Author name: Prasanna

NCERT Solutions for Class 7 English

The Rebel NCERT Solutions for Class 7 English Honeycomb Poem 2

The Rebel NCERT Text Book Questions and Answers

The Rebel  Working with the poem

Question 1.
Answer the following questions.

i. If someone doesn ‘t wear a uniform to school, what do you think the teacher will say?
Answer:
If someone doesn’t wear a uniform to school, I believe that the teacher will not appreciate it. He or she would want to meet the student’s parents, and will definitely inform the principal about this.

ii. When everyone wants a clear sky, what does the rebel want most?
Answer:
When everyone wants a clear sky, the rebel wants it to rain.

iii. If the rebel has a dog for a pet, what is everyone else likely to have?
Answer:
If the rebel has a dog for a pet, everyone else is likely to have a cat as pet.

iv. Why is it good to have rebels?
Answer:
It is good to have rebels because they challenge the beliefs of the world. They have the courage to stand out, and the world needs people who can go against it to create something new. They are the ones who bring real change in the world.

v. Why is it not good to be a rebel oneself?
Answer:
It is not good to be a rebel oneself because it would mean having less number of friends, or people who understand you. No one wants to be with someone who disagrees with them, and so the rebels are likely to be very lonely people.

vi. Would you like to be a rebel? If yes, why? If not, why not?
(Encourage the students to use their creativity and formulate their own answers.) Sample answer:
Answer:
Yes, I would like to be a rebel, as I would like to be different from the rest of the world. I want to be a harbinger of change in the world, and I believe that the rebels do just that. I would like to say what I think all the time instead of agreeing to everything that the
other person is telling me.

Question 2.
Find in the poem an antonym (a word opposite in meaning) for each of the following words.
Answer:

• long – short
• grow – cut
• quietness – disturbance
• sober – fantastic
• lost – find

Question 3.
Find in the poem lines that match the following. Read both one after the other.
Answer:

• The rebel lets his hair grow long.
• The rebel expresses a preference for cats.
• The rebel puts in a good word for dogs.
• The rebel regrets the absence of sun.
• The rebel creates a disturbance.

These NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.2

Question 1.
The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.
Answer:
Side of equilateral triangle = l
Therefore, Perimeter of equilateral triangle = 3 × side = 3l

Question 2.
The Side of a regular hexagon (see the given figure) is denoted by l. Express the perimeter of the hexagon using l.
(Hint: A regular hexagon has all its six sides equal in length.)

Answer:
Side of hexagon = l
Therefore,
Perimeter of Hexagon = 6 × side = 6l

Question 3.
A cube is a three-dimensional figure as shown in the given figure. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. Find the formula for the total length of the edges of a cube.

Answer:
Length of one edge of cube = l
Number of edges in a cube = 12
Therefore, total length = 12 × 1 = 12l

Question 4.
The diameter of a circle is a line which joins two points on the circle and also passed through the centre of the circle. (In the adjoining figure AB is a diameter of the circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius(r).

Answer:
Since, length of diameter is double the length of radius.
Therefore, d = 2r

Question 5.
To find sum of three numbers 14,27 and 13, we can have two ways:
(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54 or
(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54. Thus, (14 + 27) + 13 = 14 + (27 + 13)
This can be done for any three numbers. This property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on whole numbers, in a general way, by using variables a, b and c.
Answer:
(a + b) + c = a + (b + c )

These NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.5

Question 1.
Which of the following are models for perpendicular lines:
(a) The adjacent edges of a table top.
(b) The lines of a railway track.
(c) The line segments forming the letter ‘L’.
(d) The letter V.
Answer:
(a) The adjacent edges of a table top.

.’. They form perpendicular lines,

(b) The lines of a railway track

Here, the two adjacent sides don’t meet.
χ They do not form perpendicular lines

(c) The line segments forming the letter ‘L’.

Here, the line segments form a right angle
∴ They form perpendicular lines.

(d) Here, the angle between the lines is not a right angle.

∴ They do not form perpendicular lines.

Question 2.
Let \(\overline{\mathrm{PQ}}\) be the perpendicular to the line segment \(\overline{\mathrm{XY}}\). Let \(\overline{\mathrm{PQ}}\) and \(\overline{\mathrm{XY}}\) intersect in the point A. What is the measure of ∠PAY?
Answer:

Since, \(\overline{\mathrm{XY}}\) ⊥ PQ
Angle between them is a right angle.
.’. ∠PAY = 90°

Question 3.
There are two set-squares in your box. What are the measures of the angles that are formed at their corners? Do they have any angle measure that is common?
Answer:

The angle are 90°, 45°, 45°
The angles are 90°, 60°, 30°
So, Angle 90° is common between them

Question 4.
Study the diagram. The line l is perpendicular to line m.

(a) Is CE = EG?
(b) Does PE bisect CG?
(c) Identify any two line segments for which PE is the perpendicular bisector.
(d) Are these true?
(i) AC > FG (ii) CD = GH (iii) BC < EH
Answer:
(a) CE = CD + DE = 1 + 1=2 EG = EF + FG = 1 + 1 = 2

(b) PE & CG Intersect at point E & CE = EG
.’. PE is the bisector of CG

(c) PE is perpendicular bisector for
\(\overline{\mathrm{CG}}\)
As CE = EG = 2
& \(\overline{\mathrm{CE}}\) = EG = 2
\(\overline{\mathrm{XY}}\) ⊥ \(\overline{\mathrm{CG}}\)
\(\overline{\mathrm{BH}}\)
As \(\overline{\mathrm{BE}}=\overlline{\mathrm{BE}}\) = 3
\(\overline{\mathrm{PE}}=\overlline{\mathrm{BH}}\)

(d) (i) True
AC = AB + BC = 1 + 1 = 2
FG = 1
∴ AC > FG

(ii) True
CD = 1
GH = 1
CD > GH

(iii) True
BC = 1
EH = EF + FG + GH = 1 + 1 + 1 = 3
∴ CD > GH

These NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Exercise 8.1

Question 1.
The angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. the angles of the quadrilateral.
Solution:
Let the first angle of the quadrilateral be 3x.
Second angle of quadrilateral is 5x.
The third angle of the quadrilateral is 9x.
The fourth angle of the quadrilateral is 13x.
Therefore,
3x + 5x + 9x + 13x = 360 (sum of all four angles of a quadrilateral is 360)
⇒ 30x = 360
⇒ x = \(\frac{360}{30}\)
⇒ x = 12
First angle of quadrilateral is 3x = 3 × 12 = 36°
Second angle of quadrilateral is 5x = 5 × 12 = 60°
Third angle of quadrilateral is 9x = 9 × 12 = 108°
Fourth angle of quadrilateral is 13x = 13 × 12 = 156°

Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution:
Given: ABCD is a parallelogram, in which diagonal AC and BD are equal.
To prove that: ABCD is a rectangle.

Proof: In ∆ABC and ∆DCB.
AC = BD (Given)
BC = GB (Common)
AB = DC (Opposite sides of a parallelogram)
Therefore, by S-S-S congruency condition
∆ABC ≅ ∆DCB
So, ∠ABC = ∠BCD (By C.P.C.T)
But, ∠ABC + ∠BCD = 180 (Sum of interior angles of the same side of transversal)
⇒ 2∠ABC = 180 (∵ ∠ABC = ∠BCD)
⇒ ∠ABC = 90°
Therefore, paraileiogram ABCD is a rectangle.

Question 3.
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Solution:
Given: ABCD is a quadrilateral, in which diagonal AC and BD intersect each other at a right angle.
To prove that ABCD is a Rhombus.
Proof: In ∆AOD and ∆AOB,
AO = OA (Common)
OD = OB (Given)
∠AOD = ∠AOB (each 90°)
By S-A-S congruency condition
∠AOU ≅ ∠AQB
So, AD = AB …..(i) (By C.P.C.T)
Similarly, AD = BC ……(ii)
and AB = CD ……(iii)
By equation (i), (ii) and (iii)
AB = BC = CD = DA
Therefore, ABCD is a rhombus.

Question 4.
Show that the diagonals of a square are equal and bisect each other at right angles.
Solution:
Given: ABCD is a square, in which diagonal AC and BD intersect each other at Q.

To prove that:
(i) AC = BD
(ii) Diagonal bisects each other at right angle.
Proof:
(i) In ∆ABC and ∆BAD
AB = BA (Common)
BC = AD (sides of square)
∠ABC = ∠BAD (each 90°)
By S-A-S congruency condition
∆ABC ≅ ∆BAD
AC = BD (By C.P.C.T.)

(ii) In ∆AOB and ∆COD,
AB = CD (Sides of square)
∠AOB = ∠COD (Vertically opposite angles)
∠OBA = ∠OCD (Alternate interior angle)
Therefore, by A-S-A congruency condition
∆AOB ≅ ∆COD
So, OA = OC (By C.P.C.T.)
and OB = OD (By C.P.C.T)
Now, In ∆AOD and ∆COD
AD = CD (Sides of square)
OA = OC (Prove above)
OD = OD (Common)
By S-S-S ccngruency condition
∆AOD = ∆COD
So, ∠AOD = ∠COD (By C.P.C.T)
But, ∠AOD + ∠COD = 180 (Linear pair)
⇒ ∠AOD + ∠AOD = 180 (∵ ∠AOD = ∠COD)
⇒ 2∠AOD = 180
⇒ ∠AOD = 90° …..(ii)
Therefore, from equations (i) and (ii) it is dear that diagonal of a square bisect each other at a right angle.

Question 5.
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution:
Given: ABCD is a quadrilateral in which diagonal AC and BD bisect each other at right angles.

To prove that: ABCD is a square.
Proof: In ∆AOB and ∆COD
AO = CO (Given)
∠AOB = ∠COD (Vertically opposite angles)
OB = OD (Given)
By S-A-S Congruency Condition.
∆AOB ≅ ∆COD
So, AB = CD …..(i) (By C.P.C.T)
and ∠OAB = ∠OCD (By C.P.C.T)
But it is the pair of alternate interior angles and we know that if pair of alternate interior angles are equal then the two lines are parallel.
∴ AB || CD …..(ii)
From (i) and (ii)
ABCD is a parallelogram
Now, in ∆AOD and ∆COD
AO = OC (Given)
∠AOD = ∠COD (Each 90°)
OD = OD (Common)
By S-A-S congruency condition
∆AOD ≅ ∆COD
AD = CD …..(iii) (ByC.P.C.T.)
Again, AD = BC ……(iv)
(Opposite sides of parallel gram ABCD)
From equation (i), (iii) and (iv)
Therefore ABCD is a square.

Question 6.
Diagonal AC of a parallelogram ABCD bisects ∠A [see Fig. 8.19]. Show that
(i) it bisects ∠C also.
(ii) ABCD is a rhombus.

Solution:
(i) In ∆DAC and ∆BCA
DA = BC (Opposite sides of parallelogram)
DC = BA (Opposite sides of parallelogram)
AC = CA (Common)
By S-S-S congruency condition
∆DAC = ∆BCA
So, ∠DAC = ∠BCA …..(i) (By C.P.C.T)
and ∠ACD = ∠BAC ……(ii) (By C.P.C.T)
Add equation (i) and (ii)
∠DAC + ∠ACD = ∠BCA + ∠BAC
∠DAC + ∠ACD = ∠BCA + ∠DAC (∵ ∠DAC = ∠BAC given)
Therefore, ∠ACD = ∠BCA
or, AC bisect ∠C.

(ii) We have,
∠ACD = ∠BAC …..(iii) (From equation (ii))
But ∠BAC = ∠CAD …..(iv)
From (iii) and (iv)
∠ACD = ∠CAD
∴ DA = DC
(Side opposite to equal angles are equal)
But DA = BC (Opposite side of || gm)
∴ AB = BC = CD = DA
Therefore, ABCD is a rhombus.

Question 7.
ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Solution:
In this Fig. ABCD is a rhombus.

Therefore, AB = BC = CD = AD
In ∆ABC,
AB = BC (Because ABCD is a rhombus)
∴ ∠1 = ∠2 (Angle opposite to equal sides are equal)
But, ∠1 = ∠3 …….(ii) (Alternate interior angles)
From (i) and (ii)
∠1 = ∠3
∴ AC bisects ∠C.
Similarly, we can prove AC bisects ∠A, and BD bisects both ∠B as well as ∠D.

Question 8.
ABCD is a rectangle in which diagonal AC bisects. ∠A as well as ∠C. Show that
(i) ABCD is a square.
(ii) diagonal BD bisects ∠B as well as ∠D.

Solution:
(i) We have given ABCD is a rectangle.
∠A = ∠C (each 90°)
or, \(\frac {1}{2}\) ∠A = \(\frac {1}{2}\) ∠C
∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)
So, AD = CD ……(i) (Sides opposite to equal angles are equal)
But, AD = BC ……(ii) (Opposite sides of rectangle)
From equation (i) and (ii)
AB = BC = CD = AD
∴ ABCD is rhombus.
But, each angle of rhombus ABCD is 90°.
Therefore, ABCD is a square.

(ii) We know that diagonals of squares bisect opposite angles.
Therefore, diagonal BD bisects both ∠B and ∠D.

Question 9.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig 8.20) Show that:
(i) ∆APD ≅ ∆CQB
(ii) AP = CQ
(iii) ∆AQB = ∆CPD
(iv) AQ = CP
(v) APCQ is a parallelogram.

Solution:
Given: ABCD is a parallelogram, and two points P and Q are taken on diagonal BD such that DP = BQ.
To prove that:
(i) ∆APD ≅ ∆CQB
(ii) AP = CQ
(iii) ∆AQB = ∆CPD
(iv) AQ = CP
(v) APCQ is a parallelogram.

Construction: Draw diagonal AC of parallelogram ABCD intersecting BD at O.
Proof:
(i) In ∆APD and ∆CQB
∠ADP = ∠CBQ (Pair of alternate interior angles)
AD = BC (Opposite sides of || gm ABCD)
PD = QB (Given)
By S-A-S congruency condition.
∠APD ≅ ∠CQB

(ii) We have
∠APD ≅ ∠CQB (Prove above)
∴ AP = CQ (By C.P.C.T.)

(iii) In ∆AQB and ∆CPD
∠LBQ = ∠CDP (Pair of alternate interior angles)
AB = CD (Opposite sides of parallelogram ABCD)
BQ = DP (Given)
By S-A-S congruency condition
∆AQB ≅ ∆CPD

(iv) We have
∆AQB ≅ ∆CPD (Prove above)
∴ AQ = CP (By C.P.C.T)

(v) In ∆QCP,
AQ = CP (Prove above from (iv))
and AP = CQ (Prove above from (ii))
We know that if both opposite pairs are equal then the parallelogram.
Hence, APCQ is a parallelogram.

Question 10.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A arid C on diagonal BD respectively (see fig 8.21). Show that
(i) ∆APB ≅ ∆CQD
(ii) AP = CQ

Solution:
(i) In ∆APB and ∆CQD
∠APB = ∠CQD (Each 90°)
AB = CD (Opposite sides of || gm ABCD)
and, ∠ABP = ∠CDQ (Alternate interior angles)
By A-S-A Congruency condition
∆APB = ∆CQD

(ii) We have
∆APB = ∆CQD (Prove above)
∴ AP = CQ (By C.P.C.T)

Question 11.
In ∆ABC and ∆DEF, AB = DE and AB || DE, BC = EF and BC || EF. Vertices A, B, and C are joined to vertices D, E and F respectively (see fig 8.22). Show that
(i) quadrilateral ABED is a parallelogram.
(ii) quadrilateral BEFC is a parallelogram.
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ∆ABC ≅ ∆DEF

Solution:
Given: AB = DE and AB || DE,
BC = EF and BC || EF
To prove that:
(i) quadrilateral ABED is a parallelogram.
(ii) quadrilateral BEFC is a parallelogram.
(iii) AD || CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ∆ABC ≅ ∆DEF
Proof:
(i) We have
AB = DE and AB || DE
We know that if one opposite pair of a quadrilateral are parallel and equal, then the quadrilateral is a parallelogram.
∴ Quadrilateral ABDE is a parallelogram.

(ii) We have
BC = EF and BC || EF
Therefore, quadrilateral BCEF is a parallelogram.

(iii) From (i) we have
ABED is a parallelogram.
AD = BE and AD || BE …….(A)
(Opposite sides of parallelogram ABED)
Again, From (ii)
BEFC is a parallelogram,
BE = CF and BE || CF ……..(B)
(Opposite sides of parallelogram BEFC)
From equation (A) and (B)
AD || CF and AD = CF

(iv) We have
AD || CF and AD = CF (Prove above from part (iii))
We know that, if one opposite pair of a quadrilateral are parallel and equal, then the quadrilateral is a parallelogram.
∴ ACFD is a parallelogram.

(v) We have,
ACFD is a parallelogram
(Prove above from part IV)
∴ AC || DF
(Opposite sides of parallelogram ACFD)
and AC = DF

(vi) In ∆ABC and ∆DEF
AB = DE (Given)
BC = EF (Given)
AC = DF (Prove above)
By S-S-S Congruency condition,
∆ABC ≅ ∆DEF.

Question 12.
ABCD is, a trapezium in which AB || CD and AD = BC (see fig 8.23). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∆BAD
(iv) Diagonal AC = Diagonal BD.

Solution:
Given: ABCD is a trapezium in which AB || CD and AD = BC.
To prove that:
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∆BAD
(iv) Diagonal AC = diagonal BD.
Construction: Through C draw CE || AD which intersect AB produced at E, and join ACBD.

Proof: (i) We have,
AB || CD (Given)
AD || CE (By construction)
Therefore, AECD is a parallelogram.
So, AD = CE …..(i)
(Opposite sides of || gm)
But AD = BC ……(ii) (Given)
From (i) and (ii)
BC = CE
∴ ∠CBE = ∠CEB
(Angle opposite to equal sides are equal)
Now, ∠A + ∠CEB = 180°
(Sum of interior angles of the same side of transversal)
or, ∠A + ∠CBE = 180° ……(iii) (∵ ∠CEB = ∠CBE)
But, ∠B + ∠CBE = 180° ……(iv) (Linear pair)
From equation (iii) and (iv)
or, ∠A = ∠B

(ii) We have AB || CD
∴ ∠A + ∠D = 180° …..(v)
(Sum of an interior angle of the same side of transversal)
and, ∠C + ∠B = 180° ……(vi)
(Sum of interior angle of the same side of transversal)
From equation (v) and (vi)
∠A + ∠D = ∠C + ∠B
But, ∠A = ∠B (Prove above)
∴ ∠C = ∠D

(iii) In ∆ABC and ∆BAD
AB = BA (Common)
BC = AD (Given)
∠B = ∠A (Prove above)
By S-A-S congruency condition
∆ABC ≅ ∆BAD

(iv) We have
∆ABC ≅ ∆BAD (Prove above)
∴ AC = BD (By CPCT)
Therefore, diagonal AC = diagonal BD.

NCERT Solutions for Class 7 English

Bringing up Kari NCERT Solutions for Class 7 English An Alien Hand Chapter 2

Bringing up Kari NCERT Text Book Questions and Answers

Question 1.
The enclosure in which Kari lived had a thatched roof that lay on thick tree stumps. Examine the illustration of Kari s pavilion on page 8 and say why it was built that way.
Answer:
In the picture, the thatched roof is supported by tree stumps. Tree stumps are sturdy and rooted to the ground. They will not move or fall down when Kari bumped against them as he moved about in his retreat.

Question 2.
Did Kari enjoy his morning bath in the river? Give a reason for your answer.
Answer:
Yes, Kari enjoyed his morning bath in the river because he would squeal with pleasure after the bath. Moreover, he would often lie down in the water after the author had rubbed his body with sand to clean him.

Question 3.
Finding good twigs for Kari took a long time. Why?
Answer:
Finding good twigs for Kari took a long time because like all elephants, he would not eat tom or mutilated twigs. So, the author had to climb all kinds of trees, only find twigs that were suitably delicate and tender for Kari.

Question 4.
Why did Kari push his friend into the stream?
Answer:
Kari pushed his friend into the stream to let him know about the drowning boy who was lying at the bottom of the river.

Question 5.
Kari was like a baby. What are the main points of comparison?
Answer:
Kari was like a baby. For his daily morning bath, he would lie down on the river bank, while the author mobbed his body with sand. He would squeal a baby to let his companion know he liked it. He would eat delicate twigs and saplings, just like a baby eats soft, gooey food.

Even his call is more like a baby than an elephant, explains the author in the text. Kari also had to be trained to be good and if you did not tell him when he was naughty, he was up to more mischief than ever.

Question 6.
Kari helped himself to all the bananas in the house without anyone noticing it. How did he do it?
Answer:
Kari would slink towards the window of the dining room where the bananas’were kept, and would use his tank to steal them away from the fruit basket. Then, he would walk off with them to his pavilion to enjoy them.

Question 7.
Kari learnt the commands to sit and to walk. What were the instructions for each command?
Answer:
If the author used the word ‘dhat’ and pulled Kari by the ear, he would sit down. Similarly, if the author said ‘mali’ and pulled Kari’s tank forward, he understood that it is the signal to walk.

Question 8.
What is “the master call”? Why is it the most important signal for an elephant to learn?
Answer:
The “master call” is a strange hissing, howling sound, as if a snake and a tiger were fighting each other, and is made in an elephant’s ear.

It is a very important call because if one is lost in the jungle and is afraid of predatory animals, the master call will make the elephant pull down all the trees in front of him with his tank. This easily frightens all the animals away.

In other words, the elephant pulls down each tree in the way to make a road right through . the jungle straight to your house. A desert is generally without water and vegetation. A desert may be too hot or too cold.

An Oasis is like a green island in the middle of a desert. Desert plants and animals require less water to survive than other plants and animals. Deserts are an important part of nature’s great plan. They are not useless parts of the Earth. A desert is beautiful with desert plants and flowers and the sight can be as rewarding as that of any tropical garden whenever it rains.

NCERT Solutions for Class 7 English

The Tiny Teacher NCERT Solutions for Class 7 English An Alien Hand Chapter 1

The Tiny Teacher NCERT Text Book Questions and Answers

The Tiny Teacher Comprehension check – I

Question 1.
The story of an ant’s life sounds almost untrue.
The italicized phrase means
i. highly exaggerated.
ii. too remarkable to be true.
iii. not based on facts.
Answer:
ii. too remarkable to be true.

Question 2.
Complete the following sentences.
Answer:

• An ant is the smallest, but the wisest insect.
• We know a number of facts about an ant s life because people have kept ants as pets, and have watched their daily behaviour closely.

Question 3.
In what ways is an ant s life peaceful?
Answer:
An ant’s life is very peaceful because each ant in a colony does its share of work intelligently and bravely, and never fights with other members of the group.

The Tiny Teacher Comprehension check -II

Question 1.
How long does it take for a grub to become a complete ant?
Answer:
It takes around five to six weeks for a grub to become a complete ant.

Question 2.
Why do the worker ants carry the grubs about?
Answer:
The worker ants carry the grubs about for airing, exercise and sunshine.

Question 3.
What jobs are new ants trained for?
Answer:
New ants are trained to become workers, soldiers, builders and cleaners.

Question 4.
Name some other creatures that live in anthills.
Answer:
Some other creatures that live in anthills include beetles, some lesser breeds of ants and the greenfly.

Question 5.
Mention three things we can learn from the ‘tiny teacher Give reasons for choosing these items.
Answer:
There are many qualities that one can learn from ants. These include hard work, sense of duty and discipline, cleanliness, care for the young ones, and, above all, a firm loyalty to the land where they live.

For example, the ants work very hard to collect and store their food. Similarly, each of them does their bit of work peacefully and does not complain.

The way they care for their young ones by carrying the grub on their backs for airing and sunshine shows how caring they are. The soldiers also protect the eggs till they hatch.

The Tiny Teacher Exercises Question and Answer

Discuss the following topics in groups.

Question 1.
i. What problems are you likely to face ifyou keep ants as pets?
Answer:
There are many people who keep ant colonies as pets to study their behaviour, or to understand these creatures better. They often keep them in artificial containers called formicariums. The problems one might face as a result could be one of maintaining moisture inside the formicarium, or preventing mold in the ant habitats.

ii. When a group of bees finds nectar, it informs other bees of its location, quantity, etc. through dancing. Can you guess what ants communicate to their fellow ants by touching one another’s feelers?
Answer:
Ants communicate with each other by touching one another’s feelers. They often greet each other using their feelers, which also contain their touch and smell organs. This means that they can recognize each other through the touch and prevent intruders from entering their colony (because each ant colony has its own distinct smell).

The ants are also able to smell the food on each other so they both know the food that the other has found. As ants cannot see, speak or hear, they use the touch to alert the other of dangers, or to tell them to follow a particular trail. In fact, their feelers can help them detect if the other ant is hungry. In this case, they may also feed them with the food stored in their ‘social stomach’.

Question 2.
Complete the following poem with words from the box below. Then recite the poem.With answers
Answer:
Soldiers live in barracks
And birds in nests,
Much like a snake that rests
In a hole. No horse is able
To sleep except in a stable.
And a dog lives well,
Mind you, only in a kennel.
To say ‘hi’ to an ant, if you will,
You may have to climb an anthill.
(Encourage the students to recite the poem aloud together.)