Prasanna

MP Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.3

Question 1.
In Fig. given below, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution:
Given
∠SPR = 135°
∠PQT = 110°
To find. ∠PRQ
Calculation:
∠Q = 110°
110° + ∠PQR = 180° (LPA’s)
∠1 = 70°
∠P = ∠1 + ∠PRQ
135° = 70° + ∠PRQ
135° – 70° = ∠PRQ
65° = ∠PRQ

Question 2.
In Fig. given below, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∠XYZ, find ∠OZY and ∠YOZ.

Solution:
Given
∠X = 62°, ∠XYZ = 54°
∠OYZ = \(\frac{1}{2}\)∠XYZ
∠OZY = \(\frac{1}{2}\)∠XZY
To find ∠OZY and ∠YOZ
Calculation:
In ∠XYZ
∠X + ∠Y + ∠Z = 180° (ASP)
62° + 54° + ∠Z = 180°
116° + ∠Z = 180°
∴ ∠Z = 64°
In ∠OYZ
∠OYZ + ∠OZY + ∠O = 180° (ASP)
⇒ \(\frac{1}{2}\)∠XYZ + \(\frac{1}{2}\)∠XZY + ∠O = 180°
32° + 27° + ∠O = 180°
59° + ∠O = 180°
∠O = 121°

Question 3.
In Fig. given below, if AS ∥ DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution:
Given
AS ∥ DE,
∠BAC = 35°
∠CDE = 53°
To find. ∠DCE
Calculation:
AB ∥ DE and AE is the transversal.
∠BAE = ∠AED = 35°
In ∆DEC
∠D + ∠E + ∠C = 180°
53° + 35° + ∠C = 180°
88° + ∠C = 180°
∠C = 180° – 88° = 92°

Question 4.
In Fig. given below, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Solution:
Given
∠PRT = 40°
∠RPT = 95°
∠TSQ = 75°
To find. ∠SQT
Calculation:
In ∆PRT
∠P + ∠R + ∠T= 180° (ASP)
95° + 40 ° + ∠T= 180°
∠T = 180° – 135° = 45°
∠T = ∠STQ = 45°
In ∆TSQ
∠STQ + ∠S + ∠Q = 180° (V.O.A’S)
45° + 75° + ∠Q = 180° (ASP)
∠Q = 180° – 120° = 60°

Question 5.
In Fig. given below, if PQ ⊥ PS, PQ ∥ SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution:
Given
PQ ⊥ PS
PQ ∥ SR
∠SQR = 28°
∠QRT = 65°
To find, x and y
Calculation:
PQ ∥ SR and QR is the transversal
x + 28° = 65° (A.I.A’s)
x = 65° – 28° = 37°
In ∆PQS
x + y + 90° = 180° (ASP)
37° + y + 90° = 180°
∴ y = 180° – 127° = 53°

Question 6.
In Fig. given below, the side QR to ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = \(\frac{1}{2}\) ∠QPR.

Solution:
Given
∠TQR = \(\frac{1}{2}\)∠PQR
∠TRS = \(\frac{1}{2}\)∠PRS
To prove. ∠QTR = \(\frac{1}{2}\)∠QPR
Proof:
In ∆PQR
∠PRS = ∠P + ∠Q (EAP)
In ∆TQR, ∠TRS = ∠T + ∠TQR (EAP)
\(\frac{1}{2}\)∠PRS = ∠T + \(\frac{1}{2}\)∠PQR
Multiplying both sides by 2
∠PRS = 2∠T+ ∠PQR
From (1) and (2) we get,
∠P + ∠Q = 2∠T + ∠PQR
∠P=2∠T
∠T= \(\frac{1}{2}\)∠P
∠QTR = \(\frac{1}{2}\)∠QPR
Proved.

MP Board Class 9th Maths Solutions

MP Board Class 6th Special English Solutions Revision Exercises 1

Word Power

(a) Match the columns.

Answer:
1. → (c)
2. → (a)
3. → (b)
4. → (e)
5. → (d)

(b) Fill in the blanks with the help of given words:

(unhappy, country, riddles, busy, marbles)

1. Kassim had read a book of ………….. the day before.
2. Ali had to give Kassim ten ……………….
3. The town child is ………….. about his life in the town.
4. The old woman was too ……………. to go out and ask the chief.
5. The ………… child is unhappy in the country.

Answer:

1. riddles
2. Marbles
3. unhappy
4. busy
5. country.

(c) Make sentences with the given words and phrases.
(agreement, in search of, give up, approach, wonderful.)

Agreement  : An agreement is being made between the two countries.
In search of : The police is in search of the criminal.
Give up       : He gave up smoking.
Approach    : The lion is approaching towards the town.
Wonderful  : He has done a wonderful job.

Comprehension Questions

Answer these questions.

Question 1.
Why was Kassim eager to take a bet with Ali?
Answer:
Kassim was eager to take a bet with Ali because he was very sure of his success. Moreover his mind was fall of riddles.

Question 2.
How did Kassim get back his ten marbles from Ali?
Answer:
Kassim said that his pencil could not write any colour. Ali made a bet for ten marbles with his pencil. Kassim took out his pencil. He recited some magic words. He wrote red, blue, yellow, green, orange, on a piece of paper. Ali lost the bet. In this way Kassim got back his ten marbles from Ali.

Question 3.
Where does the town child live?
Answer:
The town child lives in a town.

Question 4.
Why does he not like his street?
Answer:
He does not like his street for its being too noisy.

Question 5.
Why had the thieves to stay in the town for the night?
Answer:
The thieves had to stay in the small town at night because their own village was far away. They had a pot of gold with them. It was very risky for them to travel at night.

Question 6.
How did Ramanna decide the case.
Answer:
Ramanna said to the thieves, “Bring your third companion and the pot will be delivered only to all three of you together. The thieves could not locate their companion. They withdrew their case. In this way he decided the case in the old woman’s favour.

Question 7.
Where is the country child’s home?
Answer:
The country child’s home is near a jungle.

Question 8.
Why does he wish to live in a town.
Answer:
He wishes to enjoy the company, the comfort and pleasure of city life.

Grammar

(a) Change these sentences into indirect speech:

1. Bholu said, “I have written the letters.”
2. Ali said, “I have done my work.”
3. “Will you have tea or coffee?” I asked the guest.
4. “Do the boys want to go to Chennai or to Delhi for excursion?” The headmaster asked the class teacher.

Answer:

1. Bholu said that he had written the letters.
2. Ali said that he had done his work.
3. I asked the guest whether he would have tea or coffee.
4. The headmaster asked the class teacher whether the boys wanted to go to Chennai or Delhi for the excursion.

Let’s Write

Question 1.
Write ten sentences about your village or town.
Answer:
My City Gwalior
Gwalior is in the northern most part of Madhya Pradesh. It offers a feast of historic sights, museums, parks, shops, etc. The city is dominated by its hill top fort, a symbol of Rajput valour and chivalry. The 15th century palace of Raja Mansingh is located in the citadel. Gwalior has also the distinction of being a centre of Indian classical music. A great musical festival is held at Gwalior every year where Tansen is lying buried. The routes to Gwalior pass through forests and fertile land.

Question 2.
Write an application to the principal of your school requesting him/her to grant you three days leave as you are ill.
Answer:
To
The Principal B-B,
Middle School,
Bhopal,
Date 18 May, 2005

Sir,
Most respectfully I beg to say that I have been suffering from high fever. It will take two or three days to recover. Therefore I request you to grant me leave for at-least three days. I shall be obliged for this act of kindness.

Thanking you
Yours faithfully
Amit Varma
Class VI-A

MP Board Class 6th English Solutions

MP Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.2

Question 1.
In Fig. given below, find the values of x and y and then showthat AB ∥ CD.

Solution:
∠y = 130° (VOA’s)
x + 50° = 180° (LPA’)
∴ x = 180° – 50° = 130°
∠x and ∠y are alternate interior angles and are equal AB ∥ CD

Question 2.
In Fig. given below, if AB ∥ CD, CD ∥ EF and y : z = 3 : 7, find x.

Solution:
Given
AB ∥ CD
CD ∥ EF
y : z = 3 : 7
∴ y = 3k
and z = 7k
To find, x
Solution:
AB ∥ EF and PQ is the transversal
x = z (A. I.A’s)
AB ∥ CD and PQ is the transversal
∴ x + y = 180° (C.I.A.’s)
z + y = 180° (∴ x = z)
7k + 3k = 180°
∴ k = 18°
y = 3 x 18° = 54°
2 = 7 x 18° = 126°
x = z = 126°

Question 3.
In Fig. given below, if AS ∥ CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Solution:
Given
AB ∥ CD
EF ⊥ CD
To find
∠AGE, ∠GEF, ∠FGE
AB ∥ CD and GE is the transversal
∴ ∠AGE = ∠GED = 126°
∠AGE + ∠FGE = 180°
126° + ∠FGE = 180°
∠FGE = 54°
∠GED = ∠GEF + ∠FED
126° = ∠GEF + 90°
126° – 90° = ∠GEF
∠GEF = 36°

Question 4.
In Fig. given below, if PQ ∥ ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

Solution:
Given
PQ ∥ ST
∠PQR = 110°; ∠TSR = 130°
To find
∠QRS Construction.
Draw a line EF ∥ PQ passing through point R.
Calculation. PQ ∥ EF (by construction)
PQ ∥ ST (Given)
EF ∥ ST (lines ∥ to the same line are parallel to each other)
PQ ∥ EF and QR is the transversal
110° + ∠1 = 180° (CIA’s)
∴ ∠1 = 70°
ST ∥ EF and SR is the transversal
130° + ∠3 = 180° (CIA’s)
∠3 = 50°
∠1 + ∠2 + ∠3 = 180° (angles on the same line)
70° + ∠2 + 50° = 180°
∠2 = 180° – 120°
∴ ∠QRS = 60°

Question 5.
In Fig. given below, if AB ∥ CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Given
AB ∥ CD
∠APQ = 50°; ∠PRD = 127°
To find, x and y
Calculation:
AB ∥ CD and PQ is the transversal
∠APQ = x = 50° (AIA’s)
y + 50° = 127°
∴ y = 77°

Question 6.
In Fig. given below, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB ∥ CD.

Solution:
Given
PQ ∥ RS
To prove:
AB ∥ CD
Construction:
Draw MB and NC normals at point B and C respectively.
Proof:
∠1 = ∠2
(Angle of incident = Angle of reflection)
and ∠3 = ∠4
∠ABC = 2∠2 and ∠BCD = 2∠3

MB ⊥ PQ and NC ⊥ RS
MB ∥ CN (Lines perpendicular to two parallel lines are parallel to each other)
MB ∥ CN and BC is the transversal
∠2 = ∠3 (AIA’s)
⇒ 2∠2 = 2∠3
(on multiplying both sides by 2)
⇒ ∠ABC = ∠BCD
∠ABC and ∠BCD are alternate interior angles and are equal AB ∥ CD

Angle Sum Property of a Triangle:

1. Triangle:
A plane figure bounded by three line segments in a plane is called a triangle. A triangle has six parts –

• Three sides – AB, BC and AC
• Three angles – ∠A, ∠B and ∠C

2. Median:
The median of a ∆ corresponding to a particular side is the line segment joining the midpoint of that side to its opposite vertex. In Fig., D, E and F are the midpoints of sides BC, AC and AB respectively. AD, BE and CF are the medians.

3. Incentre of a triangle:
The incentre of a ∆ is the point of intersection of angle bisectors of the triangle. In Fig., OA, OB and OC are the angle bisectors of ∠A, ∠B and ∠C respectively. These angle bisectors intersect at point O. Therefore, point O is the incentre of ∆ABC.

4. Circumcenter of a triangle:
The circumcenter of a triangle is the point of intersection of the perpendicular bisectors of the sides of the triangle. In Fig., O is the circumcenter of ∆ABC where the perpendicular bisectors of AB, BC and AC intersect.

Theorem 1.
The sum of the angles of a triangle is 180°.
Given
ABC is a ∆ in which ∠1, ∠2 and ∠3 are angles.
To prove:
∠1 + ∠2 + ∠3 = 180°

Construction:
Draw a line PQ passing through point A parallel to BC.
Proof:
PQ is a line.
∠4 + ∠1 + ∠5 = 180° (angles on the same line) …..(i)
PQ ∥ BC and AB is the transversal
∴ ∠4 = ∠2 (AIA’s) …..(ii)
PQ ∥ BC and AC is the transversal
∴ ∠5 = ∠3 (AIA’s) …..(iii)
Putting ∠4 = ∠2 and ∠5 = ∠3 in (i), we get
∠2 + ∠1 + ∠3 = 180°
∴ ∠1 + ∠2 + ∠3 = 180°. Proved

Exterior Angle Property:

Theorem 2.
If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two opposite interior angles. This is also known as exterior angle property (EAP).
Given
ABC is a ∆ in which side BC is produced to point D.
To prove. ∠4 = ∠1 + ∠2

Proof:
In ∆ABC,
∠1 + ∠2 + ∠3 = 180° (ASP)
∠3 + ∠4 = 180° (LPA)
From (i) and (ii), we get
∠1 + ∠2 + ∠3 = ∠3,+ ∠4
∠1 + ∠2 = ∠4. Proved

Corollary

Theorem 3.
An exterior angle of a triangle is greater than either of the opposite interior angles.

Given
In ∆ABC
∠4 = ∠1 + ∠2 (EAP)
∠4 > ∠1 and ∠4 > ∠2
Hence an exterior angle of A is greater than either of the opposite interior angle.

Lines And Angles:

Example 1.
In the Fig., AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Find the value of y.
Solution:
Let ∠DAB = x and ∠BAC = 3x
In ∆ABD
AB = BD
∠ADB = ∠DAB = x
(∴ In a ∆ angles opposite to equal sides are equal)
∠ABC = x + x = 2x
(In a ∆ Exterior angle is equal to sum of two opp. interior angles) In ∆ABC
2x + 3x + y = 180° (ASP)
5x + y = 180° …..(i)
In ∆ADC
x + y = 180° (ASP)
y = 180° – X …..(ii)

Putting y = 180° – x in (i), we get
5x + 108° – x = 180°
4x + 108° = 180°
4x = 180° – 108°
= 72°
∴ x = \(\frac{72}{4}\) = 18°
Putting x = 18° in (ii), we get
y = 108° – 18° = 90°.

Example 2.
In the Fig., given below, find the value of x.
Solution:
Given
∠ABD = 100°
∠ACE = 115°

To find. x°
∠ABC + ∠ABD = 180° (LPA’s)
∠ABC + 100° = 180°
∠ABC = 180° – 100° = 80°
∠ACB + ∠ACE = 180°
∠ACB + 115° = 180°
∠ACB = 180° – 115° = 65°
Now, ∠A + ∠ABC + ∠ACE = 180° (ASP)
⇒ x° + 80° + 65° = 180°
⇒ x° + 145° = 180°
x° = 180° – 145° = 35°

Example 3.
In Fig., PQ ⊥ PS, PQ ∥ SR, ∠SQR = 25° and ∠QRT = 65°.
Solution:
Given
∠SQR = 25°
∠QRT = 65°

To find x and y
In ∆SQR
∠QRT = ∠SQR + ∠QSR (FAP)
⇒ 65° = 25° + ∠QSR
⇒ 65° – 25° = ∠QSR
∠QSR = 40°
PQ ∥ SR and SQ is the transversal.
∴ ∠PQS = ∠QSR (AIA’s)
∴ x = 40°

Example 4.
In Fig., AM ⊥ L BC and AN is the bisector of ∠A. Find the measure of ∠MAN.

Solution:
In ∆AMB
∠ABM + ∠AMB + ∠MAB = 180°
65° + 90° + ∠MAB = 180°
155° + ∠MAB = 180°
∠MAB = 180° – 155° = 25°
In ∆BAC
∠A + ∠B + ∠C = 180° (ASP)
∠A + 65° + 30° = 180°
∠A + 95° = 180°
∠A = 180° – 95° = 85°
∠MAN = ∠BAN – ∠MAB
= \(\frac{1}{2}\) ∠A – 25°
= \(\frac{85^{\circ}}{2}\) – 25 = \(\frac{85^{\circ}-50^{\circ}}{2}\)
∠MAN = \(\frac{35^{\circ}}{2}\) = 17.5°

Example 5.
If the sides of a triangle are produced in order, prove that sum of the exterior angles so formed is equal to four right angles.
Solution:

∠EAB = ∠2 + ∠3 (EAP) …..(i)
∠FBC = ∠1 + ∠3 (EAP) …..(ii)
∠ACD = ∠1 + ∠2 (EAP) …(iii)
Adding (i), (ii) and (iii), we get
∠EAB + ∠FBC + ∠ACD = ∠2 + ∠3 + ∠1 + ∠3 + ∠1 + ∠2
= 2(∠1 + ∠2 + ∠3) = 2 x 180° = 360°. Proved

Example 6.
Bisectors of angles B and C of a triangle ABC intersect each other at the point O. Prove that ∠BOC = 90° + \(\frac{1}{2}\) ∠A.

Solution:
Given
∠1 = \(\frac{1}{2}\) ∠ABC
⇒ ∠ABC = 2∠1
∠2 = \(\frac{1}{2}\) ∠ACB
⇒ ∠ACB = 2∠2
To prove:
∠BOC = 90° + \(\frac{1}{2}\) ∠A
Proof:
In ∆BOC ∠O + ∠1 + ∠2 = 180°
∠1 + ∠2 = 180° – ∠O …..(i)
In ∆ABC
∠A + ∠ABC + ∠ACB = 180°
∠A + 2∠1 + 2∠2 = 180°
2∠1 + 2∠2 = 180° – ∠A
2(∠1 + ∠2) = 180° – ∠A
∠1 + ∠2 = \(\frac{180^{\circ}-\angle A}{2}\) = 90° – \(\frac{\angle A}{2}\) …..(ii)
From (i) and (ii) we get
180° – ∠O = 90° – \(\frac{\angle A}{2}\)
180° – 90° + \(\frac{\angle A}{2}\) = ∠O
90° + \(\frac{\angle A}{2}\) = ∠O
∴ ∠BOC = 90° + \(\frac{\angle A}{2}\)

Example 7.
In Fig. show that ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°
Solution:
In ∆ACE
∠A + ∠C + ∠E = 180° (ASP) …..(i)

In ABDF
∠B + ∠D + ∠F = 180° (ASP) …..(ii)
Adding (i) and (ii), we get
∠A + ∠C + ∠E + ∠B + ∠D + ∠F = 180° + 180°
∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°

Example 8.
In Fig., ∠Q > ∠R, PA is the bisector of ∠QPR and PM ⊥ QR. Prove that ∠APM = \(\frac{1}{2}\) (∠Q – ∠R)

Solution:
Given
∠Q > ∠R
∠QPA = ∠RPA
PM ⊥ QR
To prove:
∠APM = \(\frac{1}{2}\) (∠Q – ∠R)
Proof:
In ∆PMQ
∠Q + ∠QPM + ∠M= 180° (ASP)
∠Q + ∠QPA – ∠APM + 90° = 180°
(∴ ∠QPM = ∠QPA – ∠APM)
∠Q + ∠QPA – ∠APM = 180° – 90°
∠Q + ∠QPA – ∠APM = 90°
∠Q + ∠QPA – ∠APM = 90° …..(i)
In ∆PMR
∠R + ∠RPM + ∠PMR = 180°
∠R + ∠RPA + ∠APM + 90° = 180°
(∴ ∠RPM = ∠RPA + ∠APM)
∠R + ∠QPA + ∠APM = 180° – 90°
(∴ ∠RPA = ∠QPA)
∠R + ∠QPA + ∠APM = 90° …..(ii)
From (i) and (ii), we get
∠Q + ∠QPA – ∠APM = ∠R + ∠QPA + ∠APM
∠Q – ∠R = ∠APM – ∠QPA + ∠QPA + ∠APM
∠Q – ∠R = 2∠APM
∠APM = \(\frac{1}{2}\) (∠Q – ∠R) Proved

Example 9.
In fig., AB ∥ DC, ∠BDC = 30° and ∠BAD = 80°, find x, y and z.
Solution:
AB ∥ DC and BD is the transversal.
x = 30° (AIA’s)

In ∆ABD
x + y + 80° = 180° (ASP)
30° + y + 80° = 180°
110° + y = 180°
∴ y = 180° – 110° = 70°
y – 30° = 70 – 30° = 40°
In ∆BCD
30° + 40° + z = 180° (ASP)
[∴ y – 30° = 40°]
70° + z = 180°
∴ z = 180° – 70° = 110°

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:

1. OB = OC
2. AO bisect ∠A.

Solution:
Given
AB = AC
∠1 = ∠2, ∠3 = ∠4
To prove:

1. OB = OC
2. ∠5 = ∠6

Proof:
In ∆ABC,
AB = AC
∠B = ∠C
\(\frac{1}{2}\) ∠B = \(\frac{1}{2}\) ∠C
∠1 = ∠3 or ∠2 = ∠4
In ∆OBC
∠2 = ∠4
and so OB = OC
(In a A, sides opposite to equal angles are equal)
In ∆ABO and ∆ACO
BO = CO (proved)
∠1 = ∠3 (proved)
AB = AC (given)
∆ABO = ∆ACO (by SAS)
and so ∠5 = ∠6 (by CPCT)

Question 2.
In ∆ABC, AD is the perpendicular bisector of BC (see Fig. below). Show that AABC is an isosceles triangle in which AB =AC.

Solution:
Given
∠ABC = ∠ADC
To prove:
AB = AC
Proof:
In A ABD and A ACD
BD = CD (given)
∠ADB = ∠ADC (given each 90°)
AD = AD (common)
∴ ∆ABD = ∆ACD (BySAS)
and so AB = AC (by CPCT)

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. below). Show that these altitudes are equal.

Solution:
Given
AB = AC
∠E = ∠F (each 90°)
To prove: BE = CF
Proof:
In ∆ABE and ∆ACE
∠A = ∠A (common)
∠E = ∠F (each 90°)
AB = AC (given)
∆ABE = ∆ACE (byAAS)
and so BE = CF (by CPCT)

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. below). Show that:

1. ∆ABE ≅ ∆ACF
2. AB = AC,

i. e., ABC is an isosceles triangle.

Solution:
Given
BE = CF
∠E = ∠F (each 90°)
To prove:

1. ∆ABE = ∆ACE
2. AB = AC

Proof:
In ∆ABE and ∆ACF
∠A = ∠A (common)
BE = CF (given)
∠E = ∠F (each 90°)
∆ABE = ∆ACF (by AAS)
and so AB = AC (by CPCT)

Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see Fig. below). Show that ∠ABD = ∠ACD.

Solution:
Given
AB = AC
BD = CD
To prove
∠ABD = ∠ACD
Construction: Join AD
Proof:
In ∆ABD and ∆ACD
AD = AD (common)
AB = AC (given)
BD = CD (given)
∆ABD = ∆ACD (by SSS)
and so ∠ABD = ∠ACD (by CPCT)

Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. below). Show that ∠BCD is a right angle.
Solution:
Given: AB = AC
AD = AB
To show: ∠BCD = 90°
i.e., ∠2 + ∠3 = 90°

Proof:
AB = AC …..(1)
AB = AD …..(2)
From (1) and (2), we get
AC = AD
In ∆ABC
AB = AC
∠1 = ∠2
(In a A, angles opposite to equal sides are always equal) …p) …(3)
In ∆ACD
AC = AD
∠3 = ∠4
(In a A, angles opposite to equal sides are always equal) …(4)
In ∆BCD
∠1 + ∠2 + ∠3 + ∠4 = 180° (ASP)
∠2 + ∠2 + ∠3 + ∠3 = 180°
(∴ ∠1 = ∠2, ∠3 = ∠4)
2 (∠2 + ∠3) = 180°
(∠2 + ∠3) = 90°
∠BCD = 90°

Question 7.
ABC is a right angled triangle in which ∠A – 90° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆BAC
AB =AC
∠B = ∠C = x
∠A + ∠B + ∠C= 180°
∠B + ∠C = 180° – 90°
∠B + ∠C = 90°
2x = 90°
x = \(\frac{90^{\circ}}{2}\)

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
Given
ABC is an equilateral ∆
i. e., AB = BC = AC
To prove
∠A = ∠B = ∠C = 60°
Proof:
In ∆BAC
AB = AC
∠B = ∠C
(In a A, angles opposite to equal sides are always equal) ……(1)
AC = BC
∠A = ∠B
(In a A, angles opposite to equal sides are always equal) …..(2)
From (1) and (2), we get,

∠A = ∠B = ∠C = x (say)
∠A + ∠B + ∠C = 180° (ASP)
⇒ x + x + x = 180°
⇒ 3x = 180°
⇒ x = \(\frac{180^{\circ}}{3}\) = 60°
∴ ∠A = ∠B = ∠C = 60

Theorem 7.4.
SSS (Side-Side-Side) Congruence Theorem:
Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.
Given
In ∆s ABC and DEF we have,

AB =DE
BC = EE
and AC = DF
To prove:
∆ABC = ∆DEF
Construction:
Suppose BC is the longest side.
Draw EF such that EE = AB and FEG = ∠CBA.
Join GF and DG.

Proof:
In ∆s ABC and GEE, we have
AB = GE (Const.)
∠ABC = ∠GEF (Const.)
and BC = EF (Given)
∴ ∆ABC = ∆GEF (SAS Cong. Axiom)
∠A = ∠G (CPCT) …..(1)
and AC = GF (CPCT) …..(2)
Now AB = EG (Const.)
AB = DE (Given)
∴ DE = EG ……(3)
Similarly, DF = GF ……(4)
In ∆EDG
DE = EG (Proved above)
∴ ∠A = ∠2 (∠s opp. equal side) …..(5)
In ∆DFG
FD = FG (Proved above)
∴ ∠3 = ∠4 (∠s ppp. equal side) …..(6)
∴ ∠1 + ∠3 – ∠2 + ∠4 [From (5) and (6)]
i. e. ∠D = ∠G …..(7)
But ∠G = ∠A [From (1)]
∴ ∠A = ∠D …..(8)
In ∆s ABC and DEF,
AB – DE (Given)
AC = DF (Given)
∠A = ∠D [From (8)]
∆ABC ≅ ∆DEF (SAS Cong. Axiom)

Theorem 7.5.
RHS (Right Angle Hypotenuse Side) Congruence Theorem:
Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle.
Given
In ∆s ABC and DEF,
∠B = ∠E = 90°
AC = DF
BC = EF.

∆ABC ≅ ∆DEF
Construction:
Produce DE to M so that
EM = AB, Join ME.
Proof:
In ∆s ABC and MEF
AB = ME (Const.)
BC = EE (Given)
∠B = ∠MEF (each 90°)
∴ ∆ABC = ∆MEF (SAS Cong. Axiom)
Hence ∠A = ∠M (CPCT) …(1)
AC = MF (CPCT) …(2)
Also AC =DF (Given)
∴ DF = MF
∴ ∠D = ∠M (∠s opp. equal side of ADFM) …(3)
From (1) and (3), we have
∠A = ∠D …..(4)
Now, in ∆s ABC and DEF, we have
∠A = ∠D [From (4)]
∠B = ∠E (Given)
∴ ∠C = ∠F …..(5)
Again, in ∆s ABC and DEF, we have
BC = EF (Given)
AC = DF (Given)
∠C = ∠F [From (5)]
∴ ∆ABC = ∆DEF (SAS Cong. Axiom)

MP Board Class 9th Maths Solutions

MP Board Class 7th Science Solutions Chapter 12 Reproduction in Plants

Reproduction in Plants Intex Questions

Question 1.
Boojho wants to know if there is any advantage of vegetative propagation?
Answer:
Plants produced by vegetative propagation take less time to grow and bear flowers and fruits earlier than those produced from seeds. The new plants are exact copies of the parent plant, as they are produced from a single parent.

Question 2.
Boojho wants to know how the male gamete in the pollen grain reaches the female gamete present in the ovule?
Answer:
The ovary contains one or more ovules.

Question 3.
Boojho wants to know why flowers are generally so colorful and fragrant. Is it to attract insects?
Answer:
Yes.

Activities

Activity – 1
Take a fresh potato. Observe the scars on it with the help of a magnifying glass. You may find buds in them. These scars are also called “eyes”. Cut a few pieces of a potato, each with an eye and bury them in the soil. Water the pieces regularly for a few days and observe their progress?
Answer:

Reproduction in Plants Text Book Exercises

Question 1.
Fill in the blanks:

1. Production of new individuals from the vegetative part of parent is called ……………………….
2. A flower may have either male or female reproductive parts Such a flower is called …………………….
3. The transfer of pollen grains from the another to the stigma of the same or of another flower of the same kind is known as ………………………….
4. The fusion of male and female gametes is termed as ………………………
5. Seed dispersal takes place by means of …………………………. , ……………………… and …………………………….

Answer:

1. Vegetative propagation
2. unisexual flowers
3. pollination
4. fertilisation
5. wind, insects, water.

Question 2.
Describe the different methods of asexual reproduction. Give examples?
Answer:
The various methods of asexual reproduction are:

1. Vegetative propagation, for example stem cutting of rose or champa.
2. Budding, for example reproduction in yeast by budding.
3. Fragmentation, for example fragmentation in spirogyra.
4. Spore formation, for example reproduction in fern.

Question 3.
Explain what you understand by sexual reproduction?
Answer:
In this type of reproduction male gamete (sperm) combines tie female gamete (ovum) in animals to form zygote.

Question 4.
State the main difference between asexual and sexual reproduction?
Sexual Reproduction:

1. It occurs both in lower and the higher organisms but mostly in higher plants and animals.
2. In it, there is fusion of the two (male and female) gametes and for this the sexes are required.
3. The new individual formed is identical to the parent.

Asexual Reproduction:

1. Usually occurs in lower organisms.
2. Only one parent is sufficient in asexual reproduction.
3. The new individual formed is similar but not identical to the parents.

Question 5.
Sketch the reproductive parts of a flower?
Answer:

Question 6.
Explain the difference between self – pollination and cross – pollination?
Answer:
self – pollination:
The transfer of pollen from the another to the stigma of a flower is called pollination.
If the pollen lands on the stigma of the same flower it is called self – pollination.

cross – pollination:
When the pollen of a flower lands on the stigma of another flower of the same plant, or that of a different plant of the same kind, it is called cross – pollination.

Question 7.
How does the process of fertilisation take place in flowers?
Answer:
Fertilization is the process in which male and female gametes fuse together. In flowering plants the pollen fuses with ovules present in the overy to form a zygote, which later on develops into seed.

Question 8.
Describe the various ways by which seeds are dispersed?
Answer:
In nature same kind of plants grow at different places. This happens because seeds are dispersed to different places. Sometimes after a walk through a forest or a field or a park, we may have found seeds or fruits sticking to our clothes. Seeds and fruits of plants are carried away by wind, water and animals.

Winged seeds such as those of drumstick and maple [Fig.(a) and (b)], light seeds of grasses or hairy seeds of aak (Madar) and hairy fruit of sunflower [Fig. (a), (b)], get blown off with the wind to far away places. Some seeds are dispersed by water. These fruits or seeds usually develop floating ability in the form of spongy or fibrous outer coat as in coconut. Some seeds are dispersed by animals, especially spiny seeds with hooks which get attached to the bodies of animals and are carried to distant places. Examples are Xanthium and Urena.


Some seeds are dispersed when the fruits burst with sudden jerks. The seeds are scattered far from the parent plant. This happens in the case of castor and balsam.

Question 9.
Match items in Column I with those in Column II

Answer:

(a) – (iii)
(b) – (v)
(c) – (ii)
(d) – (i)
(e) – (iv).

Question 10.
Tick the correct answer:

Question (a)
The reproductive part of a plant is the?
(i) leaf
(ii) stem
(iii) root
(iv) flower.
Answer:
(iii) root

Question (b)
The process of fusion of the male and the female gametes is called?
(i) fertilisation
(ii) pollination
(iii) reproduction
(iv) seed formation.
Answer:
(i) fertilisation

Question (c)
Mature ovary forms the?
(i) seed
(ii) pistil
(iii) stamen
(iv) fruit.
Answer:
(iv) fruit.

Question (d)
A spore producing plant is?
(i) rose
(ii) bread mould
(iii) potato
(iv) ginger.
Answer:
(iii) potato

Question (e)
Bryophyllum can reproduce by its?
(i) stem
(ii) leaves
(iii) roots
(iv) flower.
Answer:
(ii) leaves

Extended Learning – Activities and Projects

Question 1.
Make your own cactus garden by collecting pieces cut from different kinds of cacti. Grow the variety in one single flat container or in separate pots?
Answer:
Do yourself.

Question 2.
Visit a fruit market and collect as many local fruits as possible. If many fruits are not available, you can collect tomatoes and cucumbers (these are fruits, though we use them as vegetables). Make drawings of the different fruits. Split the fruits and examine the seeds within. Look for any special characteristics in the fruits and their seeds?
Answer:
Do yourself.

Question 3.
Think of ten different fruit – bearing plants. Remember that many vegetables are also fruits of the plants. Discuss with your teacher, parents, farmers, fruit growers and agricultural experts (if available nearby) and find out the manner of their dispersal. Present your data in the form of a table as shown below:

Answer:
Do yourself

Question 4.
Suppose there is one member of a particular kind of organism in a culture dish, which doubles itself in one hour through asexual reproduction. Work out the number of members of that kind of organism which will be present in the culture dish after ten hours. Such a colony of individuals arising from one parent is called a “clone”?
Answer:
Do yourself.

Reproduction in Plants Additional Important Questions

Objective Type Questions

Question 1.
Choose the correct alternative:

Question (a)
………………………… is formed in female reproductive part of plant.
(a) pollen grain
(b) pollination
(c) female gamete
(d) stigma.
Answer:
(d) stigma.

Question (b)
The process of reaching pollen grains to stigma is called as …………………….
(a) pollination
(b) budding
(c) fission
(d) fertilisation.
Answer:
(a) pollination

Question (c)
In plants, male and female reproductive organs are found in ………………………….
(a) leaves
(b) flowers
(c) roots
(d) stem.
Answer:
(b) flowers

Question (d)
Those living beings who have male and female reproductive organs differently are called ………………………….
(a) asexual
(b) bisexual
(c) unisexual
(d) None of these.
Answer:
(c) unisexual

Question (e)
Zygote is formed:
(a) by fusion of two male gamete
(b) by fusion of two female gametes
(c) by fusion of male and female gametes
(d) by all the above methods.
Answer:
(c) by fusion of male and female gametes

Question (f)
Which one of the following organisms shows budding?
(a) Hydra
(b) Spirogyra
(c) Amoeba
(d) Paramecium.
Answer:
(a) Hydra

Question 2.
Fill in the blanks:

1. ……………………. are parts of flower.
2. Once the pollen grains reach stigma, they form …………………………..
3. ……………………… are formed in pollen grains.
4. The long thread like structure of carpel is called as ………………………
5. The grain like structure in ovary is called as …………………………
6. ………………………….. is formed in ovule.
7. Formation of a new plant from the stem of sugarcane is known as ………………………..
8. The fusion of sperm and egg is known as ………………………..
9. We can count the age of a tree through its ……………………….
10. The cells involved in sexual reproduction are called ………………………….
11. Fusion of gametes gives rise to a single cells called ……………………….
12. The process of fusion of gametes is called ………………………….. in animals and plants.
13. Animals having both reproductive organs are called ………………………….

Answer:

1. Stamen and carpel
2. Male gametes
3. Male gametes
4. Ovules
5. Female gametes
6. Zygote
7. Vegetative reproduction
8. Fertilisation
9. Annual rings
10. Gametes
11. Zygote
12. Fertilisation
13. Hermaphrodite.

Question 3.
Which of the following statements are true (T) or false (F):

1. The process of development in human and butterfly are the same because they both start from a zygote.
2. The most common type of reproduction in amoeba and paramecium is budding.
3. The common method of reproduction in yeast is budding.
4. Asexual reproduction is more common than the sexual reproduction.
5. Reproduction by spores is a method of asexual reproduction.
6. Cutting and grafting are natural means of reproduction.
7. A fertilized egg becomes a seed.
8. Plants can have indefinite growth but animals do not.
9. Insect pollinated flowers are brightly coloured.
10. Flowers which possess stamens and carpel are called unisexual.
11. Wind pollinated flowers produce pollen grains in large quantity.

Answer:

1. False
2. False
3. True
4. True
5. True
6. False
7. True
8. True
9. True
10. False
11. True

Question 4.
Match the items in Column A with Column B:

Answer:

(i) – (d)
(ii) – (c)
(iii) – (a)
(iv) – (b).

Reproduction in Plants Very short Answer type Questions

Question 1.
Name two organisms which reproduce by two types of asexual methods. What are the methods?
Answer:
Amoeba reproduces by binary fission. Hydra reproduces by budding.

Question 2.
What is meant by reproduction?
Answer:
It is a process by which an organism gives birth to produce its own kind to maintain its species.

Question 3.
What are the two methods of reproduction in plants and animals?
Answer:
The two methods of reproduction in plants and animals are:

1. Asexual reproduction.
2. Sexual reproduction.

Question 4.
What is pollination?
Answer:
It is the process of transference of pollen grains from the stamens to the stigma of the carpel of flower.

Question 5.
What do you call for the male reproductive organ of flower?
Answer:
The male reproductive organ of a flower is called ‘stamen’.

Question 6.
What do you call for the female reproductive organ of the flower?
Answer:
The female reproductive organ of the flower is called a “carpel”.

Question 7.
What is growth?
Answer:
The process in which the change in size and shape of the organism along with the increase in weight is called growth.

Question 8.
What kind of development takes place in dog and cat?
Answer:
The development in dog and cat takes place inside the body of female. The female gives birth to young ones after a definite period. This type of development is called internal development.

Question 9.
What is the difference between sperm and ovum?
Answer:
Sperm is a male gamete while ovum is a female gamete. Sperm is mobile while ovum is non-mobile, the movement of sperm is attributed to mitochondria present in sperm.

Question 10.
In which part of the flower is the ovule found?
Answer:
In the ovary part of carpel of the flower the ovule is found.

Question 11.
Where does fertilisation occur in a flowering plant?
Answer:
In flowering plants, the fertilisation occurs in the ovary of the carpel of the flower.

Question 12.
How many types of gametes are there?
Answer:
Gametes are the fundamental units of reproduction. They are of two types :

1. Male gametes or sperms
2. Female gametes or ovum.

Question 13.
What is fertilisation process?
Answer:
The fusion of male and female gametes to form a zygote during sexual reproduction is called fertilisation.

Reproduction in Plants short Answer type Questions

Question 1.
How do insects help in cross – pollination?
Answer:
Insects get attracted towards flowers either due to their bright color or fragrance or nectar. In the process of collecting nectar from flowers some of the pollens get stuck to their legs and body. If such an insect visits another flower, he transfers pollens to this flower and thus completes the process of cross-pollination.

Question 2.
What are the organs in human which produce the gametes?
Answer:
In humans the parents are separate as mother and father. Mother (female) possesses the ovary and the father (male) possesses the testes. The ovary produces the ovum (female sex cell) and the testes produces the sperm.

Question 3.
What is budding?
Answer:
Budding is a type of a sexual reproduction in which an individual is produced as an out growth called bud from the parent organism. It takes place in certain animals like Hydra and in non green plant cells of yeast.

Question 4.
How do plants developed from a seed?
Answer:
The embryo inside the seed has two distinct parts. One part produces root system, and the other shoot system under favourable conditions of water, air, light and temperature. The root system developed under the soil and shoot system above the soil. Due to cell division elongation of different parts takes place and a new plant is obtained.

Question 5.
Give some examples of growth in plants and animals?
Answer:
Examples of growth in plants:

1. A small seedling gets developed into a fully developed tree.
2. Growth of plant in length.
3. Growth of stem in thickness.

Examples of growth in animals:

1. An egg is developed into an organism by the division of cells.
2. A small child grows into a fully developed adult.
3. A wound is healed up by the division of cells in the surrounding region.

Question 6.
Explain with examples “fission”?
Answer:
When the body of an individual after a certain period of growth divides mitotically into two or more parts, it is known as fission reproduction. When a fission results in the formation of two daughters, it is known as binary fission. When fission results in the formation of two daughters, it is known as multiple fission.

Question 7.
How does reproduction take place in fungs by budding?
Answer:
A bulb like structure formed on body is known as bud. As the reproduction takes place through this bud, it is called as budding. Such types of buds are formed on the bodies of organisms like hydra, yeast etc. This bud increases gradually and detaches itself from parent body and develops in the form of new organism. In corel and sponge also reproduction takes place by budding.

Question 8.
What is called as vegetative reproduction?
Answer:
The new plant develops when branches of these plants are burried in soil during rainy season. The new saplings are produced from various parts such as root, stem, leaf of plants. This type of propagation or reproduction in plants is called as vegetative propagation.

Question 9.
What is difference between fertilisation and pollination?
Answer:
Pollination is the transference of male gametes (pollen grain) from pollen sac to stigma and the fertilisation is fusion of male gamete (pollen grain) and female gamete (egg cell) in the ovary of a flower. Fertilisation is followed by the formation of embryo. Embryo is enclosed within the seed.

Question 10.
What are the advantages of vegetative reproduction?
Answer:
Advantages of vegetative reproduction:

1. It helps in rapid propagation of plant species in some region. Examples: bamboo, sugarcane, potato crop etc.
2. Those plants, whose seeds are not capable of germination, can be reproduced by vegetative reproduction e.g. garlic,
3. There is 100 percent possibility of survival of plants by this method.
4. The species having high qualities can be conserved.
5. The plants reproduced by this method get the fruits earlier.

Reproduction in Plants Long Answer type Questions

Question 1.
What are the different ways in which reproduction in plants can take place?
Ans. Reproduction in plants can be classified in three main groups:

1. Sexual reproduction,
2. Asexual reproduction, and
3. Vegetative reproduction.

In asexual reproduction, the new individuals are reproduced from a single parent. Here one cell undergoes division into two new individuals. In sexual reproduction, two parents are needed to produce one new individual. In vegetative propagation, a few cells of plant body propagate to form a new plant.

Question 2.
Write the various steps involved in the formation of a plant seed, starting from pollination?
Answer:
There are following steps involved from pollination to the formation of the seed:

1. Fertilisation:
The process of fusion of male and female gamete in sexual reproduction is called fertilisation.

2. Pollination:
The male part of plant is stamen. It consists of the filament and the anther. The anther has pollen grains.The female part is made up of stigma, style and ovary. The ovary has ovule. The stigma receives of pollen grains. The process of transfer of pollen from anther to stigma is known as pollination.

3. Seed formation:
The whole ovary after fertilisation is converted into the fruit while the ovule is converted into the seed.

Question 3.
What is meant by the terms external fertilisation and internal fertilisation?
Answer:
Internal fertilisation:
The reproductive process remains incomplete till the male and female gamete do not fuse with each other. In some animals the process of fertilisation occurs inside the female body. This is called internal fertilization. After that female either lay eggs or given birth to young one. For example, dog, cat, bird, human etc.

External fertilisation:
In some animals the female release ovum or egg outside and the male drops sperms on these. This type of fertilization is called external fertilization. The zygote so formal undergoes regular and specific changes with rime to form a new individual.

Question 4.
What are the reproductive organs in a flower? How does pollination take place? What is a seed?
Answer:
The reproductive organs of a plant develop on a special reproductive appendage called flower. The male reproductive, organs are stamen. The stamens bear anthers where pollen grains (the male gametes) are produced. The female reproductive organ is pistil. It consists of stigma, style and ovary. Inside the ovary (the female – gametes) are developed.

Pollination is the transfer of pollen grains from another to stigma of either the same flower (self pollination) or of another flower (cross pollination). The pollen grains are carried to the stigma by air, insects water in aquatic plants or by animals. Seed is the structure formed by the fusion of ovule and pollen grains in the ovary. The seeds germinate after their dispersion, into new plants.

Question 5.
Explain reproduction by spore formation in fungus with neat diagram?
Answer:
Spore formation:
You might have seen white powder like substance on leather objects during rainy season. These small particles are spores of fungus. Small spherical structures are in yeast, fungus, moss, fern and microbes during unfavourable circumstances. They have protective layer around them and remain floating in air. During favourable conditions, the protective layer breaks open and new organism is formed.

Question 6.
What are spores? When do they form?
Answer:
Several plants produce spores which germinate to produce new individual plants e.g., rhizopus, mucus and moss etc. It is the mechanism to overcome the unfavourable conditions.

Question 7.
Draw a neat diagram to show reproduction through spore formation in fern?
Answer:

MP Board Class 7th Science Solutions

MP Board Class 9th Science Solutions Chapter 7 Diversity in Living Organisms

Diversity in Living Organisms Intext Questions

Diversity in Living Organisms Intext Questions Page No. 80

Question 1.
Why do we classify organisms?
Answer:
Classification of organism make it easy to study the millions of organisms on this earth. Similarities among them is the basis to classify them into different classes. Classification makes study easier.

Question 2.
Give three examples of the range of variations that you see in life – forms around you.
Answer:
Variations observed in life are:

1. Size: Organisms vary greatly in size – from microscopic bacteria to elephants, whales and large trees.
2. Appearance: The colour of various animals is quite different. Number of pigments are found in plants. Their body – built also varies.
3. Life time: The life span of different organisms is varied.

Diversity in Living Organisms Intext Questions Page No. 82

Question 1.
Which do you think is a more basic characteristic for classifying organisms?
(a) the place where they live.
(b) the kind of cells they are made of. Why?
Answer:
(a) Different organisms may share same habitat but may have entirely different form and structure. So, the place where they live cannot be a basis of classification.

(b) The kind of cells they are. made of. Because placement of organism to other destination can create a easy confusion.

Question 2.
What is the primary characteristic on which the first division of organisms is made?
Answer:
The primary characteristic on which the first division of organisms is made is the nature of the cell – prokaryotic or eukaryotic cell.

Question 3.
On what basis are plants and animals put into different categories?
Answer:
Plants and animals are very different from each other but main basis to differentiate is “Mode of nutrition’’. Plants are autotrophs. They can make their food own while animals are heterotrophs which are dependent on others for food. Locomotion, absence of chloroplasts etc. also make them different.

Diversity in Living Organisms Intext Questions Page No. 83

Question 1.
Which organisms are called primitive and how are they different from the so-called advanced organisms?
Answer:
A primitive organism is the one which has a simple body structure and ancient body design or features that have not changed much over a period of time. As per the body design, the primitive organisms which have simple structures are different from those so – called advanced organisms which have complex body structure and organization.

Question 2.
Will advanced organisms be the same as complex organisms? Why?
Answer:
Yes, they are developed from same ancestor once. They have relatively acquired their complexity recently. There is a possibility that these advanced or ‘younger’ organisms acquire more complex structures during evolutionary time to compete and survive in the changing environment.

Diversity in Living Organisms Intext Questions Page No. 85

Question 1.
What is the criterion for classification of organisms as belonging to kingdom Monera or Protista?
Answer:
The organisms belonging to kingdom Monera are unicellular and prokaryotic whereas the organisms belonging to Kingdom Protista are unicellular and eukaryotic. This is the main criterion of their classification.

Question 2.
In which kingdom will you place an organism which is single – celled, eukaryotic and photosynthetic?
Answer:
Kingdom Protista.

Question 3.
In the hierarchy of classification, which grouping will have the smallest number of organisms with a maximum of characteristics in common and which will have the largest number of organisms?
Answer:
In the hierarchy of classification, “species” will have the smallest number of organisms with a maximum of characteristics in common whereas “the kingdom” will have the largest number of organisms a Arthropoda.

Diversity in Living Organisms Intext Questions Page No. 88

Question 1.
Which division among plants has the simplest organisms?
Answer:
Division thallophyta.

Question 2.
How are pteridophytes different from the phanerogams?
Answer:
1. Pteridophyta: They have inconspicuous or less differentiated reproductive organs. They produce naked embryos called spores.
Examples:

• Ferns
• marsilea
• equisetum, etc.

2. Phanerogams: They have well developed reproductive organs. They produce seeds.
Example:

• Pinus
• cycas
• fir etc.

Question 3.
How do gymnosperms and angiosperms differ from each other?
Answer:
Gymnosperm:

1. They are non – flowering plants.
2. Naked seeds not enclosed inside fruits are produced.
3. Examples:
   • Pinus
   • Cedar
   • Fir
   • Cycas etc.

Angiosperm:

1. They are flowering plants.
2. Seeds are enclosed inside fruits.
3. Examples:
   • Coconut
   • Palm
   • Mango etc.

Diversity in Living Organisms Intext Questions Page No. 94

Question 1.
How do poriferan animals differ from coelenterate animals?
Answer:

Poriferan	Coelenterate
1. Mostly marine, non – motile.	1. Motile marine animals that either live in colonies or have a solitary life – span.
2. Cellular level of organisation.	2. Tissue level of organisation.
3. Spongilla, Euplectella etc.	3. Hydra, sea anemone.

Question 2.
How do annelid animals differ from arthropods?
Answer:

Annelids	Arthropods
1. Closed circulatory system	1. An open circulatory system
2. The body is divided into several identical segments	2. The body is divided into few specialized segments

Question 3.
What are the differences between amphibians and reptiles?
Answer:

Amphibian	Reptiles
1. They live at land and water both.	1. They are completely terrestrial.
2. Scales are absent.	2. Skin is covered with scales.
3. They lay eggs in water.	3. They lay eggs on land.
4. Example: frogs, toads and salamanders.	4. Example: lizards, snakes, turtles, chameleons etc.

Question 4.
What are the differences between animals belonging to the Aves group and those in the mammalia group?
Answer:
Most birds have feathers and they possess a beak.Mammals do not have feathers and the beak is also absent. Birds lay eggs. Hence, they are oviparous. Some mammals lay eggs and some give birth to young ones. Hence, they are both oviparous and viviparous.

Diversity in Living Organisms NCERT Textbook Exercises

Question 1.
What are the advantages of classifying organisms?
Answer:
Advantages of classification:

1. Better categorization of living beings based on common characters.
2. Easier study for scientific research.
3. Better understanding of human’s relation and dependency on other organisms.
4. Helps in cross breeding and genetic engineering for commercial purposes.

Question 2.
How would you choose between two characteristics to be used for developing a hierarchy in classification?
Answer:
Gross character will form the basis of start of the hierarchy and fine character will form the basis of further steps of single hierarchy.
Examples:

• Presence of vertebral column in human beings can be taken under vertebrata.
• Presence of four limbs makes them members of Tetrapoda.
• Presence of mammary glands keeps them under mammalia.

Question 3.
Explain the basis for grouping organisms into five kingdoms.
Answer:
Basis of classification:

1. Number of cells: unicellular or multicellular.
2. Complexity of cell structure: Prokaryote and Eukaryote.
3. Presence or absence of cell wall.
4. Mode of nutrition.
5. Level of organization.

Question 4.
What are the major divisions in the Plantae? What is the basis for these divisions?
Answer:
Major divisions of Kingdom Plantae:

Division	Basis for classification
1. Thallophyta or Algae	1. Thallus like body, plant body is not differentiated into roots, stems etc.
2. Bryophyta	2. Body is divided into leaf and stem, lack vascular tissue.
3. Pteridophyta	3. Body is divided into root, stem and leaf, lack seeds.
4. Gymnosperm	4. Seed bearing, naked seeds, lack flowers.
5. Angiosperm	5. Seed bearing covered seeds, produce flowers.

Question 5.
How are the criteria for deciding divisions in plants different from the criteria for deciding the subgroups among animals?
Answer:
In plants, basic structure of their body is a major criteria based on which thallophytes are different from bryophytes. Apart from this, absence or presence of seeds is another important criteria. Gymnosperms and angiosperms are further segregated based on if seeds are covered or not. It is clear that it is the morphological character which makes the basis for classification of plants.

In animal, classification is based on more minute structural variations. So in place of morphology, cytology forms the basis. Animals are classified based on layers of cells, presence or absence of coelom. Further, higher hierarchy animals are classified based on the presence or absence of smaller features, like presence or absence of four legs.

Question 6.
Explain how animals in Vertebrata are classified into further subgroups.
Answer:
Vertebrata is divided into two super classes, viz. Pisces and Tetrapoda. Animals of pisces have streamlined body with fins and tails to assist in swimming. Animals of tetrapoda have four limbs for locomotion.
Tetrapoda is further classified into following classes:

1. Amphibia: Amphibians are adapted to live in water and on land. They can breathe oxygen through kin when under water.
2. Reptilia: These are crawling animals. Skin is hard to withstand extreme temperatures.
3. Aves: Forelimbs are modified into wings to assist in flying. Beaks are present. Body is covered with feathers.
4. Mammalia: Mammary glands are present to nurture young ones. Skin is covered with hair. Most of the animals are viviparous.

Diversity in Living Organisms Additional Questions

Diversity in Living Organisms Tissues Multiple Choice Questions

Question 1.
Find out incorrect sentence.
(a) Protista includes unicellular eukaryotic organisms.
(b) Whittaker considered cell structure, mode and source of nutrition for classifying the organisms in five kingdoms.
(c) Both Monera and Protista may be autotrophic and heterotrophic.
(d) Monerans have well defined nucleus.
Answer:
(d) Monerans have well defined nucleus.

Question 2.
Which among the following has specialised tissue for conduction of water?
(i) Thallophyta
(ii) Bryophyta
(iii) Pteridophyta
(iv) Gymnosperms
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv).
Answer:
(c) (iii) and (iv)

Question 3.
Which among the following produce seeds?
(a) Thallophyta
(b) Bryophyta
(c) Pteridophyta
(d) Gymnosperms.
Answer:
(d) Gymnosperms.

Question 4.
Which one is a true fish?
(a) Jellyfish
(b) Starfish
(c) Dogfish
(d) Silverfish.
Answer:
(c) Dogfish

Question 5.
Which among the following is exclusively marine?
(a) Porifera
(b) Echinodermata
(c) Mollusca
(d) Pisces.
Answer:
(b) Echinodermata

Question 6.
Which among the following animals have pores all over their body?
(a) Porifera
(b) Aves
(c) Mollusca
(d) Pisces.
Answer:
(a) Porifera

Question 7.
Which among the following have chi tin as cell wall?
(a) Sycon
(b) Yeast
(c) Jelly fish
(d) Euplectella.
Answer:
(c) Jelly fish

Question 8.
Which among the following is not a Monocotyledonous plant?
(a) Wheat
(b) Rice
(c) Maize
(d) Gram.
Answer:
(d) Gram.

Question 9.
Which among the following is not a dicotyledonous plant?
(a) Wheat
(b) Sunflower
(c) Mango
(d) Gram.
Answer:
(a) Wheat

Question 10.
An organism with a single cell is called _______ .
(a) Thallophyta
(b) Bryophyta
(c) Unicellular
(d) Multicellular.
Answer:
(c) Unicellular

Question 11.
The amphibians of the plant is _______ .
(a) Thallophyta
(b) Bryophyta
(c) Unicellular
(d) Multicellular.
Answer:
(b) Bryophyta

Question 12.
Plant bearing naked seeds are _______ .
(a) Thallophyta
(b) Bryophyta
(c) Unicellular
(d) Gymnosperm.
Answer:
(d) Gymnosperm.

Diversity in Living Organisms Very Short Answer Type Questions

Question 1.
Name a saprophyte and also tell, why are they called so.
Answer:
Aspergillus: They are called so because they obtain their nutrition from dead and decaying matter.

Question 2.
Why are lichens called dual organisms?
Answer:
Lichens are permanent symbiotic association between algae and fungi. Therefore, they are called dual organisms.

Question 3.
State the phylum to which centipede and prawn belong.
Answer:
Arthropoda.

Question 4.
Name one reptile with four – chambered heart.
Answer:
Crocodile.

Question 5.
Identify kingdom in which organisms do not have well defined nucleus and do not show multicellular body designs.
Answer:
Monera.

Diversity in Living Organisms Short Answer Type Questions

Question 1.
Why do we differentiate organism, give two main basis?
Answer:
Due to variation in various characteristics, we differentiate organism. Two main basis are mode of nutrition and habitat.

Question 2.
Which kingdom generate food on earth and initiate food chain?
Answer:
Plantae.

Question 3.
Which kingdom do not have cell wall to their cell?
Answer:
Animalia.

Question 4.
What do you understand by biodiversity?
Answer:
Biodiversity: The variety of living beings found in a particular geographical area is called biodiversity of that area. Amazon rainforests is the largest biodiversity hotspot in the world.

Question 5.
Why classification is required?
Answer:
Classification is necessary for the study of living beings in easy way. Without proper classification, it would be impossible to study millions of organisms which exist on this earth.

Question 6.
What was the basis of classification of Ancient Greek philosopher Aristotle?
Answer:
Aristotle classified living beings on the basis of their habitat. He classified them into two groups, i.e. those living in water and those living on land.

Question 7.
How can we divide organism on the basis of mode of nutrition ?
Answer:
On this basis, organisms can be divided into two broad groups, i.e. autotrophs and heterotrophs.

Question 8.
Define Monocotyledonous plants. Give examples.
Answer:
Monocotyledonous: There is single seed leaf in a seed. A seed leaf is a baby plant.
Examples:

• wheat
• rice
• maize, etc.

Question 9.
Give example of Dicotyledonous plants.
Answer:
Dicotyledonous plants: Mustard, gram, mango etc.

Question 10.
Give one difference between prokaryotes and eukaryotes
Answer:

1. Prokaryotes: When nucleus is not organized, i.e., nuclear materials are not membrane bound; the organism is called prokaryote.
2. Eukaryotes: When nucleus is organized, i.e., nuclear materials are membrane bound; the organism is called eukaryote.

Question 11.
What is the difference between unicellular and multicellular organism?
Answer:

1. Unicellular organism: An organism with a single cell is called unicellular organism.
2. Multicellular organism: An organism with more than one cell is called multicellular organism.

Question 12.
Write short notes on the following:
(a) Thallophyta
(b) Bryophyta
Answer:
(a) Thallophyta: The plant body is thallus type. The plant body is not differentiated into root, stem and leaves. They are known as algae also.
Examples:

• Spirogyra
• chara
• volvox
• ulothtrix etc.

(b) Bryophyta: Plant body is differentiated into stem and leaf like structure. Vascular system is absent, which means there is no specialized tissue for transportation of water, minerals and food. Bryophytes are known as the amphibians of the plant kingdom, because they need water to complete a part of their life cycle.
Examples:

• Moss
• marchantia.

Question 13.
What are cryptogams and phanerogams?
Answer:
Plant body is differentiated into root, stem and leaf. Vascular system is present. They do not bear seeds and hence are called cryptogams. Plants of rest of the divisions bear seeds and hence are called phanerogams.
Examples:

• Marsilear
• ferns
• horse tails etc.

Question 14.
How gymnosperms are different from angiosperms?
Answer:

1. Gymnosperms: They bear seeds. Seeds are naked i.e., are not covered. The word ‘gymnos’ means naked and ‘sperma’ means seed.
2. Angiosperms: The seeds are covered. The word ‘angios’ means covered. There is great diversity in species of angiosperm.

Question 15.
What is porifera?
Answer:
Porifera: These animals have pores all over their body. The pores lead into the canal system. They are marine animals. Examples:

• Sycon
• Spongilla
• Euplectella, etc.

Question 16.
What is coelenterata?
Answer:
Coelenterata: The body is made up of a coelom (cavity) with a single opening. The body wall is made up of two layers of cells (diploblastic).
Examples:

• Hydra
• jelly fish
• sea anemone, etc.

Question 17.
What is Platyhelminthes?
Answer:
The body is flattened from top to bottom and hence the name platyhelminthes. These are commonly known as flatworms. The body wall is composed of three layers of cells (triploblastic).
Example:

• Planaria
• liver fluke
• tapeworm etc.

Question 18.
What is Nematohelminthes and Annelida?
Answer:
Nematohelminthes: Animals are cylindrical in shape and the body is bilaterally symmetric and there are three layers in the body wall.
Example:

• Roundworms
• pinworms
• filarial parasite (Wuchereria) etc.

Annelida: True body cavity is present in these animals. The body is divided into segments and hence the name annelida.
Example:

• Earthworm
• leech etc.

Question 19.
Explain the followings:
(a) Arthropoda
(b) Mollusca
(c) Echinodermata
(d) Protochordata
(e) Chordata.
Answer:
(a) Arthropoda: Animals have jointed appendages which gives the name arthropoda. Exoskeleton is present which is made of chitin. This is the largest group of animals; in terms of number of species.
Examples:

• cockroach
• housefly
• spider
• prawn
• scorpion etc.

(b) Mollusca: The animal has soft body; which is enclosed in a hard shell. The shell is made of calcium carbonate.
Examples:

• Snail
• mussels
• octopus etc.

(c) Echinodermata: The body is covered with spines, which gives the name echinodermata. Body is radially symmetrical. The animals have well developed water canal system, which is used for locomotion.
Examples:

• Starfish
• sea urchins etc.

(d) Protochordata: Animals are bilaterally symmetrical, triploblastic and ceolomate. Notochord is present at least at some stages of life.
Examples:

• Balanoglossus
• herdmania
• amphioxus etc.

(e) Chordata: Animals have notochord, pharyngeal gill slits and post anal tail; for at least some stages of life. Phylum chordata is divided into many sub – phyla; out of which we shall focus on vertebrata.

Diversity in Living Organisms Long Answer Type Questions

Question 1.
What is the different levels of organizations in case multicellular organism?
Answer:
Level of organization: Even in case of multicellular organisms, there can be different levels of organization:
(a) Cellular level of organization: When a cell is responsible for all the life processes, it is called cellular level of organization.

(b) Tissue level of organization: When some cells group together to perform specific function, it is called tissue level of organization.

(c) Organ level of organization: When tissues group together to form some organs, it is called organ level of organization. Similarly organ system level of organization is seen in complex organisms.

Question 2.
“Classification of living organism is based on evolution.” Explain.
Answer:
It is a well – established fact that all the life forms have evolved . from a common ancestor. Scientists have proved that the life begun on the earth in the form of simple life forms. During the course of time, complex organism evolved from them. So, classification is also based on evolution. A simple organism is considered to be primitive while a complex organism is considered to be advanced.

Question 3.
Explain five kingdom classification by Robert Whittaker (1959).
Answer:
Five Kingdom Classification by Robert Whittaker (1959):
This is the most accepted system of classification. The five kingdoms and their key characteristics are given below:

1. Monera: These are prokaryotes; which means nuclear materials are not membrane bound in them. They may or may not have cell wall. They can be autotrophic or heterotrophic. All organisms of this kingdom are unicellular. Examples: bacteria, blue green algae (cyanobacteria) and mycoplasma.

2. Protista: These are eukaryotes and unicellular. Some organisms use cilia or flagella for locomotion. They can be autotrophic or heterotrophic. Examples: unicellular algae, diatoms and protozoans.

3. Fungi: These are heterotrohic and have cell wall. The cell wall is made of chitin. Most of the fungi are unicellular. Many of them have the capacity to become multicellular at certain stage in saprophytic. Some fungi live in symbiotic relationship with other organisms, while some are parasites as well.
Examples:

• Yeast
• penicillum
• aspergillus
• mucor etc.

4. Plantae: These are multicellular and autotrophs. The presence of chlorophyll is a distinct characteristic of plants, because of which they are capable of doing photosynthesis. Cell wall is present.

5. Animalia: These are multicellular and heterotophs. Cell wall is absent. They feed on decaying organic materials.

Diversity in Living Organisms Higher Order Thinking Skills (HOTS)

Question 1.
What are the differences between Platyhelminthes and Nematohelminthes?
Answer:

Platyhelminthes	Nematohelminthes
1. Form: They are flat in shape and are called flat worms.	1. They are cylindrical in form and are called round worms.
2. Sexuality: Animals are hermaphrodite.	2. Animals are uni – sexual.
3. Coelom: Platyhelminthes are acoelomate.	3. Nematohelminthes are pseudocoelomate.
4. Digestive Tract: It is incomplete.	4. It is complete

Question 2.
Differentiate between animals belonging to the Mammalia group and those in the Aves group.
Answer:
Differences between mammals and aves.

Mammals	Aves
1. Give birth to young ones except platypus and the echidna.	1. Lay eggs.
2. Mammary glands are present.	2. Mammary glands are absent.
3. Body covered with hair.	3. Body covered with feathers.
4. Sweat and sebaceous glands are present in the skin.	4. Sweat and sebaceous glands are not present in the skin.

Diversity in Living Organisms Value Based Questions

Question 1.
Ashish, a IX class student, was studying chapter, ‘Diversity in Living Organisms’. He thought that all the fungi are harmful as these spoil food and cause various diseases. However, his elder sister Dimple told him that not all fungi are harmful as these are also used in making bread, vitamins and medicines.

1. Name any fungus which is the source of some medicine.
2. Name any fungus which is used in bread making.
3. What value are displayed by Ashish’s sister?

Answer:

1. Pencillium.
2. Yeast.
3. Dimple acted as elder sister and enhanced his younger brother’s scientific knowledge about fungi and their functions.

Question 2.
Coral is getting diminished in all the oceans due to global warming. People in Goa island protects their coral by not allowing people / tourist to take it away.

1. What is the phylum of coral?
2. What is coral made up of?
3. What value of people in Goa island is reflected here?

Answer:

1. Coelenterates is the phylum of coral.
2. It is made up of calcium carbonate.
3. They reflect the value of being responsible citizen, respecting environment and nature.

MP Board Class 9th Science Solutions

MP Board Class 9th Science Solutions Chapter 4 Structure of the Atom

Structure of the Atom Intext Questions

Structure of the Atom Intext Questions Page No. 47

Question 1.
What are canal rays?
Answer:
Canal rays are positively charged radiations which led to the discovery of positively charged sub-atomic particle called proton. These rays were discovered by E. Goldstein.

Question 2.
If an atom contains one electron and one proton, will it carry any charge or not?
Answer:
The atom will not contain any charge and will be electrically neutral because both electron and proton will balance each other.

Structure of the Atom Intext Questions Page No. 49

Question 1.
On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Answer:
According to Thomson’s model, an atom consist of a positively charged sphere and electrons are embedded in it. So, both charges are equal which makes the atom electrically neutral.

Question 2.
On the basis of Rutherford’s model of an atom, which sub – atomic particle is present in the nucleus of an atom?
Answer:
Proton is the sub – atomic particle which is present in the nucleus of an atom.

Question 3.
Draw a sketch of Bohr’s model of an atom with three shells.
Answer:

Question 4.
What do you think would be the observation if the a-particle scattering experiment is carried out using a foil of a metal other than gold?
Answer:
The observations would be same as that of gold foil.

Structure of the Atom Intext Questions Page No. 49

Question 1.
Name the three sub – atomic particles of an atom.
Answer:

1. Positively charged – Protons
2. Negatively charged – Electrons
3. No charged – Neutrons.

Question 2.
Helium atom has an atomic mass of 4u and two protons in its nucleus. How many neutrons does it have?
Answer:
Atomic mass = Number of protons + Number of neutrons
∴ 4 = 2 + Number of neutrons
∴ Number of neutrons = 4 – 2 = 2.

Structure of the Atom Intext Questions Page No. 50

Question 1.
Write the distribution of electrons in carbon and sodium atoms.
Answer:
Atomic number of Carbon = 6

Question 2.
If K and L shells of an atom are full, then what would be the total number of electrons in the atom?
Answer:
K shell is the 1 shell
So, n = 1
Then maximum electron’s = 2n² = 2 × (1)²
= 2 × 1 = 2
and L shell is the second shell.
So, n = 2
Then maximum electrons = 2(n)²
= 2 × (2)² = 8
∴ Total number of electrons = 2 + 8 = 10.

Structure of the Atom Intext Questions Page No. 52

Question 1.
How will you find the valency of chlorine, sulphur and magnesium?
Answer:
We know that valency is the number of electrons lost, gained or shared by atom to become stable or to complete 8 electrons in the shell.
Now, Chlorine,
Atomic number = 17

Then, it will take 8 – 7 = 1 electron to complete its shell.
∴ Its valency is ‘I’
Sulphur, Atomic number = 16

It will take 8-6 = 2 electrons to complete its shell.
∴ Its valency is ‘2’.
Magnesium, Atomic number = 12

It will lose 2 electrons from its outermost shell to become stable.
∴ Its valency will be ‘2’.

Structure of the Atom Intext Questions Page No. 52

Question 1.
If number of electrons in an atom is 8 and number of protons is also 8 then,
(i) What is the atomic number of the atom? and
(ii) What is the charge on the atom?
Answer:
(i) Number of electrons = 8 and,
Number of protons =8
Then, Atomic number = Number of protons = 8

(ii) Now, total electrons (-) = Total protons (+)
So, atom will be electrically neutral.

Question 2.
With the help of table 4.1 of Textbook, find out the mass number of oxygen and sulphur atom.
Answer:
From the table, we have,
Oxygen,
Mass Number = Number of protons + Number of neutrons
= 8 + 8 = 16
Sulphur,
Mass number = Number of protons + Number of neutrons
= 16 + 16 = 32.

Structure of the Atom Intext Questions Page No. 53

Question 1.
For the symbol H, D, and T tabulate three sub – atomic particles found in each of them.
Answer:
H, D, and T stand for protium, deuterium and tritium as isotopes of hydrogen atom.
Table:

Question 2.
Write the electronic configuration of any one pair of isotopes and isobars.
Answer:
Pair of isotopes: \(_{ 6 }^{ 12 }{ C }\), \(_{ 6 }^{ 14 }{ C }\)
Electronic configuration:

Structure of the Atom NCERT Textbook Exercises

Question 1.
Compare the properties of electrons, protons and neutrons.
Answer:

Question 2.
What are the limitations of J.J. Thomson’s model of the atom?
Answer:
J.J. Thomson’s model explained the existence of positive charge in the form of sphere and electrons embedded in it. But, he was unable to explain the Rutherford’s gold foil experiment in which most of positive α – particles passed straight, existence of electrons in the circular path and protons at the centre of the atom.

Question 3.
What are the limitations of Rutherford’s model of the atom?

Answer:
Rutherford explained that electrons revolve in a circular path which is found contradictory in terms of stability of atom. Because electrons are negatively charged and when they move continuously in circular paths then they should lose their energies and finally, fall into the positively charged nucleus making atoms unstable and collapse.

Question 4.
Describe Bohr’s Model of the atom.
Answer:
Neils Bohr proposed the theory for model of the atom. It is explained as:

1. Atom is made up of three sub – atomic particles as electrons, protons and neutrons.
   
2. The electrons move round the nucleus in fixed circular paths called orbits or shells.
3. The orbits are represented by the letters K, L, M, N, or the number n = 1, 2, 3, 4.
4. Centre of the atom is called the nucleus.
5. Electrons do not radiate energies while revolving in the orbits.
6. Electrons gain energy when they jump from lower shell to higher shell and lose energy when they return down from higher energy level to lower energy level.

Question 5.
Compare all the proposed models of an atom given in this chapter.
Answer:

Question 6.
Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Answer:
Rules:

(a) Maximum electrons present in a shell is given by 2n² whereas n is the number of that shell.
Like,

• K Shell, n = 1 → 2n² = 2 × (1)² = 2
• L Shell, n = 2 → 2n² = 2 × (2)² = 8
• M Shell, n = 3 → 2n² = 2 × (3)² = 18
• N Shell, n = 4 → 2n² = 2 × (4)² = 32.

(b) The outermost shell can have maximum of 8 electrons.
(c) Electrons cannot be occupied in a shell till its inner shells or orbits are completely filled.

Question 7.
Define valency by taking examples of silicon and oxygen.
Answer:
Valency is the combining capacity of an atom to become electrically stable. Or It means how many electrons are lost or gained by an atom to become stable.
In Silicon,

It has 4 valence electrons.
So, it will lose 4 electrons to become stable.
∴ Its valency is 4.
In Oxygen,

It has 6 valence electrons.
So, it will gain 8 – 6 = 2 electrons to become stable.
∴ Its valency is 2.

Question 8.
Explain with examples:
(i) Atomic number
(ii) Mass number
(iii) Isotopes
(iv) Isobars
Give any two uses of isotopes.
Answer:
(i) Atomic number: It is equal to the total number of
protons in the nucleus of its atom.
E.g.,

• Carbon has 6 protons. So, its atomic number is 6.

(ii) Mass number: It is equal to the sum of total number of protons and neutrons in the nucleus.
E.g.,

• Sodium has 11 protons and 12 neutrons. So, its mass number is 11 + 12 = 23

(iii) Isotopes: These are atoms of the same element having same atomic number, but different mass number.
E.g.

• \(_{ 35 }^{ 79 }{ Br }\), \(_{ 35 }^{ 81 }{ Br }\), \(_{ 6 }^{ 12 }{ C }\), \(_{ 6 }^{ 14 }{ C }\)

(iv) Isobars: These are the atoms of different elements having different atomic number but same mass number.
E.g.

• \(_{ 18 }^{ 40 }{ Ar }\), \(_{ 20 }^{ 40 }{ Ca }\), \(_{ 11 }^{ 24 }{ Na }\), \(_{ 12 }^{ 24 }{ Mg }\)

Use of Isotopes:

• Uranium isotope is used as a fuel in nuclear reactor for generating electricity.
• Sodium isotope is used to detect the blood clots.

Question 9.
Na+ has completely filled K and L shells. Explain.
Answer:
Atomic number of sodium (Na) is 11.

Now, if Na loses 1 electron then it will become Na⁺.

Now K shell can have maximum of 2 electrons and L shell can have maximum of 8 electrons.
Then, Na⁺ has completely filled K and L shell.

Question 10.
If bromine atom is available in the form of, say, two isotopes \(_{ 35 }^{ 79 }{ Br }\) (49.7%) and \(_{ 35 }^{ 81 }{ Br }\)Br (50.3%), calculate the average atomic mass of bromine atom.
Answer:
Average atomic mass of bromine atom
= 49.7% of atomic mass of \(_{ 35 }^{ 79 }{ Br }\) + 50.3% of atomic mass of \(_{ 35 }^{ 81 }{ Br }\)
= 49.7% of 79 + 50.3% of 81
= \(\frac { 49.7 }{ 100 }\) × 49 + \(\frac { 450.3 }{ 100 }\) × 81
= (39.263 + 40.743)u = 80.006u

Question 11.
The average atomic mass of a sample of an element X is 16.2u. What are the percentages of isotopes If \(_{ 8 }^{ 16 }{ X }\) and \(_{ 8 }^{ 18 }{ X }\) in the sample?
Answer:
Let the percentage of \(_{ 8 }^{ 16 }{ X }\) in sample be x% and percentage of \(_{ 8 }^{ 18 }{ X }\) in sample be (100 – x)%.
Now,
x% of 16 + (100 – x)% of 18 = 16.2


-2x + 1800 = 16.2 × 100 – 2x + 1800 = 1620
∴ -2x = 1620- 1800 = -180
x = \(\frac {180}{2}\) = 90.
∴ Percentage of  \(_{ 8 }^{ 16 }{ X }\) is 90% and percentage of \(_{ 8 }^{ 18 }{ X }\) is (100 – 90)% = 10%.

Question 12.
If Z = 3, what would be the valency of the element? Also, name the element.
Answer:
Z = 3
So, atomic number = 3 (∵ Z = atomic number)
∴ Electronic configuration = 2, 1
Valency = 1
The name of the element is lithium (Li).

Question 13.
Composition of the nuclei of two atomic species X and Y are given as under

Give the mass number of X and Y. What is the relation between the two species?
Answer:
Mass number of X = Protons + Neutrons = 6 + 6 = 12
And,
Mass number of Y = Protons + Neutrons = 6 + 8 = 14
Both species have same atomic number.
So, they are isotopes of the same element.

Question 14.
For the following statements, write T for True and F for False.

1. J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
2. A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
3. The mass of an electron is \(\frac {1}{2000}\) times that of proton.
4. An isotope of iodine is used for making tincture iodine which is used as a medicine.

Answer:

1. False
2. False
3. True
4. False.

Put tick (✓) against correct choice and cross (✗) against wrong choice in questions 15, 16 and 17.

Question 15.
Rutherford’s alpha – particle scattering experiment was responsible for the discovery of.
(a) Atomic nucleus
(b) Electron
(c) Proton
(d) Neutron.
Answer:
(a) Atomic nucleus

Question 16.
Isotopes of an element have.
(a) the same physical properties
(b) different chemical properties
(c) different number of neutrons
(d) different atomic numbers.
Answer:
(c) different number of neutrons

Question 17.
Number of valence electrons in Cl^– ion are:
(a) 16
(b) 8
(c) 17
(d) 18
Answer:
(b) 8

Question 18.
Which one of the following is a correct electronic configuration of sodium?
(a) 2, 8
(b) 8, 2, 1
(c) 2, 1, 8
(d) 2, 8, 1.
Answer:
(d) 2, 8, 1.

Question 19.
Complete the following table:

Atomic Number	Mass Number	Number of Neutrons	Number of Protons	Number of Electrons	Name of the Atomic Species
9	–	10	–	–	–
16	32	–	–	–	Sulphur
–	24	–	12	–	–
–	2	–	1	–	–
–	1	0	1	0	–

Answer:

Atomic Number	Mass Number	Number of Neutrons	Number of Pro-­tons	Number of Elec­trons	Name of the Atomic Species
9	19	10	9	9	Fluorine
16	32	16	6	6	Sulphur
12	24	12	12	12	Magnesium
1	2	1	1	1	Hydrogen
1	1	0	1	0	Deuterium

Structure of the Atom Additional Questions

Structure of the Atom Multiple Choice Questions

Question 1.
Which is a positive sub – atomic particle?
(a) Proton
(b) Neutron
(c) Electron
(d) None of these.
Answer:
(a) Proton

Question 2.
Electron is discovered by _____ .
(a) J.Chadwick
(b) Neils Bohr
(c) J.J Thomson
(d) Rutherford.
Answer:
(c) J.J Thomson

Question 3.
Proton is discovered by _____ .
(a) Rutherford
(b) J. Chadwick
(c) J J. Thomson
(d) E. Goldstein.
Answer:
(d) E. Goldstein.

Question 4.
Neutron is discovered by _____ .
(a) J.J. Thomson
(b) J. Chadwick
(c) Neils Bohr
(d) Rutherford.
Answer:
(b) J. Chadwick

Question 5.
Nucleus is discovered by _____ .
(a) Rutherford
(b) J. Chadwick
(c) J.J. Thomson
(d) Neils Bohr.
Answer:
(a) Rutherford

Question 6.
Mass of electron is _____ .
(a) 9 × 10^-25g
(b) 6 × 10^-28g
(c) 8 × 10^-24g
(d) 9 × 10^-28g.
Answer:
(d) 9 × 10^-28g.

Question 7.
Mass of Neutron is _____ .
(a) 1.6 × 10^-22g
(b) 1.6 × 10^-23g
(c) 1.6 × 10^-25g
(d) 1.6 × 10^-24g.
Answer:
(d) 1.6 × 10^-24g.

Question 8.
Charge on an electron is _____ .
(a) -1.8 × 10^-18C
(b) -1.7 × 10^-20C
(c) -1.6 × 10^-19C
(d) -1.5 × 10^-21C.
Answer:
(c) -1.6 × 10^-19C

Question 9.
The energy paths in an atom in which electrons revolve are called _____ .
(a) Rings
(b) Cycles
(c) Orbits
(d) Circles.
Answer:
(c) Orbits

Question 10.
ass number is the sum of _____ .
(a) Protons and Electrons
(b) Protons and Neutrons
(c) Electrons, Protons and Neutrons
(d) None of these.
Answer:
(c) Electrons, Protons and Neutrons

Question 11.
Atomic number is equal to _____ .
(a) Number of protons
(b) Number of neutrons
(c) Number of electrons
(d) Both (a) and (c).
Answer:
(d) Both (a) and (c)

Question 12.
Maximum number of electrons that can be filled in ‘M’ shell are _____ .
(a) 17
(b) 19
(c) 18
(d) 20.
Answer:
(c) 18

Question 13.
An atom has atomic number ‘17’, then its valency will be _____ .
(a) 7
(b) 2
(c) 1
(d) 8.
Answer:
(c) 1

Question 14.
Isotopes of an element have same number of _____ .
(a) Neutrons
(b) Protons
(c) Electrons
(d) Both (b) and (c).
Answer:
(c) Electrons

Question 15.
Isobars of different elements have same _____ .
(a) Atomic number
(b) Electrons
(c) Mass number
(d) Neutrons.
Answer:
(d) Neutrons

Structure of the Atom Very Short Answer Type Questions

Question 1.
Who discovered canal rays?
Answer:
E. Goldstein.

Question 2.
Name the fruit which resembles J.J. Thomson model of atom.
Answer:
Watermelon.

Question 3.
Who discovered nucleus?
Answer:
Ernest Rutherford.

Question 4.
Who discovered neutrons?
Answer:
James Chadwick.

Question 5.
Name the central part of an atom where protons and neutrons are held together.
Answer:
Nucleus.

Question 6.
What is Alpha Particle?
Answer:
It is a Helium ion (He²⁺) which has 2 units of positive charge and 4 units of mass.

Question 7.
What are cathode rays?
Answer:
Cathode rays are a beam of fast moving electrons.

Question 8.
What was the main drawback of Rutherford’s model of the atom?
Answer:
Inability to explain the stability of atom.

Question 9.
Write the symbolic representation of an element A with atomic number 10 and mass number 20.
Answer:
\(_{ 10 }^{ 20 }{ A }\)

Question 10.
Name three Isotopes of Hydrogen.
Answer:

1. Protium (\(_{ 1 }^{ 1 }{ H }\))
2. Deuterium (\(_{ 1 }^{ 2 }{ H }\))
3. Tritium (\(_{ 1 }^{ 3 }{ H }\))

Question 11.
Write the electronic configuration of potassium (K).
Answer:
Atomic number of potassium (K) =19

Question 12.
Define valency.
Answer:
It is the combining capacity of an atom to become electrically stable.

Question 13.
What is the charge of a proton?
Answer:
1.6 × 10^-19C.

Question 14.
Write the year of discoveries of these sub – atomic particles – electron, proton, neutron and neucleus.
Answer:

1. Electron – 1897
2. Proton – 1866
3. Neutron – 1932
4. Nucleus – 1911.

Question 15.
Which radioactive Isotope is used in treatment of goitre?
Answer:
Iodine – 131.

Structure of the Atom Short Answer Type Questions

Question 1.
Define:
(a) Canal rays
(b) Cathode rays
(c) Atomic number
(d) Mass number
(e) Energy shells
(f) Valency
(g) Octet
(h) Isotopes
(t) Isobars.
Answer:
(a) Canal rays: These are positively charged radiations which led to the discovery of sub – atomic positively charged particles called protons through an experiment conducted by J.J. Thomson in 1897.

(b) Cathode rays: These are negatively charged radiations which led to the discovery of sub – atomic negatively charged particles called electrons during an experiment conducted by E. Goldstein in 1866.

(c) Atomic number: It is the number of protons present in the nucleus of an atom. It is represented by the letter ‘Z’.

(d) Mass number: It is the total number of protons and neutrons present in an atom of an element.
So, Mass number = Number of protons + Number of neutrons.

(e) Energy Shells: These are fixed circular paths around the nucleus of an atom in which electrons revolve continuously with high speed. These are also called orbits. They are represented by the alphabets K, L, M, N.

(f) Valency: It is the combining capacity of an atom to become electrically stable, or it also means the number of valency electrons lost or gained by an atom to complete the eight electrons in the valence shell.

(g) Octet: The completely filled outermost shell like L, M or N with 8 electrons is called an octet. When an atom completes its octet, then it become stable.

(h) Isotopes: These are atoms of same element having same atomic number, but different mass number.

E.g.

• (\(_{ 1 }^{ 1}{ H }\)) , (\(_{ 1 }^{ 2 }{ H }\)), (\(_{ 1 }^{ 3 }{ H }\)) and \(_{ 6 }^{ 12 }{ C }\), \(_{ 6 }^{ 14 }{ C }\)are the isotopes of hydrogen and carbon respectively.

(i) Isobars: These are atoms of different elements having different atomic number but same mass number.
E.g.

• \(_{ 18 }^{ 40 }{ Ar }\), \(_{ 20 }^{ 40 }{ Ca }\) and \(_{ 11 }^{ 24 }{ Na }\), \(_{ 12 }^{ 24 }{ Mg }\).

Question 2.
Differentiate between:
(a) Electrons and protons.
(b) Atomic number and mass number.
(c) Isotopes and isobars.
(d) Valence electrons and valency.
Answer:
(a)

Electrons	Protons
(i) This is negatively charged sub – atomic particle.	(i) This is positively charged sub – atomic particle.
(ii) Its mass is 9 × 10–28gms.	(ii) Its mass is 1.6 × 10-24gms.
(iii) Its symbol is “e–”.	(iii) Its symbol is “P+”.

(b)

Atomic number	Atomic mass
(i) It is the total number of protons present in the atom.	(i) It is the sum of protons and neutrons present in the atom.
(ii) It is represented by ‘Z’.	(ii) It is represented by ‘A’.
(iii) It is written on the bottom left as a subscript with the symbol the of element.	(iii) It is written on top left as a subscript with the symbol the of element.

(c)

Isotopes	Isobars
(i) These are atoms of the same element.	(i) These are atoms of the different elements.
(ii) They have same atomic number.	(ii) They have different atomic number.
(iii) They have different mass number.	(iii) They have same mass number.
(iv) They have same chemical properties.	(iv) They have different chemical properties.

(d)

Valence Electrons	Valency
(i) These are electrons present in the outermost shell of an atom.	(i) These are electrons lost or gained through valence shell of an atom to become stable.
(ii) Valence electrons can be 1, 2, 3 …….. 8 or more.	(ii) Valency can be 0, 1, 2, 3, 4 only.

Question 3.
Draw the diagrams of:
(a) J.J. Thomson’s model of atom.
(b) Rutherford’s model of atom.
(c) Neils Bohr’s model of atom.
Answer:
(a) J.J. Thomson’s Model:

(b) Rutherford’s Model:

(c) Neils Bohr’s model of atom.

Question 4.
Write the postulates of J.J. Thomson’s model of the atom.
Answer:
J.J. Thomson’s postulates for model of the atom are as follows:

1. An atom is a positively charged sphere or ball and negatively charged electrons are embedded in it.
2. The atom is electrically neutral because negative and positive charges are equal in magnitude.

Question 5.
Write the main points of the theory given by Rutherford for model of atom.
Answer:
Main points of theory of Rutherford regarding model of atom are:

1. There is an existence of positively charged centre in the atom called as nucleus which contains all the mass of the atom.
2. The electrons revolve round the nucleus in circular paths called orbits at high speeds.
3. The size of nucleus (centre of the atom) is very small as compared to size of the atom.
4. Most of the sphere in an atom is empty.

Question 6.
Write the rules given by Bohr – Bury for arrangement of electrons in different orbits in an atom.
Answer:
Rules given by Bohr – Bury are as follows:

1. Maximum electrons present in a shell is given by 2n² where
   • n is the number of that shell.
   • Like, for first shell K, n = 1
   • For second shell L, n = 2
   • third shell M, n = 3
   • fourth shell N, n = 4 called as nucleus which contains all the mass of the atom.
2. The outermost shell can have maximum of 8 electrons.
3. Electrons cannot occupy a shell till its inner shells or orbits are completely filled.

Question 7.
An atom ‘X’ has a mass number ‘23’ and atomic number ‘11’. Find its electrons, protons and neutrons. Also, name the element
Answer:
We know,
Atomic number = Number of protons.
∴ 11 = Number of protons
And, Number of protons = Number of electrons.
∴ Number of electrons = 11
Now, Mass number = Protons + Neutrons.
23 = 11 + Neutrons
∴ Neutrons = 23 – 11 = 12
∴ Atom ‘X’ has 11 electrons, 11 protons and 12 neutrons.
The element is Sodium (Na).

Question 8.
Write the electronic configuration of neon, aluminium, sulphur, argon. Also, find valencies.
Answer:

Question 9.
What are radioactive isotopes? Write down the type of isotopes used in:
(a) Tracing blood clots and tumours in human body
(b) Treatment of cancer
(c) Treatment of Goitre
(d) Nuclear reactor as a fuel.
Answer:
Radioactive isotopes: These are unstable isotopes due to extra neutrons in their nucleus and emits different types of radiations.
Examples:

• Uranium – 235
• Cobalt – 60
• Carbon – 14.

Types of Isotopes used in:
(a) Sodium – 24 to detect blood clots and Arsenic – 72 to detect tumours.
(b) Cobalt – 60
(c) Iodine-131
(d) Uranium – 235.

Question 10.
Draw the atomic structure of:
(a) Fluorine atom (F)
(b) Sodium atom (Na)
(c) Potassium atom (K)
Answer:
(a) Fluorine atom (F):
Atomic number: 9

(b) Sodium atom (Na)
Atomic number: 11

(c) Potassium atom (K)
Atomic number: 19

Question 11.
Write the atomic number, mass number, electrons, protons and neutrons of following atoms:
(a) \(_{ 14 }^{ 24 }{ X }\)
(b) \(_{ 13 }^{ 27 }{ X }\)
Answer:
(a) \(_{ 14 }^{ 24 }{ X }\)
Atomic number = 14
Mass number = 24
Electrons = 14
Protons = 14
Neutrons = 24 – 14 = 10

(b) \(_{ 13 }^{ 27 }{ X }\)
Atomic number = 13
Mass number = 27
Electrons = 13
Protons = 13
Neutrons = 27 – 13 = 14

Question 12.
Pick out the Isotopes and Isobars from the following atoms:
\(_{ 17 }^{ 37 }{ A }\), \(_{ 18 }^{ 40 }{ A }\), \(_{ 17 }^{ 33 }{ A }\), \(_{ 20 }^{ 40 }{ A }\).
Answer:

1. Isotopes: \(_{ 17 }^{ 37 }{ A }\), \(_{ 17 }^{ 33 }{ A }\)
2. Isobars: \(_{ 18 }^{ 40 }{ A }\), \(_{ 20 }^{ 40 }{ A }\).

Question 13.
What are noble gases? Why they are stable? Give three examples.
Answer:
Noble gases are the elements which are stable and do not take part in chemical reaction.
They are stable because they have completely filled outer- most shell with 8 electrons
example:

Helium is the only noble gas which has 2 electrons in outermost shell.

Question 14.
Write all the Isotopes of:

1. Hydrogen
2. Oxygen
3. Chlorine
4. Bromine
5. Carbon
6. Neon.

Answer:

1. Hydrogen (H) – \(_{ 1 }^{ 1 }{ H }\), \(_{ 1 }^{ 2 }{ H }\)
2. Oxygen (O) – \(_{ 8 }^{ 16 }{ O }\), \(_{ 8 }^{ 17 }{ O }\), \(_{ 8 }^{ 18 }{ O }\)
3. Chlorine (Cl) – \(_{ 17 }^{ 35 }{ Cl }\), \(_{ 17 }^{ 37 }{ Cl }\)
4. Bromine (Br) – \(_{ 35 }^{ 79 }{ Br }\), \(_{ 35 }^{ 81 }{ Br }\)
5. Carbon (C) – \(_{ 6 }^{ 12 }{ C }\), \(_{ 6 }^{ 14 }{ C }\)
6. Neon (Ne) –  \(_{ 10 }^{ 20 }{ Ne }\), \(_{ 10 }^{ 21 }{ Ne }\), \(_{ 10 }^{ 22 }{ Ne }\)

Question 15.
Why is it wrong to say that atomic number of an atom is equal to its number of electrons?
Answer:
We know that in an atom number of electrons is equal to the number of protons. But, we cannot say that atomic number is equal to number of electrons because number of electrons can be changed after losing or gaining by an atom during chemical reaction. But, number of protons remain constant.

Question 16.
What explanation did Neils Bohr gave on stability of atoms?
Answer:
Neils Bohr explained the stability of atom through following points:

1. The electrons revolve around the nucleus in fixed orbits or energy levels or shells and each orbit has its fixed radius.
2. While revolving electrons do not radiate their energies, so they do not fall into the nucleus and make the atom stable.

Question 17.
What are nucleons? What is the name given to the atoms having same number of nucleons?
Answer:
Protons and neutrons together in the nucleus are called nucleons. It means number of nucleons is equal to the sum of protons and neutrons. Atoms having same number of nucleons are called isobars.

Structure of the Atom Long Answer Types Questions

Question 1.
The average atomic mass of a sample of an element X is 13u. What are the percentages of isotopes \(_{ 6 }^{ 12 }{ X }\) and \(_{ 6 }^{ 14 }{ X }\) in the sample?
Answer:
Let, the percentage of isotope \(_{ 6 }^{ 12 }{ X }\) be x%.
So, percentage of isotope \(_{ 6 }^{ 14 }{ X }\) is (100- x) %.
Now, Average atomic mass = Mass of \(_{ 6 }^{ 12 }{ X }\) + Mass of \(_{ 6 }^{ 14 }{ X }\) According to percentages,
∴ 13 = x% of 12 + (100 – x)% of 14

∴ 13 × 100 = 1400 – 2x
1300 = 1400 – 2x
1300 = 1400 – 2x
-100 = -2x
x = \(\frac { 100 }{2 }\) = 50
So, percentage of \(_{ 6 }^{ 12 }{ X }\) is 50% and percentage of \(_{ 6 }^{ 14 }{ X }\) is
= (100 – x)%
= (100 – 50)%
= 50%

Question 2.
Explain Rutherford’s Gold Foil Experiment. Also, explain its observations conclusion, theory proposed and drawback of his model.
Answer:
Ernest Rutherford performed an alpha-particles scattering experiment in which he passed a-particles on the gold foil.
Observation:

1. Most of α – particles passed straight without any deflection.
2. Some of the α – particles get deflected from their path.
3. Very few α – particles get completely bounced back.

Conclusions:

1. Maximum space in an atom is vacant as most of α – particles passed straight without any deflection.
2. Some α – particles get deflected from their paths show the existence of positive charge in the atom.
3. Very few α – particles get completely bounced back indicating the concentration of all mass with positive charge in a small volume at the centre.

Theory proposed:

1. There is an existence of positively charged centre in the atom called nucleus which contains all the mass of the atom.
2. The electrons revolve around the nucleus in circular paths called orbits at high speeds.
3. The size of nucleus (centre of the atom) is very small as compared to size of the atom.
4. Most of the space in an atom is empty.

Drawback: Rutherford’s model did not explain the stability of the atom. He proposed that electron revolves around the nucleus in circular paths. So, electrons should radiate their energies as they are continuously in circular motion. Then, they should fall into the positively charged nucleus making the atom unstable and collapse.

Question 3.
Draw the electronic structure of sodium and calcium with atomic number 11 and 20 respectively.
Answer:
Sodium has electronic distribution as 2, 8, 1
Calcium has electronic distribution as 2, 8, 8, 2
Electronic structures of sodium and calcium are given:

Question 4.
Both helium (He) and beryllium (Be) have two valence electrons. Whereas ‘He’ represents a noble gas element, ‘Be’ does not. Assign reason.
Answer:
The element He (Z = 2) has two electrons present in the only shell
i.e., K – shell. Since, this shell can have a maximum of two electrons only therefore,
‘He’ is a noble gas element.
The element ‘Be’ (Z = 4) has the electronic configuration as: 2,2.
Although, the second shell has also two electrons but it do not represent a noble gas element.

Structure of the Atom Higher Order Thinking Skills (HOTS)

Question 1.
Which isotope of hydrogen contain same number of electrons, protons and neutrons?
Answer:
Deuterium (\(_{ 1 }^{ 2 }{ D }\))
Number of electron (1) = Number of proton (1)
= Number of neutron (2 – 1 = 1)

Question 2.
Which element of these two would be chemically more reactive: element A with atomic number 18 or element B with atomic number 16 and why?
Answer:
Electric configuration of

• A – 2,8,8
• B – 2, 8, 6

Since, the outermost shell of A is complete, it would be inert and will not react. Whereas element B require two atoms to complete its octet. Therefore, B would be more reactive.

Structure of the Atom Value Based Question

Question 1.
Shivek could not solve the following question in the group. His group – mate explained him and solved his difficulty.
The question was as follows:

What information do you get from the given figure about the atomic number, mass number and valency of the given atom ‘X’:

1. What is the atomic number, the mass number and valency of the atom?
2. Name the element ‘X’.
3. What value of Shivek’s friend are reflected in this behaviour?

Answer:

1. The atomic number is 5, The mass number is 11, The valency is 3.
2. The element ‘X’ is boron.
3. Shivek’s friend showed the values of helping and caring nature.

MP Board Class 9th Science Solutions

MP Board Class 9th Maths Solutions Chapter 5 Introduction to Euclid’s Geometry Ex  5.2

Question 1.
How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?
Solution:
If a line p intersects two lines l and m such that (∠1 + ∠2) is less than 180°, then lines l and m will meet at O, as shown in Fig. below.

Question 2.
Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.
Solution:
Yes, Euclid’s fifth postulate is important to express parallel lines. Two lines will never meet if they are not according to Euclid’s fifth postulate.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 5 Introduction to Euclid’s Geometry Ex  5.1

Question 1.
Which of the following statements are true and which are false? Give reasons for your answers.

1. Only one line can pass through a single point.
2. There are infinite number of lines which pass through two distinct points.
3. A terminated line can be produced indefinitely on both the sides.
4. If two circles are equal, then their radii are equal.
5. In Fig. below, if AB = PQ and PQ = XY, then AB = XY.

Solution:

1. False, infinitely many lines can pass through a given point.
2. False, only one line can pass through two distinct points.
3. True, by postulate 2 i.e., a terminated line can be produced indefinitely.
4. True, equal circles coincide each other. Therefore their radii will be equal.
5. True, by Euclid’s axiom 1, i.e., things which are equal to the same thing are equal to one another.

Question 2.
Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them?

1. parallel lines
2. perpendicular lines
3. line segment
4. radius of a circle
5. square

Solution:
1. Parallel lines:
Two distinct lines in a plane are called parallel lines if they do not have a common point. Here the undefined terms are lines and plane.

2. Perpendicular lines:
Two lines are perpendicular to each other if they intersect each other at right angle. Here the undefined term is right angle.

3. Line segment:
A part of a line between two points on a line is called a line segment. Here the undefined term is part of a line.

4. Radius of a circle:
Radius of a circle is the distance of a point on the circle from the center of the circle.

5. Square:
A square is a rectangle having all sides equal. Here undefined term is rectangle.

Question 3.
Consider two ‘postulates’ given below:

1. Given any two distinct points A and B, there exist a third point C which is in between A and B.
2. There exist at least three points that are not on the same line.

Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain.
Solution:
Yes, these postulates contain undefined terms such as point, line, distinct points. They are consistent because they deal with two different situations:

1. Point C is lying between two distinct points A and B on a line.
2. Point C is not lying on the line through A and B.

These postulates do not follow from Euclid’s postulates. However they follow from axiom “given two distinct points, there is a unique line that passes through them.

Question 4.
If a point C lies between two points A and B such that AC = BC, then prove that AC = \(\frac{1}{2}\)AB. Explain by drawing the figure.
Solution:
Given: AC = BC
To prove: AC = \(\frac{1}{2}\)AB
Proof:
AC = BC
Adding AC on both sides
AC + AC = BC + AC
2AC = AB
AC = \(\frac{1}{2}\)AB

Question 5.
In Question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one midpoint.
Solution:
If possible, Let us assume that a line segment AB has two mid points C and D when C is the mid point of AB

AC = 1/2 AB …(i)
where D is the mid point of AB
AD = 1/2 AB …(ii)
From (i) and (ii), we get
AC = AD
(By Euclid’s axiom, things which are half of the same thing are equal to one another.). This is possible only if C and D coincides.
∴ Our assumption that C and D are two mid points of AB are wrong and hence a line segment has one and only one mid point.

Question 6.
In Fig. below, if AC = BD, then prove that AB = CD.

Solution:
Given: AC = BD
To prove: AB = CD
Proof:
AC = BD
Subtracting BC on both sides, we get
AC – BC = BD – BC (By Euclid’s axiom-3)
∴ AB = CD

Question 7.
Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate).
Solution:
We know that whole is always greater than its part.

MP Board Class 9th Maths Solutions

MP Board Class 9th Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.4

Question 1.
Give the geometric representations of y = 3 as an equation –

1. In one variable
2. in two variables.

Solution:
1. Linear equation in one variable
y = 3

2. Linear equation in two variables is
0x + y – 3 = 6

Question 2.
Give the geometric representation of 2x + 9 = 0 as an equation –

1. In one variable
2. In two variables.

Solution:
1. Linear equation in one variable
2x + 9 = 0
2x = – 9
x = – 4.5

2. Linear equation in two variables
2x + 0y + 9 = 0

Equations of a Line Parallel to the x – axis and y – axis:
This is a special case when the given point lies on the axes, either x – axis or y – axis. If the point lies on x – axis, then y – coordinate will be 0 and if the point lies on y-axis, then the x-coordinate will be 0.

Example 1.
Draw the graph of the equation represented by a straight line which is parallel to the x – axis and at a distance 3 units below it. (NCERT Exemplar)
Solution:
The equation of a line which is parallel to the x-axis and at a distance of 3 units below. It is given by
y = – 3
The solutions of the equation are:

The graph is shown below.

MP Board Class 9th Maths Solutions