Class 9

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 9th Special English Short Composition with Guidance

1. A reporter made the following notes on a theft case news item. Complete the news using the given notes. Write the answers in your answer sheet against the correct blank numbers. (5 marks)

Notes Place — dace — time Market — Raghav Colony — 12-2-20xx — 3 A.M. Items stolen — jewelry — worth 3 lacs — cash 50 thousand — electronic goods — broken window — no clue.

“Theft At Raghav Colony”

It was reported that a theft (a) ……………………. on 12-2-20xx at 3-00 A.M. near the (b) ……………………. place. The burglars (c) ……………………. Some (d) ……………………. were reported to be missing from the house. So far (e) ……………………. found about the thieves.
Answer:
(a) took place in Raghav Colony
(b) market
(c) gained access into the house through the broken window.
(d) electronic goods jewelry worth 3 lacs and 50 thousand cash
(e) no clue has been.

2. A reporter made the following notes for a news item in the newspaper. complete the news using the given notes. Write the answers in your answer sheet against the correct blank numbers. (5 marks)

Notes Train Accident — Level crossing — 2 P.M. — 1.3.20xx 12 KM — Nagpur — collides with a truck — a high speed — 4 dead — several hurts.

I Accident At Level Crossing

I At a level crossing near Nagpur (a) ……………………… away a rail accident (b) ……………………… The train was running at (c) ……………………… It was (d) ……………………… on (e) ……………………… The train driver saw the truck and gave (f) The truck was also (g) ……………………… Its driver did not (h) ……………………… and tried to (i) ……………………… but failed. The driver of the truck was (j) ……………………… of those killed. The injured were
Answer:
(a) 12 KM (kilometers)
(b) occurred
(c) a high speed
(d) a collision with a truck
(e) the level crossing
(f) a signal
(g) at high speed
(h) pay heed to the signal
(i) the level crossing
(j) among one.

3. Salim made the following notes for the report regarding ‘Onion procession’ held in the town. Making use of the notes, complete the report by filling in the blanks. Write the answers in your answer sheet against the correct blank numbers. (5 marks)

Notes Procession — Cars — Jeeps — People — Delhi Gate — Connaught Place — Leader — Open jeep Balloon — Flower garlands — Onion garlands — Onions of different sizes — Huge balloon — shape of onion.

Onions Draw Crowds

In the form of huge a (a) ………………… threaded in (b) ………………… adorning every neck which made the long (c) ………………… proceeding from Delhi Gate to Connaught atop a car, onions appeared Place, it was a great attraction. The cavalcade was led by an (d) ………………… jeeP on which Mr. Dutta stood with (e) ………………… around his neck addressing the crowd against the rising prices.
Answer:
(a) balloon in the shape of an onion
(b) the garlands
(c) procession
(d) open
(e) a garland of onions.

4. Here are a few details about Gandhiji. Use this information to make a paragraph. Write the answers in your answer sheet against the correct blank numbers. (5 marks)

Birth — 2-10-1869 Residence — Rajkot — Gujarat. Parents — Father — Diwan, Mother — religious woman Education — India and England Profession — Lawyer — South Africa

Gandhiji was (a) ………………… They lived in (b) ………………… His father (c) ………………… and his mother! He got his education (d) ………………… He became (e) ………………… and
Answer:
(a) born on 2nd October 1869.
(b) Rajkot in Gujarat.
(c) was a Diwan and his mother was a religious woman.
(d) in India and England.
(e) a lawyer and went to South Africa.

5. Karishma got the grades given below. She asked her teacher to give her a written report on her abilities and achievements. Given below is a part of the report her teacher wrote. Based on the grades complete the sentences. Do not add any new information. Write the answers in your answer sheet against the correct blank numbers. (5 marks)

Karishma is a very bright girl who shows great application and diligence in class. Her special (a) ………………… and Computer Science where (b) ………………… this-year. She should, however, (c) ………………… in the Languages, (d) ………………… and is able to work as a team. (e) ………………… in harmony. Always neatly turned out, Karishma is a pleasure to have in class.
Answer:
(a) interests lie in Mathematics, Art
(b) she has scored A+ grades
(c) improve upon her grades
(d) In her music lessons and physical education classes she can improve
(e) Her leadership qualities, application and team work are

6. Prabha is a reporter for a paper. She made these notes for a news item. Read them and complete the news item given below. Write the answers in your answer sheet against the correct blank numbers. (5 marks)

Notes local train — traveling at full speed — Canning to Talki — Gauranga — 7 yr. old — tending cattle nearby — saw a break in track — took off and waved red shorts like a flag — driver noticed, braked, stopped — avoided the accident.

Seven Year Old Averts Train Accident

New Delhi, April 18
A local train was traveling at full speed from Canning to Talki in the Sealdah South section. Gauranga (a) ………………… near the railway line, saw that (b) ………………… at one place. He immediately (c) ………………… and waved it (d) ………………… rushing towards the train. The driver noticed the boy (e) ………………… just before the point of danger. Gauranga’s alertness saved hundreds of lives.
Answer:
(a) a 7-year-old boy who was tending cattle
(b) the track had broken
(c) took off his red shorts
(d) like a flag
(e) braked and stopped the train.

7. Meera, the editor of her school magazine, has made the sched-ule given below for the school magazine. She has written a note to her friends on the editorial board based on it. Complete the gaps left in the note. Do not add any new information. Write the answers in your answer sheet against the correct blank numbers. Do not copy the whole sentence. (5 marks)

Magazine Schedule For 2007-08 ”

Collection of material : through competitions in April-Aug.’ 07 (short story, poetry, cartoons) collect articles, cartoons comics in Aug. Sept’ 07 Reporters write on school activities as they occur. Selection of material : Jul.-Sept.’ 07 Editing and Illustrations: Aug.-Nov.’ 07 – Printing : Hand material to printer Nov.’ 07 Final proofs Feb.’ 08 delivery of magazine : March 08

This yeir (a) ……………………….. printed by March 2008, so we have to start our work immediately. Different (b) ……………………….. between April and August 2007 so that we can collect enough good irìterial for inclusion in the magazine. All material to he printed in the Åfl agazine should (c) ……………………….. by September 2007. Of course, the reporters on the editorial board will (d) ……………………….. school, activities as they occur. The magazine should be rediy and printed (e) ………………………..
Answer:
(a) we intend to have the school magazine
(b) competitions will be organised
(c) be selected
(ii) give in their write-ups on
(e) by February 1997.

8. Read the information given in the columns below and then fill in gaps in the paragraph appropriately. Do not use any extra information. Write the answers against the correct blank numbers in your answer sheet. (5 Marks)

Amrita and Salma are good friends though they are very different ‘from each other. Amrita is quiet and reserved though she likes the company of friends, (a) ……………………… and likes to talk a lot Salma also likes (b) ……………………… Amrita on the other hand, prefers Indian food. Amrita (c) ……………………… very well and reads a lot while Salma is a champion (d) ……………………… to relax. Even in television programmes their tastes are quite different. However, (e) ……………………… in classical music and old film songs and often attend concerts together.
Answer:
(a) Salma, on the other hand, is cheerful and outgoing
(b) all kinds of food, especially junk food
(c) can play the sitar
(d) swimmer and loves to do painting in order
(e) their choice of music is similar and both are interested.

9. Here is a diary extract of a tourist about his tour to Rajasthan. Use the information to complete the account of his visit to Rajasthan. Do not add any new information. Write the answers in your answer sheet against the correct blank number. (5 marks)
5th March :
Arrival at Jaipur
Visit to Hawa Mahal
Man Singh Art Gallery
Amer Fort: Examples of advanced architecture

7th March :
Arrival at Ajmer
Visit to Pushkar
Dargah

8th March :
Arrival at Udaipur
Visit to Palace, Temples (incl. Jain Temples)
Lakes, Boating, Chittorgarh, .
fort mines, Pratap Memorial — all very fascinating

10th March :
To Jodhpur forts, palaces

12th March :
Visit to the Jaisalmer desert.
Evidence of bravery, generosity of people, rulers

People: Cooperative

Visit To Land Of Forts And Palaces
My visit to Jaipur, Ajmer, Udaipur, Jodhpur and Jaisalmer made me feel that Rajasthan ,(a) …………………… During my visit to Jaipur I (b) …………………… and I concluded that (c) …………………… I was extremely thrilled during my visit to Udaipur. I enjoyed (d) …………………… The story of Rana Pratap and Rani Padmavati of Chittorgarh made me feel that (e) ……………………
Answer:
(a) is a place where the generosity and co-operative attitude of the people is most evident.
(b) visited the Hawa Mahal, the Man Singh Art Gallery and Amer Fort.
(c) these were excellent examples of advanced architecture.
(d) the fascinating sights of the palaces/ the Jain Temples, Chittorgarh and Pratap Memorial.
(e) the rulers of the place were indeed brave.

10. Aparijate has been asked to write an article for the ‘Nature’ section of the School magazine. She has referred to an encyclopedia and made the following notes about the ‘golden monkey’. Use the information in the notes to complete her article. Do not add any new information. Write the answers against the correct blank number in your answer sheet. (5 Marks)
Animal : Golden Monkey
Habitat : mountain forests — China
Size : Adult male 83 cm Adult feftiale 74 cm
Weight : Male 20 kg Female 10 kg
Movement : Winter grcfcips — 70 Summer groups — 300 (food plentiful)
Endangered : hunting, golden hair — make coats
Protection : reseryes, sanctuaries

The golden monkey is mainly found (a) ……………………. An adult male is 83 cm long and weighs. 20 kg. (b) ……………………. than the male beihg 74 cm long and weighing 10 kg. In winter (c) ……………………. while in summer (d) ……………………. they form groups of 300. In the past the monkeys were hunted for their gplden hair. Now-a-days (e) …………………….
Answer:
(a) in the mountain forests of China.
(b) An adult female is comparatively smaller
(c) they form groups of 70
(d) when food is in plentiful
(e) they are protected in reserves and sanctuaries

MP Board Class 9th 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 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

MP Board Class 9th Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.2

Question 1.
Which one of the following options is true, and why? y = 3x + 5 has

(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions.

Solution:
The true option is (iii) y = 3x + 5 has infinitely many solutions. Reason. For every value of x, there is a corresponding value of y and vice-versa.

Question 2.
Write four solutions for each of the following equations:

1. 2x + y = 7
2. πx + y = 9
3. x = 4y

Solution:
1. 2x + y = 7
Take x = 0
2 x 0 + y = 7
∴ y = 7
First solution is (0, 7)

Take x = 1,
2 x 1 + y = 7
2 + y = 7
y = 7 – 2
∴ y = 5
Second solution is (1, 5)

Take x = 2,
2 x 2 + y = 7
2 + y = 7
y = 7 – 4
∴ y = 3
Third solution is (2, 3)

Take x = 3,
2 x 3 + y = 7
6 + y = 7
y = 1 – 6
∴ y = 1
Fourth solution is (3, 1)

2. πx + y = 9
Take y = 0,
πx + 0 = 9
πx = 9
∴ x = \(\frac{9}{π}\)
First solution is (\(\frac{9}{π}\), 0)

Take y = 1,
πx + 1 = 9
πx =9 – 1
∴ x = \(\frac{8}{π}\)
Second solution is (\(\frac{8}{π}\), 1)

Take y = 2,
πx + 2 = 9
πx = 9 – 2
∴ x = \(\frac{7}{π}\)
Third solution is (\(\frac{7}{π}\), 2)

Take y = 3,
πx + 3 = 9
πx = 9 – 3
∴ x = \(\frac{6}{π}\)
Fourth solution is (\(\frac{6}{π}\), 3)

3. x = 4y
Take x = 0,
0 = 4y
\(\frac{0}{4}\) = y
∴ y = o
First solution is (0, 0)

Take x = 4,
4 = 4y
y = \(\frac{4}{4}\) = 1
Second solution is (4, 1)

Take x = 8,
8 = 4y
y = \(\frac{8}{4}\) = 2
Third solution is (8, 2)

Take x = 12,
12 = 4y
y = \(\frac{12}{4}\) = 3
Fourth solution is (12, 3)

Question 3.
Check which of the following are solutions of the equation x – 2y = 4 and which are not:

1. (0, 2)
2. (2, 0)
3. (4, 0)
4. (\(\sqrt{2}\), 4\(\sqrt{2}\) )
5. (1, 1)

Solution:
x – 2y = 4
1. Putting x = 0 and y = 2, we get
0 – 2 x 2 = 4
4 = 4
∴ (0, 2) is a solution

2. Putting x = 2 and y = 0, we get
2 – 2 x 0 = 4
2 ≠ 4
∴ (2, 0) is not the solution.

3. Putting x = 4 and y = 0, we get
4 – 2 x 0 = 4
4 = 4
∴ (4, 0) is a solution.

4. Putting x = \(\sqrt{2}\) and y = 4\(\sqrt{2}\), we get
\(\sqrt{2}\) – 2 x 4\(\sqrt{2}\) = 4
\(\sqrt{2}\) – 8\(\sqrt{2}\) = 4
– 7\(\sqrt{2}\) ≠ 4
∴ (\(\sqrt{2}\), 4\(\sqrt{2}\)) is not the solution.

5. Putting x = 1 and y = 1, we get
1 – 2 x 1 = 4
1 – 2 = 4
– 1 ≠ 4
∴ (1, 1) is not the solution.

Question 4.
Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
Putting x =2 and y – 1 in 2x + 3y – k, we get
2 x 2 + 3 x 1 = k
4 + 3 = k
∴ k = 1

Graph Of A Linear Equation In Two Variables:
The graph of a linear equation in two variables is a line. To draw a line we need atleast two points. Points are the solutions of the given equation. So to find the graph of a linear equation we will first find out three solutions and then plot these points on a suitable scale to a graph to get a line.

Steps for plotting the graph of Linear Equation in Two variables:

1. Write the linear equation.
2. Express y in terms of x.
3. Choose three value of x and calculate the corresponding values of y from the given equation.
4. Tabulate the values of x and y.
5. Plot the value of x on x – axis and value of y on y – axis to a suitable scale, on a graph paper to get three points.
6. Join the three points by a straight line and extend it in both the directions.
7. The line obtained is the graph of the given equation.

Note:
To draw a graph of a linear equation in two variables, atleast, two solutions are required. In this chapter three solutions are taken to draw the graph for better result. Students can draw the graph by taking two solutions also.

MP Board Class 9th Maths Solutions