PSEB 9th Class SST Notes Geography Chapter 1a India: Size and Location

This PSEB 9th Class Social Science Notes Geography Chapter 1a India: Size and Location will help you in revision during exams.

PSEB 9th Class Social Science Notes Geography Chapter 1a India: Size and Location

→ India is a vast country. Geographically, India is the seventh-largest country in the world.

→ From the point of view of the population, after China, India is the second-largest country.

→ The extension of Himalaya mountain and sea from three sides gives India the status of Sub-continent.

→ India has 2.4% of the total land of the world.

→ India’s north-south extension is 3214 km. and the east-west extension is 2933 km. 82°30′ East is considered the standard meridian of India. Its time is 5 hours 30 minutes ahead of Greenwich’s meantime.

PSEB 9th Class SST Notes Geography Chapter 1a India: Size and Location

→ When the meridian of any country is taken as its standard meridian, its time is known as the standard time.

→ A subcontinent is a vast independent geographical unit.

→ This landmass is distinctly separated from the main continent.

→ Indian boundary touches the boundary of seven countries.

→ India has a land boundary of about 15,200 km and a total coastline of about 7516 km.

→ The meaning of SAARC is South Asian Association for Regional Co-operation.

→ India keeps the most important position among other SAARC nations.

→ India has 28 states and 8 union territories.

→ Delhi is National Capital Region (NCR). Its capital New Delhi is the capital of the country.

→ Chandigarh (Punjab, Haryana) and Hyderabad (Andhra Pradesh, Telangana) are the two cities that are the capitals of more than one state.

→ India’s eminent position in the Indian Ocean creates a quite favourable environment for international trade.

PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area Ex 11.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2

1. Estimate the area of the following figures by counting unit squares.

Question (i).
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 1
Solution:
In the given figure, number of squares covered completely = 135
Area of a square = 1 sq. unit
Area of (135 square) figure = 135 sq. units, (approx.)

Question (ii).
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 2
Solution:
In the given figure number of square covered completely = 114
Area of one square = 1 unit
∴ Area of 114 squares = 114 sq units approx
Thus area of given figure = 114 sq units approx.

PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2

2. In the following figures find the area of 

Question (i).
ΔABC
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 3
Solution:
Given length of rectangle = 15 cm
Breadth of rectangle = 8 cm
The diagonal AC divides the rectangle into two triangles ΔABC and ΔADC
So, area of ΔABC = \(\frac {1}{2}\) × Area of rectangle ABCD
= \(\frac {1}{2}\) × length × breadth
= \(\frac {1}{2}\) × 15 × 8
= 60 cm2

Question (ii).
ΔCOD
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 4
Solution:
Given side of square = 6 cm
The diagonals AC and BD divides the square into four equal posses (triangles)
So, area of ΔCOD = \(\frac {1}{4}\) × Area of square
= \(\frac {1}{4}\) × 6 × 6
= 9 cm2

3. Find the area of following parallelograms.

Question (i).
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 5
Solution:
Given base of parallelogram = 9 cm
Height of parallelogram = 6 cm
Area of parallelogram = Base × height
= 9 × 6
= 54 cm2

Question (ii).
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 6
Solution:
Given base of parallelogram = 6.5 cm
Height of parallelogram= 8.4 cm
Area of parallelogram = Base × height
= 6.5 × 8.4
= 54.6 cm2

PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2

4. Find the value of x in the following parallelograms.

Question (i).
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 7
Solution:
Given base (AD) of parallelogram = 5.6 cm
Corresponding height of parallelogram = 9 cm
Area of parallelogram = 5.6 × 9 cm2 ….(1)
Also in the paralleogram, base (AB) = x
Corresponding height of parallelogram = 7 cm
Area of parallelogram will be = x × 7 ….(2)
From (1) and (2), we get
x × 7 = 5.6 × 9
x = \(\frac{5.6 \times 9}{7}\)
= 7.2

Question (ii).
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 8
Solution:
Given base (AB) of parallelogram = 15 cm
Corresponding height = 6 cm
Area of parallelogram =15 × 6 cm2 ….(1)
Also Base (AD || BC) of parallelogram = 9 cm
Corresponding height = x
So area of parallelogram = 9 × x ….(2)
From (1) and (2)
9 × x = 15 × 6
x = \(\frac{15 \times 6}{9}\)
= 10 cm.

5. The adjacent sides of a parallelogram are 28 cm and 45 cm and the altitude on longer side is 18 cm. Find the area of parallelogram.
Solution:
Given base of the parallelogram = 45 cm
Corresponding height = 18 cm
Area of parallelogram = Base × Height
= 45 × 18
= 810 cm2

PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2

6. ABCD is a parallelogram given in figure. DN and DM are the altitudes on side AB and CB respectively. If area of the parallelogram is 1225 cm2, AB = 35 cm and CB = 25 cm, find DN and DM.
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 9
Solution:
In the given parallelogram ABCD
Base (AB) = 35 cm
Let height (DN) = x cm
So area of parallelogram = 35 × x cm2
But given area of parallelogram (ABCD) = 1225 cm2
Therefore 35x = 1225
x = \(\frac {1225}{35}\)
= 35 cm
Similarly, for base (BC) and height (DM)
1225 = BC × DM
\(\frac{1225}{\mathrm{BC}}\) = DM
or DM = \(\frac {1225}{25}\)
= 49 cm.

7. Find the area of the following triangles.

Question (i).
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 10
Solution:
Given base of triangle = 7 cm
Height of triangle = 4.8 cm
Area of triangle = \(\frac {1}{2}\) × Base × Height
= \(\frac {1}{2}\) × 7 × 4.8
= 16.8 cm2.

Question (ii).
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 11
Solution:
Given base of triangle =6 cm
Height of triangle = 9 cm
Area of triangle = \(\frac {1}{2}\) × Base × Height
= \(\frac {1}{2}\) × 6 × 9
= 27 cm2

PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2

8. Find the value of x in the following triangles.

Question (i).
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 12
Solution:
In ΔABC, BC = 8 cm, AC = 15 cm
Area of triangle ABC = \(\frac {1}{2}\) × Base × height
= \(\frac {1}{2}\) × BC × AC
= \(\frac {1}{2}\) × 8 × 15
= 60 cm2 …(1)
Also, in ΔABC, AB = 20 cm
height = x
Area of triangle ABC = \(\frac {1}{2}\) × Base × Height
= \(\frac {1}{2}\) × 20 × x ….(2)
From (1) and (2)
\(\frac {1}{2}\) × 20 × x = 60
x = \(\frac{60 \times 2}{20}\)
x = 6 cm.

Question (ii).
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 13
Solution:
In ΔABC, base (AC) = 25 cm
height = 14 cm
Area of triangle ABC = \(\frac {1}{2}\) × Base × height
\(\frac {1}{2}\) × 14 × 25 ….(1)
Also, in ΔABC, base AB = x cm
height = 20 cm
So, area of ΔABC = \(\frac {1}{2}\) × Base × Height
= \(\frac {1}{2}\) × x × 20 ….(2)
From (1) and (2) we get
\(\frac {1}{2}\) × x × 20 = \(\frac {1}{2}\) × 14 × 25
x = 17.5 cm

9. ABCD is a square, M is a point on AB such that AM = 9 cm and area of ΔDAM is 171 cm2. What is the area of the square ?
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 14
Solution:
Given area of ΔDAM = 171 cm2
Base of triangle = 9 cm
As, area of triangle ΔDAM = \(\frac {1}{2}\) × base × height
171 = \(\frac {1}{2}\) × 9 × (DA)
Hence height (DA) = \(\frac{171 \times 2}{9}\)
= 18 cm
Hence side of square (DA) = 18 cm
Therefore area of square = (side)2
= (18)2
= 324 cm2

PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2

10. ΔABC is right angled at A as shown in figure. AD is perpendicular to BC, if AB = 9 cm, BC = 15 cm and AC = 12 cm. Find the area of ΔABC, also find file length of AD.
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 15
Solution:
Given AB = 9 cm
BC = 15 cm
AC = 12 cm
Let AD = x cm
Area of triangle = \(\frac {1}{2}\) × Base × height
= \(\frac {1}{2}\) × 12 × 9 cm2.
= 54 cm2 ….(1)
Since, AD is perpendicular to BC
So, area of triangle = \(\frac {1}{2}\) × BC × AD
= \(\frac {1}{2}\) × 15 × AD ….(2)
From (1) and (2) we get
\(\frac {1}{2}\) × 15 × AD = 54
AD = \(\frac{54 \times 2}{15}\)
AD = 7.2 cm

11. ΔABC is isosceles with AB = AC = 9 cm, BC = 12 cm and the height AD from A to BC is 4.5 cm. Find the area of ΔABC. What will be the height from B to AC i.e. BN ?
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 16
Solution:
In triangle ABC, Base (BC) = 12 cm
AD = 4.5 cm
AD is perpendicular to BC
So, Area of ΔABC = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × 12 × 4.5 cm
= 27 cm ….(1)
Also, in ΔABC, Base (AC) = 9 cm
Let corresponding height (BN) = x
So area of ΔABC = \(\frac {1}{2}\) × base × height
= \(\frac {1}{2}\) × 9 × BN ….(2)
From (1) and (2)
\(\frac {1}{2}\) × 9 × BN = 27
BN = \(\frac{27 \times 2}{9}\)
= 6 cm.

PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2

12. Multiple choice questions :

Question (i).
Find the height of a parallelogram whose area is 246 cm2 and base is 20 cm.
(a) 1.23 cm2
(b) 13.2 cm2
(c) 12.3 cm2
(d) 1.32 cm2
Answer:
(c) 12.3 cm2

Question (ii).
One of the side and the corresponding height of a parallelogram are 7 cm and 3.5 cm respectively. Find the area of the parallelogram.
(a) 21 cm2
(b) 24.5 cm2
(c) 21.5 cm2
(d) 24 cm2
Answer:
(b) 24.5 cm2

Question (iii).
The height of a triangle whose base is 13 cm and area is 65 cm2 is :
(a) 12 cm
(b) 15 cm
(c) 10 cm
(d) 20 cm
Answer:
(c) 10 cm

Question (iv).
Find the area of an isosceles right angled triangle, whose equal sides are of length 40 cm each.
(a) 400 cm2
(b) 200 cm2
(c) 600 cm2
(d) 800 cm2
Answer:
(d) 800 cm2

PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2

Question (v).
If the sides of a parallelogram are increased to twice of its original length, how much will be the perimeter of the new parallelogram ?
(a) 1.5 times
(b) 2 times
(c) 3 times
(d) 4 times
Answer:
(b) 2 times

Question (vi).
In a right angled triangle one leg is double the other and area is 64 cm2 find the smaller leg.
(a) 8 cm
(b) 16 cm
(c) 24 cm
(d) 32 cm.
Answer:
(a) 8 cm

PSEB 9th Class Science Notes Chapter 15 Improvement in Food Resources

This PSEB 9th Class Science Notes Chapter 15 Improvement in Food Resources will help you in revision during exams.

PSEB 9th Class Science Notes Chapter 15 Improvement in Food Resources

→ Food is the basic requirement of all organisms for their survival.

→ Food supplies nutrients like proteins, carbohydrates, fats, vitamins, and minerals to our bodies.

→ Human beings are dependent on animals and plants for food.

→ Human beings obtain most of this food from plants and animals.

→ Therefore, it is necessary to increase production efficiency for both crops and animals (Livestock, Poultry, Fishes, Honeybees).

PSEB 9th Class Science Notes Chapter 15 Improvement in Food Resources

→ Plants provide food like cereals, pulses, oilseeds, fruits, and vegetables.

→ Animals provide milk and milk products, meat, eggs, honey, etc.

→ Agriculture is the largest industry in India, engaging about 70 percent of our population.

→ The sustainability of agriculture can be achieved by adopting various techniques such as mixed cropping, crop rotation, and crop improvement.

→ Sustainable agriculture is the successful management of resources for agriculture to satisfy the changing human needs while maintaining or enhancing the quality of the environment and conserving natural resources.

→ Mixed cropping is the practice of cultivating more than one crop, that is, growing two or more crops, simultaneously in the same field.

→ In the agroforestry system, crops are grown alongwith trees.

→ The practice of growing different types of crops alternately in the same field is called crop rotation.

→ 16 nutrients are essentially required, four are required in large quantities and are known as macronutrients whereas nine nutrients required in small quantities are known as micronutrients.

→ Manures and fertilizers are the main sources of nutrients supplied to plants.

→ Manures increase soil fertility.

PSEB 9th Class Science Notes Chapter 15 Improvement in Food Resources

→ Fertilizers are commercially manufactured materials from chemicals.

→ Fertilizers increase the yield of crops.

→ Insect pests and diseases can cause 50-70 percent damage to crops.

→ Weeds are unwanted plants that grow in fields.

→ Mineral or chemical fertilizers contain inorganic chemical substances like nitrates, phosphates, potassium, etc.

→ Chemical fertilizers are required in smaller bulk.

→ Chemical fertilizers are expensive.

→ Compost manure mainly consists of rotten vegetables and animal refuse.

→ Biofertilizers are organisms that bring about soil nutrient enrichment.

→ Soil water acts as a medium for the conveyance of salts needed in plant nutrition.

→ Breakthrough in cereal production due to the use of high-yielding varieties, higher dose of fertilisers, and irrigation is known as the green revolution.

→ Varietal improvement is required for higher yield, good quality, and changing situations.

→ Storage losses in agricultural production may be quite high due to the following biotic and abiotic factors:

  • Insects
  • Rodents
  • Fungi
  • Mites
  • Bacteria

PSEB 9th Class Science Notes Chapter 15 Improvement in Food Resources

→ Abiotic factors include excess moisture and inappropriate temperature.

→ These factors cause degradation in quality, loss in weight, poor germinability, and discoloration of produce.

→ Proper drying and fumigation using chemicals that can kill pests are some of the preventive measures.

→ Animal husbandry is the science of rearing, caring, breeding, and utilization of animals.

→ The main aim of animal husbandry is to provide proper food, water, and good shelter to animals.

→ It also involves protecting them from diseases, and their breeding for its best economic use.

→ Cows, buffaloes, fowl, fish, goats, and sheep are useful animals. They provide us food i.e., milk, egg, wool, and meat (fowl and goat).

→ Milk contains all nutrients such as carbohydrates, protein, fats, vitamins, and minerals.

→ Breeds of cattle are divided into three categories: Milch breeds (provide milk e.g. Gir, Sahiwal, Red Sindhi, etc.); draught breeds (used as work animals e.g. Malvi, Hallikar, etc.) and dual-purpose breeds (e.g. Haryana, Deoni, Tharparkar, etc.)

→ Important indigenous breeds of buffaloes are Red Sindhi, Sahiwal, Murrah, Mehsana, Surti, Nili Ravi, etc.

→ Important exotic breeds of milch cows are Holstein and Friesian (Holland), Jersey (USA), Brown-Swiss (Switzerland), etc.

→ Exotic breeds of cows have a longer lactation period and more average annual milk production than those of indigenous breeds.

→ High milk-yielding cross-breeds of cows are Karan-Swiss, Karan-Fries, Frieswal, etc. These have been developed at NDRI Kamal. Kamal-Swiss is a crossbreed of Sahiwal and Brown Swiss.

→ Two main components of cattle feed are roughage (fiber-rich but with low nutrients) and concentrate (rich in proteins, palatable, and easily digestible).

PSEB 9th Class Science Notes Chapter 15 Improvement in Food Resources

→ A good shelter improves the health and yield of animals.

→ Animal Feed: It means food provided to the animal, which are of two types:

  • Roughage: It is a coarse and fibrous food with low nutrition, such as green fodder, silage, hay, and legumes.
  • Concentrates: These are the substances that are rich in one or more nutrients. Cotton seeds, oil seeds, oil cakes, and some cereals like gram and bajra are some typical concentrates.

→ Breeding of cattle is done in two ways natural and artificial.

→ Egg production in chicken: Egg production is the most economic trait.

→ A layer hen starts laying eggs at the age of 20 weeks. The egg production period in a commercial layer is 500 days.

→ For increasing the egg production for the development of new variety with respect to quantity and quality of eggs, the following points are to be considered:

  • Egg number
  • Sexual maturity
  • Egg weight
  • Bodyweight
  • Feed efficiency
  • Egg size
  • Egg shape
  • Shell colour
  • Shell quality
  • The internal quality of egg

→ Poultry farming provides meat, eggs, and manure (poultry waste).

→ Poultry farming is raising domestic fowls, ducks, geese, turkeys, pigeons etc. Poultry production includes egg production and broiler production for poultry meat.

→ To enhance poultry production, crossbreeding is done between Indian and exotic breeds for variety improvement.

→ Fishes may be obtained from seas i.e. marine resources and rivers, lakes i.e. inland resources.

→ To increase production of a choice variety of fishes can be cultured in marine and inland ecosystems.

→ Fishes are caught by fishing nets and gears operated from fishing vessels having modern devices such as echo-sounders and navigational gadgets.

PSEB 9th Class Science Notes Chapter 15 Improvement in Food Resources

→ In freshwater, the induced breeding technique is very useful in obtaining quality fish seeds in large quantities.

→ A composite fish culture system is used commonly to produce table fish.

→ Fisheries are composed of finfishes (true fishes) and shellfish include prawns, mollusks, etc.

→ Beekeeping is done to get honey and wax.

→ The honey flow period is the total time period in which bees collect nectar and pollen.

→ Macronutrients: The mineral nutrients needed by plants in large amounts are called macronutrients e.g. Nitrogen, Phosphorus, Calcium.

→ Micronutrients: The mineral elements needed by plants in very small amounts e.g. Boron, Zinc, Copper.

→ Farm Yard Manure (FYM): Livestock farm waste i.e. cattle excreta (cow dung and urine) is stored in a pit for decomposition. After 1-2 months this is used as FYM in farming practices.

→ Compost: The process in which waste material like vegetable waste, animals refuse, domestic waste, sewage waste, straw, eradicated weeds, etc. is decomposed in pits and is known as composting. The compost is rich in organic matter and nutrients.

→ Green Manure: In the cultivation field prior to the sowing of the seeds, some crops like sun hemp, guar, etc. are grown. After some time these plants are mulched by ploughing. These green plants thus turn into green manures which helps in enriching the soil by N and P.

→ Insecticides: Chemicals that kill insects.

→ Manures: Manures are the organic substances obtained from the decomposition of vegetable and animal wastes by the action of microorganisms.

→ Weeds: The small-sized unwanted plants which grow along with a cultivated crop in a field are called weeds.

→ Weedicides: Chemicals that destroy weeds are called weedicides.

→ Humus: Manures provide humus to the soil. Humus improves the physical and chemical texture of the soil.

→ Mixed cropping is the practice of cultivating more than one crop, that is, growing two or more crops, simultaneously in the same field.

PSEB 9th Class Science Notes Chapter 15 Improvement in Food Resources

→ Crop rotation: The practice of growing different types of crops alternately in the same field is called crop rotation.

→ Animal husbandry: Science of rearing, feeding, caring, and breeding of animals.

→ Roughage: It is a coarse and fibrous food with low nutrition, such as green fodder, silage, hay, and legumes.

→ Concentrates: These are the substances that are rich in one or more nutrients. Cotton seeds, oil seeds, oil cakes, and some cereals like gram and bajra are some typical concentrates.

→ Poultry: Rearing of birds to increase the yield of meat and eggs.

→ Livestock: Domesticated animals like cattle, sheep, goats, camels, horses, pigs, etc.

→ Pisciculture: Pisciculture is the rearing and breeding of fish by a man in ponds, tanks, and artificial reservoirs.

→ Artificial insemination: Introducing the semen of a fertile male into the vagina of a fertile female.

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 5 Fractions Ex 5.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 5 Fractions Ex 5.2

1. Classify the following as proper and improper fractions:

Question (i)
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2 1
Solution:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2 2

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2

2. Express each of the following as mixed fractions, Also represent with diagrams:

Question (i)
\(\frac {27}{5}\)
Solution:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2 3
∴ Mixed fraction
= Quotient \(\frac {Remainder}{Divisor}\) = \(5\frac {2}{5}\)

Question (ii)
\(\frac {13}{4}\)
Solution:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2 4
∴ Mixed fraction
= Quotient \(\frac {Remainder}{Divisor}\) = \(3\frac {1}{4}\)

Question (iii)
\(\frac {43}{8}\)
Solution:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2 5
∴ Mixed fraction
= Quotient \(\frac {Remainder}{Divisor}\) = \(5\frac {3}{8}\)

Question (iv)
\(\frac {51}{7}\)
Solution:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2 6
∴ Mixed fraction
= Quotient \(\frac {Remainder}{Divisor}\) = \(7\frac {2}{7}\)

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2

Question (v)
\(\frac {20}{3}\)
Solution:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2 7
∴ Mixed fraction
= Quotient \(\frac {Remainder}{Divisor}\) = \(6\frac {2}{3}\)

3. Express each of the following mixed fractions as improper fractions:

Question (i)
(i) \(\)2 \frac{1}{3}\(\)
(ii) \(\)5 \frac{2}{7}\(\)
(iii) \(\)4 \frac{3}{5}\(\)
(iv) \(\)3 \frac{3}{4}\(\)
(v) \(\)9 \frac{5}{8}\(\)
Solution:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2 8

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2

4. Express the shaded portion as Improper Fraction and Mixed fraction:

Question (i)
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2 9
Solution:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.2 10

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 5 Fractions Ex 5.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 5 Fractions Ex 5.1

1. Write the fraction representing the shaded portion:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1 1
Solution:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1 2

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1

2. Colour the part according to the given fraction:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1 3
Solution:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1 4

3. Write the fraction for each of the following:

Question (i)
(i) Three-Fourth
(ii) Seven-Tenth
(iii) A Quarter
(iv) Five-Eighth
(v) Three-Twelvth.
Solution:
(i) \(\frac {3}{4}\)
(ii) \(\frac {7}{10}\)
(iii) \(\frac {1}{4}\)
(iv) \(\frac {5}{8}\)
(v) \(\frac {3}{12}\)

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1

4. Write the fraction for the followings:

Question (i)
numerator = 5
denominator = 9
Answer:
\(\frac {5}{9}\)

Question (ii)
numerator = 2
denominator = 11
Answer:
\(\frac {2}{11}\)

Question (iii)
numerator = 6
denominator = 7
Answer:
\(\frac {6}{7}\)

5. Write the numerator and the denominator for the followings:

Question (i)
\(\frac {2}{3}\)
Solution:
Given fraction is \(\frac {2}{3}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 2
and Denominator = 3

Question (ii)
\(\frac {1}{4}\)
Solution:
Given fraction is \(\frac {1}{4}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 1
and Denominator = 4

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1

Question (iii)
\(\frac {5}{11}\)
Solution:
Given fraction is \(\frac {5}{11}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 5
and Denominator = 11

Question (iv)
\(\frac {9}{13}\)
Solution:
Given fraction is \(\frac {9}{13}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 9
and Denominator = 13

Question (iv)
\(\frac {17}{16}\) = \(\frac {Numerator}{Denominator}\)
∴ Numerator = 17 and
Denominator = 16

6. Express:

Question (i)
1 day as a fraction of 1 week.
Solution:
We know 1 week = 7 days
∴ Required fraction= \(\frac {1}{7}\)

Question (ii)
40 seconds as a fraction of 1 minute.
Solution:
We know 1 minute = 60 seconds
∴ Required fraction = \(\frac {40}{60}\) or \(\frac {2}{3}\)
(Dividing both terms by 20)

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1

Question (iii)
15 hours as fraction of 1 day.
Solution:
We know 1 day = 24 hours
∴ Required fraction = \(\frac {15}{24}\) or \(\frac {5}{8}\)
(Dividing both terms by 3)

Question (iv)
2 months as a fraction of 1 year.
Solution:
We know 1 year = 12 months
∴ Required fraction = \(\frac {2}{12}\) or \(\frac {1}{6}\)
(Dividing both terms by 2)

Question (v)
45 cm as a fraction of 1 metre.
Solution:
We know 1 metre = 100 cm
∴ Required fraction = \(\frac {45}{100}\) or \(\frac {9}{20}\)
(Dividing both terms by 5)

7. Write the numbers from 1 to 25.

Question (i)
What fraction of them are even numbers?
Solution:
Numbers from 1 to 25 are:
1,2, 3, 4,5,6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25 i.e. 25 in number.
Even numbers out of these numbers are:
2, 4, 6, 8, 10, 12, 14, 16, 18; 20, 22, 24 i.e. 12 in number
∴ Required fraction = \(\frac {12}{25}\)

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1

Question (ii)
What fraction of them are prime numbers?
Solution:
Prime number out of these numbers are:
2, 3, 5, 7, 11, 13, 17, 19, 23 i.e. 9 in number
∴ Required fraction = \(\frac {9}{25}\)

Question (iii)
What fraction of them are multiples of 3?
Solution:
Multiples of 3 out of these numbers are:
3, 6, 9, 12, 15, 18, 21, 24 i.e. 8 in number
∴ Required fraction = \(\frac {8}{25}\)

8. In class 6th, there are 24 boys and 18 girls. What fraction of total students represent boys and girls?
Solution:
Boys = 24
Girls = 18
Total students = 24 + 18 = 42
Fraction which represents boys
= \(\frac {24}{42}\) or \(\frac {4}{7}\)
(Dividing both terms by 6)
Fraction which represents girls
= \(\frac {18}{42}\) or \(\frac {3}{7}\)
(Dividing both terms by 6)

9. A bag contains 6 red balls and 7 blue balls. What fraction of balls represent red and blue colour?
Solution:
Red balls = 6
Blue balls = 7
Total number of red and blue balls = 6 + 7 = 13
Fraction which represents red balls = \(\frac {6}{13}\)
Fraction which represents blue balls = \(\frac {7}{13}\)

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1

10. Sidharth has a cake. He cuts it into 10 equal parts. He gave 2 parts to Naman, 3 parts to Nidhi, 1 parts to Seema and the remaining four parts he kept for himself. Find:

Question (i)
What fraction of cake, he gave to Naman?
Solution:
Total parts = 10
Fraction of cake, he gave to Naman = \(\frac {2}{10}\) or \(\frac {1}{5}\)

Question (ii)
What fraction of cake, he gave to Nidhi?
Solution:
Fraction of cake, he gave to Nidhi = \(\frac {3}{10}\)

Question (iii)
What fraction of cake, he kept for himself?
Solution:
Fraction of cake, he kept for himself = \(\frac {4}{10}\) or \(\frac {2}{5}\)

Question (iv)
Who has more cake than others?
Solution:
Sidharth has more cake than others

11. In a box, there are 12 apples, 7 oranges and 5 guavas. What fraction of fruits in box represents each?
Solution:
Apples = 12, Oranges = 7
and Gauvas = 5
Total fruits = 12 + 7 + 5 = 24.
Fraction which represents apples
= \(\frac {12}{24}\) or \(\frac {1}{2}\)
Fraction which represents oranges
= \(\frac {7}{24}\)
Fraction which represents Gauvas
= \(\frac {5}{24}\)

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1

12. Dishmeet has 20 pens. He gives one-fourth to Balkirat. How many pens Dishmeet and Balkirat have?
Solution:
Total pens = 20
Pens Balkirat has = One-fourth of 20
= \(\frac {22}{7}\) × 20 = 5
Pens Dishmeet has = 20 – 5 = 15

13. Represent the following fraction on the number line?

Question (i)
\(\frac {2}{5}\)
Solution:
In order to represent \(\frac {2}{5}\) on number line, we divide the gap between 0 and 1 into 5 equal parts, which are, (as shown)
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1 5

Question 2.
\(\frac {5}{7}\)
Solution:
In order to represent \(\frac {5}{7}\) on number line, we divide the gap between 0 and 1 into 7 equal parts, which are, (as shown)
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1 6

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1

Question 3.
\(\frac {3}{10}\), \(\frac {5}{10}\), \(\frac {1}{10}\)
Solution:
In order to represent \(\frac {3}{10}\), \(\frac {5}{10}\), \(\frac {1}{10}\) on number line, we divide the gap between 0 and 1 into 10 equal parts, which are (as shown)
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1 7

Question 4.
\(\frac {3}{8}\), \(\frac {5}{8}\), \(\frac {7}{8}\)
Solution:
In order to represents \(\frac {3}{8}\), \(\frac {5}{8}\), \(\frac {7}{8}\) on number line, we divide the gap between 0 and 1 into 8 equal parts, which are (as shown)
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1 8

14. Find:

(i) \(\frac {3}{5}\) of 20 books
(ii) \(\frac {5}{8}\) of 32 pens
(iii) \(\frac {1}{6}\) of 36 copies 6
(iv) \(\frac {4}{7}\) of 21 apples
(v) \(\frac {3}{4}\) of 28 pencils.
Solution:
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1 9
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1 10

PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1

15. Balkirat had a box of 36 erasers. He gave \(\frac {1}{2}\) of them to Rani, \(\frac {2}{9}\) of them to Yuvraj and keeps the rest.

Question (i)
(i) How many erasers does Rani get?
(ii) How many erasers does Yuvraj get?
(iii) How many erasers does Harnik keep?
Solution:
Total erasers = 36
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1 11
(iii) Erasers Harnik gets = 36 – (18 + 8)
= 36 – 26
= 10

16. State True/False

Question (i)
PSEB 6th Class Maths Solutions Chapter 5 Fractions Ex 5.1 12
Solution:
(i) False
(ii) True
(iii) True
(iv) True

PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.5

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5

Question 1.
ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.
Answer:
PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.5 1
In ∆ ABC, draw l, the perpendicular bisector of side AB and m, the perpendicular bisector of side BC. Name the point of intersection of l and m as P.
P is a point on the perpendicular bisector of AB. Hence, P is equidistant from A and B.
∴ PA = PB
P is a point on the perpendicular bisector of BC. Hence, P is equidistant from B and C.
∴ PB = PC
Thus, PA = PB = PC
Hence, P is the required point which is equidistant from all the vertices of ∆ ABC.
Note: Since ∆ ABC given here is an acute angled triangle, point P lies in the interior of ∆ ABC. If ∆ ABC is a right angled Mangle, point P lies on the hypotenuse. Actually, in that case, point P will be the midpoint of the hypotenuse. Lastly, if ∆ ABC is an obtuse angled triangle, point P lies in the exterior of ∆ ABC. This point P is called the circumcentre of ∆ ABC.

PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.5

Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.5 2
Answer:
In ∆ ABC, draw the bisectors of ∠B and ∠C to intersect each other at point I.
I is a point on the bisector of ∠ B. Hence, I is equidistant from sides BA and BC. Similarly, I is a point on the bisector of ZC. Hence, I is equidistant from sides BC and CA.
Thus, point I is the required point which is equidistant from all the three sides AB, BC and CA of ∆ ABC.
This point I is called the incentre of ∆ ABC. It always lies in the interior of ∆ ABC irrespective of the type of ∆ ABC.

Question 3.
In a huge park, people are concentrated at three points (see the given figure):
PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.5 3

A: where there are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exit.
Where should an ice cream parlour be set up so that maximum number of persons can approach it ?
(Hint: The parlour should be equidistant from A, B and C.)
Answer:
First of all, construct ∆ ABC with the given points A, B and C as vertices. Then, as shown in example 1, draw the perpendicular s bisectors of any two sides of ∆ ABC and name their point of intersection as P.

Now, the ice cream parlour should be set up at the location given by point P as it is equidistant from all the three places (points) A, B and C.

PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.5

Question 4.
Complete the hexagonal and star shaped < Rangolies [see figure (1) and (2)] by filling them with as many equilateral triangles of > side 1 cm as you can. Count the number < of triangles in each case. Which has more triangles?
PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.5 4
Answer:
In figure (1), if we join the opposite vertices, we get three longest diagonals of hexagon ABCDEE By the intersection of these diagonals we get point O and six equilateral triangles – ∆ OAB, ∆ OBC, ∆ OCD, ∆ ODE, ∆ OEF and ∆ OFA. Each side of all these equilateral triangles will measure 5 cm. In each of these six triangles, we can fill 25 (1 + 3 + 5 + 7 + 9) equilateral triangles with side 1 cm each. Hence in the hexagonal Rangoli ABCDEF, we can fill 25 × 6 = 150 triangles.
PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.5 5
Similarly, in figure (2), we can fill 150 triangles in the inner hexagon and 150 triangles in the six triangles on the boundary of the hexagon. Thus, in figure (2), 150 + 150 = 300 triangles can be filled.
Hence, more triangles can be filled in figure (2).

PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Answer:
PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4 7
In ∆ ABC, ∠ B is a right angle.
∴ ∠ B = 90° and AC is the hypotenuse.
In ∆ ABC,
∠A + ∠B + ∠C = 180°
∴ ∠ A + 90° + ∠ C = 180°
∴ ∠ A + ∠ C = 90°
Now, ∠ A and ∠ C are both positive (in degrees) and their sum is 90°.
∴ ∠ A < 90° and ∠ C < 90°
∴ ∠ A < Z B and ∠ C < ∠ B
∴ BC < AC and AB < AC (Theorem 7.7)
Hence, hypotenuse AC is greater than each of the other two sides BC and AB. Thus, the hypotenuse is the longest side in a right angled triangle.

PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4

Question 2.
In the given figure, sides AB and AC of ∆ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠ QCB. Show that AC > AB.
PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4 1
Answer:
∠ PBC < ∠ QCB (Given)
∴ – ∠ PBC > – ∠ QCB (Multiplying an inequality by (-1), it reverses)
∴ 180° – ∠ PBC > 180° – ∠ QCB (Adding 180° on both the sides) …………. (1)
Now, ∠ ABC and ∠ PBC as well as ∠ ACB and ∠ QCB from a linear pair.
∴ ∠ ABC = 180° – ∠ PBC and
∠ ACB = 180° – ∠ QCB
Substituting these values in (1), we get
∠ ABC > ∠ ACB
Now, in ∆ ABC, ∠ ABC > ∠ ACB A
∴ AC > AB (Theorem 7.7)

PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4

Question 3.
In the given figure, ∠ B < ∠ A and ∠ C < ∠ D. Show that AD < BC.
PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4 2
Answer:
In ∆ OAB, ∠ B < ∠ A
∴ OA < OB (Theorem 7.7) …………….. (1)
In ∆ OCD, ∠ C < ∠ D
∴ OD < OC (Theorem 7.7) ……………… (2)
Adding (1) and (2),
OA + OD < OB + OC
∴ AD < BC (As O is the point of intersection of AD and BC, it lies on both the line segments.)

PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD ,(see the given figure). Show that ∠A > ∠C and ∠B > ∠D.
PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4 3
Answer:
PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4 4
Construction: In quadrilateral ABCD, draw diagonal AC.
AB is the smallest side of ABCD and CD is the longest side of ABCD.
∴ AB < BC and AD < CD.
In ∆ ABC, AB < BC
∴ ∠ ACB < ∠ BAC …… (1)
In ∆ CDA, AD < CD
∴ ∠ DCA < ∠ DAC ……… (2)
Adding (1) and (2),
∠ ACB + ∠ DCA < ∠ BAC + ∠ DAC
∴ ∠ BCD < ∠ BAD
∴ ∠ BAD > ∠ BCD
Thus, in quadrilateral ABCD, ∠ A > ∠ C. Similarly, after constructing diagonal BD and using the inequalities in A ABD and A CBD, it can be proved that ∠ B > ∠ D.

PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4

Question 5.
In the given figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.
PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4 5
Answer:
PS is the bisector of ∠QPR.
∴ ∠QPS = ∠RPS = \(\frac{1}{2}\) ∠QPR ……………. (1)
∠ PSR is an exterior angle of A PQS and ∠ PSQ is an exterior angle of A PRS.
∴ PSR = ∠ Q + ∠QPS and
∠PSQ = ∠R + ∠RPS …………….. (2)
Now, in A PQR, PR > PQ
∴ ∠ Q > ∠ R
∴ ∠Q + \(\frac{1}{2}\) ∠ QPR > ∠R + \(\frac{1}{2}\) ∠ QPR
∴ ∠Q + ∠QPS > ∠R + ∠RPS [from (1)]
∴ ∠PSR > ∠PSQ

PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4

Question 6.
Show that of all line segments drawn from a given point not on a given line, the perpendicular line segment is the shortest.
Answer:
PSEB 9th Class Maths Solutions Chapter 7 Triangles Ex 7.4 6
AB is a line and P is a point not on AB.
PM is the perpendicular line segment drawn from P to line AB.
N is any point on AB, other than M.
In ∆ PMN, ∠ M = 90°
∴ ∠ N < 90°
Thus, in ∆ PMN, ZN < ZM.
∴ PM < PN
This is true for any location of point N.
Hence, of all the line segments drawn from a point not on a given line, the, perpendicular line segment is the shortest.

PSEB 9th Class Maths MCQ Chapter 7 Triangles

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles MCQ Questions with Answers.

PSEB 9th Class Maths Chapter 7 Triangles MCQ Questions

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
In ∆ ABC, ∠A = ∠C, AC = 5 and BC = 4. Then, the perimeter of ∆ ABC is ……………. .
A. 9
B. 14
C. 13
D. 15
Answer:
C. 13

PSEB 9th Class Maths MCQ Chapter 7 Triangles

Question 2.
In ∆ PQR, PQ = PR, QR is extended to S and ∠PRS = 110°. Then, ∠PQR = ……………… .
A. 30°
B. 50°
C. 80°
D. 70°
Answer:
D. 70°

Question 3.
In ∆ ABC and ∆ DEF, AB = DE, BC = EF and ∠B = ∠E. If the perimeter of ∆ ABC is 20, then the perimeter of ∆ DEF is ……………. .
A. 10
B. 20
C. 15
D. 40
Answer:
B. 20

Question 4.
In ∆ ABC and ∆ PQR, AB = PQ, ∠A = ∠P and ∠B = ∠Q. If ∠A + ∠C = 130°, then ∠Q = …………….. .
A. 65°
B. 130°
C. 50°
D. 100°
Answer:
C. 50°

PSEB 9th Class Maths MCQ Chapter 7 Triangles

Question 5.
In ∆ PQR, ∠P = ∠Q = ∠R. If PQ = 6, then the perimeter of ∆ PQR is ………………. .
A. 12
B. 9
C. 18
D. 24
Answer:
C. 18

Question 6.
In ∆ ABC, AB < AC. Then …………… holds good.
A. ∠A < ∠B
B. ∠B < ∠C
C. ∠C < ∠A
D. ∠C < ∠B
Answer:
D. ∠C < ∠B

Question 7.
In ∆ PQR, ∠R > ∠Q. Then, ………………. holds good.
A. PQ > PR
B. QR > PQ
C. PR > PQ
D. PQ > QR
Answer:
A. PQ > PR

PSEB 9th Class Maths MCQ Chapter 7 Triangles

Question 8.
In ∆ ABC, AB > BC and BC > AC. Then, the smallest angle of ∆ ABC is …………………. .
A. ∠A
B. ∠C
C. ∠B
D. ∠A or ∠C
Answer:
C. ∠B

Question 9
………………….. cannot be the measures of the sides of a triangle.
A. 10, 12, 14
B. 2, 3, 4
C. 8, 9, 10
D. 2, 4, 10
Answer:
D. 2, 4, 10

Question 10.
In ∆ PQR, PQ = 4, QR = 6 and PR = 5. Then, …………….. is the angle with greatest measure in ∆ PQR.
A. ∠P
B. ∠Q
C. ∠R
D. ∠QRP
Answer:
A. ∠P

PSEB 9th Class Maths MCQ Chapter 7 Triangles

Question 11.
In ∆ XYZ, ∠X = 45° and ∠Z = 60°. Then, …………….. is the longest side of ∆ XYZ.
A. XY
B. YZ
C. XZ
D. XY or YZ
Answer:
C. XZ

Question 12.
In ∆ ABC, ∠B = 30° and BC is extended to D. If ∠ACD = 110°, then the longest side of ∆ ABC is ………………. .
A. AB
B. BC
C. CA
D. AB or AC
Answer:
B. BC

Question 13.
In ∆ ABC, AB = 4 and BC = 7, Then, ……………… holds good.
A. AC < 7
B. AC > 4
C. 4 < AC < 7
D. 3 < AC < 11
Answer:
D. 3 < AC < 11

PSEB 9th Class Maths MCQ Chapter 7 Triangles

Question 14.
In ∆ PQR, PQ = 3 and QR = 7. Then, …………….. holds good.
A. PR = 4
B. PR = 10
C. 10 > PR > 4
D. 7 > PR > 3
Answer:
C. 10 > PR > 4

Question 15.
In ∆ ABC, the bisectors of ∠B and ∠C intersect at I. If ∠A = 70°, then ∠BIC = ………………… .
A. 35°
B. 75°
C. 100°
D. 125°
Answer:
D. 125°

PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.1

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 11 Perimeter and Area Ex 11.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1

1. Find the perimeter and the area of a rectangle having :
(i) Length = 28 cm, Breadth = 15 cm
(ii) Length = 9.4 cm Breadth = 2.5 cm
Solution:
(i) Given length of rectangle = 28 cm
Breadth of rectangle = 15 cm
Perimeter of rectangle = 2 [length + Breadth]
= 2 [28 + 15]
= 2 × 43
= 86 cm

Area of rectangle = length × Breadth
= 28 × 15
= 420 cm2

(ii) Perimeter of rectangle = 2 [9.4 + 2.5]
= 2 × 11.9
= 23.8 cm
Area of rectangle = 9.4 × 2.5
= 23.5 cm2

2. Find the perimeter and the area of a square whose side measures
(i) 29 cm
(ii) 8.3 cm
Solution:
(i) Given side of square = 29 cm
Perimeter of square = 4 × side
= 4 × 29
= 116 cm
Area of square = (side)2
= (29)2
= 841 cm2

(ii) Perimeter of square = 4 × 8.3
= 33.2 cm
Area of square = 8.3 × 8.3
= 68.89 cm2

PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.1

3. The perimeter of a square park is 148 m. Find its area.
Solution:
Given the perimeter of square park = 148 m
Side of the square park = \(\frac{perimeter}{4}\)
= \(\frac {148}{4}\)
Area of the square park = (side)2
= (37)2
= 1369 m2

4. The area of a rectangle is 580 cm2. Its length is 29 cm. Find its breadth and also, the perimeter.
Solution:
Given area of rectangle = 580 cm2
Length of the rectangle = 29 cm
Let breadth of the rectangle = b cm
Area of the rectangle = length × breadth
580 = 29 × b
\(\frac {580}{29}\) = b
b = 20 cm
Perimeter of rectangle = 2 [length + breadth]
= 2 [29 + 20]
= 2 × 49
= 98 cm

5. A wire is in the shape of a rectangle. Its length is 48 cm and breadth is 32 cm. If the same wire is rebent into the shape of a square, what will be the measure of each side. Also, find which shape encloses more area and by how much ?
Solution:
Given length of the rectangle = 48 cm
Breadth of the rectangle = 32 cm
Perimeter of the rectangle = 2 [length + breadth]
= 2 [48 + 32]
= 2 × 80
= 160 cm
Let side of square = a cm
Perimeter of the square = 4 × a
Since wire is rebent into the shape of a square
Perimeter of square = Perimeter of rectangle
4 a = 160
Therefore, a = \(\frac {160}{4}\)
= 40 cm
Area of square = (side)2
= 40 × 40
= 1600 cm2
Area of rectangle = length × breadth
= 48 × 32
= 1536 cm2
∴ Square encloses more area by 64 cm2

PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.1

6. The area of a square park is the same as that of a rectangular park. If the side of the square park is 75 m and the length of the rectangular park is 125 m, find the breadth of the rectangular park. Also, find the perimeter of rectangular park.
Solution:
Given side of square park = 75 m
Area of square park = (75)2
= 75 × 75
= 5625 m2
Length of rectangular park = 125 m
Let breadth of rectangular park = b m
Area of rectangular park = length × breadth
= 125 × b m2
Given that
Area of rectangular park = Area of square park
125 × b = 5625
= 45 m
Perimeter of rectangular park = 2 [length + breadth]
= 2 [125 + 45]
= 2 × 170
= 340 m

7. A door of length 2.5 m and breadth 1.5 m is fitted in a wall. The length of wall, is 9 m and breadth is 6 m. Find the cost of painting the wall, if the rate of painting the wall is ₹ 30 per m2.
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.1 1
Solution:
Length of door = 2.5 m
Breadth of door = 1.5 m
Area of door = length × breadth
= 2.5 × 1.5
= 3.75 m2
Area of wall = 9 × 6
= 54 m2
Area of wall painting = Area of wall including door – Area of door
= 54 – 3.75
= 50.25 m2
Cost of painting 1 m2 of wall = ₹ 30
Cost of painting 50.25 m2 of wall = ₹ 50.25 × 30
= ₹ 1507.50

PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.1

8. A door of dimensions 3 m × 2 m and a window of dimensions 2.5 m × 1.5 m is fitted in a wall. The length of the wall is 7.8 m and breadth is 3.9 m. Find the cost of painting the wall, if the rate of painting the wall is ₹ 25 per m2.
Solution:
Area of door = 3 × 2 = 6 m2
Area of window = 2.5 m × 1.5 m
= 3.75 m2
Area of wall = 7.8 m × 3.9 m
= 30.42 m2
Area of painting the wall = Area of wall – Area of door – Area of window
= 30.42 – 6 – 3.75
= 20.67 m2
Cost of painting the wall = ₹ 25 × 20.67
= ₹ 516.75

9. Find the area and the perimeter of the following figures.
PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.1 2
Solution:
(i) Perimeter of the given figure
= AB + BC + CD + DE + EF + FG + GH + HA
= 2 + 3.5 + 3 + 2 + 5 + 3.5 + 10 + 9
= 38 cm2
Area of the figure = Area of rectangle ABCJ + Area of rectangle JDEI + Area of rectangle IFGH
= 2 × 3.5 + 5 × 2 + 10 × 3.5
= 7 + 10 + 35
= 52 cm2

(ii) Perimeter of the given figure
= 8cm + 5 cm + 1.5 cm + 2.5 cm + 2.5 cm + 1.5 cm + 1.5 cm + 1.5 cm + 2.5 cm + 1.5 cm
= 29 cm
Area of the given figure = Area of rectangle I + Area of rectangle II + Area of rectangle III
= 8 cm × 1.5 cm + 3.5 cm × 1.5 cm + 1.5 cm × 1.5 cm
= 12 cm2 + 5.25 cm2 + 2.25 cm
= 19.5 cm2

PSEB 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.1

10. Multiple Choice Questions :

Question (i).
What is the area of a rectangle of dimensions 12 cm × 10 cm ?
(a) 44 cm2
(b) 120 cm2
(c) 1200 cm2
(d) 1440 cm2
Answer:
(b) 120 cm2

Question (ii).
Find the breadth of a rectangle whose length is 12 cm and perimeter is 36 cm.
(a) 6 cm
(b) 3 cm
(c) 9 cm
(d) 12 cm
Answer:
(a) 6 cm

Question (iii).
If each side of a square is 1 m then its area is ?
(a) 10 cm2
(b) 100 cm2
(c) 1000 cm2
(d) 10000 cm2
Answer:
(d) 10000 cm2

Question (iv).
Find the area of a square whose perimeter is 96 cm.
(a) 576 cm2
(b) 626 cm2
(c) 726 cm2
(d) 748 cm2.
Answer:
(a) 576 cm2

Question (v).
The area of a rectangular sheet is 500 cm2. If the length of the sheet is 25 cm, what is its breadth ?
(a) 30 cm
(b) 40 cm
(c) 20 cm
(d) 25 cm.
Answer:
(c) 20 cm

Question (vi).
What happens to the area of a square, if its side is doubled ?
(a) The area becomes 4 times, the area of original square.
(b) The area becomes \(\frac {1}{4}\) times, the area of original square.
(c) The area becomes 16 times, the area of original square.
(d) The area becomes \(\frac {1}{6}\) times, 6 the area of original square.
Answer:
(a) The area becomes 4 times, the area of original square.

PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 4 Integers Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 4 Integers Ex 4.2

1. Using number line write the integer which is:

Question (a)
5 less than -1
Solution:
5 less than -1 = ?
We need to find the integer which is 5 less than -1.
So, we shall start with ‘-1 ’ and proceed 5 steps to the left of -1 as shown below:
PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2 1
Therefore 5 less than -1 is -6

PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2

Question (b)
5 more than -5
Solution:
5 more than -5 = ?
We need to find the integer which is 5 more than -5.
So, we shall start with -5 and proceed 5 steps to the right of -5 as shown below:
PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2 2
Therefore 5 more than -5 is zero.

Question (c)
2 less than 5
Solution:
2 less than 5 = ?
We need to find the integer which is 2 less than 5.
So, we shall start with 5 and proceed
2 steps to the left of 5 as shown below:
PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2 3
Therefore 2 less than 5 is 3.

Question (d)
3 less than -2.
Solution:
3 less than – 2 = ?
We need to find the integer which is 3 less than -2.
So, we shall start with -2 and proceed 3 steps to the left of -2 as shown below:
PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2 4
Therefore 3 less than -2 is -5.

PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2

2. Using number line, add the following integers:

Question (a)
9 + (-3)
Solution:
On number line we shall start from 0 and move 9 steps to the right of zero. Then we shall move three steps to the left of ‘+9’. We finally reach at +6. Thus, +6 is the answer.
PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2 5
Hence, 9 + (-3) = +6

Question (b)
5 + (-11)
Solution:
On number line we shall start from 0 and move 5 steps to the right of zero. Then we shall move eleven steps to the left of ‘+5’. We finally reach at ‘-6’. Thus, -6 is the answer.
PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2 6
Hence, 5 + (-11) = -6

Question (c)
(-1) + (-4)
Solution:
On number line we first move 1 step to the left of zero. Then we moved ahead 4 steps to the left of ‘-1\ We finally reaches at -5. Thus -5 is the answer.
PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2 7
Hence, (-1) + (-4) = -5

Question (d)
(-5) + 12
Solution:
On number line we shall move 5 steps to the left of zero. Then we shall move 12 steps to the right of ‘-5’ and finally reach at ‘+7’. Thus +7 is the answer.
PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2 8
Hence, (-5) + 12 = +7

PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2

Question (e)
(-1) + (-2) + (-4)
Solution:
Step I: On number line we shall move one step to the left of zero suggested by minus sign of ‘-1’.
Step II: Then we shall move 2 steps to the left of -1 suggested by minus sign of -2 and we reach at ‘-3’.
Step III: Then again we shall move 4 steps to the left of ‘-3’ suggested by minus sign of -4 and finally we reach at-7.
PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2 9
Hence, (-1) + (-2) + (-4) = -7

Question (f)
(-2) + 4 + (-5)
Solution:
Step I: On number line we shall move 2 steps to the left of zero suggested by minus sign of ‘-2’.
Step II: Then we shall move 4 steps to the right of ‘-2’ suggested by plus sign of ‘44’ and we reach at +2.
Step III: Then we shall move 5 steps to the left of ‘+2’ as suggested by minus sign of ‘-5’ and finally we reach at -3. Thus answer is ‘-3’.
PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2 10
Hence, (-2) + 4 + (-5) = -3

Question (g)
(-3) + (5) + (-4).
Solution:
Step I: On number line we shall move 3 steps to the left of zero as suggested by minus sign of ‘-3’.
Step II: Then we shall move 5 steps to the right of ‘-3’ as suggested by plus sign of ‘+5’ and we reach at ‘+2’.
Step III: Then we shall move 4 steps to the left of ‘+2’ as suggested by minus sign of ‘4’ and finally we reach at ‘-2’. Thus ‘-2’ is the answer.
PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2 11
Hence, (-3) + (5) + (-4) = -2

3. Add without using number line:

Question (a)
18 + 13
Solution:
18 + 13 = 31

PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2

Question (b)
18 + (- 13)
Solution:
18 + (- 13) = + (18 – 13) = +5

Question (c)
(-18) + 13
Solution:
(-18) + 13 = – (18 – 13) = -5

Question (d)
(-18) + (-13)
Solution:
(-18) + (-13) = -(18 + 13) = -31

Question (e)
180 + (-200)
Solution:
180 + (-200) = -(200 – 180) = -20

PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2

Question (f)
111 + (-67)
Solution:
111 + (-67) = (777 – 67) = 710

Question (g)
1262 + (-366) + (-962)
Solution:
1262 + (-366) + (-962)
= 1262 – (366 + 962)
= 1262 – 1328
= -(1328 – 1262) = -66

Question (h)
30 + (-27) + 21 + (-19) + (-3) + 11 + (-9)
Solution:
30 + (-27) + 21 + (-19) + (-3) + 11 + (-9)
= 30 + 21 + 11 + (-27) + (-19) + (-3) + (-9)
= 62 + (-58) = 62 – 58 = 4

Question (i)
(-7) + (-9) + 4 + 16
Solution:
(-7) + (-9) + 4+16 = (-16) + 20 = 4

PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2

Question (j)
37 + (-2) + (-65) + (-8).
Solution:
37 + (-2) + (-65) + (-8)
= 37 – (2 + 65 + 8)
= 37 – 75 = -38

4. Write the successor and predecessor of the following:

Question (a)
-15
Solution:
Successor of -15 = -15 + 1 = -14
Predecessor of -15 = -15 – 1 = -16

Question (b)
27
Solution:
Successor of 27 = 27 + 1 = 28
Predecessor of 27 = 27 – 1 = 26

Question (c)
-79
Solution:
Successor of -79 = -79 + 1 = -78
Predecessor of -79 = -79- 1 = -80

PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2

Question (d)
0
Solution:
Successor of 0 = 0+ 1 = 1
Predecessor of 0 = 0 – 1 = -1

Question (e)
29
Solution:
Successor of 29 = 29 + 1 = 30
Predecessor of 29 = 29 – 1 = 28

Question (f)
-18
Solution:
Successor of -18 = -18 + 1 = -17
Predecessor of -18 = -18 – 1 = -19

Question (g)
-21
Solution:
Successor of -21 = -21 + 1 = -200
Predecessor of -21 = -21 – 1 = -22

Question (h)
99
Solution:
Successor of 99 = 99 + 1 = 100
Predecessor of 99 = 99 – 1 = 98

Question (i)
-1
Solution:
Successor of-1 = -1 + 1 = 0
Predecessor of -1 = -1 – 1 = -2

Question (j)
-13.
Solution:
Successor of -13 = -13 + 1 = -12
Predecessor of -13 = -13 – 1 = -14

PSEB 6th Class Maths Solutions Chapter 4 Integers Ex 4.2

5. Complete the following addition table:

+ -3 -4 -2 +1 +2 +3
-2
-3
0
+1
+2

Solution:

+ -3 -4 -2 + 1 +2 +3
-2 -5 -6 -4 -1 0 + 1
-3 -6 -7 -5 -2 -1 0
0 -3 -4 -2 + 1 +2 +3
+ 1 -2 -3 -1 +2 +3 +4
+2 -1 -2 0 +3 +4 +5