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R.D. Sharma Solutions Class 9th: Ch 13 Linear Equations in Two Variables Exercise 13.2

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Chapter 13 Linear Equations in Two Variables R.D. Sharma Solutions for Class 9th Exercise 13.2

Exercise 13.2

1. Write two solutions for each of the following equations:
(i) 5x–2y = 7
(ii) x = 6y
(iii) x + πy = 4
(iv) 2/3x – y = 4.

Solution








2. Write two solutions of the form x = 0, y = a and x = b, y = 0 for each of the following equations:

(i) 5x − 2y = 10
(ii) -4x + 3y = 12
(iii) 2x + 3y =24

Solution


(i) We are given,

5x – 2y = 10
Substituting x = 0 in the given equation,
We get;
5×0 – 2y = 10
– 2y = 10
– y = 10/2
y = – 5
Thus x =0 and y = -5 is the solution of 5x-2y = 10
Substituting y = 0 in the given equation, we get 5x —2 x 0 = 10
5x = 10
x = 10/2
x = 2
Thus x =2 and y = 0 is a solution of 5x-2y = 10

(ii) We are given, – 4x + 3y = 12

Substituting x = 0 in the given equation,
we get;
-4 × 0 + 3y = 12
3y = 12
y = 4
Thus x = 0 and y = 4 is a solution of the -4x + 3y = 12
Substituting y = 0 in the given equation, we get;
- 4 x + 3 × 0 = 12
- 4x = 12
x = -12/4
x = -3
Thus x = -3 and y =0 is a solution of -4x + 3y = 12

(iii) We are given, 2x + 3y = 24

Substituting x = 0 in the given equation, we get;
× 0 + 3y = 24
3y =24
y = 24/3
y = 8
Thus x = 0 and y = 8 is a solution of 2x+ 3y = 24
Substituting y = 0 in the given equation, we get;
2x +3 × 0 = 24
2x = 24
x = 24/2
x = 12
Thus x =12 and y = 0 is a solution of 2x + 3y = 24

3. Check which of the following are solutions of the equations 2x - y = 6 and which are not:


Solution


We are given, 2x – y = 6

(i) In the equation 2 x – y = 6,
We have L.H.S = 2x - y and R.H.S = 6
Substituting x = 3 and y = 0 in 2x - y ,
We get L.H.S = 2 x 3 – 0 = 6
⇒ L.H.S = R.H.S
⇒ (3,0) is a solution of 2x - y = 6.

(ii) In the equation 2x - y = 6,

We have L.H.S = 2x- y and R.H.S = 6
Substituting x = 0 and y = 6 in 2x - y
We get L.H.S = 2 × 0 - 6 = -6
 L.H.S R.H.S
⇒ (0,6) is not a solution of 2x - y = 6.

(iii) In the equation 2x - y = 6,

We have L.H.S = 2x - y and R.H.S = 6
Substituting x = 2 and y = - 2 in 2x - y,
We get L.H.S = 2 × 2 - (-2) = 6
⇒ L.H.S = R.H.S
⇒ (2,-2) is a solution of 2x - y = 6.


(iv) In the equation 2x-y = 6, we have
LHS = 2x-y and RHS = 6
Substituting x = √3 and y = 0 in 2x - y, we get
LHS = 2×√3-0 = 2√3
LHS ≠ RHS
(√3, 0) is not the solution of 2x - y = 6.

(v) In the equation 2x - y = 6, we have
LHS = 2x-y and RHS = 6
Substituting x = 1/2 and y = -5 in 2x-y, we get
LHS = 2× 1/2 and y = -5 in 2x-y, we get
LHS = RHS
(1/2, -5) is the solution of 2x-y = 6.

4. If x = −1, y = 2 is a solution of the equation 3x + 4y = k, find the value of k.


Solution


We are given, 3 x + 4 y = k

Given that, (-1,2) is the solution of equation 3x + 4y = k.
Substituting x = -1 and y = 2 in 3x + 4y = k,
We get; 3x – 1 + 4 × 2 = k
k = -3 + 8
k = 5

5. Find the value of λ, if x =-λ and y=5/2 is a solution of the equation x+4y-7=0. 


  Solution



6. If x = 2α + 1 and y = α − 1 is a solution of the equation 2x − 3y + 5 = 0, find the value of α.


Solution


We are given, 2x -3y +5 = 0

(2a + 1, a – 1 ) is the solution of equation 2x – 3y + 5 = 0.
Substituting x = 2a + 1 and y = a – 1 in 2x – 3y + 5 = 0,
We get 2 x 2a + (1- 3) x a – 1 + 5 = 0
⇒ 4a + 2 – 3a + 3 + 5 = 0
⇒ a + 10 = 0
⇒ a = – 10

7If x=1 and y=6 is a solution of the equation 8x-ay+a2=0. Find the value of a.


Solution


Given,
8x - ay + a2 = 0
(1, 6) is the solution of equation 8x-ay+a2 = 0
Substituting, x = 1 and y = 6 in 8x-ay+a2 = 0, we get
8×1 - a×6 +a2 = 0
⇒ a2 - 6a + 8 = 0
Using quadratic factorization,
a2 -4a -2a + 8 = 0
⇒ a(a-4) - 2(a-4) = 0
⇒ (a-2) (a-4) = 0
Therefore, a = 2, 4

Notes of Ch 1 Business, Trade and Commerce| Class 11th Business Studies

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Notes of Ch 1 Business, Trade and Commerce| Class 11th Business Studies

• All Human beings have different types of needs. To fulfill these needs, they perform certain activities.

• Business is a major economic activity as it is concerned with the production and sale of goods and services with the purpose of earning money by meeting people’s demands for goods and services.

• The chapter is divided into two sections. 
→ Section I deals with the history of trade and commerce in ancient India. 
→ Section II deals with the concept, nature and purpose of business.

Section - I - History of trade and commerce in ancient India

History of Trade and Commerce

• Indian subcontinent has Himalayas in the North and bordered by water in the South.

• Silk Routes, a network of roads helped in establishing commercial and political contacts with
adjoining foreign kingdoms and empires of Asia and the world.

• With the help of wealth earned through trade, the chief kingdoms, important trade centres and the industrial belt flourished that helped ancient India in progress of domestic and international trade.

Indigenous Banking System

• Metals came to be used as money as the economic life of the people progressed as they are durable and divisible.
→ This accelerated economic activities.

• Documents such as Hundi and Chitti were in use for carrying out transactions in which money passed from hand to hand.
Hundi involved a contract which — (i) warrant the payment of money, the promise or order which is unconditional (ii) capable of change through transfer by valid negotiation.

• With the development of banking, people began to deposit precious metals with lending individuals functioning as bankers or Seths. 
→ Money became an instrument for supplying the manufacturers with a means of producing more goods.

• Agriculture and the domestication of animals were main part of the economic life of ancient people.
→ Also, weaving cotton, dyeing fabrics, making clay pots, utensils, and handicrafts, sculpting, cottage industries, masonry, manufacturing, transports (i.e., carts, boats and ships), etc., helped in generating surpluses and savings for further investment.

Rise of Intermediaries

• Intermediaries provided considerable financial security to the manufacturers by assuming responsibility for the risks involved, especially in foreign trade.

• During the Mughal period and the days of the East India Company, the institution of Jagat Seths also developed and exercised great influence.

• Commercial and Industrial banks later evolved to finance trade and commerce and agricultural banks to provide both short-and long-term loans to finance agriculturists.

Transport

• Transport by land and water was popular in the ancient times that helped in maintaining trade.

• Trade routes were structurally wide and suitable for speed and safety.

• Maritime trade was another important branch of global trade network.
→ Malabar Coast has a long history of international maritime trade going back to the era of the Roman Empire.

• An alternate route to India for spices that led to the discovery of America by Columbus in the closing years of 15th century and also brought Vasco da Gama to the shores of Malabar in 1498.

Trading Communities Strengthened

• In different parts of the country, different communities dominated trade.
→ Punjabi and Multani merchants handled business in the northern region.
→ The Bhats managed the trade in the states of Gujarat and Rajasthan.

• Other urban groups included professional classes, such as hakim and vaid (physician), wakil
(Lawyer), pundit or mulla (teachers), painters, musicians, calligraphers, etc.

Merchant Corporations

• Guilds were autonomous corporations formed to protect the interests of the traders through which merchant community also derived power and prestige.

• Traders had to pay octroi duties that were levied on most of the imported articles at varying rates.

• Customs duties varied according to the commodities and from province to province.

• The ferry tax was another source of income generation which had to be paid for passengers, goods,
cattle and carts.

• The guild chief dealt directly with the king or tax collectors and settled the market toll on behalf of its fellow merchants at a fixed sum of money.

Major Trade Centres

• In ancient India, the leading trade centres in ancient India were:

→ Pataliputra: Known as Patna today was a commercial town and also a major centre for export of stones.

→ Peshawar: It was an important exporting centre for wool and for the import of horses.

→ Taxila: It served as a major centre on the important land route between India and Central Asia. It was also a city of financial and commercial banks. The famous Taxila University flourished here.

→  Indraprastha: It was the commercial junction on the royal road where most routes leading to the east, west, south and north converged.

→ Mathura: Many routes from South India touched Mathura and Broach.

→ Varanasi:It grew as a major centre of textile industry and became famous for beautiful gold silk cloth and sandalwood workmanship.

→ Mithila: It established trading colonies in South China, especially in Yunnan.

→ Ujjain: Agate, carnelian, muslin and mallow cloth were exported from Ujjain to different centres.

→ Surat: Textiles of Surat were famous for their gold borders (zari).

→ Kanchi: Known as Kanchipuram today where Chinese used to come in foreign ships to purchase pearls, glass and rare stones and in return they sold gold and silk.

→ Madura: It attracted foreign merchants, particularly Romans, for carrying out overseas trade.

→ Broach: It was situated on the banks of river Narmada and was linked with all important marts by roadways.

→ Kaveripatta: It was a convenient place for trade with Malaysia, Indonesia, China and the Far
East. It was the centre of trade for perfumes, cosmetics, scents, silk, wool, cotton, corals, pearls, gold and precious stones; and also for ship building.

→ Tamralipti: It was one of the greatest ports connected both by sea and land with the West and the Far East. It was linked by road to Banaras and Taxila.

Major Exports and Imports

Exports

Spices, wheat, sugar, indigo, opium, sesame oil, cotton, parrot, live animals and animal products—hides, skin, furs, horns, tortoise shells, pearls, sapphires, quartz, crystal, lapis, lazuli, granites, turquoise and copper etc.

Imports

Horses, animal products, Chinese silk, flax and linen, wine, gold, silver, tin, copper, lead, rubies, coral, glass, amber, etc.

Position of Indians Subcontinent In World Economy (1 AD  UP to 1991)

• Between the 1st and the 7th centuries CE, India is estimated to have the largest economy of the ancient and medieval world, controlling about one-third and one-fourth of the world’s wealth.

• The 18th century India was far behind Western Europe in technology, innovation and ideas.

• In the mid-18th century, the British empire began to take roots in India and used revenues generated by the provinces under its rule for purchasing Indian raw materials, spices and goods.
→ Hence, the continuous inflow of bullion that used to come on account of foreign trade stopped.

India begins to Reindustrialise

• After Independence, India went for centralised planning.
→ In 1952, the First Five Year Plan was implemented and importance was given to the establishment of modern industries, modern technological and scientific institutes, space and nuclear programmes.
→ However, the Indian economy could not develop at a rapid pace due to lack of capital formation,
rise in population, huge expenditure on defence and inadequate infrastructure.

• Thus, India relied heavily on borrowings from foreign sources and finally, agreed to economic liberalisation in 1991.

• Today, Indian economy is one of the fastest growing economies in the world today and a preferred FDI destination.

• The recent initiatives of the Government of India such as ‘Make in India’, Skill India’, ‘Digital India’ and roll out of the Foreign Trade Policy (FTP 2015-20) is expected to help the economy in terms of exports and imports and trade balance.

Section II - Nature and Concept of Business

Concept of Business

• Business means being busy.

• Business refers to an occupation in which people regularly engage in activities related to purchase, production and/or sale of goods and services with a view to earning profits.

Characteristics of  Business Activites

• An economic activity: Business in considered as an economic activity as it is undertaken with the objective of earning money.

• Production or procurement of goods and services: Business includes all the activities concerned with the production of procurement of goods & services for sales. Services include transportation, banking, Insurance etc.

• Sale or exchange of goods and service: There should be sale or exchange of goods and service between the seller & the buyer. If goods are produced not for the purpose of sale but say for internal consumption it cannot be called a business activity.

• Dealing in goods & services on a regular basis: There should be regularity of dealings or exchange of goods & services. One single transaction of sale or purchase does not constitute business.

• Profit Earning: The main purpose of business is to earn profit. A business cannot survive without making profits. So, a businessman try to maximize profit by increasing the volume of sales or reducing costs.

• Uncertainly of return: Every business invests money with the objective of earning profit but the amount of profit earned may very also there is always a possibility of losses.

• Element of Risk: All business activities carry some elements of risk because future is uncertain and business has no control over several factors like strikes, fire, theft, change in consumer taste etc.

Comparison of Business, Profession and Employement

• Employment: It Refers to those economic activities which are connected with purchase, production or sale of goods & services with the objective of earning profit.
Examples: Fishing, Manufacturing Goods, Mining Producing or selling of electronic goods, Banking.

• Profession: It includes those activities which require special knowledge be skills in the occupation.
Examples: Medical (Doctor),  Legal (Lawyer), Accountancy (CA).

• It refers to the occupation in which people work for others and get remuneration in return.
Examples: Worker, Employee, Salesman.

Basic
Business
Profession
Employment
1. Mode of
establishment
Starts after completing some formalities if needed.Membership of a professional body and certificate of practice.Appointment letter and service agreement.
2. Nature of
work
Provision of goods and services to the public.Personalized services of expert nature.Performing work alloted by the employer according to the contract.
3. QualificationNo minimum qualification is necessary.Professional qualification and training required.Qualification and training as prescribed by the employer.
4. Reward and ReturnProfit earnedProfessional FeeSalary
5. Capital InvestmentCapital needed according to nature and sizeLimited capital for establishment.No Capital required.
6. RiskInvolves high riskDegree of riskNo risk
7. Transfer of InterestTransfer possible with some formalities.Not PossibleNot Possible
8. Code of ConductNo Code of ConductProfessional code of conduct is to be followed.Terms and conditions of services contract are to be followed.
9. ExamplesShop, FactoryLegal, Medical Profession.Jobs in Banks, insurance companies.

Classification of Business Activities

• Industry deals with the production or processing of goods and materials.

• Commerce deals with distribution of goods and services.

Industry

• Primary Industry: The primary industry includes those activities through which the natural resources are used to provide raw material for other industries. It can be classified into two types:

(i) Extractive: Industry under which something is extracted out of earth, water or air such as Coal, Iron, gas. Examples are Mining, Lumbering, Hunting.

(ii) Genetic: Industries under which the breed of animals and vegetables are improved and made more useful. Examples are Poultry Farms, dairy Farming, Fishing Fish Hatching, cattle breeding etc.

• Secondary Industry: Under this industry, new products are manufacturing by using the previously produced things e.g. producing cotton is a primary industry and manufacturing cloth out of cotton is a secondary industry. 

It is of two types.

(i) Manufacturing - These industries convert raw materials or semi-finished products e.g. paper from bamboo, sugar from Sugar cane. It is further divided into four parts.

→ Analytic: Different things are manufactured out of one material such as petrol, diesel, gasoline out of crude oil.

→ Synthetic: Many raw materials are mixed to produce more useful products such as cement.

→ Processing: Industries wherein useful things are manufactured by making raw material to pass through different production processes such as sugar and paper.

→ Assembling: Different component parts to make a new product, as in the case of television, car, computer, etc.

ii) Construction Industries - Industries that are involved in the construction of building, dams, bridge, roads as well as tunnels and canals. 

iii) Tertiary or Service Industry - This includes those services which help business to move smoothly such as transport, bank, insurance, storage and advertising.

Commerce

Commerce refers to all those activities which are concerned with the transfer of goods and services from the producers to the consumers. It embraces all those activities which are necessary for maintaining a free flow of goods and services.

The function of commerce are:

• Removing the hindrance of person by making goods available to consumers from the producers. through trade. 

• Transportation removes hindrance of place by moving goods from the place of production to the markets for sale. 

• Storage and warehousing activities remove the hindrance of time by facilitating holding of stock of goods to be sold as and when required. 

• Insurance removes hindrance of risk of loss or damage of goods due to theft. fire. accidents etc. 

• Banking removes hindrance of finance by providing funds to a businessman for acquiring assets. purchasing raw materials and meeting other expenses. 

• Advertising removes hindrance of information by informing consumers about the goods and services available in the market.

Make in India

• ‘Make in India’ is an initiative launched by the Government of India on 25 September 2014, to encourage national, as well as multinational companies to manufacture their products in India.

• Major Objective: Job creation and skill enhancement in 25 sectors of the economy.

Commerce includes two types of activities.

• Trade: Refers to buying and selling of goods and services with the objective of earning profit. It is classified into two categories :

(i) Internal Trade: Takes place within a country. Internal Trade is classified into two categories: 

→ Wholesale Trade: Refers to buying and selling o goods in large quantities. A wholesaler buys goods in large quantities from the producers and sell them to other dealers. He serves as a connecting link between the producer and retailer.

→ Retail Trade: Refers to buying of goods and services in relatively small quantities & selling them to the ultimate consumers.

(ii) External Trade: Trade between two or more countries. External trade can he classified into three categories:

→ Import trade: If goods are purchased from another country. if is called import trade. Example: Buying soft toys from China and selling here in India.

→ Export Trade: If goods are sold to other countries it is called export trade. Example: Selling Basmati Rice to USA from India.

→ Enterpot: Where goods are imported for export to other countries. Example: Indian firms may import some goods from America and export the service to Nepal.

Auxiliaries to Trade

Those activities which are meant for assisting trade are known as auxiliaries to trade. These activities help in removing various hindrances which arise in connection with the production and distribution of goods. These are:

(i) Transportation and Communication: Production of goods generally takes place in particular locations. But these goods are required for consumption in different parts of the country. The obstacle of place is removed by the transport. Along with transport communication is also an important service. It helps in exchange of information between producers. consumers and traders. The common communication services are postal service, telephone, fax, Internet etc

(ii) Banking and Finance: Business needs funds for acquiring assets, purchasing raw materials and meeting other expenses. Necessary funds can be obtained from a bank.

(iii) Insurance: It provides a cover against the loss of goods, in the process of transit, storage, theft, fire and other natural calamities.

(iv) Warehousing: There is generally a time lag between the production and consumption of goods. This problem can be solved by storing the goods in warehouses from the time of production till the time they are demanded by customers.

(v) Advertising: Advertising is one of the methods of promoting the sale of products. It helps in
providing information about available goods and services and inducing customers to buy particular items.

Objective of Business

• Market standing: Market standing refers to the position of an enterprise in relation to its competitors.

• Innovation: Innovation is the introduction of new ideas or methods in the way something is done or made.

•  Productivity: Productivity is ascertained by comparing the value of output with the value of inputs.

• Physical and financial resources: Any business requires physical resources, like plants, machines, offices and financial resources, i.e., funds to be able to produce and supply goods and services to its customers.

• Earning profits: One of the objectives of business is to earn profits on the capital employed.

• Manager performance and development: Business enterprises need managers to conduct and coordinate business activity.

• Worker performance and attitude: Workers’ performance and attitudes determine their contribution towards productivity and profitability of any enterprise.

• Social responsibility: Refers to the obligation of business firms to contribute resources for solving social problems and work in a socially desirable manner.

Business Risk

It refers to the possibility of inadequate profits or even losses due to uncertainties or unexpected events.

Nature of Business Risks

(i) Business risks arise due to uncertainties.

(ii) Risk is an essential part of every business.

(iii)  Degree of risk depends mainly upon the nature and size of business.

(iv) Profit is the reward for risk taking.

Cause of Business Risks

(i) Natural causes: Human beings have little control over natural calamities, like flood, earthquake,
lightning, heavy rains, famine, etc. property and income in business.

(ii) Human causes: Human causes include such unexpected events, like dishonesty, carelessness or
negligence of employees, stoppage of work due to power failure, strikes, riots, management inefficiency, etc.

(iii) Economic causes: These include uncertainties relating to demand for goods, competition, price,
collection of dues from customers, change of technology or method of production, etc.

(iv) Other causes - These include unforeseen events like political disturbances. fluctuation in exchange rates etc. 

Starting a Business - Basic Factors

(i) Selecting the line of business: The first thing to be decided by the entrepreneur is the line and type of business to be undertaken. 

(ii) Size of business: After deciding the line of business, the entrepreneur must decide whether he wants to set up large scale or small scale business.

(iii) Choice of form of Business organization: A business organisation may take the form of a sale proprietorship, partnership, or a joint stock company.

(iv) Location of Business Enterprise: The entrepreneur has to decide the place where the enterprise will be located. Before taking this decision he must find out availability of raw materials, power, labour, banking, transportation etc.

(v) Financing the proposition: It is concerned with providing the necessary capital for starting, as well
as, for continuing the proposed business.

(vi) Physical facilities: Availability of physical facilities, including machines and equipment, building and supportive services is an important factor to be considered at the start of the business.

(vii) Plant layout:  The entrepreneur should draw a layout plan showing the arrangement of physical facilities.

(viii) Competent and committed worked force: The entrepreneur must identify the requirement of skilled and unskilled workers and managerial staff to perform various activities.

(ix) Tax planning: The entrepreneur must try to analyze the types of taxes as there are a number of tax laws in the country which affect the functioning of business.

(x) Setting up of the Enterprise: After analyzing the above mentioned points carefully the entrepreneur can start the business which would mean mobilizing various resources and completing legal formalities.

NCERT Solutions of Chapter 1 Business, Trade and Commerce

R.D. Sharma Solutions Class 9th: Ch 13 Linear Equations in Two Variables Exercise 13.3

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Chapter 13 Linear Equations in Two Variables R.D. Sharma Solutions for Class 9th Exercise 13.3

Exercise 13.3

1. Draw the graph of each of the following linear equations in two variables:
(i) x+y = 4
(ii) x-y = 0
(iii) -x+y = 6
(iv) y = 2x
(v) 3x+5y = 15
(vi) x/2 - y/3 = 2
(vii) (x-2)/3 = y-3
(viii) 2y = -x+1

Solution












2 . Give the equations of two lines passing through (3, 12). How many more such lines are there, and why?

Solution

We observe that x = 3 and y = 12 is the solution of the following equations
4x-y = 0 and 3x-y+3 = 0
So, we get the equations of two lines passing through (3, 12) are, 4x – y = 0 and 3x – y + 3 = 0.
We know that passing through the given point infinitely many lines can be drawn. So, there are infinitely many lines passing through (3,12)

3. A three-wheeler scooter charges Rs 15 for first kilometer and Rs 8 each for every subsequent kilometer. For a distance of x km, an amount of Rs y is paid. Write the linear equation representing the above information.

Solution

Total fare of Rs y for covering the distance of x km is given by
y = 15 + 8(x − 1)
y = 15 + 8x − 8
y = 8x + 7
Where, Rs y is the total fare (x – 1) is taken as the cost of first kilometer is already given Rs 15 and 1 has to subtracted from the total distance travelled to deduct the cost of first kilometer.

4. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Aarushi paid Rs 27 for a book kept for seven days. If fixed charges are Rs x and per day charges are Rs y. Write the linear equation representing the above information.

 Solution
Total charges of Rs 27 of which Rs x for first three days and Rs y per day for 4 more days is given by 
x+y(7-3) = 27
⇒ x+4y = 27
⇒ x+y(7-3)=27
⇒ x+4y=27
Here, (7, -3) is taken as the charges for the first three days are already given at Rs x and we have to find the charges for the remaining four days as the book is kept for the total of 7 days.


5. A number is 27 more than the number obtained by reversing its digits. If its unit's and ten's digit are x and y respectively, write the linear equation representing the above statement.

Solution

The number given to us is in the form of yx,
where y represents the ten's place of the number and x represents the unit's place of the number. 
Now,
Given number is 10y + x
Number obtained by reversing the digits of the number is 10x + y
It is given to us that the original number is 27 more than the number obtained by reversing its digits
So, 
10y + x = 10x + y +27
⇒ 10y- y+x-10x = 27 
⇒ 9y-9x = 27 
⇒ 9(y- x) = 27
⇒ y-x = 27/9
⇒ y-x=3 

6. The sum of a two digit number and the number obtained by reversing the order of its digits is 121. If units and ten's digit of the number are x and y respectively, then write the linear equation representing the above statement.

Solution

7. Plot the points (3,5) and (−1, 3) on a graph paper and verify that the straight line passing through these points also passes through the point (1, 4).

Solutions


 
By plotting the given points (3, 5) and (–1, 3) on a graph paper, we get the line BC.
We have already plotted the point A (1, 4) on the given plane by the intersecting lines.
Therefore, it is proved that the straight line passing through (3, 5) and (–1, 3) also passes through A (1, 4).

8. From the choices given below, choose the equation whose graph is given in Fig. 13.13.
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x

Solution 

We are given co-ordinates (1, –1) and (–1, 1) as the solution of one of the following equations.
We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.



9. From the choices given below, choose the equation whose graph is given in Fig. 13.14.

(i) y= x + 2
(ii) y = x −2
(iii) y = x + 2
(iv) x + 2y = 6
Solution 

We are given co-ordinates (–1, 3) and (2, 0) as the solution of one of the following equations.
We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.



 


10. If the point (2, −2) lies on the graph of the linear equation 5x + ky = 4, find the value of k.

Solution 


11. Draw the graph of the equation 2x + 3y = 12. From the graph, find the coordinates of the point:
(i) whose y-coordinates is 3.
(ii) whose x-coordinate is -3.

 Solution 


12. Draw the graph of each of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes:
(i) 6x − 3y = 12
(ii) -x + 4y = 8
(iii) 2x + y = 6
(iv) 3x + 2y + 6 = 0

Solution 





13. Draw the graph of the equation 2x + y = 6. Shade the region bounded by the graph and the coordinate axes. Also, find the area of the shaded region.

Solution 



14. Draw the graph of the equation x/3 + y/4 = 1. Also, find the area of the triangle formed by the line and the coordinates axes.

 Solution 


15. Draw the graph of y = | x |.

Solution 


16. Draw the graph of y = | x | + 2.

Solution




17. Draw the graphs of the following linear equations on the same graph paper:
2x + 3y = 12, x − y = 1
Find the coordinates of the vertices of the triangle formed by the two straight lines and the area bounded by these lines and x-axis.

Solution



18. Draw the graphs of the linear equations 4x − 3y + 4 = 0 and 4x + 3y −20 = 0. Find the area bounded by these lines and x-axis.

Solution

19. The path of a train A is given by the equation 3x + 4y − 12 = 0 and the path of another train B is given by the equation 6x + 8y − 48 = 0. Represent this situation graphically. 

Solution 


20. Ravish tells his daughter Aarushi, ''Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be''. If present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically as well as graphically.

Solution 

21. Aarushi was driving a car with uniform speed of 60 km/h. draw distance-time graph. From the graph, find the distance travelled by Aarushi in
(i) 2½ Hours
(ii) 1/2 Hour

Solution 

Aarushi is driving the car with the uniform speed of 60 km/h.
We represent time on X-axis and distance on Y-axis
Now, graphically



R.D. Sharma Solutions Class 9th: Ch 1 Number System MCQ's

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Chapter 1 Number System R.D. Sharma Solutions for Class 9th MCQ's

Mark the correct alternative in each of the following:

1. Which one of the following is a correct statement?
(a) Decimal expansion of a rational number is terminating
(b) Decimal expansion of a rational number is non-terminating
(c) Decimal expansion of an irrational number is terminating
(d) Decimal expansion of an irrational number is non-terminating and non-repeating.

Solution

The decimal expansion of an irrational number is non-terminating and non- repeating. Thus, we can say that a number, whose decimal expansion is non-terminating and non- repeating, called irrational number. And the decimal expansion of rational number is either terminating or repeating. Thus, we can say that a number, whose decimal expansion is either terminating or repeating, is called a rational number. Hence the correct option is d.

2. Which one of the following statements is true? (a) The sum of two irrational numbers is always an irrational number
(b) The sum of two irrational numbers is always a rational number
(c) The sum of two irrational numbers may be a rational number or an irrational number
(d) The sum of two irrational numbers is always an integer

Solution

Since, 2 and 2 - 1 are two irrational number and -2 + (2+1) = 1
Therefore, sum of two irrational number may be rational 
Now, let 3 and 2 - 3 be two irrational numbers and 3 + (2 - 3) = 2
Therefore, sum of two irrational number may be irrational 
Hence the correct option is c.

3. Which of the following is a correct statement?
(a) Sum of two irrational numbers is always irrational
(b) Sum of a rational and irrational number is always an irrational number
(c) Square of an irrational number is always a rational number
(d) Sum of two rational numbers can never be an integer

Solution

The sum of irrational number and rational number is always irrational number .
Let a be a rational number and b be an irrational number.

Then, 
Let a be a rational number and b be an irrational number.
Then,
(a+b)2 = a2 + b2 + 2ab
= (a2 + b2) + 2ab
As 2ab is irrational number therefore (a+b)2 is irrational .
Hence (a+b) is irrational .
Therefore answer is b.

4 . Which of the following statements is true?
(a) Product of two irrational numbers is always irrational.
(b) Product of a rational and an irrational number is always irrational.
(c) Sum of two irrational numbers can never be irrational.
(d) Sum of an integer and a rational number can never be an integer.

Since, we know that the product of rational and irrational number is always an irrational.
For example : Let 1/2, 3 are rational and irrational number respectively and their product is √3/2.
Hence the correct option is b .

5. Which of the following is irrational ?
Solution



6. Which of the following is irrational?
(i) 0.14
(ii) 0.1416
(iii) 0.1416
(iv) 0.1014001400014

Solution

Given that,
0.14: 

0.1416
0.1416
• 0.1014001400014.... is not terminating or non-repeating. So, it is an irrational number.
Hence, the correct option is d.

7. Which of the following is rational ?
(a) 3
(b) π 
(c) 4/10
(d) 0/4

Solution 

Given that,
3, π , 4/10 and 0/4
Here,  0/4 = 0, this is the form of p/q . so this is a rational number 
Hence the correct option is d.

8. The number 0.318564318564318564 ........ is:
(a) a natural number
(b) an integer
(c) a rational number
(d) an irrational number 

Solution

Since, the given number 0.318564318564318564.... = 0.318564 is repeating, so it is a rational number because rational number is always either terminating or repeating. Hence, the correct option is d.

9. If n is a natural number, then √n  
(a) always a natural number
(b) always an irrational number
(c) always an irrational number
(d) sometimes a natural number and sometimes an irrational number

Solution

The term “natural number” refers either to a member of the set of positive integer. 1,2,3. And natural number starts from one of counting digit .Thus, if n is a natural number then sometimes n is a perfect square and sometimes it is not. Therefore, sometimes is a natural number and sometimes it is an irrational number  Hence the correct option is d .

10. Which of the following numbers can be represented as non-terminating, repeating decimals ?
(a) 39/24
(b) 3/16
(c) 3/11
(d) 137/25

Solution 

Given that
39/24 = 1.625
3/16 = 0.1875
3/11 = 0.272727272
1.37/25 = 0.0548
Here 3/11 is repeating but non-terminating.
Hence the correct option is c.

11. Every point on a number line represents
(a) a unique real number
(b) a natural number
(c) a rational number
(d) an irrational number

Solution

In basic mathematics, number line is a picture of straight line on which every point is assumed to correspond to real number.
Hence the correct option is a.

12. An irrational number between 2 and 2.5 is

Solution

Let
a=2
b=2.5
Here a and b are rational numbers. So we observe that in first decimal place a and b have distinct. According to question a < b.so an irrational number between 2 and 2.5 is 2.236067978 OR  √5
Hence the correct answer is b.

13. Which of the following is irrational ?

Solution

Given,
Decimal number are:
0.15, 0.1516, and 0.5015000115....
Here, the number 0.5015000115.... is non terminating or non-repeating.
Hence, the correct option is d.

14. The number of consecutive zeros in 23 ×34 ×54 ×7 is
(a) 3
(b) 2
(c) 4
(d) 5 Solution

We are given the following expression and asked to find out the number of consecutive zeros
23×34×54×7
We basically, will focus on the powers of 2 and 5 because the multiplication of these two number gives one zero. So
23×34×54×7=23×54×34×7
=23×53×5×34×7
=(2×5)3×5×34×7
=103×5×34×7
=5×81×7×1000
=2835000
Therefore the consecutive zeros at the last is 3
So the option (a) is correct


Solution


Solution


17. The smallest rational number by which 1/3 should be multiplied so that its decimal expansion terminate after one place of decimal, is
(a) 1/10
(b) 3/10
(c)  3
(d) 30

Solution



Solution



Solution



Solution



 Solution


R.D. Sharma Solutions Class 9th: Ch 4 Algebraic Identities Exercise 4.3

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Chapter 4 Algebraic Identities R.D. Sharma Solutions for Class 9th Math Exercise 4.3 

Exercise 4.3

1. Find the cube of each of the following binomials expressions:

Solution





2. Simplify each of the following:

Solution




3. If a + b = 10 and ab = 21, find the value of a3 + b3

Solution

4.  If a − b = 4 and ab = 21, find the value of a3 −b3

Solution

5. If x+1x=5x+1x=5, find the value of x3+1x3

Solution

6. If x1x=7 find the value of x31x3

Solution

Solution


Solution



Solution


10.  If 2x+3y = 13 and xy = 6, find the value of 8x3 + 27y3

Solution

11. If 3x − 2y = 11 and xy = 12, find the value of 27x3 − 8y3

Solution

12. If x4+1x4= 119x4+1x4= 119, find the value of x31x3

Solution

13.  Evaluate each of the following:
(i) (103)3
(ii) (98)3
(iii) (9.9)3
(iv) (10.4)3
(v) (598)3
(vi) (99)3

Solution







14 . Evaluate each of the following:
(i) 1113 − 893
(ii) 463+343
(iii) 1043 + 963
(iv) 933 − 1073

Solution





Solution


Solution


17. Find the value of 27x3 + 8y3, if
(i) 3x + 2y = 14 and xy = 8
(ii) 3x + 2y = 20 and xy = 149

Solution


(ii) 

18. Find the value of 64x3 − 125z3, if 4x − 5z = 16 and xz = 12.

Solution

19. If x1x=3+22x-1x=3+22, find the value of x31x3

Solution

RD Sharma Class 9 Solutions - All Chapters

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Solutions of RD Sharma for Class 9 All Chapters - Download PDF

Get Detailed Solution of RD Sharma. All the solutions of the chapters of RD Sharma is based on the latest textbook based on CBSE pattern. Our faculty has tried to make the solution in easiest and accurate way.  Below are the name of the chapters from RD Sharma Mathematics Textbooks. Click on the Exercise name to get the solution.

Chapter 1: Number System
There are total 6 exercises in this chapter and one exercise for MCQ based questions.
RD Sharma Class 9 Number System Exercise 1.1 Solutions
RD Sharma Class 9 Number System Exercise 1.2 Solutions
RD Sharma Class 9 Number System Exercise 1.6 Solutions

Chapter 2: Exponents of Real Numbers
In this chapter, there are total 2 exercise and one is MCQ's type exercise.
RD Sharma Class 9 Exponents of Real Numbers Exercise 2.1 Solutions
RD Sharma Class 9 Exponents of Real Numbers Exercise 2.2 Solutions

Ch 3: Rationalisation
RD Sharma Class 9 Rationalisation Exercise 3.1 Solutions
RD Sharma Class 9 Rationalisation Exercise 3.2 Solutions
RD Sharma Class 9 Rationalisation Exercise MCQ's Solutions

Ch 4: Algebraic Identities
RD Sharma Class 9 Algebraic Identities Exercise 4.1 Solutions
RD Sharma Class 9 Algebraic Identities Exercise 4.2 Solutions

Ch 5: Factorization of Algebraic Expressions
RD Sharma Class 9 Factorization of Algebraic Identities Exercise 5.1 Solutions
RD Sharma Class 9 Factorization of Algebraic Identities Exercise 5.2 Solutions
RD Sharma Class 9 Factorization of Algebraic Identities Exercise 5.4 Solutions

Ch 6: Factorization of Polynomials

Ch 7: Introduction to Euclid's Geometry

Ch 8: Lines and Angles

Ch 9: Triangles and its Angles

Ch 10: Congruent Triangles

Ch 11: Co-ordinate Geometry

Ch 12: Heron's Formula

Ch 13: Linear Equations in Two Variables
RD Sharma Class 9 Linear Equations in Two Variables MCQ's Solutions

Ch 14: Quadrilaterals

RD Sharma Class 9 Quadrilaterals Exercise 14.1 Solutions
RD Sharma Class 9 Quadrilaterals Exercise 14.2 Solutions
RD Sharma Class 9 Quadrilaterals Exercise 14.3 Solutions
RD Sharma Class 9 Quadrilaterals MCQ's Solutions

Ch 15: Areas of Parallelogram and Circles

RD Sharma Class 9 Areas of Parallelogram and Circles Exercise 15.1 Solutions
RD Sharma Class 9 Areas of Parallelogram and Circles Exercise 15.2 Solutions
RD Sharma Class 9 Areas of Parallelogram and Circles Exercise 15.3 Solutions
RD Sharma Class 9 Areas of Parallelogram and Circles MCQ's Solutions

Ch 16: Circles

RD Sharma Class 9 Circles Exercise 16.1 Solutions
RD Sharma Class 9 Circles Exercise 16.2 Solutions
RD Sharma Class 9 Circles Exercise 16.3 Solutions
RD Sharma Class 9 Circles Exercise 16.4 Solutions
RD Sharma Class 9 Circles Exercise 16.5 Solutions
RD Sharma Class 9 Circles MCQ's Solutions

Ch 17: Construction

RD Sharma Class 9 Construction Exercise 17.1 Solutions
RD Sharma Class 9 Construction Exercise 17.2 Solutions
RD Sharma Class 9 Construction Exercise 17.3 Solutions

Ch 18: Surface Area and Volume of a Cuboid and Cube

RD Sharma Class 9 Surface Area and Volume of a Cuboid and Cube Exercise 18.1 Solutions
RD Sharma Class 9 Surface Area and Volume of a Cuboid and Cube Exercise 18.2 Solutions
RD Sharma Class 9 Surface Area and Volume of a Cuboid and Cube MCQ's Solutions


Ch 19: Surface Area and Volume of a Right Circular Cylinder
RD Sharma Class 9 Surface Area and Volume of a Right Circular Cylinder Exercise 19.1 Solutions
RD Sharma Class 9 Surface Area and Volume of a Right Circular Cylinder Exercise 19.2 Solutions
RD Sharma Class 9 Surface Area and Volume of a Right Circular Cylinder MCQ's Solutions

Ch 20: Surface Area and Volume of a Right Circular Cone
RD Sharma Class 9 Surface Area and Volume of a Right Circular Cone Exercise 20.1 Solutions
RD Sharma Class 9 Surface Area and Volume of a Right Circular Cone Exercise 20.2 Solutions
RD Sharma Class 9 Surface Area and Volume of a Right Circular Cone MCQ's Solutions

Ch 21: Surface Area and Volume of a Sphere
RD Sharma Class 9 Surface Area and Volume of a Sphere Exercise 21.1 Solutions
RD Sharma Class 9 Surface Area and Volume of a Sphere Exercise 21.2 Solutions
RD Sharma Class 9 Surface Area and Volume of a Sphere MCQ's Solutions

Ch 22: Tabular Representation of Statistical Data
RD Sharma Class 9 Tabular Representation of Statistical Data Exercise 22.1 Solutions
RD Sharma Class 9 Tabular Representation of Statistical Data MCQ's Solutions

Ch 23: Graphical Representation of Statistical Data
RD Sharma Class 9 Graphical Representation of Statistical Data Exercise 23.1 Solutions
RD Sharma Class 9 Graphical Representation of Statistical Data Exercise 23.2 Solutions
RD Sharma Class 9 Graphical Representation of Statistical Data Exercise 23.3 Solutions
RD Sharma Class 9 Graphical Representation of Statistical Data MCQ's Solutions

Ch 24: Measures of Central Tendency


RD Sharma Class 9 Measures of Central Tendency Exercise 24.1 Solutions
RD Sharma Class 9 Measures of Central Tendency Exercise 24. 2 Solutions
RD Sharma Class 9 Measures of Central Tendency Exercise 24.3 Solutions
RD Sharma Class 9 Measures of Central Tendency Exercise 24.4 Solutions
RD Sharma Class 9 Measures of Central Tendency MCQ's Solutions

Ch 25: Probability


RD Sharma Class 9 Probability 25.1 Solutions
RD Sharma Class 9 Probability MCQ's Solutions

R.D. Sharma Solutions Class 9th: Ch 13 Linear Equations in Two Variables MCQ's

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Chapter 13 Linear Equations in Two Variables R.D. Sharma Solutions for Class 9th MCQ's

Exercise 13.3

Mark the correct alternative in each of the following:

1. If (4, 19) is a solution of the equation y = ax + 3, then a=
(a) 3
(b) 4
(c) 5
(d) 6 

Solution

We are given (4,19)as the solution of equation.
 y = ax+3
Substituting x= 4 and y= 19, we get
19 = 4a+3
⇒ 4a=19-3
⇒ 4a=16
⇒ a=4
Therefore, the correct answer is (b). 

2. If (a, 4) lies on the graph of 3x + y = 10, then the value of a is
(a) 3
(b) 1
(c) 3
(d) 4

Solution

We are given (a, 4) lies on the graph of linear equation 3x+ y= 10.
So, the given co-ordinates are the solution of the equation 3x+ y= 10.
Therefore, we can calculate the value of a by substituting the value of given co-ordinates in equation 3x+ y= 10.
Substituting x= a and y= 4 in equation 3x+ y= 10, we get
3×a + 4 = 10
⇒ 3a = 10-4
⇒ 3a=6
⇒ a=6/3
⇒ a=2
No, option is correct. 

3. The graph of the linear equation 2x − y = 4 cuts x- axis at
(a) (2, 0)
(b) (−2, 0)
(c) (0, −4)
(d) (0, 4)

Solution

Given,
2x- y = 4
we get, 
y = 2x-4
We will substitute y =0 in y = 2x-4 to get the co-ordinates for the graph of 2x- y = 4 at x axis
0 = 2x-4
⇒ 2x=0+4
⇒ 2x=4 x=2
Co-ordinates for the graph of 2x- y = 4 are (2,0). 
Therefore, the correct answer is (a). 

4. How many linear equations are satisfied by x = 2  and y = −3?
(a) Only one
(b) Two
(c) Three
(d) Infinitely many

Solution
There are infinite numbers of linear equations that are satisfied by x= 2 and y = -3 as
(i) Every solution of the linear equation represent a point on the line.
(ii) Every point on the line is the solution of the linear equation.
Therefore, the correct answer is (d).

5. The equation x − 2 = 0 on number line is represented by
(a) a line
(b) a point
(c) infinitely many lines
(d) two lines

Solution

The equation x - 2 = 0  is represented by a point on the number line.
Therefore, the correct answer is (b).

6. x = 2, y = −1 is a solution of the linear equation
(a) x + 2y = 0
(b) x + 2y = 4
(c) 2x + y = 0
(d) 2x + y = 5

Solution

We are given x=2; y = -1 as the solution of linear equation, which we have to find.
The equation is x+2y = 0 which can he proved by
Substituting x=2 and y= -1 in the equation x+2y= 0, we get
2 + 2×(-1) = 0
2-2 = 0
0 = 0
RHS = LHS
Therefore, the correct answer is (a).

7. If (2k − 1, k) is a solution of the equation 10x − 9y = 12, then k =
(a) 1
(b) 2
(c) 3
(d) 4

Solution

We are given (2k-1, k)as the solution of equation
10x - 9y = 12
Substituting x=2k-1 and y = k , we get
10 ×(2k-1)- 9×k =12
⇒ 20k - 10-9k =12
⇒ 11k = 12+10
⇒ 11k = 22
⇒ k = 22/11
⇒ k = 2
Therefore, the correct answer is (b). 

8. The distance between the graph of the equations x = −3 and x = 2 is
(a) 1
(b) 2
(c) 3
(d) 5

Solution

Distance between the graph of equations x= -3 and x = 2.
D = Distance of co-ordinate on negative side of x axis + Distance of co-ordinate on positive side of x axis
Distance of co-ordinate on negative side of x axis = x = 3 units
Distance of co-ordinate on positive side of x axis = x = 2 units 
D = 3+2
D = 5 units
Therefore, the correct answer is d 

9. The distance between the graphs of the equations y = −1 and y = 3 is
(a) 
(b) 4
(c) 3
(d) 1

Solution

Here, you can see, the distance is 4 units. The distance can be calculated by subtracting y coordinates. 3-(-1) =4. So, (b) is the correct answer.

10. If the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right triangle AOB is of length
(a) 4 units
(b) 3 units
(c) 5 units
(d) none of these

Solution



R.D. Sharma Solutions Class 9th: Ch 14 Quadrilaterals Exercise 14.1

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Chapter 14 Quadrilaterals R.D. Sharma Solutions for Class 9th Exercise 14.1

Exercise 14.1

1. Three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angles.

Solution 


2. In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measure of each angles of the quadrilateral.

Solution

3. In a quadrilateral ABCD, CO and DO are the bisectors ∠C and ∠D respectively. Prove that ∠COD = 1/2(∠A + ∠B).

Solution


4. The angles of quadrilateral are in ratio 3:5:9:13. Find all angles of the quadrilateral. 

Solution:

We know that sum of all angles of a triangle is 360°
It is given; angles are in ratio 3:5:9:13. So, we can say, the angles are 3x, 5x, 9x and 13x. 
Sum of all angles 
3x+5x+9x+13x = 360°
⇒30x=360°
⇒x = 360°/30 = 12°
So, angles are 
3x = 3×12°=36°
5x = 5×12°=60°
9x =9×12°= 108°
13x =13×12°=156°
So, angles are 36°, 60°,108°, 156°

R.D. Sharma Solutions Class 9th: Ch 14 Quadrilaterals Exercise 14.2

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Chapter 14 Quadrilaterals R.D. Sharma Solutions for Class 9th Exercise 14.2

Exercise 14.2

1. Two opposite angles of a parallelogram are (3x − 2)° and (50 − x)°. Find the measure of each angle of the parallelogram.


Solution


2. If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.


Solution



3. Find the measure of all the angles of a parallelogram, if one angle is 24° less than twice the smallest angle.

Solution


4. The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side?

Solution



5. In a parallelogram ABCD, ∠D = 135°, determine the measures of ∠A and ∠B.

Solution



6. ABCD is a parallelogram in which ∠A = 70°. Compute ∠B, ∠C and ∠D.

Solution


7. In Fig. 14.34, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet at P, prove that AD = DP, PC = BC and DC = 2AD.

Solution



8.  In Fig. 14.35, ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute ∠CDB and ∠ADB.

Solution



9. In figure 14.36, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove that AF = 2AB.


 Solution


10. Which of the following statements are true (T) and which are false (F)?
(i) In a parallelogram, the diagonals are equal.
(ii) In a parallelogram, the diagonals bisect each other.
(iii) In a parallelogram, the diagonals intersect each other at right angles.
(iv) In any quadrilateral, if a pair of opposite sides is equal, it is a parallelogram.
(v) If all the angles of a quadrilateral are equal, it is a parallelogram.
(vi) If three sides of a quadrilateral are equal, it is a parallelogram.
(vii) If three angles of a quadrilateral are equal, it is a parallelogram.
(viii) If all the sides of a quadrilateral are equal, it is a parallelogram.

Solution


(i) False
(ii) True
(iii) False
(iv) False
(v) True
(vi) False
(vii) False
(viii) True

R.D. Sharma Solutions Class 9th: Ch 14 Quadrilaterals Exercise 14.3

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Chapter 14 Quadrilaterals R.D. Sharma Solutions for Class 9th Exercise 14.3

Exercise 14.3

1. In a parallelogram ABCD, determine the sum of angles ∠C and ∠D.


Solution


2. In a parallelogram ABCD, if ∠B = 135°, determine the measure of its other angles.

Solution



3. ABCD is a square. AC and BD intersect at O. State the measure of ∠AOB.

Solution



4. ABCD is a rectangle with ∠ABD = 40°. Determine ∠DBC.

Solution


5. The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.

Solution


6. P and Q are the points of trisection of the diagonal BD of a parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ.

Solution


7. ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively. such that AE = BF = CG = DH. Prove that EFGH is a square.

Solution


8. ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.


Solution



9. ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC


Solution


R.D. Sharma Solutions Class 9th: Ch 14 Quadrilaterals Exercise 14.4

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 Chapter 14 Quadrilaterals R.D. Sharma Solutions for Class 9th Exercise 14.4

Exercise 14.4

1. In a ΔABC, D, E and F are, respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of ΔDEF.


Solution



Given that,
AB = 7cm, BC = 8 cm, AC = 9cm
In △ABC,
F and E are the mid points AB and AC.
∴ EF = 1/2 BC
Similarly,
DF = 1/2 AC and DE = 1/2 AB
Perimeter of △DEF = DE + EF + DF
= 1/2 AB + 1/2 BC + 1/2 AC
=  1/2×7 + 1/2×8 + 1/2×9
= 3.5 + 4 + 4.5 = 12 cm
Perimeter of △DEF = 12 cm.

2. In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and C =∠70°. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.

Solution



In △ABC,
D and E are mid points of AB and BC
By mid-point theorem,
DE∥AC, DE = 1/2 AC
F is the midpoint of AC.
Then, DE = 1/2 AC = CF
In quadrilateral DECF,
DE∥AC, DE = CF
Hence, DECF is a parallelogram.
∠C = ∠D = 70°  [opposites sides of parallelogram]
Similarly,
BEFD is a parallelogram, ∠B = ∠F = 60°
ADEF is a parallelogram, ∠A = ∠E = 50°
∴ Angles of △DEF are ∠D = 70°, ∠E = 50°, ∠F = 60°

3. In a triangle, P, Q and R are the mid-points of sides BC, CA and AB respectively. If AC = 21 cm, BC = 29 cm and AB = 30 cm, find the perimeter of the quadrilateral ARPQ.


Solution


In △ABC,

R and P are mid points of AB and BC
RP∥AC, RP = 1/2 AC  [By midpoint Theorem]
In a quadrilateral, a pair of side is parallel and equal.
RP∥AQ, RP = AQ
Therefore, RPQA is a parallelogram.
⇒ AR = 1/2 AB = 1/2×30 = 15 cm
AR =QP = 15 cm  [Opposites sides are equal]
⇒ RP = 1/2 AC = 1/2×21 = 10.5 cm
RP = AQ = 10.5 cm
Now,
Perimeter of ARPQ = AR + QP + RP + AQ
= 15+15+10.5+10.5 = 51 cm

4. In a ΔABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram

Solution


△ABC is given with AD as the median extended to point X such that AD = DX.


Join BX and CX.

We get a quadrilateral ABXC, we need to prove that it’s a parallelogram.
We know that AD is the median.
By definition of median we get:
BD = CD
Also, it is given that
AD = DX
Thus, the diagonals of the quadrilateral ABCX bisect each other.
Therefore, quadrilateral ABXC is a parallelogram.
Hence proved.

5. In a ABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.


Solution

ΔABC is given with E and F as the mid points of sides AB and AC.

Also, AP ⊥BC intersecting EF at Q.
We need to prove that AQ = QP
In △ABC, E and are the mid-points of AB and AC respectively.
Theorem states, the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.
Therefore, we get: EF ⊥BC
Since, Q lies on EF.
Therefore, FQ | | BC
This means,
Q is the mid-point of AP.
Thus, AQ = QP (Because, F is the mid point of AC and FQ | | BC )
Hence proved.

6. In a ΔABC, BM and CN are perpendiculars from B and C respectively on any line passing through A. If L is the mid-point of BC, prove that ML = NL.


Solution




7. In Fig. 14.95, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D, E are the mid-points of the sides AB and AC respectively, calculate

(i) The length of BC
(ii) The area of ΔADE

 Solution



8. In Fig. 14.96, M, N and P are the mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC.


Solution

Given,

MN = 3cm, NP = 3.5cm and MP = 2.5cm
To find BC, AB and AC
In △ABC,
M and N are mid-points of AB and AC.
∴ MN = 1/2 BC, MN∥BC  [By midpoint theorem]
⇒ 3 = 1/2 BC
⇒ 3×2 = BC
⇒ BC = 6 cm
Similarly,
AC = 2MP = 2×2.5 = 5 cm
AB = 2NP = 2×3.5 = 7 cm

9. ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of ΔABC


Solution




10. In Fig. 14.97, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°


Solution


11. In Fig. 14.98, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ΔABC. Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram.


Solution



12. ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.


Solution




13. Let ABC be an isosceles triangle in which AB = AC. If D, E, F be the mid-points of the sides BC, CA and AB respectively, show that the segment AD and EF bisect each other at right angles.


Solution




14. ABC is a triangle, D is a point on AB such that AD = 1/4 AB and E is a point on AC such that AE = 1/4 AC. Prove that DE = 1/2 BC.


Solution


15. In Fig. 14.99, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ=1/4 AC. If PQ produced meets BC at R, Prove that R is a midpoint of BC.



Solution



16. In Fig. 14.100, ABCD and PQRC are rectangle and Q is the midpoint of AC. Prove that
(i) DP = PC
(ii) PR = 1/2 ∠AC

Solution 



17. ABCD is a parallelogram, E and F are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC at G, P and H respectively. Prove that GP = PH.



Solution



18. RM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. IF L is the mid-point of BC, prove that LM = LN.


Solution



19. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Solution



20. Fill in the blanks to make the following statements correct:

(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is........
(ii) The triangle formed by joining the mid-points of the sides of a right triangle is ........
(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is ..........

Solution


(i) Isosceles

(ii) Right triangle
(iii) Parallelogram

R.D. Sharma Solutions Class 9th: Ch 14 Quadrilaterals MCQ's

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 Chapter 14 Quadrilaterals R.D. Sharma Solutions for Class 9th MCQ's

Multiple Choice Questions

Mark the correct alternative in each of the following:

1. The opposite sides of a quadrilateral have

(a) no common point
(b) one common point
(c) two common points
(d) infinitely many common points

Solution



2. The consecutive sides of a quadrilateral have
(a) no common point
(b) one common point
(c) two common points
(d) infinitely many common points

 Solution

3. PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following cases, PQRS is a parallelogram?
(a) ∠P = 100°, ∠Q = 80°, ∠R = 95°
(b) ∠P =85°, ∠Q = 85°, ∠R = 95°
(c) PQ = 7 cm, QR = 7 cm, RS = 8 cm, SP = 8 cm
(d) OP = 6.5 cm, OQ = 6.5 cm, OR = 5.2 cm, OS = 5.2 cm

Solution

Let us analyze each case one by one.
We have a quadrilateral named PQRS, with diagonals PR and QS intersecting at O.
(a) ∠P=100°, ∠Q = 80°, ∠R=100°
By angle sum property of a quadrilateral, we get:
∠P + ∠Q + ∠R + ∠S = 360°
⇒ 100° +80° +100° + ∠S = 360°
⇒ 280° + ∠S = 360°
⇒ ∠S - 80°
Clearly, ∠P = ∠R
And, ∠Q = ∠S
Thus we have PQRS a quadrilateral with opposite angles are equal.
Therefore, PORS is a parallelogram.

(b) ∠P = 85°, ∠Q=85°, ∠R=95°
By angle sum property of a quadrilateral, we get:
∠P + ∠Q + ∠R + ∠S = 360°
⇒ 85° +85° +95° + ∠S = 360°
⇒ 265° + ∠S = 360°
∠S = 95°
Clearly, ∠P ≠ ∠R
And ∠Q ≠ ∠S
Thus we have PQRS a quadrilateral with opposite angles are not equal. 
Therefore, PORS is not a parallelogram.

(c) PQ = 7cm, QR = 7cm, RS = 8cm, SP = 8cm
Clearly, PQ ≠ RS
And QR ≠ SP
Thus, we have PQRS a quadrilateral with opposite sides are not equal.
Therefore,
PQRS is not a parallelogram.

(d) OP = 6.5cm, OQ = 6.5cm, OR = 5.2cm, OS = 5.2cm
We know that the diagonals of a parallelogram bisect each other.
But, here we have
OP ≠ OR
And OQ ≠ OS
Therefore, PQRS is not a parallelogram.

Hence, the correct choice is (a).

4. Which of the following quadrilateral is not a rhombus?
(a) All four sides are equal
(b) Diagonals bisect each other
(c) Diagonals bisect opposite angles
(d) One angle between the diagonals is 60°

Solution



5. Diagonals necessarily bisect opposite angles in a
(a) rectangle
(b) parallelogram
(c) isosceles trapezium
(d) square

Solution



6. The two diagonals are equal in a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) trapezium

Solution



7. We get a rhombus by joining the mid-points of the sides of a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) triangle

Solution



8. The bisectors of any two adjacent angles of a parallelogram intersect at
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Solution



9. The bisectors of the angle of a parallelogram enclose a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) square

Solution



10. The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a
(a) parallelogram
(b) rectangle
(c) square
(d) rhombus

 Solution



11. The figure formed by joining the mid-points of the adjacent sides of a rectangle is a
(a) square
(b) rhombus
(c) trapezium
(d) none of these

Solution



12. The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
(a) square
(b) rectangle
(c) trapezium
(d) none of these

Solution

13.The figure formed by joining the mid-points of the adjacent sides of a square is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram

Solution



14. The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
(a) rectangle
(b) parallelogram
(c) rhombus
(d) square

Solution



15. If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of the parallelogram is
(a) 176°
(b) 68°
(c) 112°
(d) 102°

Solution


16. In a parallelogram ABCD, if ∠DAB = 75° and ∠DBC = 60°, then ∠BDC =
(a) 75°
(b) 60°
(c) 45°
(d) 55°

Solution



17. ABCD is a parallelogram and E and F are the centroids of triangles ABD and BCD respectively, then EF =
(a) AE
(b) BE
(c) CE
(d) DE

Solution



18. ABCD is a parallelogram, M is the mid-point of BD and BM bisects ∠B. Then ∠AMB =
(a) 45°
(b) 60°
(c) 90°
(d) 75°

Solution



19. ABCD is a parallelogram and E is the mid point of BC. DE and AB when produced meet at F. Then AF
(a) 3/2 AB
(b) 2 AB
(c) 3 AB
(d) 5/4 AB

Solution



20. If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is
(a) 108°
(b) 54°
(c) 72°
(d) 81°

Solution



21. If the degree measures of the angles of quadrilateral are 4x, 7x, 9x and 10x, what is the sum of the measures of the smallest angle and largest angle?
(a) 140°
(b) 150°
(c) 168°
(d) 180°


Solution 


22. In a quadrilateral ABCD, ∠A + ∠C is 2 times ∠B + ∠D. If ∠A = 140° and ∠D = 60°, then ∠B =
(a) 60°
(b) 80°
(c) 120°
(d) None of these

Solution 


23. If the diagonals of a rhombus are 18 cm and 24 cm respectively, then its side is equal to
(a) 16 cm
(b) 15 cm
(c) 20 cm
(d) 17 cm

Solution


24. The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, then ∠DPC =
(a) 70°
(b) 90°
(c) 80°
(d) 100°


Solution 



25. ABCD is a parallelogram in which diagonal AC bisects ∠BAD. If ∠BAC = 35°, then ∠ABC =
(a) 70°
(b) 110°
(c) 90°
(d) 120°

Solution



26. In a rhombus ABCD, if ∠ACB = 40°, then ∠ADB =
(a) 70°
(b) 45°
(c) 50°
(d) 60°


Solution


27. In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°. The angles of the triangle formed by joining the mid-points of the sides of this triangle are
(a) 70°, 70°, 40°
(b) 60°, 40°, 80°
(c) 30°, 40°, 110°
(d) 60°, 70°, 50°

Solution 


28.The diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90° and ∠BDC = 50°, then ∠OAB =
(a) 40°
(b) 50°
(c) 10°
(d) 90°

Solution 


29. ABCD is a trapezium in which AB || DC. M and N are the mid-points of AD and the respectively. If AB = 12 cm, MN = 14 cm, then CD =
(a) 10 cm
(b) 12 cm
(c) 14 cm
(d) 16 cm

Solution



30. Diagonals of a quadrilateral ABCD bisect each other. If ∠A= 45°, then ∠B =
(i) 115°
(ii) 120°
(iii) 125°
(iv) 135°

Solution



31. P is the mid-point of side BC of a parallelogram ABCD such that ∠BAP = ∠DAP. If AD = 10 cm, then CD =
(a) 5 cm
(b) 6 cm
(c) 8 cm
(d) 10 cm

Solution


32. In ΔABC, E is the mid-point of median AD such that BE produced meets AC at F. IF AC = 10.5 cm, then AF =
(a) 3 cm
(b) 3.5 cm
(c) 2.5 cm
(d) 5 cm

Solution

R.D. Sharma Solutions Class 9th: Ch 15 Areas of Parallelogram and Triangles Exercise 15.1

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Chapter 15 Areas of Parallelogram and Triangles R.D. Sharma Solutions for Class 9th Exercise 15.1

Exercise 15.1

1.Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels:

 Solution

(i) ΔPCB and trapezium ABCD are on the same base CD and between the same parallels AB and DC.

(ii) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.

(iii) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC but they are not on the same base.

(iv) ΔQRT and parallelogram PQRS are on the same base QR and between the same parallels QR and PS.

(v) Parallelograms PQRS and trapezium SMNR are on the same base but not between the same parallels.

(vi) Parallelograms PQRS, AQRD, BCQR are between the same parallels. Also, parallelograms PQRS, BPSC, APSD are between the same parallels.

(vi) Parallelograms PQRS, AQRD, BCQR are between the same parallels. Also, parallelograms PQRS, BPSC, APSD are between the same parallels.

R.D. Sharma Solutions Class 9th: Ch 15 Areas of Parallelogram and Triangles Exercise 15.2

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Chapter 15 Areas of Parallelogram and Triangles R.D. Sharma Solutions for Class 9th Exercise 15.2

Exercise 15.2

1. If Fig. 15.26, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm. AE = 8 cm and CF = 10 cm, find AD.
Solution

Opposite side of parallelogram are equal, therefore,
CD = AB = 16cm
Since, area of parallelogram =  Height × Base
AE is base on CD
AE×CD = 8cm×16cm
Also, area is
CF×AD = 10cm×AD
So,
Equating Both
8cm×16cm = 10cm×AD
AD = 128cm/10cm = 12.8cm

2. In Q.No. 1, if AD = 6 cm, CF = 10 cm, and AE = 8 cm, find AB.

Solution


We know that,
Area of a parallelogram ABCD = AD×CF --- (1)
Again, area of parallelogram ABCD = CD×AE --- (2)
Comparing equation (1) and (2),
AD×CF = CD×AE
⇒ 6×10 = D×8
⇒ D = 60/8 = 7.5 cm
∴ AB = DC = 7.5 cm
[∵ Opposite side of a parallelogram are equal.]

3. Let ABCD be a parallelogram of area is 124 cm. If E and F are mid points of sides AB and  CD respectively, then find the area of parallelogram AEFD.

Solution


4. If ABCD is a parallelogram, then prove that ar(△ABD) = ar(△BCD) = ar(△ABC) = 1/2 ar(∥gm ABCD)

Solution


Given,
ABCD is a parallelogram.

To prove:
ar(△ABD) = ar(△BCD) = ar(△ABC) = ar(△ACD) = 1/2 ar(∥gm ABCD)

Proof:
We know that diagonal of a parallelogram divides it into two equilaterals.
Since,
AC is the diagonal.
Then,
ar(△ABC) = ar(△ACD) = 1/2 ar(∥gm ABCD) --- (1)
Since, BD is the diagonal.
Then,
ar(△ABD) = ar(△BCD) = 1/2 ar(∥gm ABCD) --- (2)
Compare equation (1) and (2)
∴ ar△ABC) = ar(△ACD) = ar(△ABD)= ar(△BCD) = 1/2 ar(∥gm ABCD).

R.D. Sharma Solutions Class 9th: Ch 15 Areas of Parallelogram and Triangles Exercise 15.3

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Chapter 15 Areas of Parallelogram and Triangles R.D. Sharma Solutions for Class 9th Exercise 15.3

Exercise 15.3

1. In Fig. 15.74, compute the area of quadrilateral ABCD.

Solution


2. In Fig. 15.75, PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of Δ OTS if PQ = 8 cm.

Solution


3. Compute the area of trapezium PQRS in figure

Solution



4. In Fig. 15.77, ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of Δ AOB.

Solution

5. In Fig. 15.78, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

Solution



6. In Fig. 15.79, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2√5, find the area of the rectangle.

Solution


7. In Fig. 15.80, ABCD is a trapezium in which AB || DC. Prove that:
ar(Δ AOD) = ar(Δ BOC)

Solution


8. In Fig. 15.81, ABCD, ABFE and CDEF are parallelograms. Prove that
ar(Δ ADE) = ar(Δ BCF)

Solution


9. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(Δ APB) ✕ ar (Δ CPD) = ar (Δ APD) ✕ ar (Δ BPC)

Solution


10. In Fig. 15.82, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD)
Solution


11. If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.

Solution


12. If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of median AD, prove that ar(Δ BGC)= 2 ar (Δ AGC).

Solution


13. A point D is taken on the side BC of a ΔABC such that BD = 2DC. Prove that ar (Δ ABD) = 2 ar (Δ ADC).

Solution


14. ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that:
(i) ar (Δ ADO) = ar(Δ CDO)
(ii) ar (Δ ABP) = ar (Δ CBP).

Solution


15. ABCD is a parallelogram in which BC is produced to E such that CE= BC. AE intersects CD at F.
(i) Prove that ar(△ADF) = ar(△ECF)
(ii) If the area of △DFB = 3cm2, find the area of ∥gm ABCD.

Solution


16. ABCD is a parallelogram whose diagonals AC and BD intersect at O . A line through O intersects AB at P and DC at Q. Prove that ar(ΔPOA)=ar(ΔQOC).

Solution


17. ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.

Solution 



18. In a △ABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point of AP. Prove that:
(i) ar(△PBQ) = ar(△ARC)
(ii) ar(△PRQ) = 1/2 ar(△ARC)
(iii) ar(△RQC) = 3/8 ar(△ABC)

Solution 


19. ABCD is a paralleogram, G is the point on AB such that AG = 2 GB, E is a point of DC such that CE = 2 DE and F is the point of BC such that BF = 2 FC. Prove that:
(i) ar(ADEG) = ar(GBCE)
(ii) ar (△EGB) = 1/6 ar(ABCD)
(iii) ar(△EFC) = 1/2 ar(△EBF)
(iv) ar(△EBG) = ar(△EFC)
(v) Find what portion of the area of the parallelogram is the area of △EFG.

Solution


20. In Fig. 15.83, CD || AE and CY || BA.
(i) Name a triangle equal in area of ΔCBX
(ii) Prove that ar (Δ ZDE) = ar (Δ CZA)
(iii) Prove that ar (BCZY) = ar (Δ EDZ).

Solution


21. In Fig. 15.84, PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove that ar (PQE) = ar (Δ CFD).


Solution




22. In Fig. 15.85, ABCD is a trapezium in which AB∥DC and DC = 40 cm and AB = 60 cm. If X and Y are respectively the mid-points of AD and BC, prove that:
(i) XY = 50 cm
(ii) DCYX is a trapezium.
(iii) ar (trap. DCXY) = 9/11 ar (trap. XYBA)

Solution



23. In Fig. 15.86, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intesects BC in F. Prove that:
(i) ar(△BDE) = 1/4 ar(△ABC)
(ii) ar(△BDE) = 1/2 ar(△BAE)
(iii) ar(△BFE) = ar(△AFD)
(iv) ar(△ABC) = 2 ar(△BEC)
(v) ar(△FED) = 1/8 ar(△AFC)
(vi) ar(△BFE) = 2 ar(△EFD)
Solution


24. D is the mid-point of side BC of △ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar(△BOE) = 1/8 ar(△ABC)

Solution


25. In Fig. 15.87, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (Δ ABP) = ar (Δ ACQ).

Solution 


26. In Fig. 15.88, ABCD and AEFD are two parallelograms. Prove that

(i) PE = FQ
(ii) ar (Δ APE): ar (ΔPFA) = ar Δ (QFD) : ar (Δ PFD)
(iii) ar (Δ PEA) = ar (Δ QFD).

Solution



28. In a △ABC, if L and M are points on AB and AC respectively such that LM∥BC. Prove that:
(i) ar(△LCM) = ar(△LBM)
(ii) ar(△LBC) = ar(△MBC)
(iii) ar(△ABM) = ar(△ACL)
(ii) ar(△LOB) = ar(△MOC)

Solution



29. In figure, D and E are two points on BC such that BD = DE = EC. Show that:

ar(ΔABD)=ar(ΔADE)=ar(ΔAEC).

 Solution


 
30 . If Fig. 15.91, ABC is a right triangle right angled at A, BCED, ACFGand ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y.
Show that
(i) ΔMBC ≅ Δ ABD
(ii) ar (BYXD) = 2 ar (Δ MBC)
(iii) ar (BYXD) = ar (ABMN)
(iv) Δ FCB ≅ Δ ACE
(v) ar (CYXE) = 2 ar (ΔFCB)
(vi) ar (CYXE) = ar (ACFG)
(vii) ar (BCED) = ar (ABMN) + ar (ACFG)
 Solution




R.D. Sharma Solutions Class 9th: Ch 15 Areas of Parallelogram and Triangles MCQ's

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Chapter 15 Areas of Parallelogram and Triangles R.D. Sharma Solutions for Class 9th MCQ's

Exercise 15.3

1. Two parallelograms are on the same base and between the same parallels. The ratio of their areas is
(a) 1 : 2
(b) 2 : 1
(c) 1 : 1
(d) 3 : 1

Solution



2. A triangle and a parallelogram are on the same base and between the same parallels. The ratio of the areas of triangle and parallelogram is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3

 Solution


3. Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the mid-points of the sides of Δ ABC. Then the area of ΔPQR is
(a) 12 sq. units
(b) 6 sq. units
(c) 4 sq. units
(d) 3 sq. units

 Solution


4. The median of a triangle divides it into two
(a) congruent triangle
(b) isosceles triangles
(c) right triangles
(d) triangles of equal areas

 Solution

Given: A triangle with a median.
Calculation: We know that a ,”median of a triangle divides it into two triangles of equal area.”
Hence the correct answer is option (d).

5. In a △ABC, D, E, F are the mid-points of sides BC, CA and respectively. If ar(△ABC) = 16cm2, then ar(trapezium FBCE) =
(a) 4 cm2
(b) 8 cm2
(c) 12 cm2
(d) 10 cm2

Solution

6. ABCD is a parallelogram. P is ant point on CD. If ar(△DPA) = 10 cm2 and ar(△APC) 20 cm2, then ar(△APB) =
(a) 15 cm2
(b) 20 cm2
(c) 35 cm2
(d) 30 cm2

Solution


7. The area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 16 cm and 12 cm is:
(a) 28 cm2
(b) 48 cm2
(c) 96 cm2
(d) 24 cm2 

Solution 


8. A, B, C, D are mide-points of sides of parallelogram PQRS. If ar(PQRS) = 36 cm2, then ar(ABCD) = 
(a) 24 cm2
(b) 18 cm2
(c) 30 cm2
(d) 36 cm2

Solution




9. The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm is:
(a) a rhombus of area 24 cm2
(b) a rectangle of area 24 cm2
(c) a square of area 26 cm2
(d) a trapezium of area 36 cm2

Solution


10. The  mid-points of the sides of triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to
(a) ar(△ABC)
(b) 1/2 ar(△ABC)
(c) 1/3 ar(△ABC)
(d) 1/4 ar(△ABC)

Solution


11. If AD is median of △ABC and P is a point on AC such that
ar(△ADP) : ar(△ABD) = 2:3, then ar(△PDC) : ar(△ABC)
(a) 1:5
(b) 1:3
(c) 1:6
(d) 3:5

Solution




12. Medians of △ABC intersect at G. If ar(△ABC) = cm2, then ar(△BGC) = 
(a) 6 cm2
(b) 9 cm2(c) 12 cm2
(d) 18 cm2

Solution
13. In a △ABC if D and E are mid-points of BC and AD respectively such that ar(△AEC) = 4 cm2, then ar(△BEC) =
(a) 4 cm2
(b) 6 cm2
(c) 8 cm2
(d) 12 cm2

Solution

14. In Fig. 15.106, ABCD is a parallelogram. If AB = 12 cm, AE = 7.5 cm, CF = 15 cm, then AD =

Solution

15. In Fig. 15.107, PQRS is a parallelogram. If X and Y are mid-points of PQ and SR respectively and diagonal Q is joined. The ratio ar(∥gm XQRY) : ar(△QSR) =
(i) 1:4
(ii) 2:1
(iii) 1:2
(iv) 1:1

Solution


16. In Fig. 15.108, ABCD and FECG are parallelograms equal in area. If ar(△AQE) = 12 cm2, then ar(∥gm FGBQ) =
(a) 12 cm2
(b) 20 cm2
(c) 24 cm2
(d) 36 cm2

Solution


17. Diagonal AC and BD of trapezium ABCD, in which AB || DC, intersect each other at O. The triangle which is equal in area of ΔAOD is
(a) ΔAOB
(b) ΔBOC
(c) ΔDOC
(d) ΔADC

Solution


18. ABCD is a trapezium in which AB∥DC. If ar(ΔABD) = 24 cm2 and AB = 8 cm, then height of △ABC is
(a) 3 cm
(b) 4 cm
(c) 6 cm
(d) 8 cm

Solution


19. ABCD is a trapezium with parallel sides AB =a and DC = b. If E and F are mid-points of non-parallel sides AD and BC respectively, then the ratio of areas of quadrilaterals ABFE and EFCD is
(a) a : b
(b) (a + 3b): (3a + b)
(c) (3a + b) : (a + 3b)
(d) (2a + b) : (3a + b)

Solution

20. ABCD is a rectangle with O as any point in its interior. If ar(△AOD) = 3 cm2 and ar(△BOC) = 6 cm2, then area of rectangle ABCD is
(a) 9 cm2
(b) 12 cm2
(c) 15 cm2
(d) 18 cm2

Solution

R.D. Sharma Solutions Class 9th: Ch 16 Circles Exercise 16.1

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Chapter 16 Circles R.D. Sharma Solutions for Class 9th Exercise 16.1

Exercise 16.1

1. Fill in the blanks:
(i) All points lying inside/outside a circle are called ........ points /.......points.
(ii) Circles having the same centre and different radii are called ...... circles.
(iii) A point whose distance from the centre of a circle is greater than its radius lies in ......of the circle.
(iv) A continuous piece of a circle is ....... of the circle.
(v) The longest chord of a circle is a ....... of the circle.
(vi) An arc is a ..... when its ends are the ends of a diameter.
(vii) Segment of a circle is the region between an arc and ....of the circle.
(viii) A circle divides the plane, on which it lies, in ...... parts.

 Solution

(i) Interior/exterior
(ii) Concentric
(iii) The exterior
(iv) Arc
(v) Diameter
(vi) Semi-circle
(vii) Centre
(viii) Three


 2. Write the truth value (T/F) for the following with suitable reasons:
(i) A circle is a plane figure.
(ii) Line segment joining the centre to any point on the circle is a radius of the circle.
(iii) If a circle is divided into three equal arcs each is a major arc.
(iv) A circle has only finite number of equal chords.
(v) A chord of a circle, which is twice as long is its radius is a diameter of the circle.
(vi) Sector is the region between the chord and its corresponding arc.
(vii) The degree measure of an arc is the complement of the central angle containing the arc.
(viii) The degree measure of a semi-circle is 180°

 Solution

(i) True

(ii) True

(iii) True

(iv) False

(v) True

(vi) True

(vii) False

(viii) True

R.D. Sharma Solutions Class 9th: Ch 16 Circles Exercise 16.2

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Chapter 16 Circles R.D. Sharma Solutions for Class 9th Exercise 16.2

Exercise 16.2



1. The radius of a circle is 8 cm and the length of one of its chords is 12 cm. Find the distance of the chord from the centre.

Solution

2. Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm.

Solution


2.  Find the length of a chord which is at a distance of 4 cm from the centre of the circle of radius 6 cm.

Solution


3. Find the length of a chord which is at a distance of 4 cm from the centre of the circle of radius 6 cm.

Solution

4. Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel, If the distance between AB and CD is 3 cm, find the radius of the circle.

Solution

5. Give a method to find the centre of a given circle.

Solution
 Let AB and C are three distinct points on a circle .
Now join AB and BC and draw their perpendicular bisectors.
The point of intersection of the perpendicular bisectors is the centre of given circle.
Hence O is the centre of circle C(0,r).

6. Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.

Solution


7. Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.

 Solution


8. Prove that two different circles cannot intersect each other at more than two points .

Solution

We have to prove that two different circles cannot intersect each other at more than two points.
Let the two circles intersect in three points A, B and C.
Then as we know that these three points A, B and C are non-collinear. So, a unique circle passes through these three points.
This is a contradiction to the fact that two given circles are passing through A, B, C.
Hence, two circles cannot intersect each other at more than two points.
Hence, proved.

9. A line segment AB is of length 5cm. Draw a circle of radius 4 cm passing through A and B. Can you draw a circle of radius 2 cm passing through A and B? Give reason in support of your answer.

Solution

Given that, a line AB = 5 cm, one circle having radius r1 = 4 cm which is passing through point A and Band other circle of radius r2 = 2 cm. 
As we know that the largest chord of any circle is equal to the diameter of that circle. 
So, 2×r2<AB 
There is no possibility to draw a circle whose diameter is smaller than the length of the chord. 

10. An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.

Solution


11. Given an arc of a circle, complete the circle.

Solution

Let PQ be an arc of the circle.
In order to complete the circle. First of all we have to find out its centre and radius.
Now take a point R on the arc PQ and join PR and QR.
Draw the perpendicular bisectors of PR and QR respectively.
Let these perpendicular bisectors intersect at point O.
Then OP = OQ, draw a circle with centre O and radius OP = OQ to get the required circle

12. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Solution


13. Suppose you are given a circle. Give a construction to find its centre.

Solution


14. Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Solution


15. The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of the other chord from the centre?

Solution

R.D. Sharma Solutions Class 9th: Ch 16 Circles Exercise 16.3

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Chapter 16 Circles R.D. Sharma Solutions for Class 9th Exercise 16.3

Exercise 16.3

1. Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24m each, what is the distance between Ishita and Nisha.

Solution

2. A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.

Solution


R.D. Sharma Solutions Class 9th: Ch 16 Circles Exercise 16.4

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Chapter 16 Circles R.D. Sharma Solutions for Class 9th Exercise 16.4

Exercise 16.4

1. In Fig. 16.120, O is the centre of the circle. If ∠APB∠APB= 50°, find ∠AOB and ∠OAB.

Solution


2. In Fig. 16.121, it is given that O is the centre of the circle and ∠AOC = 150°. Find ∠ABC.
Solution


3. In Fig. 16.122, O is the centre of the circle. Find ∠BAC.
Solution


4. If O is the centre of the circle, find the value of x in each of the following figures.

Solution

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

5. O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A

Solution 


6. In Fig. 16.135, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.

Solution


7. In Fig. 16.136, O is the centre of the circle, prove that ∠x = ∠y + ∠z.

Solution


8. In Fig. 16.137, O and O' are centres of two circles intersecting at B and C. ACD is a straight line, find x.
Solution


9. In Fig. 16.138, O is the centre of a circle and PQ is a diameter. If ∠ROS = 40°, find ∠RTS.
Solution


10. In Fig. 16.139, if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.




Solution 


11. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Solution

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