CBSE Class 9 – Page 33

Value Based Questions in Science for Class 9 Chapter 3 Atoms and Molecules

January 19, 2023

Value Based Questions in Science for Class 9 Chapter 3 Atoms and Molecules

These Solutions are part of Value Based Questions in Science for Class 9. Here we have given Value Based Questions in Science for Class 9 Chapter 3 Atoms and Molecules

Question 1.
Kamla prepared aqueous solutions of barium chloride and sodium sulphate. She weighed them separately and then mixed them in a beaker. A white precipitate was immediately formed. She filtered the precipitate, dried it and then weighed it. After reading this narration, answer the following questions :

Will the weight of the precipitate be the same as that of the reactants before mixing ?
If not, what she should have done ?
Which law of chemical combination does this support ?
What is the value based information associated with it ?

Answer:

No, it will not be the same.
She should have weighed the total contents of the beaker after the reaction and not the precipitate alone.
It supports the law of conservation of mass.
Whenever the law of conservation of mass is to be verified in the laboratory, total mass of the reactants and also of products should be taken into account. Moreover, none of the species be allowed to leave the container.

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Question 2.
In order to verify the law of conservation of mass, a student mixed 6.3 g of sodium carbonate and 15.0 g of ethanoic acid in a conical flask. After experiment, he weighed the flask again. The weight of the residue in the flask was only 18.0 g. He approached the teacher who guided him to carry the experiment in a closed flask with a cork. There was no difference in weight of the flask before and after the experiment.

What was the mistake committed by the student ?
Why did not the two weights match earlier ?
How did the teacher help him ?
What lesson was learnt by the student ?

Answer:

He was carrying the experiment in an open flask.
CO₂ gas evolved in the reaction escaped from the flask
2CH₃COOH + Na₂CO₃ —-> 2CH₃COONa + H₂O + CO₂
Teacher asked him to cork the flask the moment the reactants were mixed.
The student learnt that in future he should not carry the experiment relating to the law of conservation of mass in an open container, particularly when one or more reactants or products are in the gaseous state.

Question 3.
A student was asked by his teacher a verify the law of conservation of mass in the laboratory. He prepared 5% aqueous solutions of NaCl and Na₂SO₄. He mixed 10 mL of both these solutions in a conical flask. He weighed the flask on a balance. He then stirred the flask with a rod and weighed it after sometime. There was no change in mass. Read this narration and answer the questions given below :

Was the student able to verify the law of conservation of mass ?
If not, what was the mistake committed by him ?
In your opinion, what he should have done ?
What is the value based information associated with this ?

Answer:

No, he could not verify the law of conservation of mass inspite of the fact that there was no change in mass.
No chemical reaction takes place between NaCl and Na₂SO₄. This means that no reaction actually took place in the flask.
He should have performed the experiment by using aqueous solutions of BaCl₂ and Na₂SO₄. A chemical reaction takes place in this case and a white precipitate of BaSO₄ is formed.
While working in the chemistry laboratory, a student must select those chemical substances which actually react with each other. Only then products will be formed.

Question 4.
Dalton was the first scientist to introduce symbols for the known elements. Modern symbols were given by J.J. Berzelius. A symbol in general may be defined as the short hand representation of the name of an element.

How do symbols help in identifying elements ?
Do we use symbols in daily life ?
What values do you attach for using symbols ?

Answer:

All the known elements are identified by their symbols.
For example, Symbol of calcium = Ca; Symbol of copper = Cu; Symbol of iron= Fe
Yes, these are very common in daily life. For example, all road signs such as diversions, dangerous, zones etc. are indicated by symbols. In playground, umpires, signify the various happenings such as ‘LBW’, ‘Out’ etc. in circket by symbols.
Symbols for road signs save many lives. The names of many complicated compounds are shown by the formulae which are collection of symbols. The chemical composition of all madicines are shown either on the strips or on the bottles by their formulae.

Question 5.
Mole concept is an important tool for dealing with chemical calculations. The elements have atomic masses while compounds have molecular masses or molar masses. Mole is in fact, a collection of Avogadro’s number (N_A) of the particles of a substance whether element or compound. The value of Avogadro’s number is 6-022 x 10²³.

Why is mole commonly called chemist’s dozen ?
What is the value associated with the term mole ?

Answer:

Just as a dozen represents 12 articles, a mole represents 6.022 x 10²³ or Avogadro’s number of particles. Therefore, it has been rightly called chemist’s dozen.
Since particles such as atoms, ions or molecules are very extremely small in size, it is very difficult to identify and express them individually. These are collectively represented as mole. For example, 3.011 x 10²³ molecules of CO₂ gas are shown as 0.5 mole which is very simple.

Hope given Value Based Questions in Science for Class 9 Chapter 3 Atoms and Molecules are helpful to complete your science homework.

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RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A

January 19, 2023

RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A.

Question 1.
Solution:
Given : In the figure, ABCD is a quadrilateral and
AB = CD = 5cm
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q1.1

Question 2.
Solution:
In ||gm ABCD,
AB = 10cm, altitude DL = 6cm
and BM is altitude on AD, and BM = 8 cm.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q2.1

Question 3.
Solution:
Diagonals of rhombus are 16cm and 24 cm.
Area = \(\frac { 1 }{ 2 } \) x product of diagonals
= \(\frac { 1 }{ 2 } \) x 1st diagonal x 2nd diagonal
= \(\frac { 1 }{ 2 } \) x 16 x 24
= 192 cm² Ans.

Question 4.
Solution:
Parallel sides of a trapezium are 9cm and 6cm and distance between them is 8cm
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q4.1

Question 5.
Solution:
from the figure
(i) In ∆ BCD, ∠ DBC = 90°
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q5.1

Question 6.
Solution:
In the fig, ABCD is a trapezium. AB || DC
AB = 7cm, AD = BC = 5cm.
Distance between AB and DC = 4 cm.
i.e. ⊥AL = ⊥BM = 4cm.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q6.1

Question 7.
Solution:
Given : In quad. ABCD. AL⊥BD and CM⊥BD.
To prove : ar(quad. ABCD)
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q7.1

Question 8.
Solution:
In quad. ABCD, BD is its diagonal and AL⊥BD, CM⊥BD
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q8.1

Question 9.
Solution:
Given : ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect each other at O.
To prove : ar(∆ AOD) = ar(∆ BOC)
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q9.1

Question 10.
Solution:
Given : In the figure,
DE || BC.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q10.1

Question 11.
Solution:
Given : In ∆ ABC, D and E are the points on AB and AC such that
ar( ∆ BCE) = ar( ∆ BCD)
To prove : DE || BC.
Proof : (∆ BCE) = ar(∆ BCD)
But these are on the same base BC.
Their altitudes are equal.
Hence DE || BC
Hence proved.

Question 12.
Solution:
Given : In ||gm ABCD, O is any. point inside the ||gm. OA, OB, OC and OD are joined.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q12.1

Question 13.
Solution:
Given : In quad. ABCD.
A line through D, parallel to AC, meets ‘BC produced in P. AP in joined which intersects CD at E.
To prove : ar( ∆ ABP) = ar(quad. ABCD).
Const. Join AC
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q13.1

Question 14.
Solution:
Given : ∆ ABC and ∆ DBC are on the same base BC with points A and D on , opposite sides of BC and
ar( ∆ ABC) = ar( ∆ DBC).
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q14.1

Question 15.
Solution:
Given : In ∆ ABC, AD is the median and P is a point on AD
BP and CP are joined
To prove : (i) ar(∆BDP) = ar(∆CDP)
(ii) ar( ∆ ABP) = ar( ∆ ACP)
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q15.1

Question 16.
Solution:
Given : In quad. ABCD, diagonals AC and BD intersect each other at O and BO = OD
To prove : ar(∆ ABC) = ar(∆ ADC)
Proof : In ∆ ABD,
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q16.1

Question 17.
Solution:
In ∆ ABC,D is mid point of BC
and E is midpoint of AD and BE is joined.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q17.1

Question 18.
Solution:
Given : In ∆ ABC. D is a point on AB and AD is joined. E is mid point of AD EB and EC are joined.
To prove : ar( ∆ BEC) = \(\frac { 1 }{ 2 } \) ar( ∆ ABC)
Proof : In ∆ ABD,
E is midpoint of AD
BE is its median
ar(∆ EBD) = ar(∆ ABE)
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q18.1

Question 19.
Solution:
Given : In ∆ ABC, D is midpoint of BC and E is die midpoint of BO is the midpoint of AE.
To prove that ar( ∆ BOE) = \(\frac { 1 }{ 8 } \) ar(∆ ABC).
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q19.1

Question 20.
Solution:
Given : In ||gm ABCD, O is any point on diagonal AC.
To prove : ar( ∆ AOB) = ar( ∆ AOD)
Const. Join BD which intersects AC at P
Proof : In ∆ OBD,
P is midpoint of BD
(Diagonals of ||gm bisect each other)
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q20.1

Question 21.
Solution:
Given : ABCD is a ||gm.
P, Q, R and S are the midpoints of sides AB, BC, CD, DA respectively.
PQ, QR, RS and SP are joined.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q21.1

Question 22.
Solution:
Given : In pentagon ABCDE,
EG || DA meets BA produced and
CF || DB, meets AB produced.
To prove : ar(pentagon ABCDE) = ar(∆ DGF)
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q22.1

Question 23.
Solution:
Given ; A ∆ ABC in which AD is the median.
To prove ; ar( ∆ ABD) = ar( ∆ ACD)
Const : Draw AE⊥BC.
Proof : Area of ∆ ABD
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q23.1

Question 24.
Solution:
Given : A ||gm ABCD in which AC is its diagonal which divides ||gm ABCD in two ∆ ABC and ∆ ADC.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q24.1

Question 25.
Solution:
Given : In ∆ ABC,
D is a point on BC such that
BD = \(\frac { 1 }{ 2 } \) DC
AD is joined.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q25.1

Question 26.
Solution:
Given : In ∆ ABC, D is a point on BC such that
BD : DC = m : n
AD is joined.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q26.1

Hope given RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2

January 19, 2023

RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2

Other Exercises

Question 1.
If f(x) = 2x³ – 13x² + 17x + 12, find
(i) f (2)
(ii) f (-3)
(iii) f(0)
Solution:
f(x) = 2x³ – 13x² + 17x + 12
(i) f(2) = 2(2)³ – 13(2)² + 17(2) + 12
= 2 x 8-13 x 4+17 x 2+12
= 16-52 + 34 + 12
= 62 – 52
= 10
(ii) f(-3) = 2(-3)³ – 13(-3)² + 17 x (-3) + 12
= 2 x (-27) – 13 x 9 + 17 x (-3) + 12
= -54 – 117 -51 + 12
= -222 + 12
= -210
(iii) f(0) = 2 x (0)³ – 13(0)² + 17 x 0 + 12
= 0-0 + 0+ 12 = 12

Question 2.
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases: [NCERT]
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q2.1
Solution:

Question 3.
If x = 2 is a root of the polynomial f(x) = 2x²-3x + la, find the value of a.
Solution:
p(x) = 2x² – 3x + 7a
∵ x = 2 is its zero, then
p(0) = 0
∴ p( 2) = 2(2)² – 3×2 + la = 0
⇒2 x 4-3 x2 + 7a = 0
⇒ 8 – 6 + 7o = 0
⇒2 + 7a = 0
⇒ 7a = -2 ⇒ a =\(\frac { -2 }{ 7 }\)
∴ Hence a = \(\frac { -2 }{ 7 }\)

Question 4.
If x = –\(\frac { 1 }{ 2 }\) is a zero of the polynomial p(x) = 8x³ – ax² – x + 2, find the value of a.
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q4.1

Question 5.
If x = 0 and x = -1 are the roots of the polynomial f(x) = 2x³ – 3x² + ax + b, find the value of a and b.
Solution:
f(x) = 2x³ – 3x² + ax + b
∵ x = 0 and x = -1 are its zeros
∴ f(0) = 0 and f(-1) = 0
Now, f(0) = 0
⇒ 2(0)³ – 3(0)² + a x 0 + b = 0
⇒ 0-0 + 0 + b= 0
∴ b = 0
and f(-1) = 0
⇒ 2(-1)³ – 3(-1)² + a(-1) + b = 0
⇒ 2 x (-1) – 3 x 1 + a x (-1) + b = 0
⇒ -2 -3-a + b = 0
⇒ -2-3-a + 0 = 0
⇒ -5- a = 0=>a =-5
Hence a = -5, b = 0

Question 6.
Find the integral roots of the polynomial f(x) = x³ + 6x² + 11x + 6.
Solution:
f(x) = x³ + 6x² + 11x + 6
Construct = 6 = ±1, ±2, +3, ±6
If x = 1, then
f(1) = (1)³ + 6(1)² + 11 x 1 + 6
= 1+ 6+11+ 6 = 24
∵ f(x) ≠ 0, +0
∴ x = 1 is not its zero
Similarly, f(-1) = (-1)³ + 6(-1)² + 11(-1) + 6
= -1 + 6 x 1-11+6
=-1+6-11+6
= 12-12 = 0
∴ x = -1 is its zero
f(-2) = (-2)³ + 6(-2)² + 11 (-2) + 6
= -8 + 24 – 22 + 6
= -30 + 30 = 0
∴ x = -2 is its zero
f(-3) = (-3)³ + 6(-3)² + 11 (-3) + 6
= -27 + 54 – 33 + 6 = 60 – 60 = 0
∴ x = -3 is its zero
x = -1, -2, -3 are zeros of f(x)
Hence roots of f(x) are -1, -2, -3

Question 7.
Find the rational roots of the polynomial f(x) = 2x³ + x² – 7x – 6.
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q7.1

Hope given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 are helpful to complete your math homework.

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NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules

January 19, 2023

NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules

These Solutions are part of NCERT Exemplar Solutions for Class 9 Science . Here we have given NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules

Question 1.
Which of the following correctly represents 360 g of water ?
(i) 2 moles of H₂O
(ii) 20 moles of water
(iii) 6.022 x 10²³ molecules of water
(iv) 1.2044 x 10²⁵ molecules of water
(a) (i)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv).
Mass of water
Correct Answer:
(d)
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 1

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Question 2.
Which of the following statements is not true about atoms ?
(a) Atoms are not able to exist independently
(b) Atoms are the basic units from which molecules and ions are formed
(c) Atoms are always neutral in nature
(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch.
Correct Answer:
(a) Atoms of inert gas elements can exist independently. However, atoms of all other elements cannot exist independently.

Question 3.
The chemical symbol for nitrogen gas is :
(a) Ni
(b) N₂
(c) N⁺
(d) N
Correct Answer:
(b).

Question 4.
The chemical symbol for sodium is :
(a) So
(b) Sd
(c) NA
(d) Na.
Correct Answer:
(d ).

Question 5.
Which of the following would weigh the maximum ?
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 2
Correct Answer:
(c).
(a) 0.2 x 342 g = 68.4 g
(b) 2 x 44 g = 88 g
(c) 2 x 100 g = 200 g
(d) 10 x 18 g = 180 g.

Question 6.
Which of the following has maximum number of atoms ?
(a) 18 g of H₂O
(b) 18 g of O₂
(c) 18 g of CO₂
(d) 18 g of CH₄.
Correct Answer:
(d).
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 3

Question 7.
Which of the following contains maximum number of molecules ?
(a) 1g CO₂
(b) 1g N₂
(c) 1g H₂
(d) 1g CH₄.
Correct Answer:
(c).
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 4

Question 8.
Mass of one atom of oxygen is :
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 5
Correct Answer:
(a).

Question 9.
3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution are
(a) 6.68 x 10²³
(b) 6.09 x 102²²(c) 6.022 x 10²³
(d) 6.022 x 10²¹.
Correct Answer:
(a).
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 6

Question 10.
A change in the physical state can be brought about :
(a) only when energy is given to the system
(b) only when energy is taken out from the system
(c) when energy is either given to, or taken out from the system
(d) without any energy change.
Correct Answer:
(c).

Short Answer Questions

Question 11.
Which of the following represents a correct chemical formula ? Name it.
(a) CaCl
(b) BiPO₄
(c) NaSO₄
(d) NaS.
Answer:
(b). Both Bi and PO₄ are trivalent ion (Bi³⁺ and PO₄^3- ).
It is known as bismuth phosphate.

Question 12.
Write the chemical formulae for the following compounds :
(a) Copper (II) bromide
(b) Aluminium (III) nitrate
(c) Calcium (II) phosphate
(d) Iron (III) sulphide
(e) Mercury (II) chloride
(f) Magnesium (II) acetate.
Answer:
(a) CuBr₂
(b) Al(NO₃)₃
(c) Ca₃(PO₄)₂
(d) Fe₂S₃
(e) HgCl₂
(f) Mg(CH₂COO)₂.

Question 13.
Write the chemical formulae of all the compounds that can be formed by the combination of following ions :
Cu²⁺, Na⁺, Fe³⁺, Cl^–, SO₄^2-, PO₄^3- .
Answer:
For Cu²⁺ ion : CuCl₂, CuSO₄, Cu₃(PO₄)₂
For Na⁺ ion : NaCl, Na₂SO₄, Na3PO₄
For Fe³⁺ um :FeCl₃, Fe₂(SO₄)₃, FePO₄.

Question 14.
Write the cations and anions present (if any) in the following compounds :
(a) CH₃COONa
(b) NaCl
(c) H₂
(d) NH₄NO₃.
Answer:
(a) CH₃O^– (anion), Na⁺ (cation)
(b) Na⁺ (cation), Cl^– (anion)
(c) Molecular compound (no ions)
(d) NH₄⁺(cation), NO₃^– (anion).

Question 15.
Give the formulae of the compounds formed from the following sets of elements :
(a) Calcium and fluorine
(b) Hydrogen and sulphur
(c) Nitrogen and hydrogen
(d) Carbon and chlorine
(e) Sodium and oxygen
(f) Carbon and oxygen.
Answer:
(a) CaF₂
(b) H₂S
(c) NH₃
(d) CCl₄
(e) Na₂O
(f) CO and CO₂.

Question 16.
Which of the following symbols of elements are incorrect ? Give their correct symbols :
(a) Cobalt Co
(b) Carbon C
(c) Aluminium Al
(d) Helium He
(e) Sodium So
Answer:
(a) Incorrect; correct symbol is Co
(b) Incorrect; correct symbol is C
(c) Incorrect; correct symbol is Al
(e) Incorrect; correct symbol is Na.

Question 17.
Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them.
(a) Ammonia
(b) Carbon monoxide
(c) Hydogen chloride
(d) Aluminium fluoride
(e) Magnesium sulphide.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 8

Question 18.
State the number of atoms present in each of the following chemical species :
(a) CO₃^2-
(b) PO₄^3-
(c) P₂O₅
(d) CO.
Answer:
(a) Four
(b) Five
(c) Seven
(d) Two.

Question 19.
What is the fraction of mass of water (H₂O) due to neutrons ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 9

Question 20.
Does the solubility of a substance change with temperature ? Explain with the help of an example.
Answer:
In most of the cases, the solubility of a substance in a particular solvent such as water increases with the rise in temperature. For example, more of sugar can dissolve in a particular volume of water by increasing the temperature. .

Question 21.
Classify each of the following on the basis of their atomicity.
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 10
Answer:
(a) 2
(b) 3
(c) 3
(d) 8
(e) 4
(f) 4
(g) 14
(h) 3
(i) 2
(j) 5
(k) 1
(l) 1.

Question 22.
You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting ?
Answer:
Dissolve both sugar and salt separately in water taken in two glass beakers. Pass electric current through both. In case, solution is conducting, it represents a salt dissolved in water. If it fails to conduct electricity, it represents a sugar solution.

Question 23.
Calculate the number of moles of magnesium present in a magnesium ribon weighing 12 g. Molar atomic mass of magnesium is 24 g mol^-1.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 11

Long Answer Questions

Question 24.
Verify by calculating that :
(a) 5 moles of CO₂ and 5 moles of H₂O do not have the same mass.
(b) 240 g of calcium and 240 g magnesium elements have mole ratio of 3 : 5.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 12

Question 25.
Find the ratio by mass of the combining elements in the following compounds :
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 13
Answer:

Question 26.
Calcium chloride when dissolved in water dissociates into its ions according to the following equation.

Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 16

Question 27.
The difference in the mass of 100 moles each of sodium atoms and sodium ions is 5.48002 g. Compute the mass of an electron.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 17

Question 28.
Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 18

Question 29.
The mass of one steel screw is 4.11 g. Find the mass of one mole of these steel screws. Compare this value with the mass of the Earth (5.98 x 10²⁴ kg). Which one of the two is heavier and by how many times ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 19

Question 30.
A sample of vitamin C is known to contain 2.58 x 10²⁴ oxygen atoms. How many moles of oxygen atoms are present in the sample ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 20

Question 31.
Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of same weight.
(a) Whose container is heavier ?
(b) Whose container has more number of atoms ?
Answer:
(a) Molar atomic mass of carbon = 12 g
No. of moles of carbon carried by Raunak = 5
Mass of 5 moles of carbon = (12 x 5) = 60 g
Molar atomic mass of sodium = 23 g
No. of moles of sodium carried by Krish = 5
Mass of 5 moles of sodium = (23 x 5) = 115 g
This shows that the container carried by Krish is heavier.
(b) Since both the containers have same number of moles, the number of atoms present in these containers are also same i.e. 5 x N_A atoms.

Question 32.
Fill in the missing data in the given table
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 21
Answer:

Question 33.
The visible universe is estimated to contain 10²² stars. How many moles of stars are present in the visible universe ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 23

Question 34.
What is the SI prefix for each of the following multiples and submultiples of a unit ?
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 42
Answer:
(a) kilo
(b) deci
(c) centri
(d) micro
(e) nano
(f) pico

Question 35.
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 25
Answer:

Question 36.
Compute the difference in masses of 10³ moles each of magnesium atoms and magnesium ions.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 27

Question 37.
Which has more number of atoms ?
100 g of N₂ or 100 g of NH₃
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 28

Question 38.
Compare the number of ions present in 5.85 g of sodium chloride.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 30

Question 39:
A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 31

Question 40.
What are ionic and molecular compounds ? Give examples.
Answer:
Simple ionic compounds are of binary nature. It means that both the positive and negative ions have one atom only. The symbols of these ions are written side by side with their valencies at their bottom. A common factor if any, is removed to get a simple ratio of the valencies of the combining atoms. The criss-cross method is then applied to arrive at the final chemical formula of the compound. Let us write the formulae of a few simple ionic compounds. For example,
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 32
The molecular compounds also called covalent compounds. The compounds listed above are heteroatomic in nature. It means that different elements participate in these compounds. They may be homoatomic also which means that these are formed from the atoms of the same element. For example, hydrogen molecule (H₂), chlorine molecule (Cl₂), oxygen molecule (0₂), nitrogen molecule (N₂), etc.
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 33

Question 41.
Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions. (Mass of an electron is 9.1 x 10^-28 g). Which one is heavier ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 34

Question 42.
A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 35

Question 43.
A sample of ethane (C₂H₆) gas has the same mass as 1.5 x 10²⁰ molecules of methane (CH₄). How many C₂H₆ molecules does the sample of gas contain ?
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 37

Question 44.
Fill in the blanks
(a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called ………….. .
(b) A group of atoms carrying a fixed charge on them is called ………….. .
(c) The formula unit mass of Ca₃(PO₄)₂ is ………….. .
(d) Formula of sodium carbonate is ………… and that of ammonium sulphate is …………… .
Answer:
(a) Law of conservation of mass
(b) Polyatomic ions
(c) 3 x Atomic mass of Ca + 2 x Atomic mass of P + 8 x Atomic mass of O.
3 x 40 + 2 x 31 + 8 x 16 = 120 + 62 + 128 = 310 u
(d) Na₂CO₃, (NH₄)₂SO₄.

Question 45.
Complete the following crossword puzzle by using the name of the chemical elements. Use the data given in table.
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 38

Answer:

Question 46.
Write the formulae for the following and calculate the molecular mass for each one of them.
(a) Custic potash
(b) Baking powder
(c) Lime stone
(d) Caustic soda
(e) Ethanol
(f) Common salt.
Answer:
(a) KOH = 39 + 16 + 1 = 56 u
(b) NaHCO₃ = 23 + 1 + 12 + 3 x 16 = 84 u
(c) CaCO₃ = 12 + 12 x 3 x 16 = 100 u
(d) NaOH = 23 + 16 + 1 = 40 u
(e) C₂H₅OH = 2 x 12 + 6 x 1 + 16 = 46 u
(f) NaCl = 23 + 35.5 = 58.5 u

Question 47.
In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex sereies of reactions to give a molecule of glucose having a molecular formula C₆H₁₂O₆ and a molecule of oxygen with molecular formula O₂. How many grams of water would be required to produce 18 g of glucose ? Compute the volume of water so consumed assuming the density of water to be 1 g cm^-3.
Answer:
NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules image - 41

Hope given NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules are helpful to complete your science homework.

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RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS

January 19, 2023

RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS

Other Exercises

Question 1.
How many least number of distinct points determine a unique line?
Solution:
At least two distinct points determine a unique line.

Question 2.
How many lines can be drawn through both of the given points?
Solution:
Through two given points, one line can be drawn.

Question 3.
How many lines can be drawn through a given point?
Solution:
Through a given point, infinitely many lines can be drawn.

Question 4.
in how many points two distinct lines can intersect?
Solution:
Two distinct lines can intersect at the most one point.

Question 5.
In how many points a line, not in a plane, can intersect the plane?
Solution:
A line not in a plane, can intersect the plane at one point.

Question 6.
In how many points two distinct planes can intersect?
Solution:
Two distinct planes can intersect each other at infinite number of points.

Question 7.
In how many lines two distinct planes can intersect?
Solution:
Two distinct planes intersect each other in one line.

Question 8.
How many least number of distinct points determine a unique plane?
Solution:
Three non-collinear points can determine a unique plane.

Question 9.
Given three distinct points in a plane, how many lines can be drawn by joining them?
Solution:
Through three given points, one line can be drawn of they are collinear and three if they are non-collinear.

Question 10.
How many planes can be made to pass through a line and a point not on the line?
Solution:
Only one plane can be made to pass through a line and a point not on the line.

Question 11.
How many planes can be made to pass through two points?
Solution:
Infinite number of planes can be made to pass through two points.

Question 12.
How many planes can be made to pass through three distinct points?
Solution:
Infinite number of planes can be made of they are collinear and only one plane, if they are non-collinear.

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RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C

January 19, 2023

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C.

Other Exercises

Question 1.
Solution:
It is clear that the given frequency distribution is in exclusive form, we represent the daily wages (in rupees) along x-axis and no. of workers along y-axis. Then we construct rectangles with class intervals as bases and corresponding frequencies as heights as shown given below:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q1.1

Question 2.
Solution:
We shall take daily earnings along x-axis and number of stores along y-axis. Then we construct rectangles with the given class intervals as bases and corresponding frequency as height as shown in the graph.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q2.1

Question 3.
Solution:
We shall take heights along x-axis and number of students along y-axis. Then we shall complete the rectangles with the given class intervals as bases and frequency as height and complete the histogram as shown below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q3.1

Question 4.
Solution:
We shall take class intervals along x-axis and frequency along y-axis Then we shall complete the rectangles with given class interval as bases and frequency as heights and complete the histogram as shown below.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q4.1

Question 5.
Solution:
The given frequency distribution is in inclusive form first we convert it into exclusive form at given below.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q5.1

Now, we shall take class intervals along x-axis and frequency along y-axis and draw rectangles with class intervals as bases and frequency as heights and complete the histogram as shown.

Question 6.
Solution:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q6.1

Question 7.
Solution:
In the given distribution class intervals are different size. So, we shall calculate the adjusted frequency for each class. Here minimum class size is 4. We know that adjusted
frequency of the class is \(\frac { max\quad class\quad size }{ class\quad size\quad of\quad this\quad class } \) x its frequency
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q7.1

Question 8.
Solution:
We shall take two imagined classes one 0-10 at the beginning with zero frequency a id other 70-80 at the end with zero frequency.
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q8.1

Now, plot the points” (5, 0), (15, 2), (25, 5), (35, 12), (45, 19), (55, 9), (65, 4), (75, 0) on the graph and join them its order to get a polygon as shown below.

Question 9.
Solution:
We take Age (in years) along x-axis and number of patients along y-axis. First we complete the histogram and them join the midpoints of their tops in order
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q9.1
We also take two imagined classes. One 0-10 at the beginning and other 70-80 at the end and also join their midpoints to complete the polygon as shown.

Question 10.
Solution:
We take class intervals along x-axis and frequency along y-axis.
First we complete the histogram and then join the midpoints of the tops of adjacent rectangles in order. We shall take two imagined classes 15-20 at the beginning and 50-55 at the end.
We also join their midpoints to complete the polygon as shown
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q10.1

Question 11.
Solution:
We represent class interval on x-axis and frequency on y-axis. First we construct the histogram and then join the midpoints of the tops of each rectangle by line segments. We shall take two imagined class i.e. 560-600 at the beginning and 840-900 at the end. Join their midpoints also to complete
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q11.1

Question 12.
Solution:
We will take the imagined class 11-0 at the beginning and 61-70 at the end. Each with frequency zero. Thus, we have,
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C Q12.1

Now we will mark the points A (-5.5, 0), B(5.5, 8), C(15.5, 3) D(25.5, 6), E(35.5, 12), F(45.5, 2), G(55.5, 7) and H(65.5, 0) on the graph and join them in order to get the required frequency polygon.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D

January 19, 2023

RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D.

Other Exercises

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C
RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D

Question 1.
Solution:
In ∆ABC,
∠B = 76° and ∠C = 48°
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> ∠A + 76° + 48° = 180°
=> ∠ A + 124° = 180°
=> ∠A= 180° – 124° = 56°

Question 2.
Solution:
Angles of a triangle are in the ratio = 2:3:4
Let first angle = 2x
then second angle = 3x
and third angle = 4x
2x + 3x + 4x = 180°
(Sum of angles of a triangle)
=> 9x = 180°
=> x = \(\frac { { 180 }^{ o } }{ 9 } \) = 20°
First angle = 2x = 2 x 20° = 40°
Second angle = 3x = 3 x 20° = 60°
and third angle = 4x = 4 x 20° = 80° Ans.

Question 3.
Solution:
In ∆ABC,
3∠A = 4∠B = 6∠C = x (Suppose)
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q3.1

Question 4.
Solution:
In ∆ABC,
∠ A + ∠B = 108° …(i)
∠B + ∠C – 130° …(ii)
But ∠A + ∠B + ∠C = 180° …(iii)
(sum of angles of a triangle)
Subtracting (i) from (iii),
∠C = 180° – 108° = 72°
Subtracting (ii) from (iii),
∠A = 180°- 130° = 50°
But ∠ A + ∠B = 108° (from i)
50° + ∠B = 108°
=> ∠B = 108° – 50° = 58°
Hence ∠A = 50°, ∠B = 58° and ∠C = 72° Ans.

Question 5.
Solution:
In ∆ABC,
∠A+∠B = 125° …(i)
∠A + ∠C = 113° …(ii)
But ∠A + ∠B + ∠C = 180° …(iii)
(sum of angles of a triangles) Subtracting, (i), from (iii),
∠C = 180°- 125° = 55°
Subtracting (ii) from (iii),
∠B = 180°- 113° – 67°
∠A + ∠B = 125°
∠ A + 67° = 125°
=> ∠ A = 125° – 67°
∠A = 58°
Hence ∠A = 58°, ∠B = 67° and ∠ C = 55° Ans.

Question 6.
Solution:
In ∆ PQR,
∠ P – ∠ Q = 42°
=> ∠P = 42°+∠Q …(i)
∠Q – ∠R = 21°
∠Q – 21°=∠R …(ii)
But ∠P + ∠Q + ∠R = 180°
(Sum of angles of a triangles)
42° + ∠Q + ∠Q + ∠Q – 21°= 180°
=> 21° + 3∠Q = 180°
=> 3∠Q = 180°- 21° = 159°
from ∠Q = \(\frac { { 159 }^{ o } }{ 3 } \) = 53°
(i)∠P = 42° + ∠Q = 42° + 53° = 95°
and from (ii) ∠R = ∠Q – 21°
= 53° – 25° = 32°
Hence ∠P = 95°, ∠Q = 53° and ∠R = 32° Ans.

Question 7.
Solution:
Let ∠ A, ∠ B and ∠ C are the three angles of A ABC.
and ∠A + ∠B = 116° …(i)
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q7.1

Question 8.
Solution:
Let ∠ A, ∠ B and ∠ C are the three angles of the ∆ ABC
Let ∠ A = ∠ B = x
then ∠C = x + 48°
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
x + x + x + 18° = 180°
=> 3x + 18° = 180°
=> 3x = 180° – 18° = 162°
x = \(\frac { { 162 }^{ o } }{ 3 } \) = 54°
∠A = 54°, ∠B = 54° and ∠C = 54° + 18° = 72°
Hence angles are 54°, 54 and 72° Ans.

Question 9.
Solution:
Let the smallest angle of a triangle = x°
their second angle = 2x°
and third angle = 3x°
But sum of angle of a triangle = 180°
x + 2x + 3x = 180°
=> 6x = 180°
=> x – \(\frac { { 180 }^{ o } }{ 6 } \) = 30°
Hence smallest angle = 30°
Second angle = 2 x 30° = 60°
and third angle = 3 x 30° = 90° Ans.

Question 10.
Solution:
In a right angled triangle.
one angle is = 90°
Sum of other two acute angles = 90°
But one acute angle = 53°
Second acute angle = 90° – 53° = 37°
Hence angle of the triangle with be 90°, 53°, 37° Ans.

Question 11.
Solution:
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q11.1
Given : In ∆ ABC,
∠ A = ∠B + ∠C
To Prove : ∆ABC is a right-angled
Proof : We know that in ∆ABC,
∠A + ∠B + ∠C = 180°
(angles of a triangle)
But ∠ A = ∠ B + C given
∠A + (∠B + ∠C) = 180°
=> ∠A + ∠A = 180°
=> 2∠A = 180°
=> ∠ A = \(\frac { { 180 }^{ o } }{ 2 } \) = 90°
∠ A = 90°
Hence ∆ ABC is a right-angled Hence proved.

Question 12.
Solution:
Given. In ∆ ABC, ∠A = 90°
AL ⊥ BC.
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q12.1
To Prove : ∠BAL = ∠ACB
Proof : In ∆ ABC, AL ⊥ BC
In right angled ∆ALC,
∠ ACB + ∠ CAL = 90° …(i)
( ∴∠L = 90°)
But ∠ A = 90° ‘
=> ∠ BAL + ∠ CAL = 90° …(ii)
From (i) and (ii),
∠BAL + ∠ CAL= ∠ ACB+ ∠CAL
=> ∠ BAL = ∠ ACB Hence proved.

Question 13.
Solution:
Given. In ∆ABC,
Each angle is less than the sum of the other two angles
∠A< ∠B + ∠C
∠B < ∠C + ∠A
and ∠C< ∠A + ∠C
Proof : ∠ A < ∠B + ∠C
Adding ∠ A both sides,
∠A + ∠A < ∠A + ∠B + ∠C => 2 ∠ A < 180°
(∴ ∠A+∠B+∠C=180°)
∠A < \(\frac { { 180 }^{ o } }{ 2 } \) => ∠A< 90
Similarly, we can prove that,
∠B < 90° and ∠C < 90°
∴ each angle is less than 90°
Hence, triangle is an acute angled triangle. Hence proved.

Question 14.
Solution:
Given. In ∆ABC,
∠B > ∠A + ∠C
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q14.1

Question 15.
Solution:
In ∆ABC
∠ ABC = 43° and Ext. ∠ ACD = 128°
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q15.1

Question 16.
Solution:
∠ ABC + ∠ ABD = 180°
(Linear pair)
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q16.1

Question 17.
Solution:
(i)In the figure, ∠BAE =110° and ∠ACD = 120°.
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q17.1

Question 18.
Solution:
In the figure,
∠A = 55°, ∠B = 45°, ∠C = 30° Join AD and produce it to E
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q18.1

Question 19.
Solution:
In the figure,
∠EAC = 108°,
AD divides ∠ BAC in the ratio 1 : 3
and AD = DB
∠EAC + ∠ BAC = 180°
(Linear pair)
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q19.1

Question 20.

Solution:
Sides BC, CA and AB
are produced in order forming exterior
angles ∠ ACD,
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q20.1

Question 21.
Solution:
Given : Two ∆ s DFB and ACF intersect each other as shown in the figure.
To Prove : ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°
Proof : In ∆ DFB,
∠D + ∠F + ∠B = 180°
(sum of angles of a triangle)
Similarly,in ∆ ACE
∠A + ∠C + ∠E = 180° …(ii)
Adding (i) and (ii), we get :
∠D + ∠F + ∠B+ ∠A+ ∠C + ∠E = 180° + 180°
=> ∠A+∠B+∠C+∠D+∠E + ∠ F = 360°
Hence proved.

Question 22.
Solution:
In the figure,
ABC is a triangle
and OB and OC are the angle
bisectors of ∠ B and ∠ C meeting each other at O.
∠ A = 70°
In ∆ ABC,
∠A + ∠B + ∠C = 180°
(sum of angles of a triangle)
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q22.1

Question 23.
Solution:
In ∆ABC, ∠ A = 40°
Sides AB and AC are produced forming exterior angles ∠ CBD and ∠ BCE
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q23.1

Question 24.
Solution:
In the figure, ∆ABC is triangle and ∠A : ∠B : ∠C = 3 : 2 : 1
AC ⊥ CD.
∠ A + ∠B + ∠C = 180°
(sum of angles of a triangle)
But ∠A : ∠B : ∠C = 3 : 2 : 1
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q24.1

Question 25.
Solution:
In ∆ ABC
AN is the bisector of ∠ A
∠NAB =\(\frac { 1 }{ 2 } \) ∠A.
Now in right angled ∆ AMB,
∠B + ∠MAB = 90° (∠M = 90°)
RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D Q25.1

Question 26.
Solution:
(i) False: As a triangle has only one right angle
(ii) True: If two angles will be obtuse, then the third angle will not exist.
(iii) False: As an acute-angled triangle all the three angles are acute.
(iv) False: As if each angle will be less than 60°, then their sum will be less than 60° x 3 = 180°, which is not true.
(v) True: As the sum of three angles will be 60° x 3 = 180°, which is true.
(vi) True: A triangle can be possible if the sum of its angles is 180°
But the given triangle having angles 10° + 80° + 100° = 190° is not possible.

Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D are helpful to complete your math homework.

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RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C

January 19, 2023

RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C.

Other Exercises

Question 1.
Solution:
Given : In trapezium ABCD,
AB || DC and E is the midpoint of AD.
A line EF ||AB is drawn meeting BC at F.
To prove : F is midpoint of BC
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C Q1.1

Question 2.
Solution:
Given : In ||gm ABCD, E and F are the mid points of AB and CD respectively. A line segment GH is drawn which intersects AD, EF and BC at G, P and H respectively.
To prove : GP = PH
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C Q2.1

Question 3.
Solution:
Given : In trapezium ABCD, AB || DC
P, Q are the midpoints of sides AD and BC respectively
DQ is joined and produced to meet AB produced at E
Join AC which intersects PQ at R.
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C Q3.1

Question 4.
Solution:
Given : In ∆ ABC,
AD is the mid point of BC
DE || AB is drawn. BE is joined.
To prove : BE is the median of ∆ ABC.
Proof : In ∆ ABC
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C Q4.1

Question 5.
Solution:
Given : In ∆ ABC, AD and BE are the medians. DF || BE is drawn meeting AC at F.
To prove : CF = \(\frac { 1 }{ 4 } \) BC.
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C Q5.1

Question 6.
Solution:
Given : In ||gm ABCD, E is mid point of DC.
EB is joined and through D, DEG || EB is drawn which meets CB produced at G and cuts AB at F.
To prove : (i)AD = \(\frac { 1 }{ 2 }\) GC
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C Q6.1

Question 7.
Solution:
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C Q7.1
Given : In ∆ ABC,
D, E and F are the mid points of sides BC, CA and AB respectively
DE, EF and FD are joined.

Question 8.
Solution:
Given : In ∆ ABC, D, E and F are the mid points of sides BC, CA and AB respectively
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C Q8.1

Question 9.
Solution:
Given : In rectangle ABCD, P, Q, R and S are the midpoints of its sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined.
To prove : PQRS is a rhombus.
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C Q9.1

Question 10.
Solution:
Given : In rhombus ABCD, P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined.
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C Q10.1

Question 11.
Solution:
Given : In square ABCD, P,Q,R and S are the mid points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C Q11.1

Question 12.
Solution:
Given : In quadrilateral ABCD, P, Q, R and S are the midpoints of PQ, QR, RS and SP respectively PR and QS are joined.
To prove : PR and QS bisect each other
Const. Join PQ, QR, RS and SP and AC
Proof : In ∆ ABC,
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C Q12.1

But diagonals of a ||gm bisect each other PR and QS bisect each other.
∴ PR and QS bisect each other

Question 13.
Solution:
Given : ABCD is a quadrilateral. Whose diagonals AC and BD intersect each other at O at right angles.
P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively. PQ, QR, QS and SP are joined.
RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C Q13.1

Hope given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS

January 19, 2023

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS

Other Exercises

Question 1.
Write the equation representing x-axis.
Solution:
The equation of x-axis is, y = 0.

Question 2.
Write the equation representing y-axis.
Solution:
The equation of y-axis is, x = 0.

Question 3.
Write the equation of a line passing through the point (0, 4) and parallel to x-axis.
Solution:
The equation of the line passing through the point (0,4) and parallel to x-axis will be y = 4.

Question 4.
Write the equation of a line passing through the point (3, 5) and parallel to x-axis.
Solution:
The equation of the line passing through the point (3, 5) and parallel to x-axis will be y = 5.

Question 5.
Write the equation of a line parallel toy-axis and passing through the point (-3, -7).
Solution:
The equations of the line passing through the point (-3, -7) and parallel to y-axis will be x = -3.

Question 6.
A line passes through the point (-4, 6) and is parallel to x-axis. Find its equation. A line passes through the point (-4, 6) and is parallel to x-axis. Find its equation.
Solution:
A line parallel to x-axis and passing through the point (-4, 6) will be y = 6.

Question 7.
Solve the equation 3x – 2 = 2x + 3 and represent the solution on the number line.
Solution:
3x – 2 = 2x + 3
⇒ 3x – 2x = 3 + 2 (By terms formation)
⇒ x = 5
∴ x = 5
Solution on the number line is
RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS Q7.1

Question 8.
Solve the equation 2y – 1 = y + 1 and represent it graphically on the coordinate plane.

Solution:
2y – 1 = y + 1
⇒ 2y – y = 1 +1
⇒ y = 2
∴ It is a line parallel to x-axis at a distance of 2 units above the x-axis is y = 2.

Question 9.
If the point (a, 2) lies on the graph of the linear equation 2x – 3y + 8 = 0, find the value of a.
Solution:
∵ Points (a, 2) lies on the equation
2x – 3y + 8 = 0
∴ It will satisfy the equation,
Now substituting the value of x = a, y = 2 in the equation
⇒ 2a – 3 x 2+ 8= 0
⇒ 2a + 2= 0
⇒ 2a = -2
⇒ a = \(\frac { -2 }{ 2 }\) = -1
∴ a = -1

Question 10.
Find the value of k for which the point (1, -2) lies on the graph of the linear equation, x – 2y + k = 0.
Solution:
∵ Point (1, -2) lies on the graph of the equation x – 2y + k = 0
∴ x = 1, y = -2 will satisfy the equation
Now substituting the value of x = 1, y = -2 in it
1-2 (-2) + k = 0
⇒ 1 + 4 + k = 0
⇒ 5+ k = 0 ⇒ k =-5
∴ k = -5

Hope given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A

January 19, 2023

RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A.

Question 1.
Solution:
Base of the triangle (b) = 24cm and height (h) = 14.5 cm
∴ Area = \(\frac { 1 }{ 2 } \) x b x h = \(\frac { 1 }{ 2 } \) x 24 x 14.5 cm²
= 174 cm² Ans.

Question 2.
Solution:
Let the length of altitude of the triangular field = x then its base = 3x.
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q2.1

Question 3.
Solution:
Sides of a triangle = 42cm, 34cm and 20cm
Let a = 42cm, b = 34cm and c = 20 cm
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q3.1

Question 4.
Solution:
Sides of the triangle = 18cm, 24cm and 30cm
Let a = 18 cm, b = 24 cm and c = 30cm
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q4.1

Question 5.
Solution:
Sides of triangular field ABC arc 91m, 98m and 105m
Let AC be the longest side
∴ BD⊥AC
Here a = 98m, b = 105m and c = 91m
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q5.1

Question 6.
Solution:
Perimeter of triangle = 150m
Ratio in the sides = 5:12:13
Let sides be 5x, 12x and 13x
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q6.1

Question 7.
Solution:
Perimeter of a triangular field = 540m
Ratio is its sides = 25 : 17 : 12
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q7.1

Question 8.
Solution:
Perimeter of the triangular field = 324 m
Length of the sides are 85m and 154m
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q8.1

Question 9.
Solution:
Length of sides are
13 cm, 13 cm and 20cm
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q9.1

Question 10.
Solution:
Base of the isosceles triangle ABC = 80cm
Area = 360 cm²
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q10.1

Question 11.
Solution:
Perimeter of the triangle
ABC = 42 cm.
Let length of each equal sides = x
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q11.1

Question 12.
Solution:
Area of equilateral triangle = 36√3 cm².
Let length of each side = a
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q12.1

Question 13.
Solution:
Area of equilateral triangle = 81√3 cm²
Let length of each side = a
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q13.1

Question 14.
Solution:
∆ ABC is a right angled triangle, right angle at B.
∴ BC 48cm and AC = 50cm
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q14.1

Question 15.
Solution:
Each side of equilateral triangle
(a) = 8cm.
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q15.1

Question 16.
Solution:
Let a be the each side of
the equilateral triangle.
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q16.1

Question 17.
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q17.1
Solution:
The given umbrella has 12 triangular pieces of the size 50cm x 20cm x 50cm. We see that each piece is of an isosceles triangle shape and we have to find firstly area of one such triangle.

Question 18.
Solution:
The given floral design is made of 16 tiles
The size of each tile is 16cm 12cm, 20cm
Now we have to find the area of firstly one tile
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q18.1

Question 19.
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q19.1
Solution:

Question 20.
Solution:
In the figure, ABCD is a quadrilateral
AB = 42 cm, BC = 21 cm, CD = 29 cm,
DA = 34 cm and ∠CBD = 90°
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q20.1

Question 21.
Solution:
from the figure
∆DAB
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q21.1

Question 22.
Solution:
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q22.1

Question 23.
Solution:
from the figure,
We know that
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q23.1

Question 24.
Solution:
RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A Q24.1

Hope given RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2

January 19, 2023

RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2

Other Exercises

Question 1.
In the figure, OA and OB are opposite rays:
(i) If x = 25°, what is the value of y?
(ii) If y = 35°, what is the value of x?
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q1.1
Solution:
(i) If x = 25°
∴ 3x = 3 x 25° = 75°
But ∠AOC + ∠BOC = 180° (Linear pair)
⇒ ∠AOC + 75° = 180°
⇒ ∠AOC = 180° – 75°
⇒ ∠AOC = 105°
∴ 2y + 5 = 105° ⇒ 2y= 105° – 5° = 100°
⇒ y = \(\frac { { 100 }^{ \circ } }{ 2 }\) = 50°
∴ If x = 25° then y = 50°
(ii) If y = 35°, then ∠AOC = 2y + 5
∴ 2y + 5 = 2 x 35° + 5 = 70° + 5 = 75°
But ∠AOC + ∠BOC = 180° (Linear pair)
⇒ 75° + ∠BOC = 180°
⇒ ∠BOC = 180°-75°= 105°
∴ 3x = 105° ⇒ x = \(\frac { { 105 }^{ \circ } }{ 3 }\) = 35°
∴ x = 35°

Question 2.
In the figure, write all pairs of adjacent angles and all the linear pairs.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q2.1
Solution:
In the given figure,
(i) ∠AOD, ∠COD; ∠BOC, ∠COD; ∠AOD, ∠BOD and ∠AOC, ∠BOC are the pairs of adjacent angles.
(ii) ∠AOD, ∠BOD and ∠AOC, ∠BOC are the pairs of linear pairs.

Question 3.
In the figure, find x, further find ∠BOC, ∠COD and ∠AOD.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q3.1
Solution:
In the figure,
AOB is a straight line
∴ ∠AOD + ∠DOB = 180° (Linear pair)
⇒ ∠AOD + ∠DOC + ∠COB = 180°
⇒ x+ 10° + x + x + 20 = 180°
⇒ 3x + 30° = 180°
⇒ 3x= 180° -30°= 150°
⇒ x = \(\frac { { 150 }^{ \circ } }{ 3 }\) = 50°
∴ x = 50°
Now ∠BOC =x + 20° = 50° + 20° = 70°
∠COD = x = 50°
and ∠AOD = x + 10° = 50° + 10° = 60°

Question 4.
In the figure, rays OA, OB, OC, OD and OE have the common end point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q4.1
Solution:
Produce AO to F such that AOF is a straight line
Now ∠AOB + ∠BOF = 180° (Linear pair)

⇒ ∠AOB + ∠BOC + ∠COF = 180° …(i)
Similarly, ∠AOE + ∠EOF = 180°
⇒ ∠AOE + ∠EOD + ∠DOF = 180° …(ii)
Adding (i) and (ii)
∠AOB + ∠BOC + ∠COF + ∠DOF + ∠EOD + ∠AOE = 180° + 180°
⇒ ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°
Hence proved.

Question 5.
In the figure, ∠AOC and ∠BOC form a linear pair. If a – 2b = 30°, find a and b.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q5.1
Solution:
∠AOC + ∠BOC = 180° (Linear pair)
⇒ a + 6 = 180° …(i)
and a – 2b = 30° …(ii)
Subtracting (ii) from (i), 3b = 150°
⇒ b = \(\frac { { 150 }^{ \circ } }{ 3 }\) = 50°
and a + 50° = 180°
⇒ a= 180°-50°= 130°
∴ a = 130°, b = 50°

Question 6.
How many pairs of adjacent angles are formed when two lines intersect in a point?
Solution:
If two lines AB and CD intersect each other at a point O, then four pairs of linear pairs are formed i.e.
∠AOC, ∠BOC; ∠BOC, ∠BOD; ∠BOD, ∠AOD and ∠AOD, ∠AOC
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q6.1

Question 7.
How many pairs of adjacent angles, in all, can you name in the figure.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q7.1
Solution:
In the given figure 10 pairs of adjacent angles are formed as given below:
∠AOB, ∠BOC; ∠AOB, ∠BOD; ∠AOC, ∠COD; ∠AOD, ∠BOE; ∠AOB, ∠BOE; ∠AOC, ∠COF; ∠BOC, ∠COD; ∠BOC, ∠COE; ∠COD, ∠DOE and ∠BOD, ∠DOE

Question 8.
In the figure, determine the value of x.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q8.1
Solution:
In the figure
∠AOC + ∠COB + ∠BOD +∠AOD = 360 (angles at a point )

⇒ 3x + 3x + x + 150° = 360°
⇒ 7x – 360° – 150° = 210°
⇒ x = \(\frac { { 210 }^{ \circ } }{ 7 }\) = 30°
∴ x = 30°

Question 9.
In the figure, AOC is a line, find x.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q9.1
Solution:
In the figure,
∠AOB + ∠BOC = 180° (Linear pair)
⇒ 70° + 2x = 180°
⇒ 2x = 180° – 70°
⇒ 2x = 110°⇒x = \(\frac { { 110 }^{ \circ } }{ 2 }\) = 55°
∴ x = 55°

Question 10.
In the figure, POS is a line, find x.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q10.1
Solution:
In the figure, POS is a line
∴ ∠POQ + ∠QOS = 180° (Linear pair)
⇒ ∠POQ + ∠QOR + ∠ROS = 180°
⇒ 60° + 4x + 40° = 180°
⇒ 4x + 100° – 180°
⇒ 4x = 180° – 100° = 80°
⇒ x = \(\frac { { 80 }^{ \circ } }{ 4 }\) =20°
∴ x = 20°

Question 11.
In the figure, ACB is a line such that ∠DCA = 5x and ∠DCB = 4x. Find the value of x.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q11.1
Solution:
ACB is a line, ∠DCA = 5x and ∠DCB = 4x
∠ACD + ∠DCB = 180° (Linear pair)
⇒ 5x + 4x = 180° ⇒ 9x = 180°
⇒ x = \(\frac { { 180 }^{ \circ } }{ 9 }\) = 20°
∴ x = 20°
∴ ∠ACD = 5x = 5 x 20° = 100° and ∠DCB = 4x = 4 x 20° = 80°

Question 12.
Given ∠POR = 3x and ∠QOR = 2x + 10°, find the value ofx for which POQ will be a line.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q12.1
Solution:
∠POR = 3x and ∠QOR = 2x + 10°
If POQ is a line, then
∠POR + ∠QOR = 180° (Linear pair)
⇒ 3x + 2x + 10° = 180°
⇒ 5x = 180° – 10° = 170°
⇒ x = \(\frac { { 170 }^{ \circ } }{ 5 }\) = 34°
∴ x = 34°

Question 13.
What value ofy would make AOB, a line in the figure, if ∠AOC = 4y and ∠BOC = (6y + 30).
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q13.1
Solution:
In the figure,
AOB is a line if
∠AOC + ∠BOC = 180°
⇒ 6y + 30° + 4y= 180°
⇒ 10y= 180°-30°= 150°
150°
⇒ y = \(\frac { { 150 }^{ \circ } }{ 10 }\) = 15°
∴ y = 15°

Question 14.
In the figure, OP, OQ, OR and OS are four rays. Prove that: ∠POQ + ∠QOR + ∠SOR + ∠POS = 360° [NCERT]
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q14.1
Solution:
In the figure, OP, OQ, OR and OS are the rays from O
To prove : ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°
Construction : Produce PO to E

Proof: ∠POQ + ∠QOE = 180° (Linear pair)
Similarly, ∠EOS + ∠POS = 180°
Adding we get,
∠POQ + ∠QOR + ∠ROE + ∠EOS + ∠POS = 180° + 180° ,
⇒ ∠POQ + ∠QOR + ∠ROS + ∠POS = 360° Hence ∠POQ + ∠QOR + ∠SOR + ∠POS = 360°

Question 15.
In the figure, ray OS stand on a line POQ. Ray OR and ray OT are angle bisectors of ∠POS and ∠SOQ respectively. If ∠POS = x, find ∠ROT. [NCERT]
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q15.1
Solution:
Ray OR stands on a line POQ forming ∠POS and ∠QOS

OR and OT the angle bisects of ∠POS and ∠QOS respectively. ∠POS = x
But ∠POS + ∠QOS = 180° (Linear pair)
⇒ x + ∠QOS = 180°
⇒ ∠QOS = 180° – x
∵ OR and OT are the bisectors of angle
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q15.3

Question 16.
In the figure, lines PQ and RS intersect each other at point O. If ∠POR : ∠ROQ = 5 : 7. Find all the angles. [NCERT]
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q16.1
Solution:
Lines PQ and PR, intersect each other at O
∵ Vertically opposite angles are equal
∴ ∠POR = ∠QOS and ∠ROQ = ∠POS
∠POR : ∠ROQ = 5:7
Let ∠POR = 5x and ∠ROQ = 7x
But ∠POR + ∠ROQ = 180° (Linear pair)
∴ 5x + 7x = 180° ⇒ 12x = 180°
⇒ x = \(\frac { { 180 }^{ \circ } }{ 12 }\) = 15°
∴ ∠POR = 5x = 5 x 15° = 75°
and ∠ROQ = 7x = 7 x 15° = 105°
But ∠QOS = POR = 75° (Vertically opposite angles)
and ∠POS = ∠ROQ = 105°

Question 17.
In the figure, a is greater than b by one third of a right-angle. Find the values of a and b.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q17.1
Solution:
In the figure,
∠AOC + ∠BOC = 180° (Linear pair)
⇒ a + b =180° …(i)
But a = b + \(\frac { 1 }{ 3 }\) x 90° = b + 30°
⇒ a – b = 30° …(ii)
Adding (i) and (ii)
210°
2a = 210° ⇒ a = \(\frac { { 210 }^{ \circ } }{ 2 }\) = 105°
and 105° + b = 180°
⇒ b = 180° – 105°
∴ b = 75°
Hence a = 105°, b = 75°

Question 18.
In the figure, ∠AOF and ∠FOG form a linear pair.
∠EOB = ∠FOC = 90° and ∠DOC = ∠FOG = ∠AOB = 30°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q18.1
(i) Find the measures of ∠FOE, ∠COB and ∠DOE.
(ii) Name all the right angles.
(iii) Name three pairs of adjacent complementary angles.
(iv) Name three pairs of adjacent supplementary angles.
(v) Name three pairs of adjacent angles.
Solution:
In the figure,
∠AOF and ∠FOG form a linear pair
∠EOB = ∠FOC = 90°
∠DOC = ∠FOG = ∠AOB = 30°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q18.2
(i) ∠BOE = 90°, ∠AOB = 30°
But ∠BOE + ∠AOB + ∠EOG = 180°
⇒ 30° + 90° + ∠EOG = 180°
∴∠EOG = 180° – 30° – 90° = 60°
But ∠FOG = 30°
∴ ∠FOE = 60° – 30° = 30°
∠COD = 30°, ∠COF = 90°
∴ ∠DOF = 90° – 30° = 60°
∠DOE = ∠DOF – ∠EOF
= 60° – 30° = 30°
∠BOC = BOE – ∠COE
= 90° – 30° – 30° = 90° – 60° = 30°
(ii) Right angles are,
∠AOD = 30° + 30° + 30° = 90°
∠BOE = 30° + 30° + 30° = 90°
∠COF = 30° + 30° + 30° = 90°
and ∠DOG = 30° + 30° + 30° = 90°
(iii) Pairs of adjacent complementaiy angles are ∠AOB, ∠BOD; ∠AOC, ∠COD; ∠BOC, ∠COE
(iv) Pairs of adjacent supplementary angles are ∠AOB, ∠BOG; ∠AOC, ∠COG and ∠AOD, ∠DOG
(v) Pairs of adjacent angles are ∠BOC, ∠COD; ∠COD, ∠DOE and ∠DOE, ∠EOF.

Question 19.
In the figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = \(\frac { 1 }{ 2 }\)(∠QOS – ∠POS). [NCERT]
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 Q19.1
Solution:
Ray OR ⊥ ROQ. OS is another ray lying between OP and OR.

To prove : ∠ROS = \(\frac { 1 }{ 2 }\)(∠QOS – ∠POS)
Proof : ∵ RO ⊥ POQ
∴ ∠POR = 90°
⇒ ∠POS + ∠ROS = 90° …(i)
⇒ ∠ROS = 90° – ∠POS
But ∠POS + ∠QOS = 180° (Linear pair)
= 2(∠POS + ∠ROS) [From (i)]
∠POS + ∠QOS = 2∠ROS + 2∠POS
⇒ 2∠ROS = ∠POS + ∠QOS – 2∠POS
= ∠QOS – ∠POS
∴ ROS = \(\frac { 1 }{ 2 }\) (∠QOS – ∠POS)

Hope given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Category: CBSE Class 9

Value Based Questions in Science for Class 9 Chapter 3 Atoms and Molecules

RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A

RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2

NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules

RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14C

RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D

RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9C

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS

RS Aggarwal Class 9 Solutions Chapter 7 Areas Ex 7A

RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2