Category: CBSE Class 9

In this page, we are providing Is Matter Around Us Pure Class 9 Extra Questions and Answers Science Chapter 2 pdf download. NCERT Extra Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 2 Extra Questions and Answers Is Matter Around Us Pure

Extra Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure with Answers Solutions

Is Matter Around Us Pure Class 9 Extra Questions Very Short Answer Type

Class 9 Science Chapter 2 Extra Questions And Answers Question 1.
What is the reason for the difference in properties of solutions, colloids and suspensions?
Answer:
Due to different particle size.

Is Matter Around Us Pure Extra Questions Question 2.
Is rain water or distilled water a pure substance?
Answer:
Yes, because it contains particles (molecules) of only water.

Is Matter Around Us Pure Class 9 Extra Questions Question 3.
Can we separate a mixture of alcohol and water by a separating funnel?
Answer:
No, the two liquids are miscible.

Ch 2 Science Class 9 Extra Questions Question 4.
Give two examples of metalloids.
Answer:
Silicon and germanium.

Class 9 Chemistry Chapter 2 Extra Questions Question 5.
Give an example of a solution in which solid is a solute as well as the solvent.
Answer:
Alloys are solid in solid solutions. For example, brass contains about 30% zinc and 70% copper. Here, zinc is the solute while copper is the solvent.

Class 9 Science Ch 2 Extra Questions Question 6.
What type of liquid mixture will kerosesne oil and water form? How will you separate it?
Answer:
Immiscible. We can separate this mixture by using a separating funnel.

Class 9 Is Matter Around Us Pure Extra Questions Question 7.
Define the term heterogeneous.
Answer:
Heterogeneous means that the substance does not have the same properties or characteristics throughout its bulk.

Chapter 2 Science Class 9 Extra Questions Question 8.
Write the constituent element of potassium hydroxide and sodium chloride.
Answer:
The constituent element of potassium hydroxide is K, H and O and sodium chloride is Na and Cl.

Ncert Class 9 Science Chapter 2 Extra Questions Question 9.
On the basis of composition, how is matter classified?
Answer:

1. Pure substance
2. Mixture.

Extra Questions Of Chapter 2 Science Class 9 Question 10.
What are different categories of pure substances?
Answer:

1. Elements
2. Compounds.

Class 9 Chapter 2 Science Extra Questions Question 11.
What are the different kinds of mixture?
Answer:

1. Homogeneous mixture
2. Heterogeneous mixture.

Is Matter Around Us Pure Class 9 Important Questions With Answers Question 12.
What are the constituents of brass?
Answer:
Brass is an alloy and is a mixture of zinc (30%) and copper (70%).

Class 9 Science Chapter 2 Extra Questions Question 13.
How are elements further classified?
Answer:
Metals, non-metals, metalloids.

Extra Questions Of Is Matter Around Us Pure Question 14.
A solution of water and alcohol contains 30 g of water and 60 g of alcohol. What is the concentration of solution?
Answer:
\(\frac { 30 }{ 30 + 60 }\) x 100 = \(\frac { 30 }{ 90 }\) x 100 = 33.3%.

Class 9th Science Chapter 2 Extra Questions Question 15.
What are aqueous solutions?
Answer:
Solutions in which water is the solvent are called aqueous solutions, e.g., sugar solution, in which sugar is dissolved in water.

Question 16.
What is an unsaturated solution?
Answer:
A solution in which some more solute can be dissolved at any fixed temperature is called an unsaturated solution.

Question 17.
How many gram of water is needed to make 8% mass by mass percentage of sodium carbonate solution if 4 g of sodium carbonate is a variable to make a solution?
Answer:
8% means 8 g in 100 g of solution. So if 4 g Na₂CO₃ is present, it means solution must be 50 g.

Question 18.
What are the conditions required to convert air into liquid air?
Answer:
200 atmospheric pressure and – 200°C.

Question 19.
Define the term inter-conversion of matter.
Answer:
The phenomenon of change of one state of matter into another and back to the original state is called inter-conversion of matter.

Question 20.
Why do fish go in deep water during day light?
Answer:
During day time, the shallow water is warmed and hence it has less dissolved oxygen. Therefore fish tend to go in deep water during day time.

Question 21.
Out of colloid, solution and a suspension which one can be separated by filtration.
Answer:
Suspension.

Question 22.
Out of colloid, solution and a suspension which has the smallest particle?
Answer:
Solution.

Is Matter Around Us Pure Class 9 Extra Questions Short Answer Type 1

Question 1.
Suggest separation technique(s) one would need to employ to separate the following mixtures. [NCERT Exemplar]
(a) Mercury and water
(b) Potassium chloride and ammonium chloride
(c) Common salt, water apd sand
(d) Kerosene oil, water and salt.
Answer:
(a) Separation by using separating funnel
(b) Sublimation
(c) Filtration followed by evaporation
Or
Centrifugation followed by evaporation/distillation
(d) Separation by using separating funnel to separate kerosene oil followed by distillation.

Question 2.
A salt can be recovered from its solution by evaporation. Suggest some other technique for the same? [NCERT Exemplar]
Answer:
Crystallisation.

Question 3.
The ‘sea water’ can be classified as a homogeneous as well as heterogeneous mixture. Comment.
Answer:

• Homogeneous – mixture of salts and water only.
• Heterogeneous – contains salts, water, mud, decayed plant, etc.

Question 4.
Fill in the following blanks:
(i) Milk is a ……………… solution but vinegar is a solution.
(ii) Milk is a colloidal solution in which ……………… is the dispersed phase and ……………… is the dispersion medium.
Answer:
(i) colloidal, true
(ii) fat, water.

Question 5.
(a) Classify Brass and Diamond as element and mixture.
(b) How is a chemical change different from a physical change?
Answer:
(a) Brass is homogeneous mixture also called alloy. The constituents are Cu and Zn. Diamond is an element. It is an allotropic form of carbon.

(b) In a chemical change, a new substance is formed as a result of chemical reaction. No new substance is formed in a physical change.

Question 6.
Identify the dispersed phase and dispersion medium in the following examples of colloids:
(a) Fog
(b) Cheese
(c) Coloured gemstone.
Answer:
(a) Fog: Liquid (water drops) acts as dispersed phase and gas (air) as the dispersion medium.
(b) Cheese: Solid (fat) acts as the dispersed phase and water (liquid) as the dispersion medium.
(c) Coloured gemstone: Solids act the dispersed phase as well as the dispersion medium.

Question 7.
Explain, why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do. [NCERT Exemplar]
Answer:
Particle size in a suspension is larger than those in a colloidal solution. Also molecular interaction in a suspension is not strong enough to keep the particles suspended and hence they settle down.

Question 8.
Smoke and fog both are aerosols. In what way are they different? [NCERT Exemplar]
Answer:
Both fog and smoke have gas as the dispersion medium. The only difference is that the dispersed phase in fog is liquid and in smoke it is a solid.

Question 9.
How will you bring about the following separation?
(i) Fine mud particles floating in water.
(ii) Carbon particles present in smoke.
Answer:
(i) By coagulation using alum and then filtering.
(ii) By passing smoke through electric plates maintained at a high potential difference. The colloidal particles of carbon get neutralised and fall down while air escapes out.

Question 10.
Classify the following as Physical or chemical properties. [NCERT Exemplar]
(a) The composition of a sample of steel is 98% iron, 1.5% carbon and 0.5% other elements.
(b) Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas.
(c) Metallic sodium is soft enough to be cut with a knife.
(d) Most metal oxides form alkalies on interacting with water.
Answer:
Physical properties: (a) and (c)
Chemical properties: (b) and (d)

Question 11.
Suggest a suitable separation technique for the following: [NCERT Exemplar]
(a) Mercury and water
(b) Coloured components from blue ink
(c) Ammonium chloride and potassium chloride
(d) Mixture of alcohol and water.
Answer:
(a) The separation can be done by the use of a separating funnel. Mercury forms the lower layer (heavier) and water the upper layer (lighter).
(b) The separation can be done with the help of chromatography.
(c) Process of sublimation can be used. Ammonium chloride collects as the sublimate while potassium chloride remains in the dish.
(d) Process of fractional distillation can be used. Alcohol (ethyl alcohol) with lower boiling point (78°C or 351 K) gets distilled leaving behind water with higher boiling point (100°C or 373 K) in the distillation flask.

Question 12.
Name the type of colloids in each of the following giving an example of each.

	Dispersed Phase	Dispersing Medium
A	Liquid	Gas
B	Liquid	Liquid
C	Liquid	Solids

Answer:

A	Aerosol	(Fog)
B	Emulsion	(Milk)
C	Gel	(Jelly)

Question 13.
You are given two samples of water labelled as ‘A’ and ‘B’ Sample ‘A’ boils at 100°C and sample ‘B’ boils at 102°C. Which sample of water will not freeze at 0°C? Comment. [NCERT Exemplar]
Answer:
Sample ‘B’ will not freeze at 0°C because it is not pure water. At 1 atm, the boiling point of pure water is 100°C and the freezing point of pure water is 0°C.

Question 14.
Identify colloids from the following: Copper sulphate solution, milk, smoke, muddy water, butter, sugar solution, face cream, lemonade.
Answer:
Colloids: milk, smoke, muddy water, butter, face cream, lemonade.

Question 15.
What are the favourable qualities given to gold when it is alloyed with copper or silver for the purpose of making ornaments? [NCERT Exemplar]
Answer:
Pure gold is very soft as compared to gold alloyed with silver or copper. Thus for providing strength to gold, it is alloyed.

Question 16.
12 mL of dettol is added to a beaker containing 500 mL of water and stirred. State four observations that you make.
Answer:
When dettol is added to a beaker containing of water, the following observations are made.

• An emulsion is formed which is of colloidal nature.
• The colour of emulsion is milky.
• It gives characteristic smell of dettol.
• The solution can pass through a filter paper.

Is Matter Around Us Pure Class 9 Extra Questions Short Answer Type 2

Question 1.
During an experiment the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution. [NCERT Exemplar]
(a) Are the two solutions of the same concentration?
(b) Compare the mass % of the two solutions.
Answer:
(a) No.
\(Mass\%=\frac{\text { Mass of solute }}{\text { Mass of solute }+\text { Mass of solvent }} \times 100\)

(b) Solution made by Ramesh
Mass % = \(\frac { 10 }{ 10 + 100 }\) = \(\frac { 10 }{ 110 }\) x 100 = 9.09%
solution made by sarika
Mass % = \(\frac { 10 }{ 100 }\) x 100 = 10%
The solution prepared by Sarixa has a higher mass % than that prepared by Ramesh.

Question 2.
State which of the following solutions exhibit Tyndall effect:
Starch solution, Sodium chloride solution, Tincture of iodine, Air.
Answer:
(i) Tyndall effect is shown both by starch solution and air which are heterogeneous mixtures and have the capacity to scatter a beam of light as it passes through them.

(ii) Sodium chloride solution and tincture of iodine (iodine crystals dissolved in ethyl alcohol) are both homogeneous in nature and do not exhibit any Tyndall effect.

Question 3.
While diluting a solution of salt in water, a student by mistake added acetone (boiling point 56°C). What technique can be employed to get back the acetone? Justify your choice. [NCERT Exemplar]
Answer:
Distillation, since acetone is more volatile it will separate out first.

Question 4.
(a) Why is crystallisation technique better than evaporation?
(b) Write any two physical properties of each of metals and non-metals.
(c) Name the technique used to separate butter from curd.
Answer:
(a) Both these techniques are used to separate solid substances from their solutions. But crystallisation is considered better because during evaporation certain solids may decompose or some of them like sugar get charred when the solution is evaporated completely to dryness. As a result of crystallisation, even the shapes of the crystals do not change.

(b) (i) Metals have a shiny surface known as lustre.
(ii) Metals are malleable and ductile.
(iii) Non-metals are mostly poor conductors of electricity.
(iv) Non-metals are generally soft.

(c) Butter can be separated from curd by the process of centrifugation. This is usually done by churning which is very common as well as convenient.

Question 5.
Name the process associated with the following:
(a) Dry ice is kept at room temperature and at one atmospheric pressure.
(b) A drop of ink placed on the surface of water contained in a glass spreads throughout the water.
(c) A potassium permanganate crystal is in a beaker and water is poured into the beaker with stirring.
(d) A acetone bottle is left open and the bottle becomes empty.
(e) Milk is churned to separate cream from it.
(f) Settling of sand when a mixture of a sand and water is left undisturbed for some time.
Answer:
(a) Sublimation
(b) Diffusion
(c) Dissolution/diffusion
(d) Evaporation, diffusion
(e) Centrifugation
(f) Sedimentation

Question 6.
On dissolving chalk powder in water, a suspension is obtained. Give any four reasons to support the fact that the mixture so obtained is a suspension only.
Answer:
It is supported by the following reasons:

• White particles of chalk powder can be seen with the naked eyes.
• The particles can be separated by ordinary filter paper.
• Upon shaking, a white turbidity reappears in solution.
• Light cannot pass through the suspension which shows that it is of opaque nature.

Question 7.
Give an example for each of the following:
(a) Solid-liquid homogeneous mixture
(b) Gas-gas homogeneous mixture
(c) Liquid-liquid heterogeneous mixture.
Answer:
(a) Mixture of sodium chloride in water.
(b) Air. It is a homogeneous mixture of a number of gases.
(c) Emulsion of oil and water.

Question 8.
What would you observe when
(a) a saturated solution of potassium chloride prepared at 60°C is allowed to cool to room temperature?
(b) an aqueous sugar solution is heated to dryness?
(c) a mixture of iron filings and sulphur powder is heated strongly?
Answer:
(a) Solid potassium chloride will separate out.
(b) Initially the water will evaporate and then sugar will get charred.
(c) Iron sulphide will be formed.

Question 9.
(a) Arrange solids, liquids and gases in increasing order of the following properties of matter
(i) rigidity
(ii) diffusion
(iii) compressibility.
(b) Write one example from your daily life which is based on diffusion of gases.
Answer:
(a) (i) Rigidity: Gases < Liquids < Solids
(ii) Diffusion: Solids < Liquids < Gases
(iii) Compressibility: Solids < Liquids < Gases.

(b) Smell of aroma or perfume released in one corner of the room soon spreads in the whole room.

Question 10.
Is air a mixture or a compound? Give three reasons.
Answer:
Air is a mixture and not a compound as discussed below:
(i) The properties of a mixture are in between those of its constituents. The two major components of air are oxygen (20.9% by volume) and nitrogen (78.1% by volume). In oxygen, any fuel bums brightly but in nitrogen it gets extinguished. In contrast, in air the fuel bums slowly.

(ii) The components of a mixture can be separated by simple physical methods. For example, the components of air can be separated by fractional distillation of liquid air.

(iii) The composition of a mixture is variable. The composition of air is also variable. It has more oxygen in the country side than in big cities.

(iv) When air is obtained by mixing its constituent gases, no heat is either evolved or absorbed.

(v) Liquid air does not have a fixed boiling point.

Is Matter Around Us Pure Class 9 Extra Questions Long Answer Type

Question 1.
Classify each of the following, as a physical or a chemical change. Give reasons.
(a) Drying of a shirt in the Sun.
(b) Rising of hot air over a radiator.
(c) Burning of kerosene in a lantern.
(d) Change in the colour of black tea on adding lemon juice to it.
(e) Churning of milk cream to get butter.
Answer:
Physical changes: (a), (b), (e)
Chemical changes: (c), (d)

Question 2.
Iron filings and sulphur were mixed together and divided into two parts, ‘A’ and ‘B’. Part ‘A’ was heated strongly while Part ‘B’ was not heated. Dilute hydrochloric acid was added to both the parts and evolution of gas was seen in both the cases. How will you identify the gases evolved?
Answer:
Part A

Part B

When dilute HCl is added to it, only the iron filings in the mixture react and sulphur remains unreacted
Fe (s) + 2 HCl (aq) → FeCl₂ + H₂ gas
H₂S gas formed has a foul smell and on passing through lead acetate solution, it turns the solution black. Hydrogen gas burns with a pop sound.

Question 3.
Fill in the blanks:
(a) A colloid is a ……………….. mixture and its components can be separated by the technique known as ………………..
(b) Ice, water and water vapour look different and display different ……………….. properties but they are the same.
(c) A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for sometime. The upper layer in the separating funnel will be of ……………….. and the lower layer will be that of
(d) A mixture of two or more miscible liquids, for which the difference in the boiling points is less than 25 K can be separated by the process called ………………..
(e) When light is passed through water containing a few drops of milk, it shows a bluish tinge. This is due to the ……………….. of light by milk and the phenomenon is called ……………….. This indicates that milk is a ……………….. solution. [NCERT Exemplar]
Answer:
(a) heterogeneous, centrifugation
(b) physical, chemically
(c) water, chloroform (hint-density of water is less than that of chloroform)
(d) fractional distillation scattering, Tyndall effect, colloidal.

Question 4.
Give an example each for the mixture having the following characteristics. Suggest a suitable method to separate the components of these mixtures. [NCERT Exemplar]
(a) A volatile and a non-volatile component.
(b) Two volatile components with appreciable difference in boiling points.
(c) Two immiscible liquids.
(d) One of the components changes directly from solid to gaseous state.
(e) Two or more coloured constituents soluble in some solvent.
Answer:
(a) Evaporation or distillation
(b) Distillation
(c) By using a separating funnel
(d) Sublimation
(e) Chromatography

Question 5.
Which separation techniques you will apply for the separation of the following mixtures?
(a) Oil from water
(b) Camphor from sand
(c) Sodium chloride from its solution in water
(d) Cream from milk
(e) Metal pieces from engine oil of a car.
Answer:
(a) By the use of a separating funnel.
(b) With the help of sublimation technique
(c) By evaporation crystallisation technique.
(d) By the use of a centrifuge
(e) By the use of filtration technique

Is Matter Around Us Pure Class 9 Extra Questions HOTS

Question 1.
Which of the tubes shown below will be more effective as a condenser in the distillation apparatus? [NCERT Exemplar]

Answer:
The presence of beads in tube (a) would provide a larger surface area for cooling. So, this tube will be more effective as a condenser.

Question 2.
Non-metals are usually poor conductors of heat and electricity. They are non-lustrous, non- sonorous, non-malleable and are coloured.
(a) Name a lustrous non-metal.
(b) Name a non-metal which exists as a liquid at room temperature.
(c) The allotropic form of a non-metal is a good conductor of electricity. Name the allotrope.
(d) Name a non-metal which is known to form the largest number of compounds.
(e) Name a non-metal other than carbon which shows allotropy.
(f) Name a non-metal which is required for combustion.
Answer:
(a) Iodine
(b) Bromine
(c) Graphite
(d) Carbon
(e) Sulphur, phosphorus
(f) Oxygen.

Question 3.
The teacher instructed three students ‘A’, ‘B’ and ‘C’ respectively to prepare a 50% (mass by volume) solution of sodium hydroxide (NaOH). ‘A’ dissolved 50 g of NaOH in 100 mL of water, ‘B’ dissolved 50 g of NaOH in 100 g of water while ‘C’ dissolved 50 g of NaOH in water to make 100 mL of solution. Which one of them has made the desired solution and why?
Answer:
‘C’ has made the desired solution.

= \(\frac { 50 }{ 100 }\) x 100
= 50% mass by volume.

Question 4.
On heating calcium carbonate gets converted into calcium oxide and carbon dioxide.
(a) Is this a physical or a chemical change?
(b) Can you prepare one acidic and one basic solution by using the products formed in the above process? If so, write the chemical equation involved.
Answer:
(a) Chemical change

(b) Acidic and basic solutions can be prepared by dissolving the products of the above process in water
CaO + H₂O → Ca(OH)₂ (basic solution)
CO₂ + H₂O → H₂CO₃ (acidic solution)

Question 5.
Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions?
(a) 1.00 g of NaCl + 100 g water
(b) 0.11 g of NaCl + 100 g of water
(c) 0.01 g of NaCl + 99.99 g of water
(d) 0.10 g of NaCl + 99.90 g of water
Answer:
(c) 0.01 g of NaCl + 99.99 g of water

= \(\frac { 0.01 }{ 0.01 + 99.99 }\) x 100
= \(\frac { 0.01 }{ 100 }\) x 100
= 0.01 g

Question 6.
A child wanted to separate the mixture of dyes constituting a sample of ink. He marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in the figure. The filter paper was removed when the water moved near the top of the filter paper. [NCERT Exemplar]

1. What would you expect to see, if the ink contains three different coloured components?
2. Name the technique used by the child.
3. Suggest one more application of this technique.

Answer:

1. Three different bands will be observed.
2. Chromatography
3. To separate the pigments present in chlorophyll.

Question 7.
A group of students took an old shoe box and covered it with a black paper from all sides. They fixed a source of light (a torch) at one end of the box by making a hole in it and made another hole on the other side to view the light. They placed a milk sample contained in a beaker/tumbler in the box as shown in the figure. They were amazed to see that milk taken in the tumbler was illuminated. They tried the same activity by taking a salt solution but found that light simply passed through it?

1. Explain why the milk sample was illuminated. Name the phenomenon involved.
2. Same results were not observed with a salt solution. Explain.
3. Can you suggest two more solutions which would show the solution?

Answer:

1. Milk is a colloid and would show Tyndall effect.
2. Salt solution is a true solution and would not scatter light.
3. Detergent solution, sulphur solution.

Question 8.
Sudha tested the solubility of four salts X, Y, Z and T at different temperatures and collected the following data. (Solubility refers to the amount in grams dissolved in 100 g of water to give a saturated solution.)

Answer the following questions from the table:

1. Which salt has the highest and lowest solubility at 323 K?
2. A student prepared a saturated solution of X at 323 K and then added 25 g water to it. What mass of X must be added to again make the solution saturated?
3. The solubility of which salt is least affected by increase in temperature?
4. What mass of T would be required to make saturated solution in 200 g of water at 290 K?

Answer:
1. At 323 K, salt Y has the highest solubility in water while salt Z has the lowest solubility.

2. By definition of saturated solution,
100 g of water at 323 K contain salt = 40 g
125 g of water at 323 K contain salt = \(\frac { 40 g }{ 100 g }\) x (125 g) = 50 g
∴ Mass of salt to be added to make the solution again saturated = (50 – 40) = 10 g

3. The data show that the solubility of the salt Y is least affected with increase in temperature.

4. At 290 K, mass of T required to make a saturated solution in 200 g of water = \(\frac { 25 g }{ 100 g }\) x (200 g) = 50 g

Is Matter Around Us Pure Class 9 Extra Questions Value Based (VBQs)

Question 1.
Mallika’s mother was suffering from cold and cough. Mallika prepared tea for her mother. She boiled water in a pan, then she added tea leaves, sugar and milk to it. She filtered the tea in a cup and served it to her mother.
(a) Explain the values shown by Mallika.
(b) Identify solute, solvent, residue and filtrate in this activity.
Answer:
(a) Mallika used the knowledge of chemistry to provide relief to her mother. Actually she prepared an extract of tea leaves which is helpful in curing cold and cough and gives warmth to the body.

(b) Solute: Tea leaves and sugar.
Solvent: water and milk.
Filtrate: homogeneous mixture of water, milk, sugar and extract of tea leaves.

Question 2.
Amit was asked by his teacher to separate a liquid mixture of acetone and ethyl alcohol. He set up a distillation apparatus and tried to distil the mixture. To his surprise, both the liquids got distilled. Teacher told Amit to repeat the experiment by using a fractionating column in the distillation flask. Amit followed the advice of the teacher and he was able to separate the two liquids.

1. Why was Amit not successful in separating the liquid mixture earlier?
2. Why did teacher ask him to use the fractionating column?
3. Which liquid was distilled first?
4. As a student of chemistry, what value based information you have gathered?

Answer:
1. The difference in boiling point temperatures of acetone (56°C) and ethyl alcohol (78°C) is only 22°C.
Therefore, process of simple distillation fails in this case.

2. Fractionating column is quite effective in this case because it obstructs the distillation of ethyl alcohol which is high boiling and at the same time helps in the distillation of acetone which is low boiling.

3. Acetone was distilled first since it has comparatively low boiling point.

4. The process of simple distillation can be used only in case, the liquids present in the mixture differ in their boiling point by 25°C or more.

In this page, we are providing Matter in Our Surroundings Class 9 Extra Questions and Answers Science Chapter 1 pdf download. NCERT Extra Questions for Class 9 Science Chapter 1 Matter in Our Surroundings with Answers will help to score more marks in your CBSE Board Exams. https://ncertmcq.com/extra-questions-for-class-9-science/

Class 9 Science Chapter 1 Extra Questions and Answers Matter in Our Surroundings

Extra Questions for Class 9 Science Chapter 1 Matter in Our Surroundings with Answers Solutions

Matter in Our Surroundings Class 9 Extra Questions Very Short Answer Type

Matter In Our Surroundings Class 9 Extra Questions With Answers Question 1.
Define the term ‘matter’.
Answer:
Matter is defined as anything that occupies some space and has definite mass.

Class 9 Science Chapter 1 Extra Questions And Answers Question 2.
What is Law of Conservation of Mass?
Answer:
Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction.

Matter In Our Surroundings Extra Questions Class 9 Question 3.
Define melting point.
Answer:
It is the temperature at which a solid becomes liquid at atmospheric pressure by absorbing heat.

Class 9 Chemistry Chapter 1 Extra Questions Class 9 Question 4.
Out of water and alcohol, which is more volatile?
Answer:
The boiling point of alcohol (78°C or 351K) is lower than that of water (100°C or 373K), therefore, alcohol is more volatile than water.

Matter In Our Surroundings Class 9 Extra Questions With Answers Pdf Question 5.
What is sublimation?
Answer:
Direct conversion of a solid into vapour and vice-versa (i.e., vapour into solid) is called sublimation.

Matter In Our Surroundings Class 9 Extra Questions Question 6.
Is dry ice the same thing as ordinary ice?
Answer:
No, dry ice is solid carbon dioxide while ordinary ice is solid water.

Extra Questions For Class 9 Science Chapter 1 Question 7.
Define latent heat of fusion.
Answer:
It is the heat energy required to convert 1 kg of solid into liquid at its melting point at atmospheric pressure.

Class 9 Science Ch 1 Extra Questions  Question 8.
Define vapourisation.
Answer:
The process of change from liquid state to gaseous (vapour) state is called vapourisation.

Class 9 Chapter 1 Science Extra Questions Question 9.
Give the important properties on the basis of which the three states of matter can be distinguished.
Answer:
The three states of matter can distinguished on the basis of shape, volume, compressibility, packing of molecules, number of free surfaces, etc.

Ncert Class 9 Science Chapter 1 Extra Questions Question 10.
Name the term used for the solid which is directly formed from the gas.
Answer:
Sublimate.

Science Class 9 Chapter 1 Extra Questions Question 11.
Define the term volatile liquid.
Answer:
Those liquids which can change into vapour easily are termed as volatile liquids.

Ch 1 Science Class 9 Extra Questions Question 12.
What is the effect of pressure on boiling point?
Answer:
Boiling point increases with increase in pressure.

Class 9 Science Chapter 1 Extra Questions Question 13.
Name any two substances which sublime.
Answer:
Camphor, napthalene, iodine, ammonium chloride.

Class 9 Science Chapter 1 Short Question Answer Question 14.
Define condensation.
Answer:
The change of a gaseous state to a liquid state on cooling is known as condensation.

Chapter 1 Science Class 9 Extra Questions Question 15.
State the effect of surface area on rate of evaporation.
Answer:
If the surface area is increased, the rate of evaporation increases.

Question 16.
Define evaporation.
Answer:
Evaporation is a physical process in which a liquid changes to its gaseous state, at a temperature lower than its boiling point.

Question 17.
What are the ways in which a gas can be liquefied?
Answer:
Applying pressure and reducing temperature can liquefy gases.

Question 18.
What is plasma?
Answer:
It is a state of matter which consists of super energetic and super excited particles. These particles are in the form of ionised gases.

Question 19.
How do solids, liquids and gases differ in shape and volume?
Answer:
Solids have a definite shape and a fixed volume, liquids have a definite volume but no fixed shape while gases neither have a definite volume nor a definite shape.

Question 20.
Kelvin scale of temperature is regarded as better scale than Celsius. Why?
Answer:
As it has a wide range of measurement and temperature in kelvin scale always has a positive sign, hence regarded as better scale than Celsius.

Matter in Our Surroundings Class 9 Extra Questions Short Answer Type 1

Question 1.
What are characteristics of particles of matter?
Answer:
The particles of matter have following characteristics:

• Particles of matter are very very small.
• Particles of matter have space between them.
• Particles of matter attract each other.
• Particles of matter are constantly moving.

Question 2.
Write four main characteristics of solid state of matter.
Answer:

• Solids have definite mass, volume and shape.
• The particles in solid state are closely packed and empty spaces in them are negligible.
• Solids are rigid.
• Solids can have a number of free surfaces.

Question 3.
Write four main characteristics of liquid state of matter.
Answer:

• Liquids have a definite mass and volume.
• A liquid can take the shape of a container.
• Liquids have only one free surface.
• Liquids show the property of diffusion.

Question 4.
Write four characteristics of gaseous state of matter.
Answer:

• A gas has definite mass but it has neither definite shape nor definite volume.
• Gases can occupy the whole of the space available to them.
• There are larger vacant spaces between the molecules of a gas.
• Gases are highly compressible.

Question 5.
Explain evaporation and its cooling effect in terms of kinetic energy of particles.
Answer:
During evaporation, the molecules which possess higher kinetic energy leave the liquid and go into the space above the liquid as vapour. The remaining molecules possessing lower kinetic energy are left in the liquid state. Consequently, the average kinetic energy decreases which results in the fall of temperature of the liquid.

Question 6.
How is heat transferred when a solid sublimes?
Answer:
Certain solids like iodine, naphthalene, solid CO₂ sublimes on heating. Heat is absorbed by the molecules of these solids rapidly which provides enough kinetic energy to show phase change into gaseous state.

Question 7.
Why do gases diffuse rapidly?
Answer:
Gases diffuse rapidly due to high speed of the particles and large space between them.

Question 8.
For any substance, why does the temperature remain constant during the change of state?
Answer:
On increasing the temperature of a substance, for example a solid, the kinetic energy of the particles increases which is used to overcome the forces of attraction between the particles therefore the temperature remains constant during the change of state.

Question 9.
Explain compressibility in gases with an example.
Answer:
Liquefied petroleum gas (LPG) cylinders are used in our homes for cooking, contains gases in the compressed state. Similarly, compressed natural gas (CNG) is used as a fuel in vehicles. Large volume of gases can be compressed in small cylinders and are transported to distant places.

Question 10.
Why solids cannot be compressed like gases?
Answer:
The particles in solids are so tightly packed that there are no or little interparticle spaces left among them. Therefore solids are not compressible like gases. Gases which have large interparticle spaces are therefore compressible.

Question 11.
Define boiling. Why is boiling considered as bulk phenomenon?
Answer:
Rapid and breaking of bubbles in the bulk of a liquid being heated is called boiling. During boiling particles from the bulk of liquid gain enough energy to get converted to vapour. Therefore it is a bulk phenomenon.

Question 12.
Why do we see water droplets on the outer surface of a glass containing ice-cold water?
Answer:
The water vapour present in air, on coming in contact with the cold glass of water, loses energy and gets converted to liquid state, which we see as water droplets.

Question 13.
Why do we sprinkle water on the roof or open ground in summer?
Answer:
During hot summer evenings, we often sprinkle water on the roof of the house or open ground in front of our house. The water evaporates by absorbing heat from the ground and the surrounding air. By losing heat, the ground becomes cool and we feel comfortable.

Question 14.
Why is ice rubbed on a burnt part of the skin?
Answer:
When a finger or some part of our body gets burnt, we rub the burnt portion with an ice cube. The reason being that due to burning, the temperature of the injured skin increases. When ice is rubbed, the excess heat from the skin is taken away by large latent heat of fusion of water. As a result, the temperature of the injured skin decreases and we feel less pain.

Question 15.
How will you demonstrate that particles of matter are continuously moving?
Answer:
When an incense stick is lit in one corner of a room, we get the smell while sitting at a distance from the stick. This is because the particles of matter are continuously moving. Because of their random motion, the particles of incense mix with the particles of air rapidly and the smell of the incense reaches us even when we are sitting at a distance from the incense stick.

Question 16.
Why do solids expand a bit on heating and contract a bit on cooling?
Answer:
The solid molecules do not have sufficient intermolecular (or interparticle) space thus its expands a bit on heating. The interparticle forces of attraction are very strong which do not let solid particles leave their mean positions. Therefore solid contracts a bit on cooling.

Question 17.
Draw diagram to shown interconversion among states of matter.
Answer:

Question 18.
Why is light not considered matter?
Answer:
Matter occupies space and has mass. Light has neither of the two and that is why it is not considered as matter. It is considers as a form of energy and electromagnetic radiation.

Question 19.
Convert the following temperatures:
(a) – 78.0°C to kelvin
(b) 775 K to °C
(c) 489 K to °C
(d) 24°C to kelvin
Answer:
(a) – 78 + 273 = 195 K
(b) 775 – 273 = 502°C
(c) 489-273 = 216°C
(d) 24 + 273 = 297 K

Question 20.
Mention the difference between gas and vapour.
Answer:
Gas – The gas is a substance which exists in the gaseous state at a temperature equal to or more than the boiling point of its liquid state. For example oxygen, hydrogen, nitrogen, etc.

Vapour – A vapour is a substance which exists in the gaseous state such that its temperature is lower than that of boiling point of its liquid state. For example, water vapour, iodine vapour, etc.

Question 21.
A sample of water under study was found to boil at 102″C at normal temperature and pressure. Is the water pure? Will this water freeze at 0°C? Comment.
Answer:
It’s freezing point will be below 0°C due to the presence of a non-volatile impurity in it.

Question 22.
Water as ice has a cooling effect, whereas water as steam may cause severe burns. Explain these observations. [NCERT Exemplar]
Answer:
In case of ice, the water molecules have low energy while in the case of steam the water molecules have high energy. The high energy of water molecules in steam is transformed as heat and may cause burns. On the other hand, in case of ice, the water molecules take energy from the body and thus gave a cooling effect.

Question 23.
It is a hot summer day, Priyanshi and Ali are wearing cotton and nylon clothes respectively. Who do you think would be more comfortable and why? [NCERT Exemplar]
Answer:
Cotton being a better absorber of water than nylon helps in absorption of sweat followed by evaporation which leads to cooling. So Priyanshi would be more comfortable than Ali.

Question 24.
You want to wear your favourite shirt to a party, but the problem is that it is still wet after a wash. What steps would you take to dry it faster? [NCERT Exemplar]
Answer:
Conditions that can increase the rate of evaporation of water are:

• An increase of surface area by spreading the shirt
• An increase in temperature by putting the shirt under the Sun
• An increase the wind speed by spreading it under the fan.

Matter in Our Surroundings Class 9 Extra Questions Short Answer Type 2

Question 1.
What is evaporation? Why does evaporation cause cooling?
Answer:
The process in which a liquid changes into its vapour state at a temperature below the boiling point is called evaporation. Evaporation is an endothermic process i.e., the liquid absorbs heat during evaporation. This heat may be provided either by the surrounding or by the liquid itself. When the evaporating liquid takes the required heat from other parts of the liquid, the rest of the liquid cools down.

On the other hand, if the liquid takes heat from the surroundings, it causes cooling of the surroundings. For example, on a hot day (sunny day) we perspire. When this sweat evaporates, it absorbs the required heat from our body, and we feel cool.

Question 2.
What factors affect the rate of evaporation?
Answer:
Factors that affect the rate of evaporation are:

• Temperature: Evaporation increases with increase in temperature.
• Humidity: Evaporation decreases with an increase in humidity.
• Wind speed: Evaporation increases with an increase in wind speed.

Question 3.
What is a dry ice and what are its properties?
Answer:
Solid carbon dioxide is known as dry ice. It is stored under high pressure. Solid CO₂ gets converted directly to gaseous state on decrease of pressure to 1 atmosphere without passing through the liquid state (i.e., sublimes). This is the reason that solid carbon dioxide is also known as dry ice.

It is mainly used as a cooling agent because its temperature is very low than ice formed from water. Dry ice is commonly used in theaters and in movies to produce the effect of fog.

Question 4.
Why should we wear cotton clothes in summer?
Answer:
During summer, we perspire more because of the mechanism of our body which keeps us cool. We know that during evaporation, the particles at the surface of the liquid gain energy from the surroundings or body surface and change into vapour. The heat energy equal to latent heat of vapourisation is absorbed from the body leaving the body cool. Cotton, being a good absorber of water helps in absorbing the sweat and exposing it to the atmosphere for easy evaporation.

Question 5.
Give the main postulates of kinetic theory of matter.
Answer:
The main postulates of kinetic theory are:

• All matter is made up of a large number of extremely small particle called molecules.
• The molecules are always in a state of rapid random motion.
• The molecules possess kinetic energy.
• There are attractive forces between the molecules.
• The kinetic energy of molecules increases with increase in temperature.

Question 6.
Identify each of the following changes of state as evaporation, boiling or condensation. Give reason for your answer.
(a) Wet clothes dry when spread on wire.
(b) After a hot shower, your bathroom mirror is covered with water.
(c) Lava flows into the ocean and forms steam.
Answer:
(a) Evaporation, because conversion of liquid water to vapour occur at room temperature.
(b) Condensation, because hot water vapour condense to form liquid water.
(c) Boiling, because heat of lava makes liquid water boil and hence steam is formed.

Question 7.
Why do surgeons often spray some ether on the skin before performing minor surgery?
Answer:
Quite often doctors spray ether on a portion of skin the before performing minor surgery. The reason being that a ether has very low boiling point (308 K). Therefore, it evaporates quite rapidly. The heat energy needed for evaporation is taken from the skin. As a result, the temperature of the skin becomes so low that it almost becomes numb. Due to this numbness, the patient does not feel much pain when a minor cut is made in the skin in order to perform surgery.

Similarly, when a player gets injured during a game, ethyl chloride on the injured portion of the body. Since the boiling point of ethyl chloiide (285.5K or 12.5°C) is very low, it quickly evaporates. The heat energy needed for evaporation is taken from the skin. By losing heat, temperature of the skin becomes so low that it almost becomes numb. Due to this numbness, the player does not feel much pain.

Question 8.
Comment upon the following:
(i) Rigidity
(ii) Compressibility
(iii) Fluidity
Answer:
(i) Rigidity means tendency to maintain shape when some outside force is applied due to strong interparticle force.

(ii) Compressibility means tendency to decrease volume when some outside force is applied. Due to large interparticle distances in gases their volume decreases when some pressure is applied on them therefore, gases possess high compressibility.

(iii) Fluidity means tendency to flow. Due to large interparticle distances and weak forces of attraction gases have highest fluidity.

Question 9.
Comment on the following statements:
(а) Evaporation produces cooling.
(b) Rate of evaporation of an aqueous solution decreases with increase in humidity.
(c) Sponge though compressible is a solid. [NCERT Exemplar]
Answer:
(а) For evaporation to occur, heat energy is needed. This heat energy is taken out from the substance or the surroundings. As a result surrounding becomes cool. Thus, evaporation causes cooling.

(b) By humidity we mean, the amount of water vapours present in the air. With increase in humidity the rate of evaporation decreases. If the humidity of air is already high, it can hold only a little more amount of water vapour to reach that optimum level, therefore the rate of evaporation decreases.

(c) Sponge has large number of minute holes in which air is trapped. When we press it, air expelled and sponge is compressed to a small amount of matter which has a definite shape as well as definite volume.

Matter in Our Surroundings Class 9 Extra Questions Long Answer Type

Question 1.
List any five physical properties of liquids.
Answer:

• Liquids do not have fixed shape or boundaries.
• They have fixed volume.
• They exhibit fluidity i.e., they can flow.
• Less compressible as compared to gases but higher than solids.
• Lower density as compared to solids.
• Compared to solids, liquids have higher kinetic energy but less than gases.
• The intermolecular forces of attraction are weaker than those of solids.
• Show the property of intermixing i.e., can diffuse.

Question 2.
Fill in the blanks:
(а) Evaporation of a liquid at room temperature leads to a ………………… effect.
(b) At room temperature the forces of attraction between the particles of solid substances are ………………… than those which exist in the gaseous state.
(c) The arrangement of particles is less ordered in the state. However, there is no order in ………………… the state.
(d) is the change of gaseous state directly to solid state without going through the ………………… state.
(e) The phenomenon of change of a liquid into the gaseous state at any temperature below its boiling point is called ………………… [NCERT Exemplar]
Answer:
(a) cooling
(b) stronger
(c) liquid, gaseous
(d) Sublimation, liquid
(e) evaporation

Question 3.
Classify the following into osmosis/diffusion [NCERT Exemplar]
(a) Swelling up of a raisin on keeping in water.
(b) Spreading of virus on sneezing.
(c) Earthworm dying on coming in contact with common salt.
(d) Shrinking of grapes kept in thick sugar syrup.
(e) Preserving pickles in salt.
Answer:
(a) Osmosis
(b) Diffusion
(c) Osmosis
(d) Osmosis
(e) Osmosis

Question 4.
You are provided with a mixture of naphthalene and ammonium chloride by your teacher. Suggest an activity to separate them with well labelled diagram. [NCERT Exemplar]
Answer:
Naphthalene is insoluble in water but soluble in ether an organic solvent. It is volatile at room temperature. Ammonium chloride is soluble in water and volatile at higher temperature. It decomposes on heating to dryness.

Question 5.
Why does the temperature of a substance remain constant during its melting point or boiling point?
Answer:
The temperature of a substance remains constant at its melting and boiling points untill all the substance melts or boils because, the heat supplied is continuously used up in changing the state of the substance by overcoming the forces of attraction between the particles. This heat energy absorbed without showing any rise in temperature is given the name latent heat of fusion/latent heat of vapourisation.

Matter in Our Surroundings Class 9 Extra Questions HOTS

Question 1.
Arrange the following substances in increasing order of force of attraction between the particles.
(a) Milk
(b) Salt
(c) Oxygen
Answer:
Oxygen < Milk < Salt.

Question 2.
Explain with an experiment to show gases do not have fixed shape or volume.
Answer:
Experiment: To show gases do not have fixed shape or volume.

Method:

• Take two balloons of different shapes. For example, one round and one heart shape or cylindrical.
• Fill the balloons with air.

Observation: Air takes up the shape of balloon.

Conclusion: This shows air has no definite shape or volume. It takes up the shape of the balloon.

Question 3.
Name the change of state during the following changes:
(a) Drying of wet clothes
(b) Melting of wax
(c) Melting of ice
(d) Formation of clouds
Answer:
(a) Liquid to gaseous state
(b) Solid to liquid state
(c) Solid to liquid state
(d) Liquid to gaseous state

Question 4.
With proper explanation, explain whether the following statements are true or false?
(a) Sublimation occurs only when the solid is heated.
(b) A lighter gas can move downwards and a heavier gas can move upwards.
(c) Interconversion of matter is a constant temperature process.
Answer:
(a) Statement is wrong. Sublimation may occur on its own or by heating, e.g., camphor, naphthalene, iodine, etc., sublime slowly at room temperature.

(b) Statement is true. Diffusion occurs against the law of gravitation. Therefore, lighter gases can also diffuse downwards and the heavier gases can also diffuse upwards. However rate of diffusion of lighter gases is faster than those of heavier gases.

(c) Statement is true. During interconversion of state of matter from solid to liquid or from liquid to gas, it tends to reach its melting point or boiling point. At this point, the temperature remains constant unit it has changed in another state.

Question 5.
What is meant by Bose-Einstein Condense?
Answer:
(a) In 1920, Indian scientist Satyendra Nath Bose did some calculations, based on which Albert Einstein predicted that a new state of matter should exist.

(b) This new state was named as Bose-Einstein Condensate (BEC). In 2001, Cornell, Ketterie and Wieman of USA received Noble Prize for actually making this state in laboratory. BEC is made by cooling gas of very low density to super low temperature.

Question 6.
A student heats a beaker containing ice and water. He measures the temperature of the content of the beaker as a function of time. Which of the following figure would correctly represent the result? Justify your choice. [NCERT Exemplar]

Answer:
Since ice and water are in equilibrium, the temperature would be zero. When we heat the mixture, energy supplied is utilised in melting the ice and the temperature does not change till all the ice melts because of latent heat of fusion. On further heating, the temperature of the water would increase. Therefore the correct option is (d).

Question 7.
Look at the figure and suggest in which of the vessels A, B, C or D the rate of evaporation will be the highest? Explain. [NCERT Exemplar]
Answer:

(c) The rate of evaporation increases with an increase in the surface area of absorption because evapo-ration in a surface phenomenon. Also, with the increase in air speed, the particles of water vapour will move away with the air, which will increase the rate of evaporation.

Matter in Our Surroundings Class 9 Extra Questions Value Based (VBQs)

Question 1.
In a hot summer day, Rajeev wants to watch a movie in a nearest cineplex. His mother wears a nylon saree so Rajeev suggest her to wear cotton saree.
(i) Why does Rajeev suggest her to wear cotton saree and not nylon saree?
(ii) Mention the values exhibited by Rajeev.
Answer:
(i) On a hot summer day, we sweat a lot. Cotton clothes absorb sweat from the body. As the sweat evaporates it results in cooling giving comfort to the body.

(ii) Caring, use of knowledge of science.

Question 2.
Mohan was getting late for school. He tried to sip tea from the cup. His father advised him to use a plate and asked him to sip the tea from the plate. Mohan followed the advice and finished his tea very easily.
(i) Why is siping off tea easier from a plate?
(ii) Mention the values exhibited by Mohan’s father.
Answer:
(i) A plate has a larger surface area than a cup. Evaporation becomes faster in this case. Since cooling is always caused during evaporation the temperature got lowered. Therefore it became easier to sip tea from a plate.

(ii) Caring, use of knowledge of science, helpful.

Here we are providing Statistics Class 9 Extra Questions Maths Chapter 14 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Statistics with Answers Solutions

Extra Questions for Class 9 Maths Chapter 14 Statistics with Solutions Answers

Statistics Class 9 Extra Questions Very Short Answer Type

Statistics Class 9 Extra Questions Question 1.
The points scored by a basketball team in a series of matches are follows:
17, 7, 10, 25, 5, 10, 18, 10 and 24. Find the range.
Solution:
Here, maximum points = 25 and
minimum points = 5
Range = Maximum value – Minimum value
= 25 – 5 = 20

Class 9 Statistics Extra Questions Question 2.
The points scored by a basketball team in a series of matches are as follows:
17, 2, 7, 27, 25, 5, 14, 18, 10. Find the median.
Solution:
Here, points scored in ascending order are 2, 5, 7, 10, 14, 17, 18, 25, 27, we have n = 9 terms

Statistics Class 9 Extra Questions Pdf Question 3.
The scores of an English test (out of 100) of 20 students are given below :
75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. .:: Find the median and mode of the data.
Solution:
The scores of an English test (out of 100) in ascending order are
44, 55, 59, 64, 67, 69, 73, 75, 75, 88, 88, 88, 88, 88, 90, 95, 95, 95, 98, 99
Here, n = 20

= Mean of 10^th and 11^th term Median
= Mean of 88 and 88 = 88
Mode = 88 [∵ 88 occured max. no. of times i.e., 5 times]

Statistics Extra Questions Class 9 Question 4.
Mean of 20 observations is 17. If in the observations, observation 40 is replaced by 12, find the new mean.
Solution:
Since mean of 20 observations is 17
∴ Sum of the 20 observations = 17 × 20 = 340
New sum of 20 observations = 340 – 40 + 12 = 312
New mean = \(\frac{312}{20}\) = 15.6

Extra Questions On Statistics Class 9 Question 5.
Mean of 36 observations is 12. One observation 47 was misread as 74. Find the correct
mean.
Solution:
Mean of 36 observations = 12
Total of 36 observations = 36 × 12 = 432
Correct sum of 36 observations = 432 – 74 + 47 = 405
Correct mean of 36 observations = \(\frac{405}{36}\) = 11.25

Class 9 Maths Chapter 14 Extra Questions Question 6.
The median of the data 26, 56, 32, 33, 60, 17, 34, 29, 45 is 33. If 26 is replaced by 62, then
find the new median.
Solution:
Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62

Hence, new median is 34.

Statistics Class 9 Extra Questions With Solutions Question 7.
There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers.
Solution:
Let x be the mean of 50 numbers.
i. Sum of 50 numbers = 50x
Since each number is subtracted from 53.
According to question, we have

Class 9 Maths Statistics Extra Questions Question 8.
To draw a histogram to represent the following frequency distribution :

Find the adjusted frequency for the class 25-45.
Solution:
Adjusted frequency of a class

Statistics Class 9 Extra Questions Short Answer Type 1

Extra Questions Of Statistics Class 9 Question 1.
For a particular year, following is the distribution of ages (in years) of primary school teachers in a district :

(i) Write the lower limit of first class interval.
(ii) Determine the class limits of the fourth class interval.
(iii) Find the class mark of the class 45 – 50.
(iv) Determine the class size.
Solution:
(i) First class interval is 15 – 20 and its lower limit is 15.
(ii) Fourth class interval is 30 – 35
Lower limit is 30 and upper limit is 35.
(iii) Class mark of the class 45 – 50 = \(\frac{45+50}{2}\) = \(\frac{95}{2}\) = 47.5
(iv) Class size = Upper limit of each class interval – Lower limit of each class interval
∴ Here, class size = 20 – 15 = 5

Statistics Class 9 Important Questions Question 2.
Find the mean of the following distribution :

Solution:

Class 9 Maths Ch 14 Extra Questions Question 3.
In figure, there is a histogram depicting daily wages of workers in a factory. Construct the frequency distribution table.

Solution:

Questions On Statistics Class 9 Question 4.
Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending
order. The median of the data is 24. Find the value of x.
Solution:
Here, the arranged data is 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43
Total number of observations = 10

But median of data is 24 (given)
⇒ \(\frac{3 x-12}{2}\) = 24
⇒ 3x – 12 = 48
⇒ 3x = 60
⇒ x = 20
∴ The value of x = 20

Statistics Class 9 Extra Questions Short Answer Type 2

Chapter 14 Maths Class 9 Extra Questions Question 1.
Draw a histogram for the given data :

Solution:
Let us represent class-intervals along x-axis and corresponding frequencies along y-axis on
a suitable scale, the required histogram is as under :

Question 2.
Given are the scores (out of 25) of 9 students in a Monday test :
14, 25, 17, 22, 20, 19, 10, 8 and 23
Find the mean score and median score of the data.
Solution:
Ascending order of scores is :
8, 10, 14, 17, 19, 20, 22, 23, 25

Question 3.
Draw a histogram of the weekly pocket expenses of 125 students of a school given below :

Solution:
Here, the class sizes are different, so calculate the adjusted frequency for each class by using the formula.
Minimum class size Frequency density or adjusted frequency for a class

Here, the minimum class size = 10 – 0 = 10

Let us represent weekly pocket money along x-axis and corresponding adjusted frequencies along y-axis on a suitable scale, the required histogram is as given below :

Question 4.
The weight in grams of 35 mangoes picked at random from a consignment are as follows:
131, 113, 82, 75, 204, 81, 84, 118, 104, 110, 80, 107, 111, 141, 136, 123, 90, 78, 90, 115, 110, 98, 106, 99, 107, 84, 76, 186, 82, 100, 109, 128, 115, 107, 115 From the grouped frequency table by dividing the variable range into interval of equal width of 20 grams, such that the mid-value of the first class interval is 70 g. Also, draw a histogram.
Solution:
It is given that the size of each class interval = 20 and the mid-value of the first class interval is 70.
Let the lower limit of the first class interval be a, then its upper limit = a + 20.
\(\frac{a+(a+20)}{2}\) = 70
⇒ a = 70 – 10 = 60
Thus, the first class interval is 60 – 80 and the other class-intervals are 80 – 100, 100 – 120, 120 – 140, 140 – 160, 160 – 180, 180 – 200 and 200 – 220.
So, the grouped frequency table is as under :


Let us represent weight (in g) along x-axis and corresponding frequencies along y-axis on a suitable scale, the required histogram is as under :

Statistics Class 9 Extra Questions Long Answer Type

Question 1.
Find the mean salary of 60 workers of a factory from the following table :

Solution:

Hence, mean salary of 60 workers is ₹5083.33.

Question 2.
In a school marks obtained by 80 students are given in the table. Draw a histogram. Also,
make frequency polygon.

Solution:
∴ Lower limit of first class interval is 305 – \(\frac{10}{2}\) = 300
Upper limit of first class interval is 305 + \(\frac{10}{2}\) = 310
Thus, first class interval is 300 – 310

Required histogram and frequency polygon is given on the graph paper.

Question 3.
The following two tables gives the distribution of students of two sections according to the marks obtained by them :

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Solution:
The class marks are as under :

Let us take class marks on X-axis and frequencies on Y-axis. To plot frequency polygon of Section-A, we plot the points (5, 3), (15, 9), (25,17), (35,12), (45, 9) and join these points by (15,19). line segments. To plot frequency polygon of Section-B, we plot the points (5, 5), (15, 19), (25, 15), (35, 10), (45, 1) on the same scale and join these points by dotted line segments.

From the above two polygons, clearly the performance of Section A is better.

Statistics Class 9 Extra Questions HOTS

Question 1.
The mean weight of 60 students of a class is 52.75 kg. If mean weight of 25 students of this class is 51 kg, find the mean weight of remaining 35 students of the class.
Solution:
Total weight of 60 students = 60 × 52.75 kg = 3165 kg
Total weight of 25 students = 25 × 51 kg = 1275 kg
∴ Total weight of 35 students = (3165 – 1275) kg = 1890 kg
∴ Mean weight of 35 students = \(\frac{1890}{35}\) = 54 kg

Question 2.
Find the missing frequencies in the following frequency distribution. If it is known that the mean of the distribution is 50.16 and the total number of items is 125.

Solution:
Since total number of items = 125
∴ 17 + f₁ + 32 + f₁ + 19 = 125
f₁ + f₂ = 125 – 17 – 32 – 19
f₁ + f₂ = 57 …..(i)
Now, mean of data = 50.16
We know that

Statistics Class 9 Extra Questions Value Based (VBQs)

Question 1.
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of the Indian society is given below :

(i) Represent the information above by a bar graph.
(ii) In the classroom, discuss what conclusions can be arrived at from the graph.
(iii) What steps should be taken to improve the situation ?
Solution:
(i) The required graph is given below :

In the graph, different sections of the society is taken on X-axis and number of girls per thousand boys is taken on the Y-axis.
[Scale : 1 cm = 10 girls.]
(ii) From the graph, the number of girls to the nearest ten per thousand boys are maximum in scheduled tribes whereas they are minimum in urban.
(iii) Pre-natal sex determination should strictly banned in urban.

Question 2.
Shimpi, a class IX student received cash award of 10000 (Ten thousand) in the singing competition. Her father advised her to make a budget plan for spending this amount. She
made the following plan :

Make a bar graph for the above data.
From above answer the following questions :
(i) Which mathematical concepts have been covered in this ?
(ii) How will you rate her budget plan ? In your opinion which head has been given
(a) more than deserved and (b) less than it deserved ?
(iii) Which values are depicted in her plan?
Solution:
The bar graph of given data is given below :

In the graph, head is taken on X-axis and amount is taken on Y-axis.
(i) Representation of data using bar graph.
(ii) Very good
(a) Picnic for family
(b) Tuition fee for needy child
(iii) Help the needy people and respect the elders.

Here we are providing Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Heron’s Formula with Answers Solutions

Extra Questions for Class 9 Maths Chapter 12 Heron’s Formula with Solutions Answers

Heron’s Formula Class 9 Extra Questions Very Short Answer Type

Heron’s Formula Class 9 Extra Questions Question 1.
Find the area of an equilateral triangle having side 6 cm.
Solutioin:
Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) × (side)² = \(\frac{\sqrt{3}}{4}\) × 6 × 6 = 9√3 cm²

Herons Formula Class 9 Extra Questions Question 2.
If the perimeter of an equilateral triangle is 90 m, then find its area.
Solutioin:

Herons Formula Extra Questions Question 3.
If every side of a triangle is doubled, then find the percent increase in area of triangle so formed.
Solutioin:
Let the sides of the given triangle be, a units, b units and c units.

Hence, percent increase = 300%

Class 9 Herons Formula Extra Questions Question 4.
If the length of a median of an equilateral triangle is x cm, then find its area.
Solutioin:

Let each equal sides of given equilateral triangle be 2y². We know that median is also perpendicular bisector.
∴ y² + x² = 4y²
⇒ x² = 3y²
⇒ x = √3y
or
⇒ y = \(\frac{x}{\sqrt{3}}\)
Now, area of given triangle = \(\frac{1}{2}\)
× 2y × X = y × x = \(\frac{x}{\sqrt{3}}\)× x = \(\frac{x^{2}}{\sqrt{3}}\)

Heron’s Formula Class 9 Extra Questions Short Answer Type 2

Class 9 Maths Chapter 12 Extra Questions Question 1.
Find the area of a triangle whose sides are 11 m, 60 m and 61 m.
Solutioin:
Let a = 11 m, b = 60 m and c = 61 m :

Chapter 12 Maths Class 9 Extra Questions Question 2.
Suman has a piece of land, which is in the shape of a rhombus. She wants her two sons to work on the land and produce different crops. She divides the land in two equal parts by drawing a diagonal. If its perimeter is 400 m and one of the diagonals is of length 120 m, how much area each of them will get for his crops ?
Solutioin:
Here, perimeter of the rhombus is 400 m.
∴ Side of the rhombus = \(\frac{400}{4}\) = 100 m
Let diagonal BD = 120 m and this diagonal divides the rhombus ABCD into two equal parts.

Hence, area of land allotted to two sons for their crops is 4800 m² each.

Extra Questions For Class 9 Maths Chapter 12 With Solution Question 3.
The perimeter of a triangular field is 144 m and its sides are in the ratio 3:4:5. Find the length of the perpendicular from the opposite vertex to the side whose length is 60 m.
Solutioin:
Let the sides of the triangle be 3x, 4x and 5x
∴ The perimeter of the triangular field = 144 m
⇒ 3x + 4x + 5x = 144
⇒ 12x = 144

Extra Questions On Herons Formula Class 9 Question 4.
Find the area of the triangle whose perimeter is 180 cm and two of its sides are of lengths 80 cm and 18 cm. Also, calculate the altitude of the triangle corresponding to the shortest side.
Solutioin:
Perimeter of given triangle = 180 cm
Two sides are 18 cm and 80 cm
∴ Third side = 180 – 18 – 80 = 82 cm

Hence, area of triangle is 720 cm² and altitude of the triangle corresponding to the shortest side is 80 cm.

Heron’s Formula Class 9 Extra Questions Long Answer Type

Extra Questions Of Herons Formula Class 9 Question 1.
Calculate the area of the shaded region.

Solutioin:

= 2 × 2 × 3 × 7 = 84 cm²
Area of shaded region = Area of ∆ABC – Area of ∆AOB
= 84 cm² – 30 cm² = 54 cm²

Class 9 Maths Herons Formula Extra Questions Question 2.
The sides of a triangular park are 8 m, 10 m and 6 m respectively. A small circular area of diameter 2 m is to be left out and the remaining area is to be used for growing roses. How much area is used for growing roses ? (use n = 3.14)
Solutioin:
The sides of the triangular park are 8 m, 10 m and 6 m.

Radius of the circle = \(\frac{2}{2}\) = 1 m
Area of the circle = πr² = 3.14 × 1 × 1 = 3.14 m²
∴ Area to be used for growing roses = Area of the park – area of the circle
= 24 – 3.14 = 20.86 m²

Heron’s Formula Class 9 Extra Questions HOTS

Herons Formula Class 9 Extra Questions With Solutions Pdf Question 1.
OPQR is a rhombus, whose three vertices P, Q and R lie on the circle with centre 0. If the radius of the circle is 12 cm, find the area of the rhombus.
Solutioin:
Since diagonals bisect each other at 90°.
∴ In right ∆QLR, (LR)² + (LQ)² = (QR)²

Class 9 Maths Ch 12 Extra Questions Question 2.
How much paper of each shade is needed to make a kite given in the figure, in which ABCD is a square with diagonal 60 cm?

Solutioin:
Since diagonals of a square are of equal length and bisect each other at right angles, therefore,
Area of ∆AOD = \(\frac{1}{2}\) × 30 × 30 = 450 cm²
Area of ∆AOD = Area of ∆DOC = Area of ∆BOC
= Area of ∆AOB = 450 cm²
[∵ ∆AOD = ∆AOB ≅ ∆BOC ≅ ∆COD, ∵ they² have equal area]
Now, area of ACEF (by Heron’s formula)
Here a = 20 cm, b = 20 cm and c = 30 cm

Now, area of orange shaded paper in kite
= Area of ∆AOD + Area of ∆CEF
= 450 cm² + 198.4 cm²
= 648.4 cm²
Area of blue shaded paper in kite
= Area of ∆AOB + Area of ∆COD
= 450 cm² + 450 cm² = 900 cm²
Area of black shaded paper in kite = Area of ∆BOC = 450 cm².

Heron’s Formula Class 9 Extra Questions Value Based (VBQs)

Heron’s Formula Class 9 Extra Questions With Solutions Question 1.
Sister Nivedita has trapezium shaped plot which she divided into three triangular portion for different purposes. I – for providing free education for orphan children, II – for providing dispensary for the needy villagers and III – for the library for villagers. Find the area of trapezium plot given in the figure. Which qualities of sister Nivedita are being depicted in question ?

Solutioin:
Here, ABCD is the trapezium with AB || DC.
Through C, Draw CF ⊥ AB
For – I: For area of ∆EBC

But, we know, Area of ∆EBC = \(\frac{1}{2}\)(base × height)
⇒ \(\frac{1}{2}\)1 × 30 × CF = 336
15 × CF = 336
⇒ CF = \(\frac{336}{15}\) = 22.4 m
Now, area of trapezium shaped plot = \(\frac{1}{2}\) (20 + 50)(22.4)
= 35 x 22.4 = 784 m²
Caring, kind, social, generous and visionary lady.

Ch 12 Maths Class 9 Extra Questions Question 2.
In an exhibition, an umbrella is made by stiching 10 triangular pieces of cloth with same message written on two triangular pieces. If each piece of cloth measures 60 cm, 60 cm and 20 cm, find how much cloth is required for each message.
Why we should respect women, educate women, save women and empower women ?

Solutioin:
Here, each triangular piece is an isosceles triangle with sides 60 cm, 60 cm and 20 cm.

Now, there are 2 triangular pieces with same message.
∴ Total area of cloth for each message = 2 × 591.61 = 1183.22 cm²
We should respect women, educate women, save women and empower women to serve humanity and give them equal opportunity so far they deprived.

Here we are providing Areas of Parallelograms and Triangles Class 9 Extra Questions Maths Chapter 9 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Areas of Parallelograms and Triangles with Answers Solutions

Extra Questions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles with Solutions Answers

Areas of Parallelograms and Triangles Class 9 Extra Questions Very Short Answer Type

Areas Of Parallelograms And Triangles Class 9 Extra Questions With Answers Question 1.
Two parallelograms are on equal bases and between the same parallels. Find the ratio of their areas.
Solution:
1:1 [∵ Two parallelograms on the equal bases and between the same parallels are equal in
area.]

Class 9 Maths Chapter 9 Extra Questions Question 2.
In ∆XYZ, XA is a median on side YZ. Find ratio of ar(∆XYA) : ar(∆XZA).

Solution:
Here, XA is the median on side YZ.
∴ YA = AZ
Draw XL ⊥ YZ

Parallelogram Questions For Class 9 Question 3.
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm (fig.). E and F are the mid-points of the non parallel sides. Find the ratio of ar(ABFE) and ar(EFCD).

Solution:

Area Of Parallelograms And Triangles Class 9 Extra Questions Question 4.
ABCD is a parallelogram and Q is any point on side AD. If ar(∆QBC) = 10 cm², find ar(∆QAB) + ar(∆QDC).

Solution:
Here, ∆QBC and parallelogram ABCD are on the same base BC and lie between the same parallels BC || AD.
∴ ar(||^gm ABCD) = 2 ar(∆QBC) ar(∆QAB) + ar(∆QDC) + ar(∆QBC) = 2 ar(∆QBC)
ar(∆QAB) + ar(∆QDC) = ar(∆QBC)
Hence, ar(∆QAB) + ar(∆QDC) = 10 cm² [∵ ar(∆QBC) = 10 cm² (given)]

Areas Of Parallelograms And Triangles Class 9 Questions With Answers Question 5.
WXYZ is a parallelogram with XP ⊥ WZ and ZQ ⊥ WX. If WX = 8 cm, XP = 8 cm and ZQ = 2 cm, find YX.

Solution:
ar(||^gm WXYZ) = ar(||^gm WXYZ)
WX × ZQ = WZ × XP
8 × 2 = WZ × 8
⇒ WZ = 2 cm
Now, YX = WZ = 2 cm [∵ opposite sides of parallelogram are equal]

Questions On Areas Of Parallelograms And Triangles Class 9 Question 6.
In figure, TR ⊥ PS, PQ || TR and PS || QR. If QR = 8 cm, PQ = 3 cm and SP = 12 cm, find ar(quad. PQRS).

Solution:
Here,
PS || QR [given]
∴ PQRS is a trapezium
Now, TR ⊥ PS and PQ || TR [given]
⇒ PQ ⊥ PS
∴ PQ = TR = 3 cm [given]
Now, ar(quad. PQRS) = \(\frac{1}{2}\) (PS + QR) × PQ = \(\frac{1}{2}\)(12 + 8) × 3 = 30 cm²

Area Of Parallelogram Class 9 Extra Questions Question 7.
In the given figure, ABCD is a parallelogram and L is the mid-point of DC. If ar(quad. ABCL) is 72 cm, then find ar(∆ADC).

Solution:
In ||^gm ABCD, AC is the diagonal

Areas Of Parallelograms And Triangles Questions With Answers Question 8.
In figure, TR ⊥ PS, PQ || TR and PS || QR. If QR = 8 cm, PQ = 3 cm and SP = 12 cm, find ar (PQRS).

Solution:
Here, PS || QR
∴ PQRS is a trapezium in which PQ = 3 cm, QR = 8 cm and SP = 12 cm
Now, TR I PS and PQ || TR
∴ PQRT is a rectangle [∵ PQ || TR, PT || QR and ∠PTR = 90°]
⇒ PQ = TR = 3 cm
Now, ar(PQRS) = \(\frac{1}{2}\)(PS + QR) × TR = \(\frac{1}{2}\)(12 + 8) × 3 = 30 cm².

Areas of Parallelograms and Triangles Class 9 Extra Questions Short Answer Type 1

Area Of Parallelogram Class 9 Questions Question 1.
ABCD is a parallelogram and O is the point of intersection of its diagonals. If ar(A AOD) = 4 cm\(²\) find area of parallelogram ABCD.

Solution:
Here, ABCD is a parallelogram in which its diagonals AC and BD intersect each other in O.
∴ O is the mid-point of AC as well as BD.
Now, in ∆ADB, AO is its median
∴ ar(∆ADB) = 2 ar(∆AOD)
[∵ median divides a triangle into two triangles of equal areas]
So, ar(∆ADB) = 2 × 4 = 8 cm²
Now, ∆ADB and ||^gm ABCD lie on the same base AB and lie between same parallels AB and CD
∴ ar(ABCD) = 2 ar(∆ADB).
= 2 × 8
= 16cm²

Areas Of Parallelograms And Triangles Question 2.
In the given figure of ∆XYZ, XA is a median and AB || YX. Show that YB is also a median.

Solution:
Here, in ∆XYZ, AB || YX and XA is a median.
∴ A is the mid-point of YZ. Now, AB is a line segment from mid-point of one side (YZ) and parallel to another side (AB || YX), therefore, it bisects the third side XZ.
⇒ B is the mid-point of XZ.
Hence, YB is also a median of ∆XYZ.

Question 3.
ABCD is a trapezium. Diagonals AC and BD intersect each other at O. Find the ratio ar (∆AOD) : ar (∆BOC).
Solution:

Here, ABCD is a trapezium in which diagonals AC and BD intersect each other at O. ∆ADC and ABCD are on the same base DC and between the same ‘parallels i.e., AB || DC.
∴ ar(∆ADC) = ar(∆BCD)
⇒ ar(∆AOD) + ar(∆ODC)
= ar(ABOC) + ar(AODC)
⇒ ar(∆AOD) = ar(∆BOC)
⇒ \(\frac { ar(∆AOD) }{ ar(∆BOC) } \) = 1

Question 4.
ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ (fig.). If AQ intersects DC at P, show that ar(∆BPC) = ar(∆DPQ).

Solution:
In ||^gm ABCD,
ar(∆APC) = ar(∆BCP) …(i)
[∵ triangles on the same base and between the same parallels have equal area]
Similarly, ar(∆ADQ) = ar(∆ADC) i …(ii)
Now, ar(∆ADQ) – ar(∆ADP) = ar(∆ADC) – ar(∆ADP)
ar(∆DPQ) = ar(∆ACP) … (iii)
From (i) and (iii), we have
ar(∆BCP) = ar(∆DPQ)
or ar(∆BPC) = ar(∆DPQ)

Question 5.
In the figure, PQRS is a parallelogram with PQ = 8 cm and ar(∆PXQ) = 32 cm². Find the altitude of gm PQRS and hence its area.

Solution:
Since parallelogram PQRS and APXQ are on the same base PQ and lie between the same
parallels PQ || SR
∴ Altitude of the ∆PXQ and ||^gm PQRS is same.
Now, \(\frac{1}{2}\)PQ × altitude = ar(∆PXQ)
⇒ \(\frac{1}{2}\) × 8 × altitude = 32
altitude = 8 cm
ar(||^gm PQRS) = 2 ar(∆PXQ)
= 2 × 32 = 64 cm²
Hence, the altitude of parallelogram PQRS is 8 cm and its area is 64 cm².

Question 6.
In ∆ABC. D and E are points on side BC such that CD = DE = EB. If ar(∆ABC) = 27 cm, find ar(∆ADE)

Solution:
Since in ∆AEC, CD = DE, AD is a median.
∴ ar(∆ACD) = ar(∆ADE)
[∵ median divides a triangle into two triangles of equal areas]
Now, in ∆ABD, DE = EB, AE is a median
ar(∆ADE) = ar(∆AEB)… (ii)
From (i), (ii), we obtain
ar(∆ACD) = ar(∆ADE) = ar(∆AEB)\(\frac{1}{3}\)ar(∆ABC)
∴ ar(∆ADE) = \(\frac{1}{3}\) × 27 = 9 cm²

Areas of Parallelograms and Triangles Class 9 Extra Questions Short Answer Type 2

Question 1.
For the given figure, check whether the following statement is true or false. Also justify your answer. PQRS is a trapezium with PQ || SR, PS || RU and ST || RQ, then ar(PURS) = ar(TQRS)

Solution:
Since ST || RQ and SR || TQ [given]
⇒ STQR is a ||^gm
Similarly, PS || UR and SR || PU [given]
⇒ PSRU is a ||^gm
Also, similarly, ||^gm STQR and ||^gm PSRU lie on same base SR and between same parallels PQ and SR.
∴ ar(||^gm STQR) = ar(||^gm PSRU)
Hence, the given statement is true.

Question 2.
In the given figure, WXYZ is a quadrilateral with a point P on side WX. If ZY || WX, show that :
(i) ar(∆ZPY) = ar(∆ZXY)
(ii) ar(∆WZY) = ar(∆ZPY)
(iii) ar(∆ZWX) = ar(∆XWY)

Solution:
∆ZPY and ∆ZXY lie on same base ZY and between same parallels ZY and WX
∴ ar(∆ZPY) = ar(∆ZXY)
Again, (∆WZY) and (∆ZPY) lie on same base ZY and between same parallels ZY and WX
∴ ar(∆WZY) = ar(∆ZPY)
Also, ∆zwX and ∆XWY lie on same base XW and between same parallels ZY and WX
∴ ar(∆ZWX) = ar(∆XWY)

Question 3.
In ∆ABC ; D, E and F are mid-points of sides BC, AC and AB respectively. A line through C drawn parallel to DE meets FE produced to G. Show that ar(∆FDE) = ar(∆EGC).

Solution:
Here, in ∆ABC; D, E and F are the mid-points of sides BC, AC and AB respectively. A line through C is drawn parallel to DE meets FE produced at G.
Since a line segment drawn through the mid-points of two sides, is parallel to third side and is half of it.
∴ DE || AB, EF || BC and FD || AC
⇒ AEDF, EFDC and EFDB are parallelograms
Also, a diagonal of a parallelogram divides it into two congruent triangles
∴ ∆AFE ≅ ∆DEF
∆DEF ≅ ∆FDB and
∆DEF ≅ ∆EDC
∴ ar(∆FDE) ≅ ar(∆EDC) …(i)
Again, in quad. EGCD, we have
CG || DE and DC || EG [given]
∴ EGCD is a parallelogram .
∴ ∆EDC = ∆CGE
⇒ ar(AEDC) = ar(ACGE) … (ii)
From (i) and (ii), we obtain
ar(∆FDE) = ar(∆EGC)

Question 4.
In ∆PQR, A and B are points on side QR such that they trisect QR. Prove that ar(∆PQB) = 2ar(∆PBR).
Solution:
Here, in ∆PQR, A and B are points on side QR
such that QA = AB = BR.

Through P, draw a line / parallel to QR
Now, APQA, APAB and APBR on the equal bases
and between the same parallels l || QR
⇒ ar(∆PQA) = ar(∆PAB) = ar(∆PBR) …. (i)
Now, ar(∆PQB) = ar(∆PQA) + ar(∆PAB)
= 2ar(∆PQA)[using (i)]
= 2ar(∆PBR) [using (i)]

Question 5.
The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). Show that ar(||^gm ABCD) = ar(||^gm PBQR).

Solution:

Join AC and QP, also it is given that ∆Q || CP
∴ ∆ACQ and ∆APQ are on the same base ∆Q and lie between the same parallels ∆Q || CP.
∴ ar(∆ACQ) = ar(∆APQ)
or ar(∆ABC) + ar(∆ABQ) = ar(ABPQ) + ar(∆ABQ)
ar(∆ABC) = ar(ABPQ)
or \(\frac{1}{2}\)ar(||^gm ABCD) = \(\frac{1}{2}\)ar (||^gm PBQR)
or ar(||^gm ABCD) = ar(||^gm PBQR)

Areas of Parallelograms and Triangles Class 9 Extra Questions Long Answer Type

Question 1.
EFGH is a parallelogram and U and T are points on sides EH and GF respectively. If ar(∆EHT) = 16 cm, find ar(∆GUF).

Solution:
∴ ar(∆EHT) = \(\frac{1}{2}\) ar(||^gm EFGH) …..(i)
Similarly, ∆GUF and parallelogram EFGH are on the same base GF and lie between the same parallels GF and HE
∴ ar(∆GUF) = \(\frac{1}{2}\) ar(||^gm EFGH) …..(ii)
From (i) and (ii), we have
ar(∆GUF) = ar(∆EHT)
= 16 cm² [∵ ar(∆EHT) = 16 cmcm²] [given]

Question 2.
ABCD is a parallelogram and P is any point in its interior. Show that :
ar(∆APB) + ar(∆CPD) = ar(∆BPC) + ar(∆APD)

Solution:
Through P, draw a line LM || DA and EF || AB
Since ∆APB and ||^gm ABFE are on the same base AB and lie between the same parallels AB and EF.
∴ ar(∆APB) = \(\frac{1}{2}\) ar(||^gm ABFE) … (i)

Similarly, ACPD and parallelogram DCFE are on the same base DC and between the same parallels DC and EF.
∴ ar(∆CPD) = \(\frac{1}{2}\) ar(||^gm DCFE) … (ii)
Adding (i) and (ii), we have
ar(∆APB) + ar(∆CPD) = \(\frac{1}{2}\) ar (||^gm ABFE) + ar(||^gm DCFE)
= \(\frac{1}{2}\) ar(|lgm ABCD) … (iii)
Since ∆APD and parallelogram ADLM are on the same base AB and between the same parallels AD and ML
∴ ar(∆APD) = \(\frac{1}{2}\) ar(||^gm ADLM) …..(iv)
Similarly, ar(∆BPC) = \(\frac{1}{2}\) = arc||^gm BCLM) ….(v)
Adding (iv) and (u), we have
ar(∆APD) + ar (∆BPC) = \(\frac{1}{2}\) ar(||^gm ABCD) ….(vi)
From (iii) and (vi), we obtain
ar(∆APB) + ar(∆CPD) = ar(∆APD) + ar(ABPC)

Question 3.
In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP.

Solution:
In ∆PAD, ∠A = 90° and DA = PA = AB
⇒ ∠ADP = ∠APD = \(\frac{90^{\circ}}{2}\) = 45°
Similarly, in ∆QBC, ∠B = 90° and BQ = BC = AB
⇒ ∠BCQ = ∠BQC = \(\frac{90^{\circ}}{2}\) = 45°
In ∆PAD and ∆QBC, we have
PA = BQ [given]
∠A = ∠B [each = 90°]
AD = BC [sides of a square]
⇒ ∠PAD ≅ ∆QBC [by SAS congruence rule]
⇒ PD = QC [c.p.c.t.]
Now, in APDC and ∆QCD
DC = DC [common]
PD = QC [prove above]
∠PDC = ∠QCD [each = 90° + 45° = 135°]
⇒ ∆PDC = ∆QCD [by SAS congruence rule]
⇒ PC = QD or DQ = CP

Areas of Parallelograms and Triangles Class 9 Extra Questions HOTS

Question 1.
In the given figure, PQRS, SRNM and PQNM are parallelograms, Show that :
ar(∆PSM) = ar(∆QRN).

Solution:
Since PQRS is a parallelogram.
∴ PS = QR and PS || QR
Since SRNM is also a parallelogram.
∴ SM = RN and SM || RN
Also, PQNM is a parallelogram
∴ PM || QM and PM = QM
Now, in APSM and ∆QRN
PS = QR
SM = RN
PM = QN
∆PSM ≅ ∆QRN [by SSS congruence axiom]
∴ ar (∆PSM) = ar (∆QRN) [congruent triangles have same areas)

Areas of Parallelograms and Triangles Class 9 Extra Questions Value Based (VBQs)

Question 1.
Naveen was having a plot in the shape of a quadrilateral. He decided to donate some portion of it to construct a home for orphan girls. Further he decided to buy a land in lieu of his donated portion of his plot so as to form a triangle.
(i) Explain how this proposal will be implemented?
(ii) Which mathematical concept is used in it?
(iii) What values are depicted by Naveen?
Solution:

(i) Let ABCD be the plot and Naveen decided to donate some portion to construct a home for orphan girls from one corner say C of plot ABCD. Now, Naveen also purchases equal amount of land in lieu of land CDO, so that he may have triangular form of plot. BD is joined. Draw a line through C parallel to DB to meet AB produced in P.
Join DP to intersect BC at 0.
Now, ABCD and ABPD are on the same base and between same parallels CP || DB.
ar(∆BCD) = ar(∆BPD) ar(∆COD) + ar(∆DBO) = ar(∆BOP) + ar(∆DBO)
ar(ACOD) = ar(ABOP) ar(quad. ABCD)
= ar(quad. ABOD) + ar(∆COD)
= ar(quad. ABOD) + ar(∆BOP)
[∵ ar(ACOD) = ar(ABOP)] (proved above]
= ar(∆APD)
Hence, Naveen purchased the portion ABOP to meet his requirement.
(ii) Two triangles on the same base and between same parallels are equal in area.
(iii) We should help the orphan children.

Question 2.
A flood relief camp was organized by state government for the people affected by the natural calamity near a city. Many school students volunteered to participate in the relief work. In the camp, the food items and first aid centre kits were arranged for the flood victims. The piece of land used for this purpose is shown in the figure.
(a) If EFGH is a parallelogram with P and Q as mid-points of sides GH and EF respectively, then show that area used for first aid is half of the total area.
(b) What can you say about the student volunteers working for the relief work?

Solution:
(a) Here, EFGH is a ||^gm
∴ EF = GH and EF || GH

Hence, area used for first aid is half of the total area.

(b) Students working for the noble cause show compassion towards the affected people. They also realize their social responsibility to work for helping the ones in need.

Here we are providing Quadrilaterals Class 9 Extra Questions Maths Chapter 8 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Quadrilaterals with Answers Solutions

Extra Questions for Class 9 Maths Chapter 8 Quadrilaterals with Solutions Answers

Quadrilaterals Class 9 Extra Questions Very Short Answer Type

Quadrilateral Class 9 Extra Questions Question 1.
If one angle of a parallelogram is twice of its adjacent angle, find the angles of the parallelogram.
Solution:
Let the two adjacent angles be x and 2x.
In a parallelogram, sum of the adjacent angles are 180°
∴ x + 2x = 180°
⇒ 3x = 180°
⇒ x = 60°
Thus, the two adjacent angles are 120° and 60°. Hence, the angles of the parallelogram are 120°, 60°, 120° and 60°.

Class 9 Quadrilaterals Extra Questions Question 2.
If the diagonals of a quadrilateral bisect each other at right angles, then name the
quadrilateral.
Solution:
Rhombus.

Quadrilaterals Class 9 Extra Questions With Solutions Question 3.
Three angles of a quadrilateral are equal and the fourth angle is equal to 144o. Find each of the equal angles of the quadrilateral.
Solution:
Let each equal angle of given quadrilateral be x.
We know that, sum of all interior angles of a quadrilateral is 360°
∴ x + x + x + 144° = 360°
3x = 360° – 144°
3x = 216°
x = 72°
Hence, each equal angle of the quadrilateral is of 72o measures.

Extra Questions On Quadrilaterals Class 9 Question 4.
If ABCD is a parallelogram, then what is the measure of ∠A – ∠C ?
Solution:
∠A – ∠C = 0° (opposite angles of parallelogram are equal]

Quadrilaterals Class 9 Extra Questions Question 5.
PQRS is a parallelogram, in which PQ = 12 cm and its perimeter is 40 cm. Find the length of each side of the parallelogram.
Solution:

Here, PQ = SR = 12 cm
Let PS = x and PS = QR
∴ x + 12 + x + 12 = Perimeter
2x + 24 = 40
2x = 16
x= 8
Hence, length of each side of the parallelogram is 12 cm, 8 cm, 12 cm and 8 cm.

Class 9 Maths Quadrilaterals Extra Questions Question 6.
Two consecutive angles of a parallelogram are (x + 60)° and (2x + 30)°. What special name can you give to this parallelogram?
Solution:
We know that consecutive interior angles of a parallelogram are supplementary.
∴ (x + 60° + (2x + 30)° = 180°
⇒ 3x° + 90° = 180°
⇒ 3x° = 90°
⇒ x° = 30°
Thus, two consecutive angles are (30 + 60)°, 12 x 30 + 30)”. i.e., 90° and 90°.
Hence, the special name of the given parallelogram is rectangle.

Class 9 Quadrilaterals Extra Questions Pdf Question 7.
ONKA is a square with ∠KON = 45°. Determine ∠KOA.
Solution:
Since ONKA is a square
∴ ∠AON = 90°
We know that diagonal of a square bisects its ∠s
⇒ ∠AOK = ∠KON = 45°
Hence, ∠KOA = 45°

Quadrilaterals Extra Questions Class 9 Question 8.
In quadrilateral PQRS, if ∠P = 60° and ∠Q : ∠R : ∠S = 2 : 3 : 7, then find the measure of ∠S.
Solution:
Let ∠Q = 2x, ∠R = 3x and ∠S = 7x
Now, ∠P + ∠Q + ∠R + ∠S = 360°
⇒ 60° + 2x + 3x + 7x = 360°
⇒ 12x = 300°
x = \(\frac{300^{\circ}}{12}\) = 25°
∠S = 7x = 7 x 25° = 175°

Quadrilaterals Class 9 Extra Questions Short Answer Type 1

Questions On Quadrilaterals For Class 9 Question 1.
ABCD is a parallelogram in which ∠ADC = 75° and side AB is produced to point E as shown in the figure. Find x + y.
Solution:

Here, ∠C and ∠D are adjacent angles of the parallelogram.
∴ ∠C + ∠D = 180°
⇒ x + 75° = 180°
⇒ x = 105°
Also, y = x = 105° [alt. int. angles]
Thus, x + y = 105° + 105° = 210°

Class 9 Maths Chapter 8 Extra Questions With Solutions Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:

Given: A parallelogram ABCD, in which AC = BD.
To Prove: ΔBCD is a rectangle.
Proof : In ΔABC and ΔBAD
AB = AB (common]
AC = BD (given]
BC = AD(opp. sides of a ||^gm]
⇒ ΔABC ≅ ΔBAD
[by SSS congruence axiom]
⇒ ∠ABC = ∠BAD (c.p.c.t.)
Also, ∠ABC + ∠BAD = 180° (co-interior angles)
∠ABC + ∠ABC = 180° [ ∵ ∠ABC = ∠BAD ]
2∠ABC = 180°
∠ABC = 1/2 x 180° = 90°
Hence, parallelogram ABCD is a rectangle.

Quadrilateral Extra Questions Class 9 Question 3.
In the figure, ABCD is a rhombus, whose diagonals meet at O. Find the values of x and y.

Solution:
Since diagonals of a rhombus bisect each other at right angle.
In ∴ ΔAOB, we have
∠OAB + ∠x + 90° = 180°
∠x = 180° – 90° – 35°
= 55°
Also,
∠DAO = ∠BAO = 35°
∠y + ∠DAO + ∠BAO + ∠x = 180°
⇒ ∠y + 35° + 35° + 55° = 180°
⇒ ∠y = 180° – 125o = 55°
Hence, the values of x amd y are x = 55°, y = 55°.

Extra Questions For Class 9 Maths Quadrilaterals With Solutions Question 4.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see fig.). Show that :
(i) ΔAPB = ΔCQD
(ii) AP = CQ
Solution:

Given : In ||gm ABCD, AP and CQ are perpendiculars from the
vertices A and C on the diagonal BD.
To Prove: (i) ΔAPB ≅ ΔCQD
(ii) AP = CQ
Proof : (i) In ΔAPB and ΔCQD
AB = DC (opp. sides of a ||gm ABCD]
∠APB = ∠DQC (each = 90°)
∠ABP = ∠CDQ (alt. int. ∠s]
⇒ ΔAPB ≅ ΔCQD[by AAS congruence axiom]
(ii) ⇒ AP = CQ [c.p.c.t.]

Quadrilaterals Class 9 Extra Questions Short Answer Type 2

Extra Questions Of Quadrilaterals Class 9 Question 1.
The diagonals of a quadrilateral ABCD are perpendicular to each other. Show that the quadrilateral formed by joining the mid-points of its sides is a rectangle.

Solution:
Given: A quadrilateral ABCD whose diagonals AC and BD are perpendicular to each other at O. P, Q, R and S are mid-points of side AB, BC, CD and DA respectively are joined are formed quadrilateral PQRS.
To Prove: PQRS is a rectangle.
Proof : In ∆ABC, P and Q are mid-points of AB and BC respectively.
∴ PQ || AC and PQ = \(\frac{1}{2}\) AC … (i) (mid-point theorem]
Further, in SACD, R and S are mid-points of CD and DA respectively.
SR || AC and SR = \(\frac{1}{2}\) AC … (ii) (mid-point theorem]
From (i) and (ii), we have PQ || SR and PQ = SR
Thus, one pair of opposite sides of quadrilateral PQRS are parallel and equal.
∴ PQRS is a parallelogram.
Since PQ|| AC PM || NO
In ∆ABD, P and S are mid-points of AB and AD respectively.
PS || BD (mid-point theorem]
⇒ PN || MO
∴ Opposite sides of quadrilateral PMON are parallel.
∴ PMON is a parallelogram.
∠ MPN = ∠ MON (opposite angles of ||gm are equal]
But ∠MON = 90° [given]
∴ ∠MPN = 90° ⇒ ∠QPS = 90°
Thus, PQRS is a parallelogram whose one angle is 90°
∴ PQRS is a rectangle.

Class 9 Maths Chapter 8 Extra Questions Question 2.
In the fig., D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. Prove that DEF is also an equilateral triangle.

Solution:
Since line segment joining the mid-points of two sides of a triangle is half of the third side.
Therefore, D and E are mid-points of BC and AC respectively.
⇒ DE = \(\frac{1}{2}\)AB …(i)
E and F are the mid-points of AC and AB respectively.
∴ EF = \(\frac{1}{2}\)BC … (ii)
F and D are the mid-points of AB and BC respectively.
∴ FD = \(\frac{1}{2}\) AC … (iii)
Now, SABC is an equilateral triangle.
⇒ AB = BC = CA
⇒ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)BC = \(\frac{1}{2}\)CA
⇒ DE = EF = FD (using (i), (ii) and (iii)]
Hence, DEF is an equilateral triangle.

Quadrilateral Class 9 Questions Question 3.
In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY is a parallelogram. Show that ABCD is a parallelogram.

Solution:
Since BXDY is a parallelogram.
XO = YO
DO = BO
[∵ diagonals of a parallelogram bisect each other]
But AX = CY …. (iii) (given]
Adding (i) and (iii), we have
XO + AX = YO + CY
⇒ AO = CO …. (iv)
From (ii) and (iv), we have
AO = CO and DO = BO
Thus, ABCD is a parallelogram, because diagonals AC and BD bisect each other at O.

Class 9 Maths Ch 8 Extra Questions Question 4.
ABCD is a quadrilateral in which the bisectors of ∠A and ∠C meet DC produced at Y and BA produced at X respectively. Prove that
∠X +∠Y = \(\frac{1}{2}\)(∠A + ∠C)
Solution:

Here, ∠1 = ∠2 and ∠3 = ∠4
In ΔXBC, we have
∠X + ∠B + ∠4 = 180°
∠X + ∠B + \(\frac{1}{2}\)∠C = 180
In ΔADY, we have
∠2 + ∠D + ∠Y= 180°
\(\frac{1}{2}\) ∠A + ∠D + ∠Y = 180°
Adding (i) and (ii), we have
∠X + ∠Y + ∠B + ∠D + \(\frac{1}{2}\) ∠C + \(\frac{1}{2}\) ∠A = 360°
Also, in quadrilateral ABCD
∠A + ∠B + ∠C + ∠D = 360°
∠X + ∠Y + ∠B + ∠D + \(\frac{1}{2}\) ∠C + \(\frac{1}{2}\) ∠A = ∠A + ∠B + ∠C + ∠D
∠X + ∠Y = ∠A – \(\frac{1}{2}\) ∠A + \(\frac{1}{2}\) ∠C – \(\frac{1}{2}\) ∠C
∠X+ ∠Y = \(\frac{1}{2}\) (∠A + ∠C)

Quadrilaterals Class 9 Extra Questions Long Answer Type

Quadrilateral Questions For Class 9 Question 1.
In the figure, P, Q and R are the mid-points of the sides BC, AC and AB of ΔABC. If BQ and PR intersect at X and CR and PQ intersect at Y, then show that XY = \(\frac{1}{4}\) BC.

Solution:
Here, in ΔABC, R and Q are the mid-points of AB and AC respectively.
∴ By using mid-point theorem, we have
RQ || BC and RQ = \(\frac{1}{2}\) BC
∴ RQ = BP = PC [∵ P is the mid-point of BC]
∴ RQ || BP and RQ || PC
In quadrilateral BPQR
RQ || BP, RQ = BP (proved above]
∴ BPQR is a parallelogram. [∵ one pair of opp. sides is parallel as well as equal]
∴ X is the mid-point of PR. [∵ diagonals of a ||gm bisect each other]
Now, in quadrilateral PCQR
RQ || PC and RQ = PC [proved above)
∴ PCQR is a parallelogram [∵ one pair of opp. sides is parallel as well as equal]
∴ Y is the mid-point of PQ [∵ diagonals of a ||gm bisect each other]
In ΔPQR
∴ X and Y are mid-points of PR and PQ respectively.

Class 9 Quadrilaterals Important Questions Question 2.
In the given figure, AE = DE and BC || AD. Prove that the points A, B, C and D are concyclic. Also, prove that the diagonals of the quadrilateral ABCD are equal.

Solution:
Since AE = DE
∠D = ∠A …. (i) [∵ ∠s opp. to equal sides of a Δ]
Again, BC || AD
∠EBC = ∠A …. (ii) (corresponding ∠s]
From (i) and (ii), we have
∠D = ∠EBC …. (iii)
But ∠EBC + ∠ABC = 180° (a linear pair]
∠D + ∠ABC = 180° (using (iii)]
Now, a pair of opposite angles of quadrilateral ABCD is supplementary
Thus, ABCD is a cyclic quadrilateral i.e., A, B, C and D’are concyclic. In ΔABD and ΔDCA
∠ABD = ∠ACD [∠s in the same segment for cyclic quad. ABCD]
∠BAD = ∠CDA [using (i)]
AD = AD (common]
So, by using AAS congruence axiom, we have
ΔABD ≅ ΔDCA
Hence, BD = CA [c.p.c.t.]

Question 3.
In ΔABC, AB = 8 cm, BC = 9 cm and AC = 10 cm. X, Y and Z are mid-points of AO, BO and CO respectively as shown in the figure. Find the lengths of the sides of ΔXYZ.

Solution:
Here, in ΔABC, AB = 8 cm, BC = 9 cm, AC = 10 cm.
In ΔAOB, X and Y are the mid-points of AO and BO.
∴ By using mid-point theorem, we have
XY = \(\frac{1}{2}\) AB = \(\frac{1}{2}\) x 8 cm = 4 cm
Similarly, in Δ𝜏BOC, Y and Z are the mid-points of BO and CO.
∴ By using mid-point theorem, we have
YZ = \(\frac{1}{2}\) BC = \(\frac{1}{2}\) x 9cm = 4.5 cm
And, in Δ𝜏COA, Z and X are the mid-points of CO and AO.
∴ ZX = \(\frac{1}{2}\) AC = \(\frac{1}{2}\) x 10 cm = 5 cm
Hence, the lengths of the sides of ΔXYZ are XY = 4 cm, YZ = 4.5 cm and ZX = 5 cm.

Question 4.
PQRS is a square and ∠ABC = 90° as shown in the figure. If AP = BQ = CR, then prove that ∠BAC = 45°

Solution:
Since PQRS is a square.
∴ PQ = QR … (I) [∵ sides of a square are equal]
Also, BQ = CR … (ii) [given]
Subtracting (ii) from (i), we obtain
PQ – BQ = QR – CR
⇒ PB = QC … (iii)
In Δ𝜏APB and Δ𝜏BQC
AP = BQ[given
∠APB = ∠BQC = 90°](each angle of a square is 90°)
PB = QC (using (iii)]

So, by using SAS congruence axiom, we have
ΔAPB ≅ ΔBQC
∴ AB = BC [c.p.c.t.]
Now, in ΔABC
AB = BC [proved above]
∴ ∠ACB = ∠BAC = x° (say) [∠s opp. to equal sides]
Also, ∠B + ∠ACB + ∠BAC = 180°
⇒ 90° + x + x = 180°
⇒ 2x° = 90°
x° = 45°
Hence, ∠BAC = 45°

Question 5.
ABCD is a parallelogram. If the bisectors DP and CP of angles D and C meet at P on side AB, then show that P is the mid-point of side AB.
Solution:

Since DP and CP are angle bisectors of ∠D and ∠C respectively.
: ∠1 = ∠2 and ∠3 = ∠4
Now, AB || DC and CP is a transversal
∴ ∠5 = ∠1 [alt. int. ∠s]
But ∠1 = ∠2 [given]
∴ ∠5 = ∠2

Now, in ABCP, ∠5 = ∠2
⇒ BC = BP … (I) [sides opp. to equal ∠s of a A]
Again, AB || DC and DP is a transversal.
∴ ∠6= ∠3 (alt. int. Δs]
But ∠4 = ∠3 [given]
∴ ∠6 = ∠4
Now, in ΔADP, ∠6 = ∠4
⇒ DA = AP …. (ii) (sides opp. to equal ∠s of a A]
Also, BC = DA… (iii) (opp. sides of parallelogram)
From (i), (ii) and (iii), we have
BP = AP
Hence, P is the mid-point of side AB.

Question 6.
In the figure, ΔBCD is a trapezium in which AB || DC. E and F are the mid-points of AD and BC respectively. DF and AB are produced to meet at G. Also, AC and EF intersect at the point O. Show that :
(i) EO || AB
(ii) AO = CO

Solution:
Here, E and F are the mid-points of AD and BC respectively.
In ΔBFG and ΔCFD
BF = CF [given]
∠BFG = ∠CFD (vert. opp. ∠s]
∠BGF = ∠CDF (alt. int. ∠s, as AB || DC)
So, by using AAS congruence axiom, we have
ΔBFG ≅ ΔCFD
⇒ DF = FG [c.p.c.t.)
Now, in ΔAGD, E and F are the mid-points of AD and GD.
∴ By mid-point theorem, we have
EF || AG
or EO || AB
Also, in ΔADC, EO || DC
∴ EO is a line segment from mid-point of one side parallel to another side.
Thus, it bisects the third side.
Hence, AO = CO

Here we are providing Number Systems Class 9 Extra Questions Maths Chapter 1 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Number Systems with Answers Solutions

Extra Questions for Class 9 Maths Chapter 1 Number Systems with Solutions Answers

Number Systems Class 9 Extra Questions Very Short Answer Type

Number System Class 9 Extra Questions With Solutions Question 1.
Simplify: (√5 + √2)².
Solution:
Here, (√5 + √2² = (√5² + 2√5√2 + (√2)²
= 5 + 2√10 + 2 = 7 + 2√10

Question 2.
Find the value of √(3)^-2.
Solution:

Number System Class 9 Extra Questions Question 3.
Identify a rational number among the following numbers :
2 + √2, 2√2, 0 and π
Solution:
O is a rational number.

Class 9 Maths Chapter 1 Extra Questions Question 4.
Express 1.8181… in the form \(\frac{p}{q}\) where p and q are integers and q ≠ 0.
Solution:
Let x =1.8181… …(i)
100x = 181.8181… …(ii) [multiplying eqn. (i) by 100]
99x = 180 [subtracting (i) from (ii)]
x = \(\frac{180}{99}\)
Hence, 1.8181… = \(\frac{180}{99}\) = \(\frac{20}{11}\)

Class 9 Number System Extra Questions Question 5.
Simplify : √45 – 3√20 + 4√5
Solution:
√45 – 3√20 + 4√5 = 3√5 – 6√5 + 4√5 = √5.

Class 9 Maths Number System Extra Questions With Solutions Question 6.
Find the value of’

Solution:

Class 9 Maths Ch 1 Extra Questions Question 7.
Find the value of

Solution:

Number Systems Class 9 Extra Questions Short Answer Type 1

Extra Questions For Class 9 Maths Chapter 1 With Solution Pdf Question 1.
Evaluate : (√5 + √2² + (√8 – √5)²
Solution:
(√5 + √2)² + (√8 – √5² = 5 + 2 + 2√10 + 8 + 5 – 2√40
= 20 + 2√10 – 4√10 = 20 – 2√10

Extra Questions On Number System Class 9 Question 2.
Express \(23 . \overline{43}\) in \(\frac{p}{q}\) form, where p, q are integers and q ≠ 0.
Solution:
Let x = \(23 . \overline{43}\)
or x = 23.4343…         ….(i)
100x = 2343.4343…    …(ii) [Multiplying eqn. (i) by 100]
99x = 2320 [Subtracting (i) from (ii)
⇒ x = \(\frac{2320}{99}\)
Hence, \(23 . \overline{43}\) = \(\frac{2320}{99}\)

Number System Extra Questions Question 3.
Let ‘a’ be a non-zero rational number and ‘b’ be an irrational number. Is ‘ab’ necessarily an irrational ? Justify your answer with example.
Solution:
Yes, ‘ab’ is necessarily an irrational.
For example, let a = 2 (a rational number) and b = √2 (an irrational number)
If possible let ab = 2√2 is a rational number.
Now, \(\frac{ab}{a}\) = \(\frac{2 \sqrt{2}}{2}\) = √2 is a rational number.
[∵ The quotient of two non-zero rational number is a rational]
But this contradicts the fact that √2 is an irrational number.
Thus, our supposition is wrong.
Hence, ab is an irrational number.

Chapter 1 Maths Class 9 Extra Questions Question 4.
Let x and y be a rational and irrational numbers. Is x + y necessarily an irrational number? Give an example in support of your answer.
Solution:
Yes, x + y is necessarily an irrational number.
For example, let x = 3 (a rational number) and y = √5 (an irrational number)
If possible let x + y = 3 + √5 be a rational number.
Consider \(\frac{p}{q}\) = 3 + √5, where p, q ∈ Z and q ≠ 0.
Squaring both sides, we have

∵ \(\frac{p}{q}\) is a rational
⇒ √5 is a rational
But this contradicts the fact that √5 is an irrational number.
Thus, our supposition is wrong.
Hence, x + y is an irrational number.

Number Systems Class 9 Extra Questions Short Answer Type 2

Ch 1 Maths Class 9 Extra Questions Question 1.
Represent √3 on the number line.
Solution:


On the number line, take OA = 1 unit. Draw AB = 1 unit perpendicular to OA. Join OB.
Again, on OB, draw BC = 1 unit perpendicular to OB. Join OC.
By Pythagoras Theorem, we obtain OC = √3. Using
compasses, with centre O and radius OC, draw an arc, which intersects the number line at point
D. Thus, OD = √3 and D corresponds to √3.

Class 9 Chapter 1 Maths Extra Questions Question 2.
Represent √3.2 on the number line.
Solution:
First of all draw a line of length 3.2 units such that AB = 3.2 units. Now, from point B, mark a distance of 1 unit. Let this point be ‘C’. Let ‘O’ be the mid-point of the distance AC. Now, draw a semicircle with centre ‘O’ and radius OC. Let us draw a line perpendicular to AC passing through the point ‘B’ and intersecting the semicircle at point ‘D’.
∴ The distance BD = √3.2

Now, to represent √3.2 on the number line. Let us take the line BC as number line and point ‘B’ as zero, point ‘C’ as ‘1’ and so on. Draw an arc with centre B and radius BD, which intersects the number line at point ‘E’.
Then, the point ‘E’ represents √3.2.

Class 9 Ch 1 Maths Extra Questions Question 3.
Express 1.32 + 0.35 as a fraction in the simplest form.
Solution:
Let . x = 1.32 = 1.3222…..(i)

Multiplying eq. (i) by 10, we have
10x = 13.222…
Again, multiplying eq. (i) by 100, we have
100x = 132.222… …(iii)
Subtracting eq. (ii) from (iii), we have
100x – 10x = (132.222…) – (13.222…)
90x = 119
⇒ x = \(\frac{119}{90}\)
Again, y = 0.35 = 0.353535……
Multiply (iv) by 100, we have …(iv)
100y = 35.353535… (v)
Subtracting (iv) from (u), we have
100y – y = (35.353535…) – (0.353535…)
99y = 35
y = \(\frac{35}{99}\)

Extra Questions For Class 9 Maths Ch 1 Question 4.
Find the square root of 10 + √24 + √60 + √40.
Solution:

Question 5.
If x = 9 + 4√5, find the value of √x – \(\frac{1}{\sqrt{x}}\).
Solution:
Here,
x = 9 + 4√5
x = 5 + 4 + 2 x 2√5
x = (√5² + (2² + 2 x 2x √5).
x = (√5 + 2)²
√x = √5 + 2
Now, \(\frac{1}{\sqrt{x}}\) = \(\frac{1}{\sqrt{5}+2}\)

Question 6.
If x = \(\frac{1}{\sqrt{5}-2}\) , find the value of x³ – 3² – 5x + 3
Solution:

∴ x – 2 = √5
Squaring both sides, we have
x² – 4x + 4 = 5
x² – 4x – 1 = 0 …(i)
Now, x³ – 3² – 5x + 3 = (x² – 4x – 1) (x + 1) + 4
= 0 (x + 1) + 4 = 4 [using (i)]

Question 7.
Find ‘x’, if 2^x-7 × 5^x-4 = 1250.
Solution:
We have 2^x-7 × 5^x-4 = 1250
⇒ 2^x-7 × 5^x-4 = 2 5 × 5 × 5 × 5
⇒ 2^x-7 × 5^x-4 = 21 × 54
Equating the powers of 2 and 5 from both sides, we have
⇒ x – 7 = 1 and x – 4 = 4
⇒ x = 8 and x = 8
Hence, x = 8 is the required value.

Question 8.
Evaluate:

Solution:

Number Systems Class 9 Extra Questions Long Answer Type

`Question 1.
If x = \(\frac { \sqrt { p+q } +\sqrt { p-q } }{ \sqrt{p+q}-\sqrt{p-q} }\), then prove that q² – 2px + 9 = 0.
Solution:


Squaring both sides, we have
⇒ q²x² + p² – 2pqx = p² – q²
⇒ q²x² – 2pqx + q² = 0
⇒ q(q² – 2px + q) = 0
⇒ qx² – 2px + q = 0 (∵ q ≠ 0)

Question 2.
If a = \(\frac{1}{3-\sqrt{11}}\) and b = \(\frac{1}{a}\), then find a² – b²
Solution:

Question 3.
Simplify:

Solution:

Question 4.
Prove that:

Solution:

Question 5.
Find a and b, if

Solution:

Number Systems Class 9 Extra Questions HOTS

Question 1.
If x^a = y, y^b = z and z^c = x, then prove that abc = 1.
Solution:
We have x^a = y, y^b = z and z^c = x

Question 2.
Prove that:

Solution:
Taking L.H.S., we have

Question 3.
Show that:

Solution:

Number Systems Class 9 Extra Questions Value Based (VBQs)

Question 1.
Sudhir and Ashok participated in a long jump competition along a straight line marked as a number line. Both start the jumps one by one but in opposite directions. From ‘O’ Ashok jumps one unit towards the positive side while Sudhir jumps double in units as Ashok jumps, along negative side. After jumping 4 jumps each, at which point Ashok and Sudhir reached. What is the distance between their final positions ? Ashok argue that he is the winner since Sudhir is at negative side. Who do you think is winner and why? What is the value of the competition ?

Solution:
After jumping four jumps each, Ashok reached at 4 in positive direction and Sudhir reached at -8 i.e., in negative direction. Distance between their final positions is 12 units. Here, distance covered by Sudhir is 8 units and distance covered by Ashok is 4 units. Thus, Sudhir is the winner. Competition inculcate spirit of performance.

Question 2.
Manu went to his mathematics teacher and asked him “Sir, I want some chocolates to distribute among my classmates for my birthday but I have no money. Can you provide me some chocolates”. Teacher told Manu, I am giving you two numbers \(\frac{1}{3+2 \sqrt{2}}\) and \(\frac{1}{3-2 \sqrt{2}}\) and if you can find the value of sum of their squares, then I will provide you as many chocolates as the resulting value of sum of squares of given numbers. Find the number of chocolates. What value is depicted from this action?
Solution:

= (3 – 2√2)² + (3 + 2√2²
= 9 + 8 – 2 × 3 × (2√2) + 9 + 8 + 2 × 3 × 2√2 = 34.
Hence, resulting value of sum of squares of numbers = number of chocolates = 34. By doing this, teacher motivates the students to use their knowledge and apply it in day to day life with caring and kindness.

Here we are providing Coordinate Geometry Class 9 Extra Questions Maths Chapter 3 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Coordinate Geometry with Answers Solutions

Extra Questions for Class 9 Maths Chapter 3 Coordinate Geometry with Solutions Answers

Coordinate Geometry Class 9 Extra Questions Very Short Answer Type

Coordinate Geometry Class 9 Extra Questions Question 1.
Write the signs convention of the coordinates of a point in the second quadrant.
Solution:
(-ve, +ve)

Class 9 Coordinate Geometry Extra Questions Question 2.
Write the value of ordinate of all the points lie on x-axis.
Solution:
0

Coordinate Geometry Extra Questions Class 9 Question 3.
Write the value of abscissa of all the points lie on y-axis.
Solution:
0

Class 9 Maths Chapter 3 Extra Questions Question 4.
If in coordinates of a point B(3, -2), signs of both coordinates are interchanged, then it will lie in which quadrant ?
Solution:
When signs of both coordinates of B(3, -2) are interchanged, then coordinates of new point are B'(-3, 2) and it will lie in second quadrant.

Extra Questions On Coordinate Geometry Class 9 Question 5.
Find distances of points C(-3, -2) and D(5, 2) from x-axis and y-axis.
Solution:
Distances of point C(-3, -2) from x-axis is 2 units in the negative direction and from y-axis is 3 units in the negative direction. Distances of point D(5, 2) from x-axis is 2 units and from y-axis is 5 units.

Extra Questions Of Coordinate Geometry Class 9 Question 6.
Find the values of x and y, if two ordered pairs (x – 3, – 6) and (4, x + y) are equal.
Solution:
Here, two ordered pairs are equal.
⇒ Their first components are equal and their second components are separately equal.
⇒ x – 3 = 4 and x + y = -6
⇒ x = 7 and 7 + y = -6 ⇒ y = – 13
Hence, x = 7 and y = – 13.

Questions On Coordinate Geometry Class 9 Question 7.
In which quadrant does the point (-1, 2) lie ?
Solution:
(-1, 2) lie in second quadrant.

Coordinate Geometry Class 9 Important Questions Question 8.
Find the distance of the point (0, -5) from the origin.
Solution:
5 units.

Class 9 Maths Coordinate Geometry Extra Questions Question 9.
Write the shape of the quadrilateral formed by joining (1, 1), (6, 1), (4, 5) and (3, 5) on graph paper.
Solution:
Trapezium.

Coordinate Geometry Class 9 Extra Questions Short Answer Type 1

Coordinate Geometry Class 9 Questions With Solutions Question 1.
In the given figure, ABCD is a rectangle with length 6 cm and breadth 3 cm. O is the mid-point of AB. Find the coordinates of A, B, C and D.

Solution:

We have taken 1 cm = 1 unit and origin O is the mid-point of AB
∴ OA = OB = 3 cm
and BC = AD = 3 cm
Thus, the coordinates of A are (-3, 0)
the coordinates of B are (3, 0)
the coordinates of C are (3, 3)
the coordinates of D are (-3, 3)

Ch 3 Maths Class 9 Extra Questions Question 2.
Write the coordinates of A, B, C and D from the figure given alongside.

Solution:
Coordinates of the point A are (5, 0)
Coordinates of the point B are (5, 3)
Coordinates of the point C are (-2, 4)
Coordinates of the point D are (0, -2)

Class 9 Maths Ch 3 Extra Questions Question 3.
A point lies on x-axis at a distance of 9 units from y-axis. What are its coordinates? What will be the coordinates of a point, if it lies on y-axis at a distance of -9 units from x-axis ?
Solution:
As shown in graph, the coordinates of a point which lies on x-axis at a distance of 9 units from y-axis are (9, 0) and the coordinates of a point which lies at a distance of -9 units from x-axis are (0, -9).

Coordinate Geometry Class 9 Hots Questions Question 4.
Plot the point P(2, -6) on a graph paper and from it draw PM and PN perpendiculars to x-axis and y-axis respectively. Write the coordinates of the points M and N.
Solution:

As shown in graph, coordinates of M are (2, 0) and coordinates of N are (0, -6).

Coordinate Geometry Class 9 Extra Questions Short Answer Type 2

Chapter 3 Maths Class 9 Extra Questions Question 1.
Without plotting the points indicate the quadrant in which they lie, if :
(i) ordinate is 5 and abscissa is – 3
(ii) abscissa is -5 and ordinate is – 3
(iii) abscissa is – 5 and ordinate is 3
(iv) ordinate is 5 and abscissa is 3
Solution:
(i) Clearly, point (-3, 5) lies in 2nd quadrant.
(ii) Clearly, point (-5, – 3) lies in 3rd quadrant.
(ii) Clearly, point (-5, 3) lies in 2nd quadrant.
(iv) Clearly, point (3, 5) lies in 1st quadrant.

Coordinate Geometry Class 9 Extra Questions With Solutions Question 2.
Plot the points A(1, 4), B(-2, 1) and C(4, 1). Name the figure so obtained on joining them in order and also, find its area.
Solution:

Triangle.
Area of ∆ABC = \(\frac{1}{2}\) × BC × Height
= \(\frac{1}{2}\) × 6 × 3
= 9 sq. units

Coordinate Geometry Questions Class 9 Question 3.
Plot the following points, join them in order and identify the figure thus formed : A(1, 3), B(1, -1), C(7, -1) and D(7, 3)
Write the coordinates of the point of intersection of the diagonals.
Solution:

ABCD is a rectangle.
Point of intersection of the diagonals AC and BD is (4, 1).

Class 9 Chapter 3 Maths Extra Questions Question 4.
Plot the points A(2, 5), B(8,5) and C(5, -3) and join AB, BC and CA. What figure do you obtain ?
Solution:

We obtain an isosceles triangle in which AC = BC.

Extra Questions For Class 9 Maths Chapter 3 With Solution Question 5.
(i) Plot the points M(4, 3), N(4, 0), 0(0, 0), P(0, 3).
(ii) Name the figure obtained by joining MNOP.
(iii) Find the perimeter of the figure.
Solution:
(i)

(ii) As shown in graph, the figure obtained by joining MNOP is rectangle.
(iii) Perimeter of rectangle MNOP = 2 (ON + OP) = 2 (4 + 3) = 2 × 7 = 14 units.

Question 6.
Plot D(-2, -3) on the graph paper. Also, plot reflections of D in x-axis and y-axis.
Solution:

Reflection of D(-2, -3) in x-axis is D'(-2, 3) and reflection of D(-2, -3) in y-axis is D'(2, -3).

Question 7.
If the coordinates of a point M are (-2, 9) which can also be expressed as (1 + x, y) and y > 0, then find in which quadrant do the following points lie : Ply, x), Q(2, x), R(x, y − 1), S(2x, -3y).
Solution:
Here, M(-2, 9) can also be expressed as (1 + x, y²)
∴ (1 + x, y²) ⇔ (-2, 9) and y > 0
⇒ 1 + x = -2 and y² = 9
⇒ x = -3 and y = √9 = 3 (∵ y > 0)
P (y, x) = P(3, -3), it lies in IV quadrant
Q(2, x) = Q(2, -3), also lies in IV quadrant
R(x², y – 1) = R((-3)², 3 – 1) = R(9, 2), it lies in I quadrant
S(2x, -3y) = S(2 × (-3), -3 × 3) = S(-6, -9), it lies in III quadrant

Question 8.
In the given figure, PQR is an equilateral triangle with coordinates of Q and R as (-2, 0) and (2, 0) respectively. Find the coordinates of the vertex P.
Solution:

Here,
QR = OQ + OR.
= 2 + 2 = 4 units
∴ ∆PQR is an equilateral triangle.
∴ PQ = PR = QR = 4 units
In right-angled ∆OPQ, ∠POQ = 90°
∴ We have PQ² = OP² + OQ²
⇒ OP² = PQ² – OQ²
= 4² – 2² = 16 – 4 = 12
⇒ OP = √12 = 2√3 units
∴ The coordinates of P are (0, 2√3).

Coordinate Geometry Class 9 Extra Questions HOTS

Question 1.
Plot the points A(3, 2), B(-2, 2), C(-2, -2) and D(3, -2) in the cartesian plane. Join these points and name the figure so formed.
Solution:

Figure so formed is ABCD a rectangle.

Question 2.
Write the coordinates of two points on X-axis and two points on Y-axis which are at equal distances from the origin. Connect all these points and make them as vertices of quadrilateral. Name the quadrilateral thus formed.
Solution:
Let a be the equal distance from origin on both axes. Now, the coordinates of two points on equal distance ‘a’on x-axis are Pla, 0) and R(-a, 0). Also, the coordinates of two points on equal distance ‘a’ on Y-axis are Q(0, a) and S(0, -a). Join all the four points on the graph. Now, PQRS, thus formed is a square.

Coordinate Geometry Class 9 Extra Questions Value Based (VBQs)

Question 1.
On environment day, class-9 students got five plants of mango, silver oak, orange, banyan and amla from soil department. Students planted the plants and noted their locations as (x, y).

Plot the points (x, y) in the graph and join them in the given order. Name the figure you get. Which social act is being done by students of class-9 ?
Solution:

The given trees (points) are Mango (2, 0), Silver Oak (3, 4), Orange (0,7), Banyan (-3, 4) and Amla (-2, 0). The location of these trees are Orange (0,7) shown in the graph.
On joining the points of mango, silver oak, orange, banyan and amla in order, the figure so formed is a regular pentagon.
Planting more trees helpful in reducing pollution and make the environment clean and green for the coming generations.

Here we are providing Linear Equations for Two Variables Class 9 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Linear Equations for Two Variables with Answers Solutions

Extra Questions for Class 9 Maths Chapter 4 Linear Equations for Two Variables with Solutions Answers

Linear Equations for Two Variables Class 9 Extra Questions Very Short Answer Type

Linear Equations In Two Variables Class 9 Extra Questions Question 1.
Linear equation x – 2 = 0 is parallel to which axis ?
Solution:
Here, linear equation is x – 2 ⇒ 0 x = 2
Thus, it is parallel to the y-axis.

Linear Equations Class 9 Extra Questions Question 2.
Express x in term of y: \(\frac{x}{7}\) + 2y = 6
Solution:
Given equation is
\(\frac{x}{7}\) + 2y = 6
⇒ \(\frac{x}{7}\) = 6 – 2y
Thus, x = 7(6 – 2y).

Linear Equations In Two Variables Extra Questions Class 9 Question 3.
If we multiply or divide both sides of a linear equation with a non-zero number, then what will happen to the solution of the linear equation ?
Solution:
Solution remains the same.

Class 9 Maths Chapter 4 Extra Questions Question 4.
Find the value of k for which x = 0, y = 8 is a solution of 3x – 6y = k.
Solution:
Since x = 0 and y = 8 is a solution of given equation
3x – 6y = k
3(0) – 6(8) = k
⇒ k = – 48

Class 9 Linear Equations In Two Variables Extra Questions Question 5.
Write the equation of a line which is parallel to x-axis and is at a distance of 2 units from the origin.
Solution:
Here, required line is parallel to x-axis and at a distance of 2 units from the origin.
∴ Its equation is
y + 2 = 0
or y – 2 = 0

Extra Questions For Class 9 Maths Chapter 4 With Solution Question 6.
Find ‘a’, if linear equation 3x – ay = 6 has one solution as (4, 3).
Solution:
Since (4, 3) is a solution of given equation.
∴ 3(4) – a(3) = 6
⇒ 12 – 3a = 6
⇒ a = \(\frac{-6}{-3}\)
Hence, a = 2

Linear Equations In Two Variables Class 9 Extra Questions With Solutions Question 7.
Cost of a pen is two and half times the cost of a pencil. Express this situation as a linear equation in two variables.
Solution:
Let cost of a pen be ₹ x and cost of a pencil be ₹ y.
According to statement of the question, we have
x = 2\(\frac{1}{2}\) y
⇒ 2x = 5y or 2x – 5y = 0

Extra Questions On Linear Equations In Two Variables Class 9 Question 8.
In an one day international cricket match, Raina and Dhoni together scored 198 runs. Express the statement as a linear equation in two variables.
Solution:
Let runs scored by Raina be x and runs scored by Dhoni be y.
According to statement of the question, we have
x + y = 198
x + y – 198 = 0

Extra Questions For Class 9 Maths Linear Equations In Two Variables Question 9.
The cost of a table is 100 more than half the cost of a chair. Write this statement as a linear equation in two variables.
Solution:
Let the cost price of a table be ₹ x and that of a chair be ₹ y.
Since the cost price of a table is 100 more than half the cost price of a chair.
∴x = \(\frac{1}{2}\)y + 100
⇒ 2x = y + 200 or 2x – y – 200 = 0.

Linear Equations for Two Variables Class 9 Extra Questions Short Answer Type 1

Class 9 Maths Ch 4 Extra Questions Question 1.
Write linear equation representing a line which is parallel to y-axis and is at a distance of 2 units on the left side of y-axis.
Solution:
Here, required equation is parallel to y-axis at a distance of 2 units on the left side of y-axis.
x = -2 or x + 2 = 0

Chapter 4 Maths Class 9 Extra Questions Question 2.
In some countries temperature is measured in Fahrenheit, whereas in countries like India it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius :
F = \(\left(\frac{9}{5}\right)\)C + 32°
If the temperature is – 40°C, then what is the temperature in Fahrenheit?
Solution:
Given linear equation is
F = \(\left(\frac{9}{5}\right)\)C + 32°
Put C = -40°, we have
F = \(\frac{9}{5}\)(-40°) + 32°
F = – 72° + 32°
F= -40°

Ch 4 Maths Class 9 Extra Questions Question 3.
Give equation of two lines on same plane which are intersecting at the point (2, 3).
Solution:
Since there are infinite lines passing through the point (2, 3).
Let, first equation is x + y = 5 and second equation is 2x + 3y = 13.
Clearly, the lines represented by both equations intersect at the point (2, 3).

Class 9 Maths Linear Equations In Two Variables Extra Questions Question 4.
If ax + 3y = 25, write y in terms of x and also, find the two solutions of this equation.
Solution:

Linear Equations In 2 Variables Class 9 Extra Questions Question 5.
Find the value of k, if (1, -1) is a solution of the equation 3x – ky = 8. Also, find the coordinates of another point lying on its graph.
Solution:
Since (1, -1) is a solution of the equation 3x – ky = 8
∴ 3(1) – k(-1) = 8
⇒ k = 8 – 3 = 5
Thus, the given equation is
3x – 5y = 8

Hence, the coordinates of another point lying on the graph of 3x – 5y = 8 is (6, 2).

Linear Equations In Two Variables Class 9 Important Questions Question 6.
Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is the value of y, when x = 5 ?
Solution:
Given y varies directly as x implies y = kx
But y = 12 for x = 4
⇒ 4k = 12 = k = 3
Put k = 3 in y = kx, we have
y = 3x
Now, when x = 5, y = 3 x 5 = y = 15 …(i)

Linear Equations for Two Variables Class 9 Extra Questions Short Answer Type 2

Linear Equations In Two Variables Class 9 Extra Questions Pdf Question 1.
A fraction becomes , when 2 is subtracted from the numerator and 3 is added to the denominator. Represent this situation as a linear equation in two variables. Also, find two solutions for this.
Solution:
Let numerator and denominator of the given fraction be respectively x and y. According to the statement, we obtain
\(\frac{x-2}{y+3}\) = \(\frac{1}{4}\)
⇒ 4x – 8 = y + 3
⇒ 4x – y – 11 = 0
Which is the required linear equation. When y = 1, then x = 3. When y = 5, then x = 4. Hence, the two solutions are (3, 1) and (4, 5).

Extra Questions For Class 9 Maths Chapter 4 Question 2.
The path of an aeroplane is given by the equation 3x – 4y = 1 2: Represent the path graphically. Also, show that the point (-4,-6) lies on the graph.
Solution:
Given equation is 3x – 4y = 12
∴ y = \(\frac{3 x-12}{4}\).

When x = 0, then y = -3
When x = 4, then y = 0
Table of solutions is

Plot the points (0, – 3), (4, 0) on the graph and join them to get the required graph. From the graph, we see, when
x = -4, then y = -6.
Therefore, (-4,- 6) lies on the graph of given equation.

Class 9 Chapter 4 Maths Extra Questions Question 3.
Express y in terms of x for the equation 3x – 4y + 7 = 0. Check whether the points (23, 4) and \(\left(0, \frac{7}{4}\right)\) lie on the graph of this equation or not.
Solution:
Given equation is 3x – 4y + 7 = 0
∴y = \(\frac{3 x+7}{4}\)
When x = -1, then y = 1
When x = 3, then y = 4
When x = 7, then y = 7
Table of solutions is :


Plot the points (-1, 1), (3, 4), (7, 7). and join them to get the given graph. From the graph, we see that, when x = 0, then y \(\frac{7}{4}\) = 5 and when y = 4,
then x = 3. Thus, (0, \(\frac{7}{4}\)) lies on the graph, whereas (23, 4) does not lie on the graph.

Extra Questions Of Chapter 4 Class 9 Maths Question 4.
The following observed values of x and y are thought to satisfy a linear equation. Write the linear equation :

Draw the graph using the values of x, y as given in the above table. At what points the graph of the linear equation :
(i) cuts the x-axis (ii) cuts the y-axis
Solution:
Let ax + by + c = 0 ……(i)
be the linear equation in two variables. From the table, we have two points A(6, -2) and B(-6, 6) which lie on the graph of the linear equation
⇒ 6a – 2b + c = 0
and -6a + 6b + c = 0
Adding the above two equations, we obtain

Which is the required linear equation in two variables. Plots the points A(6, -2) and B(-6, 6) on the graph. Join them to get line AB.
From the graph, we see that the graph cuts the x-axis at (3, 0) and the y-axis at (0, 2).

Linear Equations for Two Variables Class 9 Extra Questions Long Answer Type

Question 1.
Write the equations of the lines drawn in following graph :

Also, find the area enclosed between these lines.
Solution:
Equations of the lines drawn in the graph are as :
x = -1 or x + 1 = 0,
x = 2 or x – 2 = 0,
y = 1 or y – 1 = 0 and
y = 3 or y – 3 = 0
Figure formed by these lines is a rectangle of dimensions 3 units by 2 units.
Hence, the area enclosed between given lines = 6 sq. units.

Question 2.
If (2, 3) and (4, 0) lie on the graph of equation ax + by = 1. Find the value of a and b. Plot the graph of equation obtained.
Solution:
(2, 3) and (4, 0) lie on the graph of equation
ax + by = 1 …(i)
∴ We have 2a + 3b = 1 … (ii)
and 4a + 0 = 1

Which is required linear equation.
Put x= 0 in eq. (iii), we have
⇒ 3(0) + 2y = 12
⇒ 2y = 12
⇒ y = 6
Put x = 2 in eq. (iii), we have
⇒ 3(2) + 2y = 12
⇒ 2y = 6
⇒ y = 3
Put x = 4 in eq. (iii), we have
⇒ 3(4) + 2y = 12
⇒ 2y = 0
⇒ y = 0
We have the following table :

By plotting the points (0, 6), (2, 3) and (4, 0). Joining them, we obtained the graph of 3x + 2y = 12.

Question 3.
Draw the graphs of the following equations on the same graph sheet :
x = 4, x = 2, y = 1 and y – 3 = 0.
Solution:
Graphs of the given equations are drawn on graph sheet.

Question 4.
Cost of 1 pen is ₹ x and that of 1 pencil is ₹ y. Cost of 2 pens and 3 pencils together is ₹ 18. Write a linear equation which satisfies this data. Draw the graph for the same.
Solution:
Here, cost of 1 pen is ₹x and that
of 1 pencil is ₹ y According to the statement of the question, we have
2x + 3y = 18
⇒ x = \(\frac{18-3 y}{2}\)
When y = 0, x = 9
When y = 4, x= 3.
When y = 6, x = 0
Table of solutions is :

Plot the points (0, 6), (3, 4) and (9, 0). Join them in pairs to get the required line.

Question 5.
Sum of two numbers is 8. Write this in the form of a linear equation in two variables. Also, draw the line given by this equation. Find graphically the numbers, if difference between them is 2.
Solution:
Let the two numbers be x and y.
It is given that sum of two numbers is 8.
∴ x + y = 8
y = 8 – x
When x = 0,
When x = 4, y = 4
When x = 8, y = 0

Table of solutions is :

Plot the points (0, 8), (4, 4), (8, 0) and join them in pairs, we get the required graph.
When difference between two number is 2, then
x – y = 2, x > y
⇒ x = y + 2

When x = 0, y = -2
When x = 2, y = 0
When x = 4, y = 2
Table of solutions is :

Plot these poitns (0, -2), (2, 0), (4, 2) and join them to get the required line.
Graphically, the numbers are : (-2; 4), (-1, -3), (0, -2), (1, -1), (2, 0), (3, 1), (4, 2), (5, 3), (6, 4), (7,5) etc.

Linear Equations for Two Variables Class 9 Extra Questions HOTS

Question 1.
Solve for y :

Solution:

Question 2.
A and B are friends. A is elder to B by 5 years. B’s sister C is half the age of B while A’s father D is 8 years older than twice the age of B. If the present age of D is 48 years, find the present ages of A, B and C.
Solution:
Let the age of B’s sister i.e., C be x years.
∴Age of B be 2x years, age of A be (2x + 5) years.
And age of A’s father i.e., D be 2(2x) + 8 = 4x + 8 years
According to the statement of the question, we have
4x + 8 = 48
⇒ 4x = 48 – 8 = 40
⇒ x = \(\frac{40}{4}\) = 10
Age of A = (2x + 5) years i.e., (2 × 10 + 5) years = 25 years
Age of B = 2x years = 2 × 10 years = 20 years
and Age of C = \(\frac{1}{2}\) of B’s age = \(\frac{1}{2}\) × 20 years = 10 years

Linear Equations for Two Variables Class 9 Extra Questions Value Based (VBQs)

Question 1.
Mrs Sharma lost her purse containing 50 rupee and 100 rupee notes amount to 1500 in a shop. Next day shopkeeper found the purse during dusting. He immediately went to Mrs Sharma’s house and returned the purse and rupees. Mrs Sharma appreciates the shopkeeper for his act.
(i) Represent the situation as an equation and draw the graph.
(ii) What value do you learn from shopkeeper’s act ?
Solution:
(i) Let the number of 50 rupee notes be x
And the number of 100 rupee notes be y
∴ We have 50x + 100y= 1500
⇒ x + 2y = 30
(dividing each term by 50)
2y = 30 – x

We have the following table :

By plotting the points (8, 11), (10, 10) and (12, 9) on the graph and by joining them, we obtain the straight line represented by equation
(i) as shown in graph.
(ii) We should be always honest to feel good.

Question 2.
In an election, a good candidate may lose because 40% of voters do not cast their votes due to various reasons. Form an equation and draw the graph with data. From the graph, find :
(i) The total number of voters, if 720 voters cast their votes.
(ii) The number of votes cast, if the total number of voters are 1000.
(iii) What message did you get from above information ?
Solution:
(i) We have, total number of voters who do not cast their votes = 40%
⇒ Total number of voters who cast their votes = 60%
Let the total number of voters be x and number of voters cast their votes be y

Thus, we have the following table :

By plotting the points (100, 60), (200, 120), (300, 180) on the graph and by joining them, we get the graph of equation (i) as shown in figure.

From the graph, we see that:
(i) When total votes polled = 720
i.e., y = 720, the total number of voters i.e., x = 1200
Hence, total number of voters = 1200.
(ii) When total number of voters x = 1000
Number of votes cast is 600.
(iii) Every voters should cast vote to elect an honest candidate.

Here we are providing Introduction to Euclid’s Geometry Class 9 Extra Questions Maths Chapter 5 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers. https://ncertmcq.com/extra-questions-for-class-9-maths/

Extra Questions for Class 9 Maths Introduction to Euclid’s Geometry with Answers Solutions

Extra Questions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry with Solutions Answers

Introduction to Euclid’s Geometry Class 9 Extra Questions Very Short Answer Type

Class 9 Maths Chapter 5 Extra Questions With Solutions Question 1.
Give a definition of parallel lines. Are there other terms that need to be defined first ? What are they and how might you define them?
Solution:
Two coplanar lines in a plane) which are not intersecting are called parallel lines. The other term intersecting is undefined.

Introduction To Euclidean Geometry Class 9 Extra Questions Question 2.
Give a definition of perpendicular lines. Are there other terms that need to be defined first ? What are they and how might you define them?
Solution:
Two coplanar (in a plane) lines are perpendicular if the angle between them at the point
of intersection is one right angle. The other terms point of intersection and one right angle are undefined.

Euclid’s Geometry Class 9 Extra Questions Question 3.
Give a definition of line segment. Are there other terms that need to be defined first ? What are they and how might you define them ?
Solution:
A line segment PQ of a line ‘l is the continuous part of the line I with end points P and Q.

Here, continuous part of the line ‘l is undefined.

Euclid’s Geometry Class 9 Extra Questions With Solutions Question 4.
Solve the equation a – 15 = 25 and state which axiom do you use here.
Solution:
a – 15 = 25
On adding 15 to both sides, we obtain
a – 15 + 15 = 25 + 15 [using Euclid’s second axiom]
a = 40

Questions On Euclid Geometry Class 9 Question 5.
Ram and Ravi have the same weight. If they each gain weight by 2 kg, how will their new weights be compared ?
Solution:
Let x kg be the weight each of Ram and Ravi.
On adding 2 kg,
Weight of Ram and Ravi will be (x + 2) kg each.
According to Euclid’s second axiom, when equals are added to equals, the wholes are equal.
So, weight of Ram and Ravi are again equal.

Chapter 5 Maths Class 9 Extra Questions Question 6.
If a point C be the mid-point of a line segment AB, then write the relation among AC, BC and AB.
Solution:
Here, C is the mid-point of AB
⇒ AC = BC
⇒ AC = BC = \(\frac{1}{2}\)AB

Euclid Geometry Class 9 Questions And Answers Question 7.
If a point P be the mid-point of MN and C is the mid-point of MP, then write the relation between MC and MN.
Solution:
Here, P is the mid-point of MN and C is the mid-point of MP.

∴ MC = \(\frac{1}{4}\)MN

Extra Questions For Class 9 Maths Chapter 5 Question 8.
How many lines does pass through two distinct points ?
Solution:
One and only one.

Euclid Geometry Class 9 Questions Question 9.
In the given figure, if AB = CD, then prove that AC = BD. Also, write the Euclid’s axiom used for proving it.

Solution:
Here, given that
AB = CD
By using Euclid’s axiom 2, if equals are added to equals, then the wholes are equal, we have
AB + BC = CD + BC
⇒ AC = BD

Introduction to Euclid’s Geometry Class 9 Extra Questions Short Answer Type 2

Euclids Geometry Class 9 Extra Questions Question 1.
Define :
(a) a square (b) perpendicular lines.
Solution:
(a) A square : A square is a rectangle having same length and breadth. Here, undefined terms are length, breadth and rectangle.
(b) Perpendicular lines : Two coplanar (in a plane) lines are perpendicular, if the angle between them at the point of intersection is one right angle. Here, the term one right angle is undefined.

Class 9 Maths Chapter 5 Important Questions Question 2.
In the given figure, name the following :
(i) Four collinear points
(ii) Five rays
(iii) Five line segments
(iv) Two-pairs of non-intersecting line segments.

Solution:
(i) Four collinear points are D, E, F, G and H, I, J, K
(ii) Five rays are DG, EG, FG, HK, IK.
(iii) Five line segments are DH, EI, FJ; DG, HK.
(iv) Two-pairs of non-intersecting line segments are (DH, EI) and (DG, HK).

Class 9 Maths Chapter 5 Extra Questions Question 3.
In the given figure, AC = DC and CB = CE. Show that AB = DE. Write the Euclid’s axiom to support this.

Solution:
We have
AC = DC
CB = CE
By using Euclid’s axiom 2, if equals are added to equals, then wholes are equal.
⇒ AC + CB = DC + CE
⇒ AB = DE.

Question 4.
In figure, it is given that AD=BC. By which Euclid’s axiom it can be proved that AC = BD?

Solution:
We can prove it by Euclid’s axiom 3. “If equals are subtracted from equals, the remainders are equal.”
We have AD = BC
⇒ AD – CD = BC – CD
⇒ AC = BD

Question 4.
If A, B and C are any three points on a line and B lies between A and C (see figure), then prove that AB + BC = AC.

Solution:
In the given figure, AC coincides with AB + BC. Also, Euclid’s axiom 4, states that things
which coincide with one another are equal to one another. So, it is evident that:
AB + BC = AC.

Question 5.
In the given figure, AB = BC, BX = BY, show that
AX = CY.

Solution:
Given that AB = BC
and BX = BY
By using Euclid’s axiom 3, equals subtracted from equals, then the remainders are equal, we have
AB – BX = BC – BY
AX = CY

Question 6.

In the above figure, if AB = PQ, PQ = XY, then AB = XY. State True or False. Justify your answer.
Solution:
True. ∵ By Euclid’s first axiom “Things which are equal to the same thing are equal to one another”.
∴ AB = PQ and XY = PQ ⇒ AB = XY

Question 7.
In the given figure, if ∠1 = ∠3, ∠2 = ∠4 and ∠3 = ∠4, write the relation between ∠1 and ∠2, using an Euclid’s axiom.

Solution:
Here, ∠3 = ∠4, ∠1 = ∠3 and ∠2 = ∠4. Euclid’s first axiom says, the things which are equal to equal thing are equal to one another. So ∠1 = ∠2.,

Question 8.
In the given figure, we have ∠1 = ∠2, ∠3 = ∠4. Show that ∠ABC = ∠DBC. State the Euclid’s Axiom used.
Solution:
Here, we have 1 = ∠2 and ∠3 = ∠4. By using Euclid’s Axiom 2. If equals are added to
equals, then the wholes are equal..
∠1 + ∠3 = ∠2 + ∠4
∠ABC = ∠DBC.

Question 9.
In the figure, we have BX and \(\frac{1}{2}\) AB =\(\frac{1}{2}\) BC. Show that BX = BY.

Solution:
Here, BX = \(\frac{1}{2}\) AB and BY = \(\frac{1}{2}\) BC …(i) [given]
Also, AB = BC [given]
⇒ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)BC …(ii)
[∵ Euclid’s seventh axiom says, things which are halves of the same thing are equal to one another]
From (i) and (ii), we have BX = BY

Question 10.
In the given figure, AC = XD, C is mid-point of AB and D is mid-point of XY. Using an Euclid’s axiom, show that AB = XY.

Solution:
∵ C is the mid-point of AB
AB = 2AC
Also, D is the mid-point of XY
XY = 2XD
By Euclid’s sixth axiom “Things which are double of same things are equal to one another.”
∴ AC = XD = 2AC = 2XD ⇒ AB = XY

Introduction to Euclid’s Geometry Class 9 Extra Questions HOTS

Question 1.
For given four distinct points in a plane, find the number of lines that can be drawn through :
(i) When all four points are collinear.
(ii) When three of the four points are collinear.
(iii) When no three of the four points are collinear.

Solution:

(i) Consider the points given are A, B, C and D.
When all the four points are collinear :
One line \(\overrightarrow{\mathrm{AD}}\).

(ii) When three of the four points are collinear :
4 lines
Here, we have four lines \(\stackrel{\leftrightarrow}{A B}, \stackrel{\leftrightarrow}{B C}, \stackrel{\leftrightarrow}{B D}, \stackrel{\leftrightarrow}{A D}\) (four).

(iii) When no three of the four points are collinear :
6 lines Here, we have
\(\stackrel{\leftrightarrow}{A B}, \stackrel{\leftrightarrow}{B C}, \stackrel{\leftrightarrow}{A C}, \stackrel{\leftrightarrow}{A D}, \stackrel{\leftrightarrow}{B D}, \stackrel{\leftrightarrow}{C D}\) (six).

Question 2.
Show that : length AH > sum of lengths of AB + BC + CD.

Solution:
We have
AH = AB + BC + CD + DE + EF + FG + GH
Clearly, AB + BC + CD is a part of AH.
⇒ AH > AB + BC + CD
Hence, length AH > sum of lengths AB + BC + CD.

Introduction to Euclid’s Geometry Class 9 Extra Questions Value Based (VBQs)

Question 1.
Rohan’s maid has two children of same age. Both of them have equal number of dresses. Rohan on his birthday plans to give both of them same number of dresses. What can you say about the number of dresses each one of them will have after Rohan’s birthday? Which Euclid’s axiom is used to answer this question? What value is Rohan depicting by doing so ? Write one more Euclid’s axiom.
Solution:
Here, Rohan’s maid has two children of same age group and both of them have equal number of dresses. Rohan on his birthday plans to give both of them same number of dresses.
∴ By using Euclid’s Axiom 2, if equals are added to equals, then the whole are equal. Thus, again both of them have equal number of dresses. Value depicted by Rohan are caring and other social values. According to Euclid’s Axiom 3, if equals are subtracted from equals, then the remainders
are equal.

Question 2.
Three lighthouse towers are located at points A, B and C on the section of a national forest to protect animals from hunters by the forest department as shown in figure. Which value is department exhibiting by locating extra towers ? How many straight lines can be drawn from A to C? State the Euclid Axiom which states the required result. Give one more. Postulate.

Solution:
One and only one line can be drawn from A to C. According to Euclid’s Postulate, “A straight line may be drawn from any point to any other point:” An other postulate : “A circle may be described with any centre and any radius.” Wildlife is a part of our environment and conservation of each of its element is important for ecological balance.

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Class 9 English MCQs Multiple Choice Questions with Answers Beehive, Moments

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MCQ Questions for Class 9 English Language and Literature

MCQ Questions for Class 9 English with Answers Beehive

MCQ Questions for Class 9 English with Answers Beehive Prose

1. The Fun They Had Class 9 MCQ Questions
2. The Sound of Music Class 9 MCQ Questions
3. The Little Girl Class 9 MCQ Questions
4. A Truly Beautiful Mind Class 9 MCQ Questions
5. The Snake and the Mirror Class 9 MCQ Questions
6. My Childhood Class 9 MCQ Questions
7. Packing Class 9 MCQ Questions
8. Reach for the Top Class 9 MCQ Questions
9. The Bond of Love Class 9 MCQ Questions
10. Kathmandu Class 9 MCQ Questions
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MCQ Questions for Class 9 English with Answers Beehive Poems

1. The Road Not Taken Class 9 MCQ Questions
2. Wind Class 9 MCQ Questions
3. Rain on the Roof Class 9 MCQ Questions
4. The Lake Isle of Innisfree Class 9 MCQ Questions
5. A Legend of the Northland Class 9 MCQ Questions
6. No Men are Foreign Class 9 MCQ Questions
7. The Duck and the Kangaroo Class 9 MCQ Questions
8. On Killing a Tree Class 9 MCQ Questions
9. The Snake Trying Class 9 MCQ Questions
10. A Slumber did my Spirit Seal Class 9 MCQ Questions

MCQ Questions for Class 9 English with Answers Moments

1. The Lost Child Class 9 MCQ Questions
2. The Adventures of Toto Class 9 MCQ Questions
3. Iswaran the Storyteller Class 9 MCQ Questions
4. In the Kingdom of Fools Class 9 MCQ Questions
5. The Happy Prince Class 9 MCQ Questions
6. Weathering the Storm in Ersama Class 9 MCQ Questions
7. The Last Leaf Class 9 MCQ Questions
8. A House is not a Home Class 9 MCQ Questions
9. The Accidental Tourist Class 9 MCQ Questions
10. The Beggar Class 9 MCQ Questions

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