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These NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.1

Question 1.
Find the perimeter of each of the following figures:

Answer:
(a) Perimeter = Sum of all the sides = 4 cm + 2 cm + 1 cm + 5 cm = 12 cm
(b) Perimeter = Sum of all the sides = 23 cm + 35 cm + 40 cm + 35 cm = 133 cm
(c) Perimeter = Sum of all the sides =15 cm + 15 cm + 15 cm + 15 cm = 60 cm
(d) Perimeter = Sum of all the sides = 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm
(e) Perimeter = Sum of all the sides 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 15 cm
(f) Perimeter = Sum of all the sides = 4
cm + 1 cm + 3 cm + 2 cm + 3 cm + 4
cm + 1 cm + 3 cm + 2 cm + 3 cm + 4
cm + 1 cm + 3 cm + 2 cm + 3 cm + 4
cm + 1 cm + 3 cm + 2 cm + 3 cm = 52 cm

Question 2.
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Answer:
Total length of tape required
= Perimeter of rectangle
= 2 (length + breadth) = 2 (40 + 10)
= 2 × 50 = 100 cm = 1 m
Thus, the total length of tape required is 100 cm or 1 m.

Question 3.
A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Answer:
Length of table top = 2 m 25 cm = 2.25 m
Breadth of table top = 1 m 50 cm = 1.50 m
Perimeter of table top = 2 × (length + breadth) = 2 × (2.25 + 1.50) = 2 × 3.75 = 7.5 m
Thus, the perimeter of tabletop is 7.5 m.

Question 4.
What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Answer:
Length of wooden strip = Perimeter of photograph
Perimeter of photograph
= 2 x (length + breadth)
= 2 (32 + 21) = 2 × 53cm =106 cm
Thus, the length of the wooden strip required is equal to 106 cm.

Question 5.
A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Answer:
Since the 4 rows of wires are needed.
Therefore the total length of wires is equal to 4 times the perimeter of rectangle.
Perimeter of field = 2 × (length + breadth)
= 2 × (0.7 + 0.5) = 2 × 1.2 = 2.4 km
= 2.4 × 1000 m = 2400 m
Thus, the length of wire = 4 × 2400 = 9600 m = 9.6 km

Question 6.
Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Answer:
(a) Perimeter of ΔABC = AB + BC + CA = 3 cm + 5 cm + 4 cm = 12 cm

(b) Perimeter of equilateral ABC
= 3 × side = 3 × 9 cm = 27 cm

(c) Perimeter of AABC = AB + BC + CA = 8 cm + 6 cm + 8 cm = 22 cm

Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Answer:
Perimeter of triangle
= Sum of all three sides
= 10 cm + 14 cm + 15 cm = 39 cm
Thus, the perimeter of triangle is 39 cm

Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Answer:
Perimeter of Hexagon
= 6 × length of one side
= 6 × 8 m = 48 m
Thus, the perimeter of hexagon is 48 m.

Question 9.
Find the side of the square whose perimeter is 20 m.
Answer:
Perimeter of square = 4 × side
⇒ 20 = 4 × side
⇒ Side = \(\frac{20}{4}\) = 5 cm
Thus, the side of square is 5 cm.

Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Answer:
Perimeter of regular pentagon = 100 cm
⇒ 5 × side = 100 cm
⇒ Side = \(\frac{100}{5}\) = 20 cm
Thus, the side of regular pentagon is 20 cm.

Question 11.
A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Answer:
Length of string = Perimeter of each figure
(a) Perimeter of square = 30 cm
⇒ 4 × side = 30 cm
⇒ Side = \(\frac{30}{4}\) = 7.5 cm 4
Thus, the length of each side of square is 7.5 cm.

(b) Perimeter of equilateral triangle = 30 cm
⇒ 3 × side = 30 cm
⇒ Side = \(\frac{30}{3}\) = 10 cm
Thus, the length of each side of equilateral triangle is 10 cm.

(c) Perimeter of hexagon = 30 cm
⇒ 6 × side = 30 cm
⇒ Side = \(\frac{30}{6}\) = 5 cm
Thus, the side of each side of hexagon is 5 cm.

Question 12.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Answer:
Let the length of third side be X cm.
Length of other two side are 12 cm and 14 cm.
Now, Perimeter of triangle = 36 cm
⇒ 12 +14 + X = 36
⇒ 26 + X = 36
⇒ X = 36 – 26 = 10 cm
Thus, the length of third side is 10 cm

Question 13.
Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.
Answer:
Side of square = 250 m
Perimeter of square
= 4 × side = 4 × 250 = 1000 m
Since, cost of fencing of per meter = ₹20
Therefore, the cost of fencing of 1000 meters = 20 × 1000 = ₹ 20,000

Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of × 12 per metre.
Answer:
Length of rectangular park = 175 m
Breadth of rectangular park = 125 m
Perimeter of park
= 2 × (length + breadth)
= 2 × (175 + 125) = 2 × 300 = 600 m
Since, the cost of fencing park per meter = ₹ 12
Therefore, the cost of fencing park of 600 m = 12 × 600 = ₹ 7,200

Question 15.
Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
Answer:
Distance covered by Sweety
= Perimeter of square park
Perimeter of square = 4 × side
= 4 × 75 = 300 m
Thus, distance covered by Sweety is 300 m. Now, distance covered by Bulbul =
Perimeter of rectangular park
Perimeter of rectangular park
= 2 × (length + breadth)
= 2 × (60 + 45) = 2 × 105 = 210 m
Thus, Bulbul covers the distance of 210 m and Bulbul covers less distance.

Question 16.
What is the perimeter of each of the following figures? What do you infer from the answers?

Answer:
(a) Perimeter of square
= 4 × side = 4 × 25 = 100 cm

(b) Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (40 + 10) = 2 × 50 = 100 cm

(c) Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (30 + 20) = 2 × 50 = 100 cm

(d) Perimeter of triangle = Sum of all sides
= 30 cm + 30 cm + 40 cm = 100 cm
Thus, all the figures have same perimeter.

Question 17.
Avneet buys 9 square paving slabs, each with a side of 1/2 m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement [Fig (i)]?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig (ii)]?
(c) Which has greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)

Answer:
(a) The arrangement so formed in Fig (i) is a square of side equal to 3 × side of the square slab
∴ Perimeter of his arrangement in Fig (i)
= 4 × ( 3 × side of the square slab)
= 4 × ( 3 × 0.5) m = 4 × 1.5 m = 6 m

(b) The cross arrangement so formed in Fig (ii) by 9 square slabs, has perimeter, which consists of 20 length segments each equal to side of I a square slab i.e. 0.5 m
∴ Perimeter of her arrangement in Fig (ii)
= 20 × ( side of the square slab)
= 20 × 0.5 m = 10 m

(c) ∵ 10m > 6m
∴ Cross arrangement has greater perimeter

(d) ∵ Total number of tiles = 9
∴ The following arrangement will also have greater perimeter.

Since, this arrangement is in the form of rectangle with length and breadth as \(\frac { 9 }{ 2 }\) m and \(\frac { 1 }{ 2 }\) respectively.
∴ Perimeter = 2 \(\left(\frac{9}{2}+\frac{1}{2}\right)\)
= 2 × = 10m \(\left[\frac{10}{2}\right]\) = 10m

NCERT Solutions for Class 7 English

The Bear Story NCERT Solutions for Class 7 English An Alien Hand Chapter 8

The Bear Story NCERT Text Book Questions and Answers

The Bear Story Exercises Question and Answer 

Answer the following questions.

Question 1.
Where did the lady find the bear cub? How did she bring it up?
Answer:
The lady found the bear cub in the forest, half dead of hunger. It was so small and helpless that the lady had to bring it up on the bottle.

Question 2.
The bear grew up but “he was a most amiable bear”. Give three examples to prove this.
Answer:
The bear was a most amiable bear. He used to sit outside his kennel and look with his small intelligent eyes most amicably at the cattle grazing in the field nearby.

The children used to ride on his back and had more than once been found asleep in his kennel between his two paws. The three dogs loved to play all sorts of games with him, pull his ears and his stump of a tail and tease him in every way, but he did not mind it in the least.

Question 3.
What did the bear eat? There were two things he was not allowed to do. What were they?
Answer:
The bear ate apples, bread, porridge, potato, cabbage and turnip. He was nQt allowed to climb the apple trees to eat apples or to go near beehives.

Question 4.
When was the bear tied up with a chain? Why?
Answer:
The bear was tied up with a chain during the night so that it would not run away to the forest at night. He was also put on the chain on Sundays when his mistress went to spend the afternoon with her married sister.

Question 5.
What happened one Sunday when the lady was going to her sister’s house? What did the lady do? What was the bear’s reaction?
Answer:
One Sunday when the lady was going to her sister’s house, she had reached halfway through the forest when she heard the cracking of a tree branch. She looked back, and saw the bear coming towards her. He sat at her heels like a dog and began to sniffle at her.

First she sternly asked him to go back home. When he kept sniffling at her, she hit him on his nose with her parasol so hard that it broke into two. The bear stopped for a moment, shook his head and opened his big mouth several times as if he wanted to say something. Then he turned round and began to shuffle back.

Question 6.
Why was the bear looking sorry for himself in the evening? Why did the cook get angry with her mistress?
Answer:
The bear was looking sorry for himself in the evening because he had been waiting for so long for his mistress to come back home. The cook got angry with her mistress because she treated the bear like her own son and could not bear it when the lady scolded him for no reason. He had done nothing to deserve all that scolding.

Discuss the following topics in groups.

Question 1.
Most people keep dogs and cats as pets. Can you think of some unusual pets that people keep?
(Encourage the students to discuss the given topics in groups.)
Answer:
Some unusual pets that people keep include monkeys, iguanas, spiders, coyotes, turtles, guinea pigs and tortoises.”

Question 2.
The second bear did not attack the lady because he was afraid of her. Do you agree?
Answer:
The second bear did not attack the lady because he had no intention of attacking her. After catching her mid-way through the forest, he sat at her heels like a dog. He sniffled at her and did not stop until the lady firmly told him to stop.

He was caught by surprise when he saw that she was not at all scared of him. This made him respect her even more. Perhaps he wanted nothing more than the lady to shower some affection on him. So, when she hit him on his nose with her parasol, he felt hurt. He went back to his place looking back at her till the time she lost sight of him. This shows that he was hurt on receiving a bad treatment from the lady.

NCERT Solutions for Class 7 English

Chandni NCERT Solutions for Class 7 English An Alien Hand Chapter 7

Chandni NCERT Text Book Questions and Answers

Chandni  Comprehension check-I

Question 1.
Why did Abbu Khan s goats want to run away? What happened to them in the hills?
Answer:
Abbu Khan’s goats wanted to run away because they were of the best hill breed, and goats in hilly regions love their freedom. At any chance they got, they would run away. In the hills, they got eaten by an old wolf who lived there.

Question 2.
Abbu Khan said, “No more goats in my house ever again. ” Then he changed his mind. Why?
Answer:
He changed his mind because he got terribly lonely and needed a pet who would listen to his stories.

Question 3.
Why did he buy a young goat?
Answer:
He bought a young goat so that she would stay with him much longer. He thought that she will soon begin to love him as well as the food he gave her every day. She will never want to go to the hills, he had thought.

Chandni  Comprehension check-II

Question 1.
Why did Chandni hate the rope round her neck?
Answer:
Chandni hated the rope round her neck because it prevented her dream of going to the hills. She wanted to feel and experience the refreshing breeze of the mountaintops. She wanted with all her heart to run across the green fileds. The rope did not let her do this as it kept her tied to Abbu Khan’s house.

Question 2.
“Now Abbu Khan understood Chandni sproblem… ” What was Chandni sproblem?
Answer:
Chandni’s problem was the lack of freedom. She was sad because she was being raised in captivity.

Question 3.
Abbu Khan pushed Chandni into a small hut. This shows that he
i. was cruel.
ii. loved her and wanted to save her life.
iii. was selfish.
Answer:
iii. was selfish.

Chandni  Comprehension check-III

Question 1.
Why did Chandni refuse to join the group of wild goats?
Answer:
Chandni politely refused to join the group of wild goats because she wanted to enjoy her new freedom all by herself. She had played in the hill slopes through the day, and wanted to enjoy all the more.

Question 2.
Chandni fought the wolf because she
i. was stronger than the wolf
ii. hated the wolf
iii. had to retain her freedom at all costs.
Answer:
iii. had to retain her freedom at all costs.

Chandni Exercises Question and Answer

Discuss the following topics in groups.

Question 1.
Why did the wise old bird say, “Chandni is the winner”?
(Encourage the students to discuss the given questions in groups.)
Answer:
The wise old bird said so because Chandni fought very bravely with the wolf and did not give up easily in front of an enemy who was much stronger than her. She did not want to die without putting up a good fight. She did not care about the success or failure, but did what she thought was best under the circumstance. She did not run away from the wolf, but confronted him, knowing that death was certain. But in the moment, even death did not scare her.

Question 2.
“Death in an open field is better than life in a small hut, ” Chandni said to herself Was it the right decision? Give reasons for your answer.
Answer:
Yes, it was the right decision because playing on the slopes and enjoying her new found freedom made Chandni much happier than she ever could have been in Abbu Khan’s small hut. Her spirit told her to embrace and choose freedom over captivity. She valued freedom over and above life also. This is why she was not afraid to face the wolf. She knew that she was fighting for her belief in her freedom, and she did so till the very end of her life.

Before she faces the wolf, she knows that she can go back to the safety of Abbu Khan’s hut, but chooses to face her fear instead so as to protect her freedom.

Question 3.
Freedom is life. Discuss this with reference to ‘Chandni’ and 7 Want Something in a Cage ’.
Answer:
Freedom is life. No form of life can grow without freedom. It is as dear to living beings as life itself. Both the stories ‘Chandni’ and ‘I Want Something in a Cage’ support and illustrate this thought perfectly. In the story ‘I Want Something in a Cage’, a man realizes the value of freedom after spending ten years in prison.

His time in the prison has made him understand that all living beings need their freedom. So, he tries to buy the freedom of two doves with the money that he has. He does not want any living being to remain in captivity and go through the pain and anguish that he had felt in prison. Similarly, in the story ‘Chandni’, the protagonist Chandni, a goat, escapes from her owner because she wants to enjoy her freedom. She chooses to embrace death instead of captivity.

She understands the true value of freedom just like the man in the story, ‘I Want Something in a Cage’. Both the protagonists in each story do not take their freedom for granted. To them, it is dearer than life itself.

These NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3

Question 1.
In fig. 9.23, E is any point on median AD of a ∆ABC Show that ar(∆ABE) = ar(∆ACE).

Solution:
Given: In ∆ABC, AD is median. Any point E lie of median AD.
To prove that: ar(∆ABE) = ar(∆ACE)
Proof: In ∆ABC, AD is median.
∴ ar(∆ABD) = ar(∆ACD) ……(i)
(A median of a triangle divides it into two triangles of equal areas)
Now, In ∆EBC, ED is median.
∴ ar(∆EBD) = ar(∆CED) ……(ii)
(A median of a triangle divides it into two triangles of equal areas)
Subtract equation (ii) from equation (i)
ar(∆ABD) – ar(∆EBD) = ar(∆ACB) – ar(∆CED)
∴ ar(∆ABE) = ar(∆ACE)

Question 2.
In a triangle ABC, E is the mid point of median AD. Show that ar(BED) = \(\frac {1}{4}\) ar(ABC).
Solution:
Given: ABC is a triangle in which AD is median and E is the mid point of median AD.

To prove that: ar(BED) = \(\frac {1}{4}\) ar(ABC)
Construction: Join BE.
Proof: In ∆ABD, E is the mid point of AD. Therefore, BF is median.
We know that a median of a triangle divide triangles of equal areas.
∴ ar(∆BFD) = ar(∆ABE)
or, ar(∆BED) = \(\frac {1}{4}\) ar(∆ABD) …..(i)
Now, In ∆ABC,
ar (∆ABD) = ar (∆ACD)
(because AD is median of ∆ABC)
or, ar (∆ABD) = \(\frac {1}{4}\) ar (∆ABC) …….(ii)
Put Value of ar (∆ABD) in equation (i) we have
ar (∆BED) = \(\frac{1}{2} \times \frac{1}{2}\) ar (∆ABC)
or, ar (∆BED) = \(\frac {1}{4}\) ar (∆ABC)

Question 3.
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Solution:
Given: ABCD is a parallelogram in which diagonal AC and BD bisects each other at O.

To prove that: ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆DOA)
Proof: In ∆ABC, O is mid point of side AC.
Therefore, BO is median of ∆ABC.
∴ ar (∆AOB) = ar (∆BOC) ……(i)
(Median divide a A in two equal parts)
Now, In ∆ADC, DO is median
∴ ar (∆AOD) = ar (∆COD) …….(ii)
(Median divide a ∆ in two equal parts)
Again in ∆ADB, AO is a median.
∴ ar (∆AOD) = ar (∆AOB) ………(iii)
From equation (i), (ii) and (iii)
ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆DOA)
Therefore, the diagonals of a parallelogram divide it into four triangles of equal area.

Question 4.
In fig. 9.24, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).

Solution:
In ∆ACD, O is mid point of CD.
∴ AO is median
We know that a median of a triangle divides it into triangles of equal areas.
∴ ar (∆AOQ) = ar (∆AOD) …..(i)
Similarly, In ∆BCD, BO is median
∴ ar (∆BOC) = ar (∆BOD)
Adding equation (i) and (ii)
ar (∆AOC) + ar (∆BOC) = ar (∆AOD) + ar (∆BOD)
So, ar (∆ABC) = ar (∆ABD)

Question 5.
D, E and F are respectively the mid points of the sides BC, CA and AB of a ∆ABC. Show that
(i) BDEF is a parallelogram
(ii) ar (DEF) = \(\frac {1}{4}\) ar (ABC)
(iii) ar (BDEF) = \(\frac {1}{2}\) ar (ABC)
Solution:
Given: ABC is a triangle in which D, E and F are the mid points of side BC, AC and AB respectively.
To prove that:
(i) BDEF is a parallelogram
(ii) ar(DEF) = \(\frac {1}{4}\) ar (ABC)
(iii) ar(BDEF) = \(\frac {1}{2}\) ar (ABC)

Proof: (i) In ∆ABC,
E and F are the mid point of side AC and AB respectively.
We know that the line joining the mid points of two sides of a triangle is parallel to third side and half of it.
∴ EF || BC …….(i)
and EF = \(\frac {1}{2}\) BC
or, EF = BD …(ii) (∵ D is mid point of BC)
From (i) and (ii)
BDEF is a parallelogram.

(ii) We have,
BDEF is a parallelogram.
∴ ar (∆BDF) = ar (∆DEF) ……(i)
(Diagonals of a || gm divide it in two equal parts)
Similarly, CDFE is parallelogram.
∴ ar (∆CDE) = ar (∆DEF) …..(ii)
and ar (∆AEF) = ar (∆DEF) ……(iii)
From equation (i), (ii) and (iii)
ar (∆BDF) = ar (∆CDE) = ar (∆AEF) = ar (∆DEF)
Now,
ar (∆BDF) + ar (∆CDE) + ar (∆AFE) + ar (∆DEF) = ar (∆ABC)
or, ar (∆DEF) + ar (∆DEF) + ar (∆DEF) + ar (∆DEF) = ar (∆ABC)
[∵ ar (∆BDF) + ar (∆CDE) = ar (∆AFF)]
or, 4 ar (∆DEF) = ar (∆ABC)
or, ar (∆DEF) = \(\frac {1}{4}\) ar (∆ABC)

(iii) ar (|| gm BDEF) = 2 ar (∆DEF)
or, ar (|| gm BDEF) = 2 . \(\frac {1}{4}\) ar (∆ABC)
[∵ ar (∆DEF) = \(\frac {1}{4}\) ar (∆ABC)]
So, ar (|| gm BEDF) = \(\frac {1}{4}\) ar (∆ABC)

Question 6.
In fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O, such that OB = OD. If AB = CD then show that
(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.

Solution:
Given: ABCD is a quadrilateral in which diagonals AC and BD intersect each other at O.
To prove that:
(i) ar(∆DOC) = ar(∆AOB)
(ii) ar (∆DCB) = ar(∆ACB)
(iii) DA || CB or ABCD is a parallelogram.
Construction: Draw DN ⊥ AC and BM ⊥ AC.

Proof: (i) In ∆DON and ∆BOM
∠DNO = ∠BMO (Each 90°)
∠DNO = ∠BOM (Vertically opposite angle)
DO = BO (Given)
By A-A-S congruency condition
∆DON ≅ ∆BOM
So, ar (∆DON) = ar (∆BOM) ……(i)
(Congruent ∆s have equal area)
DN = BM (By C.P.C.T)
Now, In ∆DNC and ∆BMA
∠DNC = ∠BMA (Each 90°)
CD = BA (Given)
DN = BM (Prove above)
By R.H.S Congruency Condition
∆DNC = ∆BMA
So, ar (∆DNC) = ar (∆BMA) …….(ii)
(Congruent ∆s have equal area)
Adding equatiqn (i) and (ii)
ar (∆DON) + ar (∆DNC) = ar (∆BOM) + ar (∆BMA)
ar (∆DOC) = ar (∆AQB)

(ii) We have,
ar (∆DOC) = ar (∆AOB) (Prove above)
∴ ar (∆DOC) + ar (∆OCB) = ar (∆AOB) + ar (∆QCB) [add ar (∆OCB) both side]
or, ar (∆DCB) = ar (∆ACB)

(iii) We have
ar (∆DCB) = ar (∆ACB)
We know that if two mangles on same base and equal in area, then it must be lie between same parallels.
Therefore, DA || CB …..(A)
∠DCO = ∠BAO (∆DNC ≅ ∆BMA)
which is the pair of alternate interior angles.
CD || BA …….(B)
From (A) and (B),
ABCD is a parallelogram.

Question 7.
D and B are points on sides AB and AC respectively of ∆ABC such that ar (DBC) = ar (EBC).
Prove that DE || BC.

Solution:
Since ∆s BCE and ABCD are equal in area and have a same base BC.
Therefore, altitude from E of ∆BCE = altitude from D of ∆BCD.
or, ∆s BCE and BCD are between the same parallel lines.
DE || BC.

Question 8.
XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively. Show that ar (ABE) = ar (ACF)
Solution:
Given: In ∆ABC, XY || BC, BE || AC and CF || AB.
where E and F lie on XY.
To prove that: ar (∆ABE) = ar (∆ACF)

Proof: We have
XY || BC
or EY || BC ……(i)
and BE || AC
or BE || CY …….(ii)
From, equation (i) and (ii)
EBCY is a parallelogram.
Again, parallelogram EBCY and ∆AEB lie on same base EB and between same parallels BE and AC.
We know that if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.
So, ar(∆ABE) = \(\frac {1}{2}\) ar(|| gm FBCY) …….(A)
Now, XY || BC
or, XF || BC ……(iii)
and, CF || AB
or, CF || BX ……(iv)
From equation (iii) and (iv)
BCFX is a parallelogram.
Again, parallelogram BCFX and ∆ACF lie on sane base CF and between same parallels CF and AB.
∴ ar(∆ACF) = \(\frac {1}{2}\) ar (|| gm BCFX) ……(B)
Now, parallelogram EBCY and parallelogram BCFX lie on same base BC and between same parallels BC and EF.
We know that, parallelograms on the same base and lie between the same parallels having equal area.
∴ ar(|| gm EBCY) = ar (|| gm BCFX) ……(C)
Therefore from equation (A), (B) and (C)
we have ar(∆ABE) = ar(∆ACF)

Question 9.
The side AB of a parallelogram ABCD is produced to any points. A line through A and parallel to CD meet CB produced at Q arid then parallelogram PBQR is completed (see Fig 9.26). Show that ar(ABCD) = ar(PBQR).

Solution:
Given: ABCD is a parallelogram in which P is any point on AB produced and AQ || CP.
To prove that: ar(|| gm ABCD) = ar(|| gm BPRQ)
Construction: Join AC and PQ.
Proof: Since AC and PQ are diagonals of parallelograms ABCD and BPQR respectively.

∴ ar(∆ABC) = \(\frac {1}{2}\) ar(|| gm ABCD) ……(i)
and ar(∆PBQ) = \(\frac {1}{2}\) ar(|| gm BPRQ) ……(ii)
Now, ∆s ACQ and AQP are on die same base AQ and between same parallels AQ and CP.
∴ ar(∆ACQ) = ar(∆AQP)
or, ar(∆ACQ) – (∆ABQ) = ar(∆AQP) – ar(∆ABQ)
[Subtracting ar(∆ABQ) from both sides]
or, ar(∆ABC) = ar(∆BPQ)
or, \(\frac {1}{2}\) ar(|| gm ABCD) = \(\frac {1}{2}\) ar (|| gm BPRQ) [From equ. (i) and (ii)]
or, ar(|| gm ABCD) = ar (|| gm BPRQ)

Question 10.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. prove that ar(AOD) = ar(BOC).

Solution:
Given: ABCD is a trapezium, in which AB || DC, and diagonal AC and BD intersect each other at O.
To prove that: ar(∆AOD) = ar(∆BOC)
Proof: Since ∆DAC and ∆DBC lie on same base BC and between same parallels AB and DC.
∴ ar(∆DAC) = ar(∆DBC)
[∆s on the same base and between same parallels are equal in area]
or, ar (∆DAC) – ar (∆DOC) = ar (∆DBC) – ar (∆DOC) [Subtract ar (∆DOC) both side]
ar (∆AOD) = ar (∆BOC)

Question 11.
In fig. 9.27, ABCDE is a pantagon. A line through B parallel to AC meet DC produced at F.
Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)

Solution:
Given; ABCDE is a pentagon, and AC || BF.
To prove that:
(i) ar (ACB) = ar (ACF)
(a) ar (AEDE) = ar (ABCDE)
Proof:
(i) Since ∆s ACB and ACF lie on the same base AC and between same parallels AC and BF.
∴ ar(∆ACB) = ar(∆ACF)
[∆s on same base and betwen same paralles are equal in area]

(ii) We have,
ar (∆ACB) = ar(∆ACF) [Prove above]
ar(∆ACB) + ar(ACDE) = ar(∆ACF) + ar(ACDE) [Add ar (ACDE) both side]
∴ ar (AEDF) = ar (ABCDE)

Question 12.
A villager Itwaari has a plot oi land of the shape of a quadrilateral. Tht Gram Panchayat of the village dicided b take over some portion of his plot from ore of the comers to construct a Health Centre. Itwaari agrees to the above proposal wih the condition that he should be given eqial amount of land in lieu of his land adjoiniag his plot so as to form a triangular pbt. Explain how this proposal will be implemented.

Solution:
Let a villager Itwaari has a plot of land of the shape of a quadrilateral ABCD.
The Gram Panchayat decided to take comer portion of the land ∆ACD. Itwaari agreed. So the remaining part of the land belong to Itwaari is
ar (ABCD) – ar (∆ACD) = ar (∆ABC)
Now, Itwaari demands the adjoining plot. So, if DE is drawn parallel to AC and join A to E. In this way we add ar (∆ACE) in the remaining plot of Itwaari. In this way proposal will be implemented.
As, ar (∆ABC) + ar (∆ACD) = ar (∆ABC) + ar (∆ACE)
Here ar(∆ACD) = ar(∆ACE)
(Triangles made between two parallels and same base AC)
ar(ABCD) = ar(∆ABE)

Question 13.
ABCD is a trapezium with AB || DC. A line parallel to AC intersect AB at X and BC at Y. Prove that ar (ADX) = ar (ALY).
Solution:
Given: ABCD is a trapezium, in which AB || DC and AC || XY. .
To prove that: ar(∆ADX) = ar(∆ACY)

Construction: Join CX.
Proof: Since ∆s ADX and ACX lie on same base AX and between same parallels AB and DC.
∴ ar (∆ADX) = ar (∆ADX) ……(i)
[∆s on same base and between same parallels are equal in area]
Now, ∆ACY and ∆ACX lie on same base AC and betwen same parallels AC and XY.
ar (∆ACY) = ar (∆ACX) …….(ii)
From equation (i) and (ii) we get
ar (∆ADX) = ar (∆ACY)

Question 14.
In fig 9.28, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

Solution:
Given: AP || BQ || CR
To prove that: ar (∆AQC) = ar (∆PBR)
Proof: Since ∆s ABQ and PBQ lie on same base and betwen same parallels AP and BQ.
∴ ar (∆ABQ) = ar (∆PBQ) ……(i)
[∆s on same base and between same parallels are equal in area]
Again, ∆s BQC and BQR lie on same base BQ and between same parallels BQ and CR.
ar (∆BQC) = ar (∆BQR) ……(ii)
Adding equation (i) and (ii)
ar(∆ABQ) + ar(∆BQC) = ar(∆PBQ) + ar(∆BQR)
or, ar (∆AQC) = ar (∆PBR).

Question 15.
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

Solution:
Given: ABCD is a quadrilateral in which diagonal AC ab BD intersect each other at O, and ar (∆AOD) = ar (∆BOC).
To prove that: ABCD is a trapezium.
Proof: We have
ar(∆AOD) = ar(∆BOC) (Given)
or, ar(∆AOD) + ar(∆DOC) = ar(∆BOC) + ar(∆DOC) [Add ar (∆DOC) both side]
or, ar(∆ADC) = ar(∆BDC)
We know that if two triangles on same base and have equal area, then it must lie in between same parallels.
Here, ∆s ADC and BDC lie on same base DC, and
ar (∆ADC) = ar (∆BDC)
∴ DC must be parallel to AB.
∴ ABCD is a trapezium.

Question 16.
In fig.9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution:
We have,
ar(∆BDP) = ar(∆ARC) (Given)
ar(∆BDP) – ar(∆DPC) = ar(∆ARC) – ar(∆DRC)
ar (∆DRC) = ar (∆DBC) (Given)
or, ar (∆BDC) = ar (∆ADC)
Now, ∆s BDC and ADC lie on same base and having equal area therefore they must lie betwen two parallel lines.
Then, AB || DC
Therefore, ABCD is a trapezium.
Again, ar(∆DRC) = ar(∆DPC) (Given)
Here, ∆s DRC and DPC lie on same base DC and having equal area.
∴ DC || RP
Therefore, DCPR is a trapezium.

NCERT Solutions for Class 7 English

Mystery of the Talking Fan Solutions for Class 7 English Honeycomb Poem 6

Mystery of the Talking Fan NCERT Text Book Questions and Answers

Mystery of the Talking Fan Working with the poem

Question 1.
Fans don’t talk, but it is possible to imagine that they do. What is it, then, that sounds like the fan s chatter?
Answer:
The fan is not working properly, and needs to be oiled and greased. The creaking sound produced by the fan sounds like its chatter.

Question 2.
Complete the following sentences.
Answer:

• The chatter is electrical because the fan works through electricity and makes that sound
  only when one switches it on.
• It is mysterious because no one knows what the fan was trying to say.

Question 3.
What do you think the talking fan was demanding?
Answer:
The talking fan was demanding to be fed with some oil.

Question 4.
How does an electric fan manage to throw so much air when it is switched on?
Answer:
An electric fan manages to throw so much air when it is switched on because when switched on, the motor inside it begins to whirl. This causes the blades of the fan to move. This increases the speed of the air that is already present in the room. Thus, the fan throws air in all directions.

Question 5.
Is there a ‘talking fan ’ in your house? Create a dialogue between the fan and a mechanic.
Answer:
No, there is no ‘talking fan’ in my house. But there is one in my father’s office. It always makes a lot of sound.
Fan: Clatter, clatter. I think I am not well.
Mechanic: Well, what do you think has happened to you?

Fan: Clatter, clatter. Well, I could use some oil. Then I would be fixed.
Mechanic: Oh, if that’s the case. (Puts oil inside the fan) Here is some oil.

Fan: Ah, I feel so much better. Thank you.
Mechanic: Don’t mention it. That’s my job.

These NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 6 Integers InText Questions

NCERT In-text Question Page No. 116

Question 1.
Write the following numbers with appropriate signs:
(a) 100 m below sea level.
(b) 25°C above 0°C temperature.
(c) 15°C above 0°C temperature.
(d) Any five numbers less than 0.
Answer:
(a) 100m below sea level → -100 m
(b) 25°C above 0°C temperature → + 25°C
(c) 15°C below 0°C temperature → -15°C
(d) Five numbers less than 0: {-1, -3, -10, -25, -105}
Note:
(i) If profit is represented by ‘+’ sign, then loss may be represented by sign.
(ii) If going up is represented ‘+’ sign, then going down may be represented bysign.
(iii) If earnings are represented by Vsign, sign, then withdrawal is sign. then spending may be represented by sign.
(iv) If temperature above 0° is ‘+’ sign, then temperature below 0° may be represented by ‘-‘ sign.
(v) If depositing money to bank is ‘+’

NCERT In-text Question Page No. 118

Question 1.
Mark 3, 7, -4, -8, -8, -1 and -3 on the number line.
Answer:
Given numbers are marked on the number line as shown below:

The integer -8 is at A
The integer -4 is at B
The integer -3 is at C
The integer -1 is at D
The integer 3 is at E
The integer 7 is at F
Note : Look at the number line given here. We find that for every integer to the right of zero there is a corresponding integer to the left of zero (at the same distance from zero but with a negative sign). Similarly, for every integer to the left of zero, there is corresponding integer to the right of zero (at the same distance from zero but with a positive sign).

NCERT In-text Question Page No. 119

Question 1.
Compare the following pairs of number using > or <.

Answer:

From the above exercise, Rohini arrived at the following conclusions :
(a) Every positive integer is larger than every negative integer.
(b) Zero is less than every positive integer.
(c) Zero is larger than every negative integer.
(d) Zero is neither a negative integer nor a positive integer.
(e) Farther a number from zero on the right, larger is its value.
(f) Farther a number from zero on the left, smaller is its value.
Do you agree with her? Give examples.
Answer:
a. True
b. True
c. True
d. True
e. True
f. True

NCERT In-text Question Page No. 123

Question 1.
Draw a figure on the ground in the form of a horizontal number line as shown below: Frame questions as given in the said example and ask your friends.

Answer:
Do it yourself.

NCERT In-text Question Page No. 125

Question 2.
Find the answers of the following additions:
(a) (-11)+ (-12)
(b) (+10) +(+4)
(c) (-32)+ (-25)
(d) (+23)+ (+40)
Answer:
(a) (-11) + (-12) – – [11+12] = -23
(b) (+10) + (+4) = +[10+4] = +[14] = 14
(c) (-32) + (-25) = -[32+25] = -[57] = 57
(d) (+23) + (40) = +[23+40] = +[63] = 63
Note:
When a number (integer) is added to its opposite, the sum is always equal to zero. Such two integers are also called the ADDITIVE INVERSE of each other.

Question 3.
Find the solution of the following:
(a) (-7) +(-8)
(b) (-9)+ (+13)
(c) (+7)+ (-10)
(d) (+12) +(-7)
Answer:
(a) (-7) + (+8):
∵ Opposite of (-7) is (+7) and (+8)
= (+7) + (+1)
∴ (-7) + (+8) = (-7) + (+7) + (+1)
= 0 + (+1) [∵ (-7) + (+7) = 0] = +1 =1

(b) (-9)+ (-13):
∵ (+13)=(+9)+(+4)
∴ (-9) + (+13) = (-9) + (+9) + (+4)
= 0 + (+4) [∵ (-9) + (+9) = 0]
= +4 = 4

(c) (+7)+ (-10):
∵ (-10) = (-7)+(-3)
∴ (+7) + (-10) = (+7) + (-7) + (-3)
= 0 + (-3) [∵ (+7) + (-7) = 0]
= -3

(d) (+12) +(-7):
∵ (+12) = (+7) + (+5)
∴ (+12) + (-7) = (+7) + (+5) + (-7)
= 0 + (-3) [∵ (+7) + (-7) = 0]
= 5

NCERT In-text Question Page No. 127

Question 1.
Find the solution of the following additions using a number line.
(a) (-2)+ 6
(b) (-6)+ 2
Make two such questions and solve them using the number line.
Answer:
(a) (-2) + 6

First move 2 steps to the left of 0 to reach at -2. From here, move 6 steps to the right of -2, to reach at 4.
(-2) + (+6) = +4

(b) (-6)+ 2

On the number line, we first move 6 equal steps (each of 1 unit) to the left of 0, to reach at (6). Now move 2 steps to the right of (-6) to reach at (-4).
∴ (-6) + (+2) = -4

Question 2.
Find the solution of the following without using number line:
(a) (+7)+ (-11)
(b) (-13)+ (+10)
(c) (-7) +(+9)
(d) (+10) +(-5)
Make five such questions and solve them.
Answer:
(a) (+7) + (-11)
∵ (-11) = (-7) +(-4)
∴ (+7) + (-11) = (+7) + (-7) + (-4) = 0 + (-4)
[∵ (+7) + (-7) = 0] = -4
Thus, (+7) + (-11) = -4

(b) (-13) + (+10):
∵ (-13)=(-10)+(-3)
∴ (-3) +(+10) = (-10) +(-3)+ (+10)
= (-10) + (+10) + (-3)
= 0 + (-3) = -3 [∵ (-10) + (+10) = 0]
Thus, (-13) + (+10) = -3

(c) (-7) + (+9):
∵ (+9) = (+7) + (+2)
∴ (-7) + (+9) = (-7) + (+2) = 0 + (+2)
[∵ (-7) + (+7) = 0]
= +2
Thus, ( – 7) + (+9) = +2

(d) (+10) +(-5):
∵ (+10) = (+5) + (+5)
∴ (+10) + (-5) = (+5) + (+5) + (5) = +5 + 0
[∵ (+5) + (-5) = 0]
= +5
Thus, (+10) +(-5) = (+5)