा, रिमझिम पावसात भिजण्याचा आम्ही अनुभव घेतला आहे.

प्रश्न 3.
निसर्गाचे संतुलन राखण्यासाठी तुम्ही काय करता?
उत्तर:
निसर्गाचे संतुलन राखण्यासाठी आम्ही झाडे लावतो, पक्ष्यांचे जीव वाचवतो, पाणी जपून वापरतो.

प्रश्न 4.
‘कोणापासून काय घ्यावे’ या कवितेच्या कवयित्री कोण आहेत?
उत्तर:
‘कोणापासून काय घ्यावे’ या कवितेच्या कवयित्री ‘निलम माणगावे’ आहेत.

प्रश्न 5.
प्रभातकाळी काय शिकूया?
उत्तरः
प्रभातकाळी जीव जगवत जगूया.

प्रश्न 6.
संध्यासमयी काय करूया?
उत्तर:
संध्याकाळी सारे मित्र एक होऊन हसूया.

प्रश्न 7.
रिकाम्या जागी कवितेतील योग्य शब्द लिहा.
1. …………………………. ” ताऱ्यापासून
……………………………… घेऊया कांती.
2. ……………………… एक होऊनी.
…………………………… मित्र हसूया.
उत्तर:
1. चमचमणाऱ्या, दिव्य
2. संध्यासमयी, सारे

Maharashtra Board Class 5 Marathi Solutions Chapter 8 कोणापासून काय घ्यावे

प्रश्न 8.
खालील कवितेच्या ओळी पूर्ण करा.
फुलापासून ……………………….
……………………………… तराणे
उत्तर:
फुलापासून गंध घेऊया .
कोकिळाकडून गाणे
झुळझुळणाऱ्या झऱ्यापासुनी
घेऊ नवे तराणे

व्याकरण व भाषाभ्यास:

प्रश्न 1.
विरुद्धार्थी शब्द लिहा.

  1. शांती
  2. गंध
  3. नवे
  4. ऊन
  5. जीवन
  6. प्रभात
  7. हसणे
  8. एक
  9. सकाळ

उत्तर:

  1. अशांती
  2. दुर्गंध
  3. जुने
  4. सावली
  5. मरण
  6. संध्या
  7. रडणे
  8. अनेक
  9. संध्याकाळ

Maharashtra Board Class 5 Marathi Solutions Chapter 8 कोणापासून काय घ्यावे

प्रश्न 2.
वचन बदला

  1. तारा
  2. ढग
  3. फूल
  4. मित्र
  5. सावली
  6. झाड
  7. झरा
  8. गाणे

उत्तर:

  1. तारे
  2. ढग
  3. फुले
  4. मित्र
  5. सावल्या
  6. झाडे
  7. झरे
  8. गाणी

प्रश्न 3.
लिंग बदला.

  1. मित्र
  2. कोकीळा
  3. मुले

उत्तर:

  1. मैत्रीण
  2. कोकीळ
  3. मुली

कोणापासून काय घ्यावे? Summary in Marathi

पदयपरिचय:

संपर्कात येणाऱ्या प्रत्येक व्यक्तीकडून आपण काही ना काही घेत असतो (शिकत असतो.) त्याचप्रमाणे निसर्गातील घटकांकडूनही घेण्यासारखे काहीना काही असते. हाच संदेश कवयित्री आपल्याला या कवितेद्वारे देत आहे.

Maharashtra Board Class 5 Marathi Solutions Chapter 8 कोणापासून काय घ्यावे

शब्दार्थ:

  1. सूर्य – एक ग्रह (the Sun)
  2. रंग – वर्ण (colour)
  3. दिव्य – दैवी (devine)
  4. कांती – लकाकी (shine)
  5. फुले – पुष्पे (flowers)
  6. गंध – सुवास (smell)
  7. झरा – निर्झर (spring)
  8. गडगडणे – मेघगर्जना (thundering)
  9. तराणे – मंजूळ गाणे (a meldious songs)
  10. प्रभात – सकाळ (morning)
  11. संध्या – संध्याकाळ (evening)
  12. कोकिळ – एक गाणारा पक्षी (the cuckoo)
  13. माती – मृत्तिका (soil)
  14. शांती – शांतता (calmness)
  15. सावली – छाया (shadow)
  16. जीव – प्राण (living being)

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

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

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

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

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

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

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

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

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

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.
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 49 1

(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:
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 49 6

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

(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

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

(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

Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 49 2

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.

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

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

Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 49 3

(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.

Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 49 4

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

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

(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.

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

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.

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

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

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

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

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

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

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

(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.

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

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.

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

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.
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 1
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

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

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.
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 2

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.

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

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

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

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.
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 3
Answer:
24 cm

Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 4
Answer:
18 cm

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

Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 5
Answer:
21 cm

Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 6
Answer:
20 cm

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

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?
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 7
Answer:
20 cm

Question 5.
Match the columns ‘A’ and ‘B’
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 8
Answer:
(1- d),
(2- a),
(3 – b),
(4 – c)

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

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

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

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.
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 9
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

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

Question 8.
Find the area of the. following figures. (All small squares are having side 1 cm)
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 10
Answer:
5 sq.cm

Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 11
Answer:
4 sq.cm

Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 12
Answer:
9 sq.cm

Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 13
Answer:
16 sq.cm.

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

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.
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 14
Answer:
(1) 9
(2) 9

Question 10.
Fill in the blanks:
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 50 15
Answer:
(1) Area = 24 sq.cm., Perimeter = 22 cm
(2) Breadth = 4 cm, Perimeter = 20 cm
(3) Length = 2 cm, Perimeter = 8 cm

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

Question 1.
For his birthday, Ajay gave 20 l 450 ml of milk to the children in an Ashramshala and 28 l 800 ml to the children in an orphanage. How much milk did Ajay donate?
Solution:

lml
1
2 0
+ 2 8
4 5 0
8 0 0
4 92 5 0

450 ml + 800 ml
= 1250 ml
= 11 + 250 ml

∴ Ajay donated 49 l 250 ml milk

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

Question 2.
Under the Rural Cleanliness Mission, college students cleaned 1 km 750m of a village road that is 2 km 575m long. How much remained to be cleaned?
Solution:

kmm
11 5 7 5
2
– 1
5 7 5
7 5 0
08 2 5

750 m cannot be subtracted from 575 m. So, convert 1 km = 1000 m.
∴ 825 m remained to be cleaned

Question 3.
Babhulgaon used 21,250 liters of treated waste water in the fields. Samvatsar used 31,350 litres of similar water. How much treated waste water was used in all?
Solution:
2 1 2 5 0 litres Babhulgaon used
+ 3 1 3 5 0 litres Samvatsar used
___________
5 2 6 0 0
___________

∴ 52,600 litres of waste water used in all

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

Question 4.
If half a litre of milk costs 22 rupees, how much will 7 litres cost?
Solution:
\(\begin{array}{l}\frac{1}{2}+\frac{1}{2}=\frac{1+1}{2}=\frac{2}{2}=1 \text { litre } \\ 22+22=₹ 44\end{array}\)
That is, 1 litre cost ₹ 44
∴ 7 litres costs 44 x 7 = ₹ 308
∴ 7 litres costs ₹ 308

Question 5.
If the speed of a motorcycle is 40 km per hour, how far will it travel in an hour and a quarter?
Solution:
Hour and quarter = 1 + \(\frac{1}{4}\) hours
= 40 km + \(\frac{1}{4}\) x 40 km
= 40 km + 10 km
= 50 km
∴ Motorcycle will travel in a hour and a quarter 50 km

Question 6.
If a man walks at a speed of 4 kmph, how long will it take him to walk 3 km?
Solution:
1 km = 1000 m
4 km in 1 hour, 4 km in 60 minutes
That is
2 km in 30 minutes
+ 1 km in 15 minutes
_______________________
3 km in 45 minutes

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 1
That is 1 km in 15 minutes Hence, 3
km in 15 x 3 = 45 min

∴ 3 km in 15 x 3 = 45 min

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

Question 7.
If a rickshaw travels at a speed of 30 kmph, how far will it travel in three quarters of an hour?
Solution:
30 kmph means
In 60 minutes 30 km and 30 minutes 15 km
and 15 minutes \(\frac{15}{2}=\frac{15 \times 5}{2 \times 5}=\frac{75}{10}\) = 7.5 km
∴ In 45 minutes 15 km + 7.5 km = 22.5 km

Question 8.
During Cleanliness Week, children cleaned the public park in their town. They collected three quarter kilograms of plastic bags and five and a half kilograms of other garbage. How much garbage did they collect in all?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 2

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

Question 9.
If one shirt needs 2 m 50cm of cloth, how much cloth do we need for 5 shirts?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 3
∴ 12 m 50 cm cloth needs

Question 10.
If a car travels 60 km in an hour, how far will it travel in
(1) 2 hours?
(2) 15 minutes?
(3) half an hour?
(4) three and a half hours?
Solution:
60 kmph
In 60 minutes 60 km
Hence, 1 minute 1 km
(1) 2 hours = 2 x 60 = 120 km
(2) In 15 minutes = 15 km
(3) In half an hour 60 ÷ 2 = 30 km
(4) In three and half hours
= 3 x 60 + 30
= 180 + 30
= 210 km

∴ (1) 120 km
(2) 15 km
(3) 30 km
(4) 210 km

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

Question 11.
If one gold bangle is made from 12 grams 250 milligrams of gold, how much gold will be needed to make 8 such bangles? (1000mg = 1 g)
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 4
∴ 98 grams gold needed

Question 12.
How many pouches of 20g cloves each can be made from 1 kg 240g of cloves?
Solution:
1 kg 240 gm
= 1000 gm + 240 gm
= 1240 gm
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 5
∴ pouches can be made

Question 13.
Seema’s mother bought 2m 70cm of cloth for a kurta and 2 m 40cm for a shirt. How much cloth did she buy in all?
Solution:
70 cm + 40 cm
= 110 cm
= 1 m 10 cm

mcm
1
2
+ 2
7 0
4 0
51 0

cloth for Kurta
cloth for Shirt

∴ 5 m 10 cm cloth in all

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

Question 14.
A water tank holds 125 l of water. If 97 l 500 ml of the water is used, how much water remains in the tank?
Solution:
1 litre = 1000 ml
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 6
water tank holds
water used
water remain

∴ 27 l 500 ml water remain in tank

Question 15.
Harminder bought 57 kg 500g of wheat from one shop and 36 kg 800 g of wheat from another shop. How much wheat did he buy altogether?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 7
bought from 1 shop
bought from another shop
500 + 800 = 1300 gm
= 1000 + 300
= 1 kg 300 gm

∴ 94 kg 300 gm bought altogether

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

Question 16.
Renu took part in a 100m race. She tripped and fell after running 80 m 50 cm. How much distance did she have left to run?
Solution:

mcm
9 91 0 0
1 0 0
– 8 0
0 0
5 0
1 95 0

Borrow l m = 100 cm
So, 100 m = 99 m + 100 cm
Total distance to run
Distance covered
Distance left to run

∴ 19 m 50 cm distance left to run

Question 17.
A sack had 40kg 300 grams of vegetables. There were 17kg 700 g potatoes, 13 kg 400g cabbage and the rest were onions. What was the weight of the onions?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 8
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 9

∴ Weight of onions is 9 kg 200 gm

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

Question 18.
One day, Gurminder Singh walked 3 km 750m and Parminder Singh walked 2km 825m. Who walked farther and by how much?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 11

∴ Gurminder walked more by 925 metres

Question 19.
Suresh bought 3kg 250g of tomatoes, 2 kg 500g of peas and 1kg 750g of cauliflower. How much was the total weight of the vegetables he bought?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 12

∴ Total weight 7 kg 500 gm

Question 20.
Jalgaon, Bhusawal, Akola, Amravati and Nagpur lie serially on a certain route. The distances between Akola and these other places are given below.

Use them to make word problems and solve the problems.
Amravati – 95 km, Bhusawal – 154 km,
Nagpur – 249 km, Jalgaon – 181 km
Solution:
(1) What is the distance between Bhusaval and Nagpur?
249 km – 154 km = 95 km

∴ The distance between Bhusaval and Nagpur is 95 km

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

(2) What is the distance between Amravati and Jalgaon?
181 km – 95 km = 86 km

∴ The distance between Amravati and Jalgaon is 86 km.

Question 21.
Complete the following table and prepare the total bill.
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 16
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 15

Activity

  • You have 1 kg of potatoes. Find out which other ingredients you will need to make potato vadas and approximately how much of each ingredient you will need. Also find out approximately how much each ingredient will cost and how many vadas you will be able to make.
  • Fix a 1 m long stick in an open field. Measure the shadow of the stick at 9:00 in the morning, at 12:00 noon, at 3:00 in the afternoon and at 5:00 in the evening. Observe at which time of the day the shadow is shortest and at what time, it is longest.
  • Measure the length of a pen refill.

Problems on Measurement Problem Set 47 Additional Important Questions and Answers

Question 1.
One can contains 30 l 560 ml of milk, while second contains 251890 ml of milk and third one contains 20 l 760 ml of milk. How much milk is there in the three cans together?
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47 14
∴ 77 l 210 ml total milk

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

Question 2.
Add the following:
(1) ₹ 13, 85 paise + ₹ 16, 40 paise
(2) 15 kg 280 gm + 18 kg 920 gm
(3) 24 l 690 ml + 25 l 780 ml
(4) 22 km 750 m + 27 km 500 m
(5) 17 m 40 cm + 19 m 85 cm
(6) 38 cm 8 mm + 17 cm 2 mm
(7) 10 km 950 m + 15 km 125 m
(8) 83 kg 468 gm + 109 kg 532 gm
Answer:
(1) ₹ 30, 25 paise
(2) 34 kg 200 gm
(3) 50 1 470 ml
(4) 50 km 250 m
(5) 37 m 25 cm
(6) 56 cm
(7) 26 km 75 m
(8) 193 kg

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

Question 3.
Subtract the following:
(1) ₹ 21, 30 paise – ₹ 13, 80 paise
(2) 16 kg 130 gm – 9 kg 250 gm
(3) 9 l 350 ml – 5 l 470 ml’
(4) 41 m 10 cm – 14 m 40 cm
(5) 38 km 175 m – 20 km 365 m
(6) 27 cm 5 mm – 11 cm 8 mm
(7) 28 km 725 m – 13 km 590 m
(8) 380 kg – 232 kg 730 gm
Answer:
(1) ₹ 7, 50 paise
(2) 6 kg 880 gm
(3) 51 880 ml
(4) 26 m 30 cm
(5) 17 km 810 m
(6) 15 cm 7 mm
(7) 15 km 135 m
(8) 147 kg 270 gm

Question 4.
Fill in the blanks:
(1) 1250 m = …………………… km …………………… m
(2) 2.5 m = …………………… m …………………… cm
(3) 3 l 50 ml = …………………… ml
(4) ₹ 2.5 = …………………… paise
Answer:
(1) 1 km 250 m
(2) 2 m 50 cm
(3) 3050 ml
(4) 250 paise

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

Question 5.
(A) Match the following:

‘A’‘B’
(1) Potato 3.5 kg, rate per kg ₹ 12(a) ₹ 40
(2) Onion 2 kg, rate per kg ₹ 20.50(b) ₹ 42
(3) Vegetables 2.5 kg, rate per kg ₹ 16(c) ₹ 39
(4) Others 6.5 kg, rate per kg ₹ 6(d) ₹ 41

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

Maharashtra Board Class 5 Maths Solutions Chapter 11 Problems on Measurement Problem Set 47

(B) Match the following:

‘A’‘B’
(1) Half metre(a) 5 cm
(2) Half kilometre(b) 50 cm
(3) 50 millimetre(c) 500 cm
(4) 5 kilometre(d) 500 m
(5) 5 metre(e) 5000 m

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

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

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)
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 1
Solution:
Perimeter [ ] DABCD
= 20 + 16 + 7 + 14
= 57 cm

∴ 57 cm

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

(2)
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 3
Solution:
Perimeter of the figure
= 12 + 18 + 8 + 8 + 18
= 64m

∴ 64m

(3)
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 2
Solution:
Perimeter of the figure
= 10 + 6 + 6 + 10 + 8 + 8
= 48 cm

∴ 48 cm

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

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?
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 4Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 5
Solution:
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 13
Perimeter
= 2 + 3 + 3 + 2 + 1 + 1
= 12 cm

∴ 12 cm

Formula for the perimeter of a rectangle

Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 6
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

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

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.
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 7

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

Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 8 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.
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 9
Perimeter of a square = 4 × length of one side
= 4 × 6
= 24

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

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.

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

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.
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 10
Solution:
Perimeter of the figure
= 2 + 6 + 2 + 6
= 16 cm

∴ 16 cm

Question 2.
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 11
Solution:
Perimeter of the figure
= 3 + 3 + 3 + 3
= 12 cm

∴ 12 cm

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

Question 3.
Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 12
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?

Maharashtra Board Class 5 Maths Solutions Chapter 12 Perimeter and Area Problem Set 48 14
Solution:
Perimeter
= 2 + 1 + 1 + 2 + 1 + 1 + 2 + 1 + 1 + 2 + 1 + 1
= 16 cm

∴ 16 cm

Maharashtra Board Class 5 Marathi Solutions Chapter 9 सिंह आणि बेडूक

Balbharti Maharashtra State Board Class 5 Marathi Solutions Sulabhbharati Chapter 9 सिंह आणि बेडूक Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Marathi Sulabhbharati Solutions Chapter 9 सिंह आणि बेडूक

5th Standard Marathi Digest Chapter 9 सिंह आणि बेडूक Textbook Questions and Answers

प्रश्न 1.
ऐका, वाचा, लक्षात घ्या.
एका …………………. एक राहायचा. त्याला त्या ……………………. कंटाळा आला आणि तो नव्या …………………. राहायला गेला. एकदा ……………… मोठ्याने गरजला. त्या …………….. अनेक ………………… तसेच काही बेडूकही राहात होते. त्यांनी यापूर्वी …………….. पाहिले नव्हते. त्यांचा ……………….. पुढारी म्हणाला, कोणीतरी मोठा आवाज काढत आहे. आता मीही मोठा आज काढतो. ..मोठा आवाज काढला. …………… हा आवाज नवीन होता …………………. वाटले, कोणीतही आपल्याला आव्हान देत आहे, आपण सावध राहिले पाहिजे. ……….. आपली गर्जना थांबवली. तो शांत उभा राहिला ……………………. मोठमोठ्याने ओरडत पुढे पुढे सरकू लागला. ……………….. त्याला पाहिले व त्याचा आवाज ऐकला. ………………. वेगाने पुढे सरकला ……………….. डोक्यावर पाय दिला गयावया करू लागला. ……………………….. त्याला सोडून दिले.
उत्तर:
एका जंगलात एक सिंह राहायचा. त्याला त्या जंगलाचा कंटाळा आला आणि तो नव्या जंगलात राहायला गेला. एकदा सिंह मोठ्याने गरजला. त्या जंगलात अनेक प्राणी व पक्षी तसेच काही बेडूकही राहात होते. त्यांनी यापूर्वी सिंहाला पाहिले नव्हते. त्यांचा बेडूक पुढारी म्हणाला, कोणीतरी मोठा आवाज काढत आहे. आता मीही मोठा आज काढतो.

बेडकाने मोठा आवाज काढला. सिंहाला हा आवाज नवीन होता सिंहाला वाटले, कोणीतही आपल्याला आव्हान देत आहे, आपण सावध राहिले पाहिजे. सिंहाने आपली गर्जना थांबवली. तो शांत उभा राहिला. बेडूक मोठमोठ्याने ओरडत पुढे पुढे सरकू लागला. सिंहाने त्याला पाहिले व त्याचा आवाज ऐकला. सिंह वेगाने पुढे सरकला सिंहाने बेडकाच्या डोक्यावर पाय दिला बेडूक गयावया करू लागला. सिंहाने त्याला सोडून दिले.

Maharashtra Board Class 5 Marathi Solutions Chapter 9 सिंह आणि बेडूक

1. या गोष्टीतील प्राण्यांची नावे लिहा.

प्रश्न 1.
या गोष्टीतील प्राण्यांची नावे लिहा.
उत्तरः
सिंह, बेडूक.

2. सिंह व बेडूक यांमध्ये हुशार कोण ते सांगा.

प्रश्न 1.
सिंह व बेडूक यांमध्ये हुशार कोण ते सांगा.
उत्तरः
सिंह

3. खालील प्राण्यांच्या ओरडण्याला काय म्हणतात ते कंसातून शोधून लिहा.

प्रश्न 1.
गोष्टीतील प्राण्यांचा आवाज कसा आहे ते माहीत करून घ्या. आवाज काढून दाखवा.
Maharashtra Board Class 5 Marathi Solutions Chapter 9 सिंह आणि बेडूक 1
उत्तरः

  1. डरकाळी
  2. भुंकणे
  3. चीत्कार
  4. म्याँव – म्याँव
  5. हंबरणे
  6. बें-बें
  7. खिंकाळणे
  8. कुईकुई

Maharashtra Board Class 5 Marathi Solutions Chapter 9 सिंह आणि बेडूक

खालील प्राण्यांच्या ओरडण्याला काय म्हणतात ते वाचा.
(अ) वाघाची – डरकाळी
(आ) हत्तीचा – चीत्कार
(इ) गाईचे – हंबरणे
(ई) बकरीचे – बें-बें
(उ) घोड्याचे – खिंकाळणे
(ऊ) कुत्र्याचे – भुंकणे

चित्रसंदेश:

1. ऐका, वाचा.
Maharashtra Board Class 5 Marathi Solutions Chapter 9 सिंह आणि बेडूक 2

2. वाचा. लक्षात ठेवा.
वरील संदेशात ‘पाटी’ हा शब्द दोन अर्थांनी आला आहे.
पाटी – 1. टोपली. 2. ज्यावर लिहिले जाते ती.
संदेश – मुलामुलींच्या डोक्यावर पाटी नको, हातात पाटी दया, म्हणजे मुलामुलींना शिकवा.

Marathi Sulabhbharati Class 5 Solutions Chapter 9 सिंह आणि बेडूक Additional Important Questions and Answers

प्रश्न 1.
एका शब्दात उत्तरे लिहा.

  1. सिंहाच्या राहण्याच्या ठिकाणाला काय म्हणतात?
  2. सिंहाच्या ओरडण्याला काय म्हणतात?
  3. सिंहाच्या पिल्लाला काय म्हणतात?
  4. पाठातील सिंह कोठे राहत होता?
  5. बेडूक कसा ओरडतो?
  6. कोणी आपली गर्जना थांबवली?
  7. सिंहाने कोणाला सोडून दिले?

उत्तरः

  1. गुहा
  2. गर्जना
  3. छावा
  4. जंगलात
  5. डराव डराव
  6. सिंहाने
  7. बेडकाला

Maharashtra Board Class 5 Marathi Solutions Chapter 9 सिंह आणि बेडूक

2. एका वाक्यात उत्तरे लिहा.

प्रश्न 1.
सिंह नव्या जंगलात राहायला का गेला?
उत्तरः
सिंहाला जुन्या जंगलात राहायचा कंटाळा आला होता.

प्रश्न 2.
जंगलात कोण-कोण राहत होते?
उत्तर:
जंगलात पशु-पक्षी तसेच काही बेडूकही राहत होते.

प्रश्न 3.
बेडकांचा पुढारी काय म्हणाला?
उत्तर:
कुणीतरी मोठा आवाज काढत आहे. मीही त्याच्याप्रमाणेच मोठा आवाज काढतो, असे बेडकांचा पुढारी म्हणाला.

Maharashtra Board Class 5 Marathi Solutions Chapter 9 सिंह आणि बेडूक

प्रश्न 4.
बेडूक गयावया का करू लागला?
उत्तर:
सिंहाने बेडकाच्या डोक्यावर पाय दिला, म्हणून बेडूक गयावया करू लागला.

प्रश्न 5.
सिंहाने आपली गर्जना का थांबवली?
उत्तर:
कोणीतरी आपल्याला आव्हान देत आहे. आपण वेळीच सावध व्हावं, या विचाराने सिंहाने आपली गर्जना थांबवली.

प्रश्न 6.
बेडकाने या अगोदर कोणाला पाहिले नव्हते?
उत्तर:
बेडकाने या अगोदर सिंहाला पाहिले नव्हते.

3. थोडक्यात उत्तर लिहा.

प्रश्न 1.
सिंह व बेडकाच्या गोष्टीतून तुम्ही काय शिकलात ते लिहा.
उत्तरः
स्वत:ची श्रेष्ठता सिद्ध करण्यासाठी कोणालाही आव्हान देऊ नये, त्याने आपलेच नुकसान होते हे सिंह व बेडकाच्या गोष्टीतून शिकलो.

Maharashtra Board Class 5 Marathi Solutions Chapter 9 सिंह आणि बेडूक

प्रश्न 2.
खाली दिलेल्या चित्रांसाठी कंसातील शब्द वापरूनकथा पूर्ण करा. (मुलगा, म्हाताऱ्या आजीबरोबर, झोपडीत, वर्तमानपत्रे, बागेला, म्हातारी आजी, भांडी, पाकीट, वर्गशिक्षकांकडे, पोलीस स्टेशनला, पोलीसांनी, फोन, पाकिट मालकाने, बक्षीस)
सदा नावाचा एक होता. तो एका राहत होता. त्याला अभ्यास करणे खूप आवडायचे. काही झाले तरी भरपूर शिकायचे, असे त्याने ठरवले. सदा रोज सकाळी टाकायला जायचा. संध्याकाळी दोन घरी पाणी घालायचा. चार घरची घासायची. एक दिवस सदा दुपारी शाळेत जात असताना त्याला रस्त्यावर एक सापडले. सदाने ते पाकीट जसेच्या तसे दिले. शिक्षकाने ते पाकीट बघितले. त्यावर मालकाचा पत्ता, नाव होते. शिक्षक सदाबरोबर गेले व ते पाकीट पोलिसांना दिले. लगेच फोन करून मालकाला बोलावले. व त्यांचे पाकीट त्यांना परत देऊन टाकले. सदाला शाबासकी दिली. त्याला दिले व सदाची शिक्षणाची सर्व जबाबदारी धनिकाने स्विकारली. त्यामुळे सदाचे स्वप्न पूर्ण झाले.
उत्तरः
सदा नावाचा एक मुलगा होता. तो म्हाताऱ्या आजीबरोबर एका झोपडीत राहत होता. त्याला अभ्यास करणे खूप आवडायचे. काही झाले तरी भरपूर शिकायचे, असे त्याने ठरवले. सदा रोज सकाळी वर्तमानपत्रे टाकायला जायचा. संध्याकाळी दोन घरी बागेला पाणी घालायचा.म्हातारी आजी चार घरची भांडी घासायची. एक दिवस सदा दुपारी शाळेत जात असताना त्याला रस्त्यावर एक पाकीट सापडले. सदाने ते पाकीट जसेच्या तसे वर्गशिक्षकांकडे दिले.

शिक्षकाने ते पाकीट बघितले. त्यावर मालकाचा पत्ता, नाव होते. शिक्षक सदांबरोबर पोलीस स्टेशनला गेले व ते पाकीट पोलिसांना दिले. पोलीसांनी लगेच फोन करून मालकाला बोलावले. व त्यांचे पाकीट त्यांना परत देऊन टाकले. पाकिट मालकाने सदाला शाबासकी दिली. त्याला बक्षीस दिले व सदाची शिक्षणाची सर्व जबाबदारी धनिकाने स्विकारली. त्यामुळे सदाचे स्वप्न पूर्ण झाले.

प्रश्न 3.
खालील चित्र पाहा व कंसात दिलेल्या योग्य शब्दांची जोडी वापरून चित्राखाली दिलेले संदेश पूर्ण करा.
(वृक्ष – शान, प्लॅस्टीकची पिशवी – कापडी पिशवी, वेगाला – जीवाला, इंधनाची – देशाची, कचरा – आरोग्याची)
Maharashtra Board Class 5 Marathi Solutions Chapter 9 सिंह आणि बेडूक 3
उत्तरः

  1. प्लॅस्टीकची पिशवी – कापडी पिशवी
  2. इंधनाची – देशाची
  3. कचरा – आरोग्याची
  4. वृक्ष – शान (५) वेगाला – जीवाला

Maharashtra Board Class 5 Marathi Solutions Chapter 9 सिंह आणि बेडूक

व्याकरण व भाषाभ्यास:

प्रश्न 1.
समानार्थी शब्द लिहा.

  1. जंगल
  2. सिंह
  3. नवे
  4. पशू
  5. पक्षी
  6. आवाज
  7. शांत
  8. कंटाळा
  9. पुढारी
  10. पाय
  11. सावध
  12. बेडूक

उत्तर:

  1. वन, रान
  2. वनराज
  3. नूतन
  4. प्राणी, जनावरे
  5. खग, विहंग
  6. ध्वनी
  7. निमूट
  8. आळस
  9. नेता
  10. चरण
  11. सज्ज
  12. मंडूक

प्रश्न 2.
विरुद्धार्थी शब्द लिहा.

  1. कंटाळा
  2. नव्या
  3. पूर्वी
  4. सावध
  5. एक
  6. पुढारी
  7. शांत
  8. वेगाने
  9. पुढे
  10. मोठा
  11. उत्साह

उत्तरः

  1. उत्साह
  2. जुन्या
  3. आता
  4. बेसावध
  5. अनेक
  6. जनता
  7. अशांत
  8. सावकाश
  9. मागे
  10. लहान
  11. निरुत्साह

प्रश्न 3.
वचन बदला.

  1. जंगल
  2. एक
  3. आव्हान
  4. गर्जना

उत्तर:

  1. जंगले
  2. अनेक
  3. आव्हाने
  4. गर्जना
  5. पाव

सिंह आणि बेडूक Summary in Marathi

पाठ्यपरिचय:

आपली मर्यादा ओळखून आपण पुढे सरकावे. स्वत:ची श्रेष्ठता सिद्ध करण्यासाठी कोणालाही आव्हान देऊ नये. त्याने आपलेच नुकसान होते, या अर्थाची कथा या पाठात सांगितली आहे.

Maharashtra Board Class 5 Marathi Solutions Chapter 9 सिंह आणि बेडूक

शब्दार्थ:

  1. जंगल – वन (forest)
  2. कंटाळा – निरसता (bore)
  3. गरजणे – गर्जना करणे (To roar)
  4. बेडूक – मंडूक (a frog)
  5. आव्हान – मुकाबला करण्यासाठी आमंत्रण देणे (a challenge)
  6. पुढारी – नेता (a leader)
  7. संतुष्ट – समाधानी (satisfied)
  8. सरकणे – पुढे जाणे (To move on)
  9. आवाज – ध्वनी (sound)
  10. सावध – जागरुक (alert)
  11. शांत – शांतता (calm)
  12. गयावया – दीनवाणी प्रार्थना (to plead)
  13. वेग – गती (speed)
  14. नवीन – नूतन (recent, new)