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Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 1.1 Geometry 10th Class Maths Part 2 Answers Solutions Chapter 1 Similarity.

10th Standard Maths 2 Practice Set 1.1 Chapter 1 Similarity Textbook Answers Maharashtra Board

Class 10 Maths Part 2 Practice Set 1.1 Chapter 1 Similarity Questions With Answers Maharashtra Board

Question 1.
Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.
Solution:
Let the base, height and area of the first triangle be b₁, h₁, and A₁ respectively.
Let the base, height and area of the second triangle be b₂, h₂ and A₂ respectively.

[Since Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ The ratio of areas of the triangles is 3:4.

Question 2.
In the adjoining figure, BC ± AB, AD _L AB, BC = 4, AD = 8, then find \(\frac{A(\Delta A B C)}{A(\Delta A D B)}\)

Solution:
∆ABC and ∆ADB have same base AB.

[Since Triangles having equal base]

Question 3.
In the adjoining figure, seg PS ± seg RQ, seg QT ± seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.

Solution:
In ∆PQR, PR is the base and QT is the corresponding height.
Also, RQ is the base and PS is the corresponding height.
\(\frac{A(\Delta P Q R)}{A(\Delta P Q R)}=\frac{P R \times Q T}{R Q \times P S}\) [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
∴ \(\frac{1}{1}=\frac{P R \times Q T}{R Q \times P S}\)
∴ PR × QT = RQ × PS
∴ 12 × QT = 6 × 6
∴ QT = \(\frac { 36 }{ 12 } \)
∴ QT = 3 units

Question 4.
In the adjoining figure, AP ⊥ BC, AD || BC, then find A(∆ABC) : A(∆BCD).

Solution:
Draw DQ ⊥ BC, B-C-Q.

AD || BC [Given]
∴ AP = DQ   (i)  [Perpendicular distance between two parallel lines is the same]
∆ABC and ∆BCD have same base BC.

Question 5.
In the adjoining figure, PQ ⊥ BC, AD ⊥ BC, then find following ratios.

Solution:
i. ∆PQB and tPBC have same height PQ.

ii. ∆PBC and ∆ABC have same base BC.

iii. ∆ABC and ∆ADC have same height AD.

Question 1.
Find \(\frac{A(\Delta A B C)}{A(\Delta A P Q)}\)

Solution:
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.

Class 10 Maths Digest

• Practice Set 1.1 Class 10 Answers
• Practice Set 1.2 Class 10 Answers
• Practice Set 1.3 Class 10 Answers
• Practice Set 1.4 Class 10 Answers
• Problem Set 1 Class 10 Answers

Balbharti Maharashtra State Board Class 10 English Solutions Unit 1.3 On Wings of Courage Notes, Textbook Exercise Important Questions and Answers.

Class 10 English Chapter 1.3 Question Answer Maharashtra Board

On Wings of Courage Poem 10th Std Question Answer

Question 1.
The ranks of officers in Indian Army, Navy and Air Force are jumbled up. Discuss with your group and put them in the appropriate boxes.

Commander, Brigadier, Wing-Commander, Vice-Admiral, Squadron-Leader, Major, Colonel, Field Marshal, Air Marshal, Admiral of Fleet, Lieutenant-General, Flying Officer, Commodore, Rear Admiral, Air-Commodore.

ARMY	NAVY	AIR FORCE

Answer:

Army	Navy	Air Force
Brigadier,	Commander,	Wing-
Major, Colonel,	Vice-Admiral,	Commander,
Field Marshal,	Admiral	Squadron-
Lieutenant-	of Fleet,	Leader, Air
General	Commodore,	Marshal, Flying
	Rear Admiral	Officer, Air-Commodore

Question 2.
Homophones/ Homographs
(A) Make sentences to bring out the difference between-
(1) (a) wear ……………………………………..
(b) ware ……………………………………..
(2) (a) here ……………………………………..
(b) hear ……………………………………..
(3) (a) there ……………………………………..
(b) their ……………………………………..
(4) (a) cell ……………………………………..
(b) sell ……………………………………..
Answer:
(1) (a) wear: The little girl wanted to wear a pink, frilly dress.
(b) ware: The silver ware laid out on the King’s table was exquisite.

(2) (a) here: “You must sit here,” said the man to his guest.
(b) hear: The children could hear the sound of the planes quite clearly.

(3) (a) there: “I had kept my bag there,” said the woman to the policeman.
(b) their: The girls picked up their bags and went home.

(4) (a) cell: The prisoner sat in the dark cell without talking.
(b) sell: The hawker wanted to sell all his wares before evening.

(B) Write what the underlined Homographs in the following sentences mean.
(1) (a) A bear is an omnivorous animal. ……………………………………..
(b) She could not bear the injustice. ……………………………………..
(2) (a) A bat is the only bird which is a mammal. ……………………………………..
(b) His bat broke as it struck the ball. ……………………………………..
(3) (a) He had to pay a fine for breaking the traffic signal. ……………………………………..
(b) Use a fine cloth for the baby’s clothes. ……………………………………..
(4) (a) We enjoyed a lot at the temple fair. ……………………………………..
(b) She has a fair complexion. ……………………………………..
Answer:
(1) (a) A bear is an omnivorous animal.
bear – a large, heavy animal
(b) She could not bear the injustice,
bear – to tolerate

(2) (a) A bat is the only bird which is a mammal.
bat – a mammal that flies
(b) His bat broke as it struck the ball.
bat – a wooden implement used for hitting the ball in many games.

(3) (a) He had to pay a fine for breaking the traffic signal.
fine – penalty
(b) Use a fine cloth for the baby’s clothes,
fine – delicate, soft

(4) (a) We enjoyed a lot at the temple fair.
fair – a gathering of stalls and amusements for public entertainment
(b) She has a fair complexion, fair – light, not dark

Maharashtra Board Class 10 English Kumarbharati Unit 1.3 Questions and Answers

Question 1.
Read the text and fill in the flow chart of the promotions received by Arjan Singh.

Answer:

Question 2.
With the help of facts given in the text prepare a Fact file of Air Marshal Arjan Singh.
(a) Date of Birth
(b) Place of Birth
(c) Education
(d) First Assignments
(e) Important posts held
(a) In Air Force
(b) After retirement
(f) Awards
(g) Most outstanding contribution in IAF
(h) Retirement
Answer:
(a) Date of birth: April 15, 1919
(b) Place of birth: Lyalpur
(c) Education: at Montgomery; Empire Pilot Training Course at RAF (Cranwell)
(d) First Assignment: to fly Westland Wapiti biplanes in the North-Western Frontier Province as a member of the No. 1 RIAF Squadron
(e) Important posts held:
(1) In Air Force: Member of No. 1. RIAF, Flying Officer, Squadron Leader, Wing Commander, Group Captain, Air Commodore, Air Officer Commanding, Air Vice Marshal, Air Officer Commanding-in-Chief, Deputy Chief of Air Staff, Vice Chief of Air Staff, Chief of Air Staff, Air Chief Marshal.
(2) After retirement: Ambassador to Switzerland Lieutenant Governor of Delhi
(f) Awards: Distinguished Flying Cross (1944); Padma Vibhushan
(g) Most outstanding contribution in IAF: Transforming the IAF into one of the most potent air forces globally and the fourth biggest in the world.
(h) Retirement: in August 1969.

Question 3.
Fill in the web.

Answer:
(1) Singh had successfully led a young IAF during the 1965 Indo-Pak war.
(2) Singh played a major role in transforming the IAF into one of the most potent air forces globally and the fourth biggest in the world.
(3) Singh was honoured with the rank of Marshal on the Republic Day in 2002.
(4) Singh’s contribution was most outstanding during the 1965 Indo-Pak war.

Question 4.
Say what actions preceded the following promotions of Arjan Singh in his career in the IAF.
(a) Selected for Empire Pilot training course at RAF
(b) Promoted to Squadron Leader
(c) Leader of a flypast of over 100 aircraft at Red Fort, Delhi
(d) Awarded Padma Vibhushan
(e) First Air Chief Marshal of Indian Air Force
Answer:
(a) The authorities selected Singh for the Empire Pilot training course.
(b) He flew against the tribal forces and moved back to No. 1 Squadron as a Flying Officer to fly the Hawker Hurricane.
(c) On 15th August 1947, Arjan Singh achieved the unique honour of leading a fly-past of over a hundred IAF aircraft over the Red Fort in Delhi.
(d) He was awarded the Padma Vibhushan for his astute leadership of the Air Force and for inspiring the IAF to victory in the 1965 Indo-Pak war.
(e) He was a source of inspiration to all the personnel of the Armed Forces through the years.

Question 5.
Replace the underlined words/phrases with the appropriate ones, to retain the proper meaning.
(be the epitome of, gear up, a brief stint, play a major role, in recognition of, take over reins)
(a) He contributed notably in bringing up the school.
(b) Our school cricket team got ready for the final match against P. Q. R. High School.
(c) After a short period of working as a lecturer, Ravi took up an important post in a multi-national company.
(d) Our class monitor is a perfect symbol of duty and discipline.
(e) Accepting the great value of his research; they awarded him with a Ph.D. (degree)
(f) After the murder of King Duncan, Macbeth took over the control of Scotland.
Answer:
(a) He played a major role in bringing up the school.
(b) Our school cricket team geared up for the final match against P.Q.R.High School.
(c) After a brief stint as a lecturer, Ravi took up an important post in a multinational company.
(d) Our class monitor is the epitome of duty and discipline.
(e) In recognition of his research, they awarded him with a Ph.D. (degree)
(f) After the murder of King Duncan, Macbeth took over the reins of Scotland.

Question 6.
Build the word wall with the words related to ‘Military’.

Answer:

Question 7.
(A) State the different meanings of the following pairs of Homophones and make sentences of your own with each of them.

Word	Meaning	Sentence
(a) led lead(b) role roll(c) air heir(d) feat feet(e) reign rein rain	………………………….. ………………………………………………………… ……………………………………………………… ………………………………………………………. ………………………………………………………. …………………………..	………………………….. ………………………………………………………. ………………………………………………………. ………………………………………………………. ………………………………………………………. …………………………..

Answer:

Word	Meaning	Sentence
(a) led	past participle of lead (to guide or conduct)	The captain led his team to safety.
lead	graphite used as part of a pencil	Do you have a lead pencil?
(b) role	a part (in a play, film, etc.)	Marie got the leading role in the new movie.
roll	move in a particular direction by turning over and over	The boy wanted to roll in the mud while playing.
(c) air	the invisible gaseous substance surrounding the earth	There Is a lot of humidity in the air during the monsoon.
heir	successor or inheritor	The family did not know who the heir to the property was.
(d) feat	a great achievement	Climbing Mt. Everest is a feat.
feet	a unit of measurement	The girl saw to her shock that the lion was only a few feet away.
(e) reign	rule as king or queen	Queen Elizabeth’s reign has been a long one.
rein	a restraining influence	The new manager kept a tight rein on her employees.
rain	water that falls In drops from clouds in the sky	Children love to play in the rain.

(B) The following Homographs have the same spelling and pronunciation but can have different meanings. Make sentences of your own to show the difference.

Answer:
(a) firm: (i) My neighbour recently Joined an electronics firm as Sales Executive.
(ii) Many people feel that they must be firm with their children when they are growing.

(b) train: (i) The train left from platform 2 at seven p.m. sharp.
(ii) You must always train your pets to obey you.

(c) type: (i) The man asked his secretary to type the letter immediately.
(ii) Cows eat only a particular type of grass.

(d) post: (i) My aunt quit her job because she felt that the post was not suitable for her.
(ii) The little boy ran to the post office to post the letter to Santa Claus.

(e) current : (i) The minister was disturbed when he read about the current situation of unrest In the country.
(ii) It is a difficult task to row against the current in a river.

Question 8.
Glance through the text and prepare notes from the information that you get. Take only relevant points. Don’t use sentences. Arrange the points in the same order. You may use symbols or short forms. Present the points sequentially. Use highlighting techniques.
Answer:
Air Force Marshal Arjart Singh—Icon of India’s Military History

1. Date of Birth: 15 April, 1919
2. Qualifications: Empire Pilot Training Course at RAF (Cranwell)
3. Responsibilities:

• first assignment to fly Westland Wapiti biplanes in No.l RIAF Squadron
• brief stint in No.2 RIAF Squadron; moved back to No. 1 RIAF Squadron as Flying Officer
• overall commander of ‘Shiksha’
• led the IAF during the 1965 Indo-Pak war
• led a squadron against the Japanese during the Arakan Campaign; assisted the advance of Allied Forces to Yangoon
• led a fly-past on August 15, 1947
• commanded Ambala in the rank of Group Captain; took over as AOC of an operational command
• took over reins of the IAF
• ambassador to Switzerland; Lieutenant Governor of Delhi

(4) Achievements:

• selected for the Empire Pilot Training Course at RAF (Cranwell) in 1938, at age 19
• promoted to the rank of Squadron Leader in 1944
• led a fly-past over the Red Fort on August 15, 1947
• promoted to the rank of Wing Commander; promoted to the rank of Air Commodore in 1949
• longest tenure as AOC (1949-1952 and 19571961)
• appointed as Deputy Chief of Air Staff at the end of the 1962 war; appointed as Vice Chief of Air Staff in 1963
• rank of Air Marshal in August 1964; took over reins of IAF
• successfully led the IAF in 1965 Indo-Pak war
• promoted as Air Vice Marshal; appointed as AOC-in-C of an operational command
• first Air Chief to keep his flying currency till his CAS rank; has flown more than 60 different types of * aircraft
• first and only Air Chief Marshal of the IAF

(5) Awards:

– Distinguished Flying Cross (1944)
– Padma Vibhushan

(6) After retirement: Ambassador to Switzerland; Lieutenant Governor of Delhi
(Students can put these points attractively in boxes and use highlighting techniques.)

Question 9.
Develop a story suitable to the conclusion/end given below. Suggest a suitable title.
………………………………………………….. (Title)
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
…………………………………………………………………………………………………………………..
………………………………………………………….. and so, with tears of joy and pride, the 10 year old Sanyogita More received the National Bravery Award from the Prime Minister.
Answer:
A WONDERFUL ACT OF BRAVERY
It was the 26th of July in Mumbai. It was raining cats and dogs. Ten-year-old Sanyogita More stood at the door of her hut. The street was flooded with water. Sanyogita was frightened. Her parents had not returned from work and she was all alone.

Suddenly, she saw two little boys, Rohan and Sohan, come out from the neighbouring hut to play in the water. As Sanyogita watched, there came a sudden gush of water and the boys were dragged towards an open manhole, which had been marked with a pole. They caught hold of the pole, but the pole began to tilt. It would soon fall—and the boys would go down the manhole!

Sanyogita ran as fast as she could towards the boys. Pulling a rope from a nearby door, she looped it around a large stone. She held onto the rope and extended her hand towards the boys. “Catch my hand, Sohan, Rohan,” she shouted. “Catch! Catch soon!”

The boys were in a panic but they did as they were told. Sohan held Rohan’s leg, Rohan held Sanyogita’s hand, and Sanyogita held onto the rope.

“Help! Help! she shouted, knowing that if the rope broke or the stone was dislodged, they would all go into the manhole.

She stood there shivering, her arms numb, for nearly 15 minutes before help arrived. Sanyogita collapsed after the incident. The news of her brave deed spread far and wide, and reached the ears of 1 the Prime Minister, who decided to honour her with an award. And so, with tears of joy and pride, the 10- I year-old Sanyogita More received the National Bravery ‘ Award from the Prime Minister.

Question 10.
You wish to join any one of the Indian Armed Forces. Fill in the following application form.
To
The Advertiser
N/AF Recruitment Service
Purangaon – 456 789

Affix recent
passport size
photograph

Application For Recruitment
Rect notice No 1234

1. Post applied for
2. Name and surname of Candidate (in Block letters)
3. Father’s Name ………………………………… Mother’s Name …………………………………
4. Date of Birth
5. Contact details :
Tel. No. (Res) ………………….. . Mobile No.
Email ID ………………….. .
6. Permanent Address :
House No./Street/Village ………………….. .
Post Office ………………….. .
District ………………….. State ………………….. .
Pincode ………………….. .
7. Educational Qualifications :

Serial Number	Qualification	Name of School/College	Name of Board/University	Percentage obtained

8. Whether registered at any employment exchange Yes/ No ………………….. (If yes, mention registration number and the name of the Employment Exchange.)

9. Outstanding achievements in extra-curricular activities/ sports/ games, etc.
………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………….

10. Why you wish to join Armed Forces. …………………………………………………………………
………………………………………………………………………………………………………………………………………….
………………………………………………………………………………………………………………………………………….

Read More:

• Where the mind is Without fear Question Answers
• The Thief’s Story Question Answers
• On Wings of Courage Question Answers
• All the World’s a Stage Question Answers
• Joan of Arc Question Answers
• The Alchemy of Nature Question Answers

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 25 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 6 Angles Problem Set 25

Measure the angles given below and write the measure in the given boxes.

Answer:
(1) 40°
(2) 120°
(3) 90°
(4) 85°

Drawing an angle of the given measure
Example Draw ∠ABC of measure 70°.
B is the vertex of∠ABC and BA and BC are its arms.

1. First draw arm BC with a ruler.

2. Since B is the vertex, we must draw a 70° angle at that point.

Put the centre of the protractor on B. Place the protractor so that the baseline lies on arm BC. Count the divisions starting from the 0 near point C. Mark a point with your pencil at the division that shows 70°. Lift the protractor.

Draw a line from vertex B through the point marking the 70° angle. Name the other end of the line A.

∠ABC is an angle of measure 700.
Rahul and Sayali drew ∠PQR of measure 800 as shown below.

Teacher : Have Rahul and Sayali drawn the angles correctly?
Shalaka : Sir, Rahul’s angle is wrong. Sayali’s angle is correct.
Teacher : Why is Rahul’s angle wrong?
Rahul : I counted 10, 20, 30…from the left and drew the angle at 80.
Teacher : Rahul measured the angle from the left. Under the baseline on the left of Q, there is nothing. The arm of the angle is on the right of Q. Therefore, the point should have been marked 80° counting from the right side, that is, on the side on which point R lies.

Angles Problem Set 25 Additional Important Questions and Answers

Answer:
60°

Answer:
110°

Answer:
100°

Answer:
90°

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 16 Preparation for Algebra Problem Set 55 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 16 Preparation for Algebra Problem Set 55

Question 1.
Say whether right or wrong.

(1) (23 + 4) = (4 + 23)
Answer:
27 = 27 is right

(2) (9 + 4) > 12
Answer:
13 > 12 is right

(3) (9 + 4) < 12
Answer:
13 < 12 is wrong

(4) 138 > 138
Answer:
Wrong

(5) 138 < 138
Answer:
Wrong

(6) 138 = 138
Answer:
right

(7) (4 × 7) = 30 – 2
Answer:
28 = 28 is right

(8) \(\frac{25}{5}\) > 5
Answer:
5 > 5 is wrong.

(9) (5 × 8) = (8 × 5)
Answer:
40 = 40 is right

(10) (16 + 0) = 0
Answer:
16 + 0
= 16
16 = 0 is wrong

(11) (16 + 0) = 16
Answer:
16 = 16 is right.

(12) (9 + 4) = 12
Answer:
13 = 12 is wrong.

Question 2.
Fill in the blanks with the right symbol from <, > or =.

(1) (45 ÷ 9) [ ] (9 – 4)
Answer:
45 ÷ 9 = 5,
9 – 4 = 5 5
= 5
so, (45 + 9) = (9 – 4)

(2) (6 + 1) [ ] (3 × 2)
Answer:
6 + 1 = 7,
3 x 2 = 6
7 > 6
so, (6 + 1) > (3 x 2)

(3) (12 × 2) [ ] (25 + 10)
Answer:
12 x 2 = 24,
25 + 10 = 35
24 < 35
so, (12 x 2) < (25 + 10)

Question 3.
Fill in the blanks in the expressions with the proper numbers.

(1) (1 × 7) = ( [ ] × 1)
Answer:
1 x 7 = 7,
7 x 1 = 7
so, (1 x 7) = ( 7 x 1)

(2) (5 × 4) > (7 × [ ] )
Answer:
5 x 4 = 20, 7 x ………… must be less than 20.
7 x 2 = 14
so, (5 x 4) > ( 7 x 2)

(3) (48 ÷ 3) < ( [ ] × 5)
Answer:
48 – 3 = 16,
5 x 4 = 20
5 x 3 = 15
16 > 15 and 16 < 20 so, (48 + 3) <(4 x 5)

(4) (0 + 1) > (5 × [ ] )
Answer:
0 + 1 = 1,
5 x 1 = 5
5 x 0 = 0
1 < 5 and 1 > 0 so, (0 + 1) > (5 x Q)

(5) (35 ÷ 7) = ( [ ] + [ ] )
Answer:
35 ÷ 7 = 5,
3 + 2 = 5 so, (35 + 7) = (3 + 2)

(6) (6 – [ ] ) < (2 + 3)
Answer:
6 – < 2 + 3 = 5
5 > 6 – 2
so, (6 – 2) < (2 + 3)

Using letters
Symbols are frequently used in mathematical writing. The use of symbols makes the writing very short. For example, using symbols, ‘Division of 75 by 15 gives us 5’ can be written in short as ‘75 ÷ 15 = 5’. It is also easier to grasp.

Letters can be used like symbols to make our writing short and simple.

While adding, subtracting or carrying out other operations on numbers, you must have discovered many properties of the operations.

For example, what properties do you see in sums like (9 + 4), (4 + 9)?

The sum of any two numbers and the sum obtained by reversing the order of the two numbers is the same.

Now see how much easier and faster it is to write this property using letters.

• Let us use a and b to represent any two numbers. Their sum will be ‘a + b’.

Changing the order of those numbers will make the addition ‘b + a’. Therefore, the rule will be : ‘For all values of a and b, (a + b) = (b + a).’

Let us see two more examples.

• Multiplying any number by 1 gives the number itself. In short, a × 1 = a.
• Given two unequal numbers, the division of the first by the second is not the same as the division of the second by the first.

In short, if a and b are two different numbers, then (a ÷b) ≠ (b ÷a).

Take the value of a as 8 and b as 4 and verify the property yourself.

Preparation for Algebra Problem Set 54 Additional Important Questions and Answers

Say whether right or wrong.

(1) (15 ÷ 3) =5
Answer:
5 = 5 is right.

(2) (2 x 1) = 1
Answer:
2 = 1 is wrong.

(3) (16 ÷ 8) = (2 x 2)
Answer:
2 = 4 is wrong.

(4) (13 – 7) = 6
Answer:
6 = 6 is right.

(5) (1 x 0) = 1
Answer:
1 = 1 is wrong.

(6) (1 + 0) = 1
Answer:
1 = 1 is right.

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

(1) (12 + 6) (10 X 2)
Answer:
12 + 6 = 18,
10 x 2 = 20
18 < 20
so, (12 + 6) < (10 x 2)

(2) (4 X 5) (10 X 2)
Answer:
4 x 5 = 20,
10 x 2 = 20
20 = 20
so, (4 x 5) = (10 x 2)

(3) (7 + 3) ………….. (3 X 3)
Answer:
7 + 3 = 10,
3 x 3 = 9
10 > 9
so, (7+ 3) > (3 x 3)

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

(1) (8 + ………….. ) = (8 x 1)
Answer:
8 + ……………. = 8,
8 x 1 = 8
8 + 0 = 8
so, (8 + 0) = (8 x 1)

(2) (5 x 6) > (14 x ……….. )
Answer:
5 x 6 = 30,
14 x 1 = 14
14 x 2 = 28
14 x 3 = 42
30 >28
so, (5 x 6) >(14 x 2)

(3) (6 X 7) < ( x 5)
Ans.
6 x 7 = 42,
9 x 5 = 45
42 < 45, 50, 55
so, (6 x 7) < (9 x 5)

Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 16 Preparation for Algebra Problem Set 54 Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 5 Maths Solutions Chapter 16 Preparation for Algebra Problem Set 54

Question 1.
Using brackets, write three pairs of numbers whose sum is 13. Use them to write three equalities.
Answer:
(7 + 6), (8 + 5), (9 + 4). since 7 + 6
= 13,8 + 5
= 13, 9 + 4
= 13.

(7 + 6)
= (8 + 5), (7 + 6)
= (9 + 4) or (8 + 5)
= (9 + 4).

Question 2.
Find four pairs of numbers, one for each of addition, subtraction, multiplication and division that make the number 18. Write the equalities for each of them.
Answer:
(9 + 9), (20 – 2), (9 x 2), (36 ÷ 2).
since 9 + 9
= 18, 20 – 2
= 18, 9 x 2
= 18 and 36 + 2
= 18, so (9 + 9)
= (20 – 2)
= (9 x 2)
= (36 ÷ 2).

Inequality
The values of 7 + 5 and 7 × 5 are 12 and 35 respectively. It means that they are not equal. To represent ‘not equal’, the symbol ‘≠’ is used.

To show that (7 + 5) and (7 × 5) are not equal, we write (7 + 5) ≠ (7 × 5) in short.

This kind of representation is called an ‘inequality’.

(9 – 5) ≠ (15 ÷ 3) means that the expressions (9 – 5) and (15 ÷ 3) are not equal.

If two expressions are not equal, one of them is greater or smaller than the other.

To show greater or lesser values, we use the symbols ‘<’ and ‘>’. Therefore, these symbols can also be used to show inequalities.

The value of (9 – 5) is 4 and the value of (15 ÷ 3) is 5. 4 < 5, so the relation between (9 – 5) and (15 ÷ 3) can be shown as (9 – 5) < (15 ÷ 3) or (15 ÷ 3) > (9 – 5).

Fill in the boxes between the expressions with <, = or > as required.

(1) (9 + 8) [ ] (30 ÷ 2)
9 + 8 = 17,
30 ÷ 2 = 15
17 > 15
Therefore (9 + 8) [ > ] (30 ÷ 2)

(2) (16 × 3) (4 × 12)
16 × 3 = 48,
4 × 12 = 48,
48 = 48
Therefore (16 × 3) [ = ] (4 × 12)

(3) (16 – 5) [ ] (2 × 7)
16 – 5 = 11,
2 × 7 = 14,
11 < 14
Therefore (16 – 5) [ < ] (2 × 7)

Write a number in the box that will make this statement correct.
(1) (7 × 2) = ( [ ] – 6)

The value of the expression 7 × 2 is 14, so the number in the box has to be one that gives 14 when 6 is subtracted from it. Subtracting 6 from 20 gives us 14.

Therefore (7 × 2) = ( [ 20 ] – 6 )
(2) (24 ÷ 3) < (5 + [ ] )
The value of the expression 24 ÷ 3 is 8, so the number in the box has to be such that when it is added to 5, the sum is greater than 8.

Now, 5 + 1 = 6, 5 + 2 = 7, 5 + 3 = 8. So the number in the box has to be greater than 3.

Therefore, writing any number like 4, 5, 6 … onwards will do. It means that this problem has several answers. (24 ÷ 3) < (5 + [ 4 ] ) is one among many answers. Even if that is true, writing only one answer will be enough to complete this statement.

Preparation for Algebra Problem Set 54 Additional Important Questions and Answers

Question 1.
Fill in the blanks.
(1) 7 + 3 = …………….. – ……………..
(2) 7 + 3 = …………….. x ……………..
(3) 7 + 3 = …………….. + ……………..
Answer:
(1) 7 + 3 = 10 and 12 – 2 = 10 or 15 – 5 = 10
(2) 7 + 3 = 10 and 10 x 1 = 10 or 5 x 2 = 10
(3) 7 + 3 = 10 and 20 + 2 = 10 or 30 + 3 = 10

Question 2.
Write the proper number in the box.
(1) 7 + 8 = 10 + [ ]
(2) 7 + 8 = 20 – [ ]
(3) 7 + 8 = 30 + [ ]
(4) 7 + 8 = 5 x [ ]
Answer:
(1) 7 + 8 = 15 so, 10 + [ ] = 15.
∴ [ ] = 15 – 10 = 5

(2) 7 + 8 = 15 s0, 20 – [ ] = 15.
∴[ ] = 20 – 15 = 5

(3) 7 + 8 = 15 so, 30 + [ ] = 15.
∴ [ ] = 30 + 15 = 2

(4) 7 + 8 = 15 so, 5 x [ ] = 15.
∴[ ] = 15 + 5 = 3

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 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: