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Punjab State Board PSEB 7th Class English Book Solutions English Letter Writing Exercise Questions and Answers, Notes.

PSEB 7th Class English Letter Writing

1. Application for Leave (For going to a doctor)

Write an application to your Headmaster to grant you leave as you are going to the doctor.

The Headmaster
Govt. High School
Ludhiana
12 March 20…….
Sir

I wish to inform you that my mother is ill. I have an appointment with the doctor at 12 for her medical check up. Kindly allow me to leave the school at 11 a.m.

Thanking you
Yours obediently
Baldev Singh
Class VII B

Word-Meanings : Appointment- मिलने का समय, Allow- अनुमति देना।

2. Application for School Leaving Certificate

Imagine you are Baldev Oberoi. You are a student of Govt. High School, Ropar. Your father has been transferred to Sangrur. Write an application to the Headmaster of your school, requesting him to issue you the school leaving certificate.
Or
Write an application to the headmaster of your school requesting him to issue you the school living certificate.

The Headmaster
Govt. High School
Ropar
12 March 20……
Sir

With due respect I wish to say that I am a student of VII A of your school. My father has been transferred to Sangrur. He is to report on duty there within three days. All the members of our family are leaving for Sangrur.

I will have to join some school at Sangrur. So you are requested to issue my school leaving certificate and oblige.

Yours obediently
Baldev Oberoi
Class VII A

Word-Meanings : Transferred- बदली हो गई, Oblige- कृतार्थ करें।

3. Application for Sick Leave

Suppose you are Harjit Singh. You are a student of Govt. High School, Balachaur. Write an application to the Headmaster to grant you leave for four days because you are ill.

The Headmaster
Govt. High School
Balachaur
March 13, 20……..
Sir

I have been ill since yesterday. The doctor has advised me complete rest. So I cannot attend the school.
Kindly grant me leave for four days. I shall be highly thankful to you for this.

Yours obediently
Harjit Singh
VII C

Word-Meanings : Advised- परामशं दिया हे, Highly thankful- अति धन्यवादी

4. Application for Full Fee Concession

You are a student of Khalsa Model School, Patiala. Write an application to the Headmaster requesting him to grant you full fee concession. Give reasons like father’s income, family conditions, academic and sports merit.

The Headmaster
Khalsa Model School
Patiala
8 March 20……
Sir

I am a student of VII B of your school. My father is a clerk. His income is very small but our family is very large. So he is unable to pay my school fee.

I am a good student. I always get good marks. I am a good player of hockey also. I was a free student last year. Kindly grant me full fee concession this year also. I shall be thankful to you for this.

Yours obediently
Sham Singh
Roll No. 25
VIIB

Word-Meanings : Unable- असमर्थ, Free student- फीस मुक्त विद्यार्थी

5. Application for Remission of Fine Write an application to the Headmistress of your school for remission of fine.

The Headmistress
GGS High School
X.Y.Z.
10 March 20 ………
Madam

I beg to say that our English teacher gave us a test on Sunday. My father was seriously ill on that day. I was to attend upon him. So I could not take the test. My teacher fined me ten rupees.

My father is a poor man. He cannot pay my fine. Besides, I am a good student. I always stand first in English. Kindly remit my fine.
I shall be very thankful to you.

Yours obediently
Simaran
VIID

Word-Meanings : Attend upon- देखरेख करना, Besides- इसके अतिरिक्त

6. Encouraging Brother / Sister on Losing a Match

Your brother/sister is the captain of his/her school’s hockey team. The school lost the match and could not qualify for the zonals. He/she is sad about it. Write a letter encouraging him/her and giving some advice.

42 Court Road
Amritsar
April 2, 20……
Dear Kamal

Through a newspaper, I have come to know that your school has lost the hockey match. It has failed to qualify for the zonals. Since you are the captain of the team, you must be very upset. But there is nothing to be sad about it. Victory and defeat go side by side. Moreover, failures are the stepping stones to success. So have courage and work harder. One day the success will be yours.
With lots of love

Your loving
brother Balvinder

Word-Meanings : Qualify- स्थान बनाना, Defeat- हार, Side by side- साथ साथ, Stepping stones to success- सफलता की सीढियां

7. Letter to a Sportsperson

Write a letter to a sportsperson telling him/her why you admire him/her.

215 Model Town
Khanna
18 Feb. 20 ………
Dear Sachin

I am very fond of sports. I like cricket the most and you are my favourite sportsperson. Your style of playing is really admirable. I feel thrilled when you are at the crease. The way you bowl and field is superb. So I consider you the best all-rounder of today. I wish you set more and more records.
With heartiest well-wishes

Your fan
Navjat

Word-Meanings : Sportsperson- खिलाड़ी, Admirable- प्रशंसनीय, Thrilled- रोमांचित Superb- शानदार।

8. Letter about School

Write a letter to your cousin in U.S.A. telling him about your school.

105 The Mall
Amritsar
October 1, 20……
Dear Sanju

You will be glad to know that I have joined a new school. It is a government school. It is outside the city. The school building is very beautiful. There are 40 rooms in it. It has two large playgrounds. It has a garden also. The library is very big.

The school has 40 teachers. They are well qualified. The headmaster is a very able person. He is an M.A., B.T. The school has 1500 students. The examination results are always good. The school is really a model school.
With love

Yours affectionately
Raman

Word-Meanings : Qualified- योग्य, Model- अदर्श

9. Inviting a Friend to a Birthday Party

Write a letter to your friend Vinod inviting him to your birthday party.

Kailash Colony
……….. City
March 12, 20 ……..
My dear Vinod

You will be glad to know that my birthday falls on March 20. I am planning to celebrate it at home. There will be a programme of dance and music. A dinner will also be served. All our friends are coming. I hope you will also join us. I am sure you will enjoy yourself.

Yours sincerely
Prem

Word-Meanings : Celebrate- मनाना, Falls- को पड़ता है।

10. Mistakes in the English Companion

Write a letter to Raman Publishers about some mistakes in their English Companion of Class VII.

Dev Kumar
H.No. 315 Mithu Basti
Jalandhar
M/s Raman Publishers
Lal Chowk
Ludhiana
March 10, 20 …….
Sir

It is to bring to your notice that there are some mistakes in the English Companion Class VII. – The errors are as under:
Page No. Incorrect Correct rain
reign 21 Rami
Rani 24
Ambrella Umbrella 60 Jansi
Jhansi There can be more errors. Kindly revise the book and rectify the mistakes. Thanking you Yours truly Dev Kumar

Word-Meanings : Errors- गन्तियां, Rectily- सुधारना

11. Extension of Holidays

Write a letter to your Principal for extension of your holidays.

The Principal
Guru Ramdas High School
Khanna
11 March 20 ……..
Sir

Our summer vacation are coming to a close. The sun is still very hot because it did not rain during the holidays. The heat is unbearable. This is the hottest month breaking all previous (fted) records. I would, therefore, request you to extend (Q@FTI) the holidays for at least ten days.

Thanking you
Yours obediently
Simranjeet
VII-B

Word-Meanings : Unbearable- असहनीय, Previous- पिछले, Extend- बढ़ाना, At least- कम से कम

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 13 Direct and Inverse Proportions Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.1

1. Following are the car parking charges near a railway station upto:

Check if the parking charges are in direct proportion to the parking time.
Solution:
Here, ratio of parking charges and parking time are as follow:

Parking time	Parking charges	Parking charge / Parking time
4 hours	₹ 60	\(\frac{60}{4}=\frac{15}{1}\)
8 hours	₹ 100	\(\frac{100}{8}=\frac{25}{2}\)
12 hours	₹ 140	\(\frac{140}{12}=\frac{35}{3}\)
24 hours	₹ 180	\(\frac{180}{24}=\frac{15}{2}\)

Here, \(\frac {15}{1}\) ≠ \(\frac {25}{2}\) ≠ \(\frac {35}{3}\) ≠ \(\frac {15}{2}\)
Thus, the parking charges are not in direct proportion to the parking time.

2. A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added:
Solution:

Parts of red pigment	1	4	7	12	20
Parts of base	8	…	…	…	…

If parts of red pigment are x₁, x₂, x₃, x₄ and x₅ respectively, then parts of base are y₁, y₂, y₃, y₄ and y₅ respectively. Here, it is clear that mixture preparation is in direct proportion.

Thus, the table is

Parts of red pigment	1	4	7	12	20
Parts of base	8	32	56	96	160

3. In Question 2 above, if 1 part of a red pigment requires 75 ml of base, how much red pigment should we mix with 1800 ml of base?
Solution:
See as per question 2 –
x₁ = 1, y₁ = 75, x₂ = ? and yx₂ = 1800
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{1}{75}=\frac{x_{2}}{1800}\)
∴ x₂ = \(\frac{1 \times 1800}{75}\)
∴ x₂ = 24
Thus, 24 ml of red pigment should be mixed with 1800 ml of base.

4. A machine in a soft drink factory fills 840 bottles in six hours. How many bottles will it fill in five hours?
Solution:
Let the number of bottles filled by machine in 5 h be x.

Number of hours (x)	6	5
Number of bottles filled (y)	840	?

Here, as the number of hours decreases, the number of bottles filled will also decrease.
∴ It is case of direct proportion.
Here, x₁ = 6, y₁ = 840, x₂ = 5 and y₂ = ?
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{6}{840}=\frac{5}{y_{2}}\)
∴ y₂ = \(\frac{5 \times 840}{6}\)
∴ y₂ = 700
Thus, 700 bottles will be filled in 5 hours.

5. A photograph of a bacteria enlarged 50,000 times attains a length of 5 cm as shown in the diagram. What is the actual length of the bacteria? If the photograph is enlarged 20,000 times only, what would be its enlarged length?
Solution:

Enlargement in picture of bacteria	Length (cm)
50,000 times enlarged (x1)	5 (y1)
1 (x2)	? (y2)

Here, length of bacteria increases as picture of bacteria enlarges.
∴ It is case of, direct proportion.

Hence, the actual length of bacteria is 10^-4 cm.
Now, the photograph is enlarged 20,000 times.

Enlargement in picture of bacteria	Length (cm)
50,000 times enlarged (x1)	5 (y1)
20,000 times enlarged (x1)	? (y2)

∴ \(\frac{50,000}{5}=\frac{20,000}{y_{2}}\)
∴ y₂ = \(\frac{20,000 \times 5}{50000}\)
∴ y₂ = 2
Thus, its enlarged length would be 2 cm.

6. In a model of a ship, the mast is 9 cm high, while the mast of the actual ship is 12 m high. If the length of the ship is 28 m, how long is the model ship?
Solution:

–	Actual ship	Model ship
Length of the ship x	28 m	?
Height of mast y	12m	9 cm

This is a case of direct proportion.
x₁ = 28, y₁ = 12, x₂ = ?, y₂ = 9
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{28}{12}=\frac{x_{2}}{9}\)
∴ x₂ = \(\frac{28 \times 9}{12}\)
∴ x₂ = 21
Thus, the length of model ship is 21 cm.

7. Suppose 2 kg of sugar contains 9 × 10⁶ crystals. How many sugar crystals are there in

Question (i)
5 kg of sugar?
Solution:

Weight of sugar (kg) x	Number of sugar crystals y
x1 = 2	y1 = 9 × 106
x2 = 5	y2 = (?)

This is a case of direct proportion.

Thus, there are 2.25 × 10⁷ crystals of sugar in 5 kg of sugar.

Question (ii)
1.2kg of sugar?
Solution:

Weight of sugar (kg) x	Number of sugar crystals y
x1 = 2	y1 = 9 × 106
x2 = 1.2	y2 = (?)

Here. x₁ = 2, x₁ = 9 × 10⁶, x² = 1.2, y² = ?
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{2}{9 \times 10^{6}}=\frac{1.2}{y_{2}}\)
∴ y₂ = \(\frac{1.2 \times 9 \times 10^{6}}{2}\)
∴ y₂ = 0.6 × 9 × 10⁶
∴ y₂ = 5.4 × 10⁶
Thus, there are 5.4 × 10⁶ crystals of sugar in 1.2 kg of sugar.

8. Rashmi has a road map with a scale of 1 cm representing 18 km. She drives on a road for 72 km. What would be her distance covered in the map?
Solution:

Actual distance (km) x	Distance on the map (cm) y
x1 = 18	y1 = 1
x2 = 72	y2 = (?)

This is a case of direct proportional.
Here, x₁ = 18 km, y₁ = 1 cm, x₂ = 72 km, y₂ = ?
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{18}{1}=\frac{72}{y_{2}}\)
∴ y₂ = \(\frac{72 \times 1}{18}\)
∴ y₂ = 4
Thus, the distance covered by her on the map is 4 cm.

9. A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time

Question (i)
the length of the shadow cast by another pole 10 m 50 cm high
Solution:

Height of vertical pole x	Length of shadow y
x1 = 5 m 60 cm = 560 cm	y1 = 3 m 20 cm = 320 cm
x2 = 10 m 50 cm = 1050 cm	y2 = (?)

This is a case of direct proportionality.

Thus, the length of the shadow cast by another pole is 6 m.

Question (ii)
the height of a pole which casts a shadow 5 m long.
Solution:

Height of vertical pole x	Length of shadow y
x1 = 560 cm	y1 = 320 cm
x2 = (?)	y2 = 5 m = 500 cm

x₁ = 560, y₁ = 320, x₂ = ?, y₂ = 500
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{560}{320}=\frac{x_{2}}{500}\)
∴ x₂ = \(\frac{560 \times 500}{320}\)
∴ x₂ = 875 cm
∴ x₂ = 8.75 cm
Thus, the height of the pole is 8,75 m.

10. A loaded truck travels 14 km in 25 minutes. If the speed remains the same, how far can it travel in 5 hours?
Solution:

Distance (km) x	Time (minute) y
x1= 14	y1 = 25
x2 = (?)	y2 = 5 hours = 300

This is a case of direct proportion.
∴ Here, x₁ = 14, y₁ = 25, x₂ = ?, y₂ = 300
\(\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}\)
∴ \(\frac{14}{25}=\frac{x_{2}}{300}\)
∴ x₂ = \(\frac{14 \times 300}{25}\)
∴ x₂ = 168
Thus, loaded truck can travel 168 km in 5 h.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

Question1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of it.
Solution:
Radius of cone = Radius of hemisphere = 1 cm

R = 1 cm
Height of cone (H) = 1 cm
Volume of solid = volume of cone + volume of hemisphere
= \(\frac{1}{3}\) πR²H + \(\frac{2}{3}\) πR³
= \(\frac{1}{3}\) πR²2 [H + 2R]
= \(\frac{1}{3}\) π × 1 × 1 [1 + 2 × 1]
= \(\frac{1}{3}\) π × 3 = π cm³
Hence, Volume of solid = π cm³.

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model the Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:

Radius of cone = Radius of cylinder (R) = \(\frac{3}{2}\) cm
∴ R = 1.5 cm
Height of eah cone (h) = 2 cm
∴ Height of cylinder = 12 – 2 – 2 = 8 cm
Volume of air in cylinder = volume of cylinder + 2 (volume of cone)

Volume of air in cylinder = \(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{28}{3}\)
= 22 × 3 = 66 cm³

Hence, Volume of air in cylinder =66 cm³.

Question 3.
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Solution:
Gulab Jamun is in the shape of cylinder
Diameter of cylinder = Diameter of hemisphere = 2.8 cm

Radius of cylinder = Radius of hemisphere (R)
= \(\frac{28.2}{2}\) = 1.4 cm
R = 1.4 cm
Height of cylindrical part = 5 – 1.4 – 1.4
= (5 – 2.8) cm = 2.2 cm.
Volume of one gulab Jamun = Volume of cylinder + 2 [Volume of hemisphere]
= πR²H + 2 \(\frac{2}{3}\) πR³

= πR² H + \(\frac{4}{3}\) R

= \(\frac{4}{4}\) × 1.4 × 1.4 2.2 + \(\frac{4}{3}\) × 1.4

= \(\frac{22}{7}\) × \(\frac{14}{10}\) × \(\frac{14}{10}\) 2.2 + \(\frac{4}{3}\) × 1.4

= \(\frac{22}{7}\) × \(\frac{196}{100}\) [2.2 + 1.86]

= \(\frac{22 \times 28}{100}\) [4.06]

Volume of one gulab Jamun = 25.05 cm³
Now volume of 45 gulab Jamuns = 45 × 25.05 cm³ = 1127.25 cm³
Volume of sugar syrup = 30% volume of 45 gulab Jamuns
= \(\frac{30 \times 1127.25}{100}\) = 338.175 cm3
Hence, Approximately sugar syrup = 338 cm³.

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution. Length of cuboid (L) = 15 cm
Width of cuboid (B) = 10 cm
Height of cuboid (H) = 3.5 cm
Radius of conical cavity (r) = 0.5 cm
Height of conical cavity (h) = 1.4 cm
Volume of wood in Pen stand = volume of cuboid – 4 [volume of cone]
= LBH – 4 \(\frac{1}{3}\) πr²h

= 15 × 10 × 3.5 – \(\frac{4}{3}\) × \(\frac{22}{7}\) × 0.5 × 0.5 × 1.4

= \(\frac{15 \times 10 \times 35}{10}-\frac{4}{3} \times \frac{22}{7} \times \frac{5}{10} \times \frac{5}{10} \times \frac{14}{10}\)

= \(15 \times 35-\frac{22}{3 \times 5}\)
= 525 – 1.466 = 523.534 cm³
Hence, Volume of wood in Pen stand = 523.53 cm³.

Question 5.
A vessel is in the form of an inverted cone. Its height ¡s 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:

Radius of cone (R) = 5 cm
Height of cone (H) = 8 cm
Radius of each spherical lead shot (r) = 0.5 cm
Let number of shot put into the cone = N
According to Question,
N [Volume of one lead shot] = \(\frac{1}{4}\) Volume of water in cone

= 10 × 10 = 100
Hence, Number of lead shots = 100.

Question 6.
A solid iron pole consists of a cylinder of height 220 cfi and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of Iron has approximately 8 g mass. (Use n = 3.14)
Solution:
Diameter of lower cylinder = 24 cm
Radius of lower cylinder (R) = 12 cm
Height of lower cylinder (H) = 220 cm
Radius of upper cylinder (r) = 8 cm
Height of upper cylinder (h) = 60 cm
Volume of pole = Volume of Lower cylinder + volume of upper cylinder
= πR²H + πr²h
= 3.14 × 12 × 12 × 220 + 3.14 × 8 × 8 × 60
= 99475.2 + 12057.6

Volume of pole = 111532.8 cm³
Mass of 1 cm³ = 8 gm
Mass of 111532.8 cm³ = 8 × 111532.8 = 892262.4 gm
= \(\frac{892262.4}{1000}\) kg = 892.2624 kg
Hence, Mass of Pole = 892.2624 kg.

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Radius of cone = Radius of hemisphere = Radius of cylinder

Height of cone (h) = 120 cm
Height of cylinder (H) = 180 cm
Volume of cylindrical vessel = πR²H
= \(\frac{22}{7}\) × 60 × 60 × 180 = 2036571.4 cm³
Volume of solid inserted in cylinder = Volume of hemisphere + Volume of cone
= \(\frac{2}{3}\) πR³ + \(\frac{1}{3}\) πR²h

= \(\frac{1}{3}\) πR² [2R + h]

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 60 × 60 [2 × 60 + 120]

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3600 [120 + 120]

= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 3600 × 240 = 905142.86 cm³
Volume of water flows out = 90514186 cm³
∴ Volume of water left in cylinder = Volume of cylinder – Volume of solid inserted in th’e vessel
= (2036571.4 – 905142.86) cm³ = 1131428.5 cm³
= \(\frac{1131428.5}{100 \times 100 \times 100}\) m³ = 1.131 m³
Hence, Volume of water left in cylinder = 1.131 m³.

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Diameter of neck (cylindrical Portion) = 2 cm

Radius of neck (r) = 1 cm
Height of cylindrical portion (H) = 8 cm
Diameter of spherical portion = 8.5 cm
Radius of spherical portion (R) = \(\frac{8.5}{2}\) cm = 4.25 cm
Volume of water in vessel = Volume of sphere + Volume of cylinder
= \(\frac{4}{3}\) πR³ + πR²h
= \(\frac{4}{3}\) × 3.14 × 4.25 × 4.25 × 4.25 × 3.14 × 1 × 1 × 8
= 321.39 + 25.12 = 346.51 cm³
Hence, Volume of water in vessel = 346.51 cm³ and She is wrong.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 14 Factorization InText Questions

Try These : [Textbook Page No. 219]

1. Factorise:

Question (i)
12x + 36
Solution:
12x = 2 × 2 × 3 × x and
36 = 2 × 2 × 3 × 3
Common factors = 2 × 2 × 3
∴ 12x + 36 = (2 × 2 × 3 × x) + (2 × 2 × 3 × 3)
= (2 × 2 × 3) (x + 3)
= 12 (x + 3)

Question (ii)
22y – 33z
Solution:
22y = 2 × 11 × y and 33z = 3 × 11 × z
Common factor =11
∴ 22y – 33z = (2 × 11 × y) – (3 × 11 × z)
= (11) × (2 × y – 3 × z)
= 11 (2y – 3z)

Question (iii)
14pq + 35pqr
Solution:
14pq = 2 × 7 × p × q and
35pqr = 7 × 5 × p × q × r
Common factor = 7pq
∴ 14pq + 35pqr = (2 × 7 × p × q) + (7 × 5 × p × q × r)
= 7 × p × q (2 + 5 × r)
= 7pq (2 + 5r)

Try These : [Textbook Page No. 225]

1. Divide:

Question (i)
24xy²z³ by 6yz²
Solution:
= \(\frac{24 x y^{2} z^{3}}{6 y z^{2}}\)
= \(\frac{2 \times 2 \times 2 \times 3 \times x \times y \times y \times z \times z \times z}{2 \times 3 \times y \times z \times z}\)
= \(\frac{2 \times 2 \times x \times y \times z}{1}\) = 4xyz
∴ 24xy²z³ ÷ 6yz²
= 4xyz

Question (ii)
63a²b⁴c⁶ by 7a²b²c³
Solution:
= \(\frac{63 a^{2} b^{4} c^{6}}{7 a^{2} b^{2} c^{3}}\)
= \(\frac{3 \times 3 \times 7 \times a^{2} \times b^{4} \times c^{6}}{7 \times a^{2} \times b^{2} \times c^{3}}\)
= 3 × 3 × \(\frac{a^{2}}{a^{2}} \times \frac{b^{4}}{b^{2}} \times \frac{c^{6}}{c^{3}}\)
= 9 × a^2-2 × b^4-2 × c^6-3
= 9 × a⁰ × b² × c³
= 9 × 1 × b² × c³
= 9b²c³
∴ 63a²b⁴c⁶ ÷ 7a²b²c³
= 9b²c³

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 14 Factorization Ex 14.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.4

1. Find and correct the errors in the following mathematical statements:

Question 1.
4 (x – 5) = 4x – 5
Solution:
Error: 4 × – 5 = (- 20) and not (- 5)
Correct statement: 4 (x – 5) = 4x – 20

Question 2.
x (3x + 2) = 3x² + 2
Solution:
Error: x × 2 = 2x
Correct statement: x (3x + 2) = 3x² + 2x

Question 3.
2x + 3y = 5xy
Solution:
Error: 2x and 3y are unlike terms.
So their sum is not possible.
Correct statement: 2x + 3y = 2x + 3y

Question 4.
x + 2x + 3x = 5x
Solution:
Error: x, 2x and 3x are like terms. So sum of their coefficient =1 + 2 + 3 = 6.
Correct statement: x + 2x + 3x = 6x

Question 5.
5y + 2y + y – 7y = 0
Solution:
Error : 5y, 2y, y and – 7y all are like terms here. So sum of their coefficient = 5 + 2 + 1 – 7 = 1.
Correct statement: 5y + 2y + y – 7y = y

Question 6.
3x + 2x = 5x²
Solution:
Error: When like terms are added or subtracted their exponents do not change.
Correct statement: 3x + 2x = 5x

Question 7.
(2x)² + 4 (2x) + 7 = 2x² + 8x + 7
Solution:
Error: (2x)² = 2x × 2x = 4x²
Correct statement:
(2x)² + 4 (2x) + 7 = 4x² + 8x + 7

Question 8.
(2x)² + 5x = 4x + 5x = 9x
Solution:
Error : (2x)² = (2x × 2x) = 4x²
Correct statement: (2x)² + 5x = 4x² + 5x

Question 9.
(3x + 2)² = 3x² + 6x + 4
Solution:
Error : (3x + 2)²
= (3x)² + 2 (3x)(2) + (2)²
= 9x² + 12x + 4
Correct statement:
(3x + 2)² = 9x² + 12x + 4

10. Substituting x = – 3 in

Question (a)
x² + 5x + 4 gives (- 3)² + 5 (- 3) + 4 = 9 + 2 + 4 = 15
Solution:
Error : 5 (- 3) = – 15 and not 2
Correct statement:
Substituting x = (- 3) in, x² + 5x + 4
= (- 3)² + 5 (-3) + 4
= 9 – 15 + 4 = 9 + 4 – 15
= 13 – 15
= (-2)

Question (b)
x² – 5x + 4 gives (- 3)² – 5 (- 3) + 4 = 9 – 15 + 4 = -2
Solution:
Error: -5 (-3) = + 15 and not (-15)
Correct statement:
Substituting x = (- 3) in,
x² – 5x + 4
= (- 3)² – 5 (- 3) + 4
= 9 + 15 + 4
= 28

Question (c)
x² + 5x gives (- 3)² + 5 (- 3) = – 9 – 15 = – 24
Solution:
Error : (- 3)² = + 9
Correct statement:
Substituting x = (- 3) in, x² + 5x
= (- 3)² + 5 (- 3)
= 9 – 15 = – 6

Question 11.
(y – 3)² = y² – 9
Solution:
Error : (y – 3)²
= (y)² – 2 (y)(3) + (- 3)²
= y² – 6y + 9
Correct statement: (y – 3)² = y² – 6y + 9.

Question 12.
(z + 5)² = z² + 25
Solution:
Error : (z + 5)²
= (z)² + 2 (z)(5) + (5)²
= z² + 10 z + 25
Correct statement: (z + 5)²
= z² + 10z + 25

Question 13.
(2a + 3b) (a – b) = 2a² – 3b²
Solution:
Error : (2a + 3b) (a – b)
= 2a (a-b) + 3b (a-b)
= 2a² – 2ab + 3ab – 3b²
= 2a² + ab – 3b²
Correct statement: (2a + 3b) (a – b)
= 2a² + ab – 3b²

Question 14.
(a + 4) (a + 2) = a² + 8
Solution:
Error : (a + 4) (a + 2) = a (a + 2) + 4 (a + 2)
= a² + 2a + 4a + 8
= a² + 6a + 8
Correct statement: (a + 4) (a + 2)
= a² + 6a + 8

Question 15.
(a – 4) (a – 2) = a² – 8
Solution:
Error : (a – 4) (a – 2) = a (a – 2) – 4 (a – 2)
= a² – 2a – 4a + 8
= a² – 6a + 8
Correct statement: (a – 4) (a – 2)
= a² – 6a + 8

Question 16.
\(\frac{3 x^{2}}{3 x^{2}}\) = 0
Solution:
Error: Numerator and denominator, both are same. So their division is 1.
Correct statement:\(\frac{3 x^{2}}{3 x^{2}}\) = 1

Question 17.
\(\frac{3 x^{2}+1}{3 x^{2}}\) = 1 + 1 = 2
Solution:
Error: \(\frac{3 x^{2}+1}{3 x^{2}}=\frac{3 x^{2}}{3 x^{2}}+\frac{1}{3 x^{2}}\)
= 1 + \(\frac{1}{3 x^{2}}\)
Correct statement: \(\frac{3 x^{2}+1}{3 x^{2}}\) = 1 + \(\frac{1}{3 x^{2}}\)

Question 18.
\(\frac{3 x}{3 x+2}=\frac{1}{2}\)
Solution:
Error: Here, simplification of LHS is not possible.
Correct statement: \(\frac{3 x}{3 x+2}=\frac{3 x}{3 x+2}\)

Question 19.
\(\frac{3}{4 x+3}=\frac{1}{4 x}\)
Solution:
Error: Here, simplification of LHS is not possible.
Correct statement: \(\frac{3}{4 x+3}=\frac{3}{4 x+3}\)

Question 20.
\(\frac{4 x+5}{4 x}\) = 5
Solution:
Error: \(\frac{4 x+5}{4 x}\)
= \(\frac{4 x}{4 x}+\frac{5}{4 x}\)
= 1 + \(\frac{5}{4 x}\)
Correct statement: \(\frac{4 x+5}{4 x}\) = 1 + \(\frac{5}{4 x}\)

Question 21.
\(\frac{7 x+5}{5 x}\) = 7x
Error: \(\frac{7 x+5}{5 x}\)
= \(\frac{7 x}{5}+\frac{5}{5}\)
= \(\frac{7 x}{5}\) + 1
Correct statement: \(\frac{7 x+5}{5}=\frac{7 x}{5}\) + 1

Punjab State Board PSEB 7th Class English Book Solutions English Message Writing Exercise Questions and Answers, Notes.

PSEB 7th Class English Message Writing

Type – I

1. Read the following telephonic conversation between Kavita and Karan. Karan will not be able to meet Avik. He leaves a message for him. Write this message by using not more than 50 words.

Kavita : Hello ! Hello ! I am Kavita from Indore. Can I speak to Avik? I am his sister.
Karan : Hello ! Kavita I am Karan, Avik’s colleagues. Avik is on leave today. Can I take a message ?
Kavita : Yes, Karan, I am coming to Mumbai tomorrow. Ask him to pick me up at the
airport. I have an interview for the post of Scientist at NPL on the day after tomorrow.
Karan : Which flight are you coming on ?
Kavita : It is the Jetline flight which arrives there at 7.15 p.m. I am bringing along with me that big box which contains his books. I hope it won’t be any trouble for you coming.
Karan : Not at all, I will leave a message on his table. Okay, Kavita.
Kavita : Thank you, Karan. Bye.

Message

Avik

Today, there was a telephonic message from your sister, Kavita. She is coming to Mumbai tomorrow as she has some interview here. She is coming by the Jetlnte flight which will arrive here at 7.15 p.m. She says she will be bringing with her a big box containing your books. Please pick her up at the airport.

Karan

2. Read the following telephone conversation which took place when Suresh was staying with his uncle. Write the message from Suresh to his maid, using not more than 50 words.

Seshu : Hello ! Hello ! This is Sheshu from Lucknow. Can I speak to Mrs. Rao, please ? I am a friend of his son, Madhav.
Suresh : This is Suresh Rao. My uncle is not here at the moment. We heard about the earthquake. Is Madhav all right ?
Seshu : Yes, yes. He’s okay now. He had a bad fall during the earthquake and he broke his left leg. It was a multiple fracture, but there’s nothing to worry about now.
Suresh : Is he in hospital ?
Sheshu : Yes, he’s at the Tata Memorial Hospital here. Would you please inform his family ?
Suresh : Of course I will.

Message

Dear Uncle

There was a telephonic call for you from one Mrs. Sheshu. He is our Madhav’s friend, from Lucknow. He had a fall during the earthquake and he broke his leg. He got a multiple fracture and has been in the Tata Memorial Hospital there. But he added that there was nothing to worry.

Suresh

3. Here is telephonic talk between Gurbani and Jaspreet. Gurbani give her a message , Write the message on behalf of Gurmeet not more than 50 words.

Gurbani : Hi ! Gurmeet.
Jaspreet : Sorry, I’m not Gurmeet. I’m her elder sister Jaspreet. Can I know who is calling ?
Gurbani : I Gurbani, her friend, Is it not her contact number ? I have some urgent message for her.
Jaspreet : Yes, it is but she has gone to the market to by some fruit and her mobile, is with me.
Gurbani : Would you please convey my message to her ? .
Jaspreet : But she is not coming back for about two hours. I am also going.to the hospital to see one of our neighbours. Would you gave me the message. I’ll leave it on her table before I go.
Gurbani : Sure ! Please note our family is going to Hazoor Sahib on Sunday. She can accompnay as if her parents allow. It will be a good company for me.
Jaspreet : All right, Don’t worry. The message will reach her.
Gurbani : Thank you very much.

Message

22.06.2020
Dear Gurmeet,

There was a call from your friend Gurbani in your absence. Their family is going to Hazoor Sahib on Sunday. You can accompany them if our parents permit. She will be feeling good in your company. Talk to her for confirmation.

Jaspreet.

4. Read the conversation between Mrs. Singh and the Principal of G.S.S. Modern School. Write the message on behalf of the Principal that he will send to the Preeti’s class-teacher.

Mrs. Singh : Hello ! Is that G.S.S. Modem School ?
Principal : Yes, what do you want ?
Mrs. Singh : I would like to speak to the school Principal.
Principal : yes, speaking. What can I do for you.
Mrs. Singh : My daughter, Preeti is a student of VII A of your school. Today was the last day for the payment of her school fee. I have deposited it the school account.
Principal : What is the problem in that ? It was your duty.
Mrs. Singh : Madam, she was worried about fine before she left for school and looked sad. Send this message to her in class, to make her tension free.
Principal : Of course ! The message will be sent to her through her class teacher.
Mrs. Singh : Thank you, Madam.

Message

12.05.2020
Dear Preeti

You looked worried and sad before you left for school. It was natural because your school fee was not paid and you could be fined or punished in some other way for that. Now, don’t take any tension as your fee has been deposited.

Mummy

5. Read the following telephonic conversation between Ravinder and Ranjit about to leave for her coaching soon and his mother is not at home for the moment. Write this message on behalf of Ranjit.

Ravinder : Hello, is that Amit ?
Ranjit : No I am her elder brother, Ranjit speaking May I know who is speaking ?
Ravinder : I am Ravinder speaking. I wanted to speak to your brother for an important message.
Ranjit : He is not at home now. He has just gone to visit one of two friends. I am also leaving for my office. If there is some message I shall give him.
Ravinder : OK, then Please tell Amit all about me. We are to play a friendly cricket match tomorrow morining. I am one of the players in his team front. I will not be able to take part in it as I am suddenly fell ill and the doctor has advised me complete rest. He can take any other player with him.

Message

23 March, 2020
Dear Amit

In your absence there was a telephone from your friend, Ravinder. He is one of the players of your cricket team for the tomorrow match.

But he will not be able to come, as he has suddenly fallen ill and the doctor has advised him complete rest. You can take any other friend with you.

Ranjit

6. Here is telephonic talk between Gurbani and Jaspreet Gurmeet gives her message for her brother. Write the message on behalf of Gurmeet not more than 50 words.

Gurmeet : Hi ! Gurmeet.
Jaspreet : Sorry, I’m not Gurmeet. I am his elder sister, Jaspreet. Can I know who is calling ?
Gurbani : I am Gurbani, Bedi, her tutor. Is it not his contact number ? I have some urgent message for her.
Jaspreet : Yes it is. But she has gone to the market to buy fruit and her mobile is with me.
Gurbani : Would you please, convey my message to him
Jaspreet : But he is not coming back for about two hours, I’m also going to the hospital to see one of our neighbours. Would you give me the message. I’ll leave it on his table before I go.
Gurbani : Sure ! Please note this. I shall not be able ‘to come for coaching as I have sprained my ankle. Therefore she should not write for me in the evening.
Jaspreet : All right. Don’t worry the message will reach him.
Gurbani : Thank you very much.

Message

April 10, 2020
Dear Sarbjit

There was a call from you tutor when you were not at home. She will not be coming today for coaching because she has sprained her ankle. Therefore don’t wait for her in the evening.

Mother/Mummy

7. Read the following telephone conversation between kamal and Hardeep from a hospital. Hardeep wants to talk to kamlesh sharma but she is at present not at home. She will be back after an hour. Thinking yourself as Kamal and using the telephone conversation as the subject, write a message to Kamlesh Sharma in not more than 50 words.

Hardeep : Is that Kamlesh Sharma ?
Kamal : No, we are her tanents.
Hardeep : May I speak to Mrs Sharma ? I have to talk to her urgently.
Kamal : She is not here at present. She has gone out and will return after an hour.
Hardeep : Then, would you give a message to her as soon as she returns ?
Kamal : Yes, of course. But I am also going to my friend’s home for his birthday party this evening. However, you needn’t worry I shall leave your message on table for her. What’s it ?
Hardeep : Kindly tell her that her grandmother has met with an accident and she is at present, in the civil hospital here in Khanna. She has broken her leg and is in plaster. Now she is feeling easy. Please tell kamlesh sharma to reach the hospital at her earliest with her husband.
Kamal : O.M.G. Please worry not. I shall leave the message for her before I leave Hardeep. Thank you very much.
Kamal : By, Who speaks on the other side ?
Hardeep : I am her neighbour, Hardeep Sodhi.

Message

May 5, 2020.
Dear Sharma Aunty

There was a telephone call for you from Khanna. I am sorry to inform that your grandmother has broken her leg and is in the Civil Hospital there. Don’t worry she is feeling easy now though she is still in plaster. Reach the hospital with your husband at your earliest.

Kamal

8. Read the following telephonic conversation between Ravinder and Shilpa. Shilpa is about to leave for her coaching within five minutes and her sister is not at home for the moment. Write this message conveyed on behalf of Shilpa.

Mrs. Ravinder : Hello, is that Jaspreet ?
Shilpa : No, I am her sister, Shilpa speaking. May I know who speaks on the other side ?
Mrs. Ravinder : I am Mrs. Ravinder, her friend Sonum’s mother speaking from bus stand. I wanted to speak to your sister urgently.
Shilpa : She is not at home now aunty. She has just gone to her college if there is some message I shall give her.
Mrs. Ravinder : Ok, then please note down I have returned from Amritsar. I have brought some holy books for her. I wanted her to collect the packet at bus stand as I am already late for home. However, she can collect it from there in the evening today urgently. I will not be availabe for the night as we have to attend a marriage party.
Shilpa : You needn’t worry aunty. As I am also going to the Gurudwara, I shall leave this message on table before leave.
Mrs. Ravinder : Thank you very much.

Message

March 5, 2020
Didi

There was a telephone from Mrs. Ravinder, Sonum’s mother. She has bought some holy books for you from Amritsar. Collect the packet from her home in the evening today, urgenlty as she will not be available at night as they are going to attend a marriage party.

Shilpa.

TYPE – II

1. You want to send a message to your niece on her birthday as you are unable to attend it because of an urgent meeting in the office. Write the S.M.S. is not more than 50 words.

May 15, 2020
Dear Vani

Many many happy returns of the day. Stay blessed and in high spirits. Don’t mind my absence as I am unable to attend your birthday party due to an urgent meeting in the office. You will soon receive a lovely gift from me through courier.

Your loving uncle
Gurnarn

2. You were to attend the marriage of your friend but you suddenly fell ill that night. Send your friend an S.M.S. informing your friend about your disability to reach and giving him congratulations and expressing your good wishes for the wedding couple.

Dear Madhur

Congratulations on the wedding of your elder brother. Please pardon my absence as I am unable to attend the marriage ceremony because of sudden illness. I had got my briefcase ready to take a bus, but I was forced it lie in bed. Let me convey my hearty wishes for the happy life of the wedding couple.

Sharan

Punjab State Board PSEB 7th Class English Book Solutions English Notice Writing Exercise Questions and Answers, Notes.

PSEB 7th Class English Notice Writing

Notice लोगों को किसी घटना की जानकारी देने का संदेश अथवा समाचार होता है। उदाहरण के लिए किसी स्कूल के Notice-board पर विद्यार्थियों को स्कूल की विभिन्न गतिविधियों की जानकारी दी जा सकती है। Notice लिखते समय अग्रलिखित बातों का उल्लेख अवश्य करें:

1. शीर्षक-इसमें स्पष्ट किया जाना चाहिए कि Notice किसके लिए है।
2. Notice लिखे जाने की तिथि
3. संबंधित घटना का दिन, समय तथा स्थान
4. Notice के नीचे लिखने वाले का नाम, पद, पता आदि।
5. शब्द-सीमा लगभग 30 शब्दों तक।

1. You are the PTI of your school. Write a notice asking the students to enrol for free yoga classes.

Free Yoga Classes

Attention!

20 March 20 ……

All students interested in attending free yoga classes from 10th April every. morning from 6 a.m. to 7 a.m. should contact the undersigned before 7th April.

Sd/
B.S. Bedi
PTI

2. You are the librarian of your school. Write a notice asking the students to return borrowed books before the school closes down.

Attention!

10 March 20…….

The school is going to be closed down for the summer vacation next week. All students who have borrowed any book from the library must return it before the school closes down.

Raman Kumar
Librarian

3. You are Anupam, the editor of the school magazine, and want to hold an interclass competiton to collect poems and cartoons for the magazine before Sept. 9. Draft a notice for the students ‘Notice-board inviting entires.

Interesting Contest, Amazing Prizes

We are going to hold an inter-class compition to collet poems and cartoons for the school magazine. Entries for the same are invited to reach the under signed befors 9th September. The result of the contest will be declared on 15th. The best poem and cartoon shall win a free school blazer. There will be some consolation prizes to.

4. You are Sangeets, the secretary of the school quiz club. You want to hold an inter-class competition to decide on entires for an inter-school competition 2 weeks from now. Draft a notice for the students notice-board inviting participants.

Prizes Through Quizzes

An inter-school quiz-competition is going to be held two weeks from now. In order to decide entreis for the same, we have decided to hold on inter-class competition on Monday, the 15th. All those who desire to participets schould give their name to the undersigned by tomorrow. The three best participants shall be awarded free school blazers Sangeeta.

(Secretary, School Quiz Club.)

5. You are Kulbir Singh of Class VII. You have lost your new water bottle. Write the notice that you would like to put up on the school notice-board.

Lost ! Lost ! Lost !

12 March 20……..

I have lost my new water bottle somewhere in the school garden. The bottle is of Eagle make with blue colour. The finder is requested to return it to me or deposit it with the school office.

Kulbir Singh
Roll No. 2
VII B.

6. You are the Sarpanch of your village. Write a notice inviting adults to donate blood at the blood donation camp to be held at the community centre.

Blood Donation Camp

8 March 20 ………

The village Panchayat is going to organise a blood donation camp in the community centre on 8th April from 9 a.m. to 11 a.m. All the adults of the village should come forward and donate blood to save lives of many.

Balwan Singh
Village Sarpanch.

7. You have misplaced a library book ‘Panchtantra Tales’. Write a notice that you would like to put up in the classroom.

Book Misplaced

10 March 20 …….

I have misplaced my book “Panchtantra Tales’ somewhere in the classroom. I borrowed it from the library yesterday. The finder is requested to return it to me or deposit it with the class teacher.

Rajni
Roll No. 5
VII A.

8. You are the sports captain of your school. Write a notice to all participants to submit their names and event in which they are taking part.

Attention ! Sports Participants

02 May 20 …….

All the participants in the school sports are requested to submit their names with the undersigned before Sunday, the 6th May. They must mention the event they are taking part in.

Geeta Sharma
Sports Captain
Khalsa Girls High School, Moga.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

Question 1.
2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Let side of cube = x cm
Volume of cube = 64 cm³
[volume of cube = (side)³]
x³ = 64
x = \(\sqrt[3]{4 \times 4 \times 4}\)
x = 4 cm
∴ side of cube = 4 cm.

When cubes are joined end to end and cuboid is formed
whose Length = 2x cm = 2(4) = 8 cm
Width = x cm = 4 cm
Height = x cm = 4 cm

Surface area of cuboid = 2[LB + Bh + hL]
= 2 [8 × 4 + 4 × 4 + 4 × 8]
= 2 [32 + 16 + 32]
= 2 [80]
∴ Surface area of cuboid = 160 cm².

Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere ¡s 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:

Diameter of hemisphere = Diameter of cylinder
= 14 cm
2R = 14 cm
Radius of hemisphere (R) = 7 cm
Total height of vessel = 13 cm
∴ Height of cylinder = (13 – 7) = 6 cm
Inner surface area of vessel = inner surface area of cylinder + Inner surface area of hemisphere
= 2πRH + 2πR²
= 2πR [H + R]
= 2 × \(\frac{22}{7}\) × 7(16 + 7)
= 44 × 13 = 572 cm²
Hence, Inner surface area of vessel = 572 cm²

Question 3.
A toy is ¡n the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Radius of cone = Radius of hemisphere (R) = 3.5 cm
Total height of toy = 15.5 cm
∴ Height of cone (H) = (15.5 – 3.5) = 12 cm

Slant height of cone = \(\sqrt{\mathrm{R}^{2}+\mathrm{H}^{2}}\)

= \(\sqrt{(3.5)^{2}+(12)^{2}}\)

= \(\sqrt{12.25+144}\) = \(\sqrt{156.25}\)
Slain height conk (l) = 12.5 cm
Total surface area of toy = Surface area of cone + Surface area of hemisphere
= πRL + 2πR²
= πR[L + 2R]
= \(\frac{22}{7}\) × 3.5 [12.5 + 2 (3.5)1 cm²
= \(\frac{22}{7}\) × 3.5 [19.5] cm²
= \(\frac{1501.5}{7}\) = 214.5 cm²
∴ Total surface area of toy = 214.5 cm²

Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have?
Find the surface area of the solid.
Solution:
Side of cubical box = 7 cm
Diameter of hemisphere = Side of cubical box = 7 cm
2R = 7
R = \(\frac{7}{2}\) cm

Surface area of solid = (surface area of the cube) – (area of base of hemisphere) + curved surface area of hemisphere)
= 6l² – πR² + 2πR²
= 6l² + πR²
= 6(7)² + \(\frac{22}{7}\) \(\frac{7}{2}\)²
= [6 × 49 + 11 × \(\frac{7}{2}\)]cm²
= 294 + 38.5 = 332.5 cm²

Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of (he hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Let each side of cube = a
∴ Diameter of hemisphere = Side of cube
2R = a
R = \(\frac{a}{2}\)

Surface area of remaining solid = Total surface area of euboid – Area of the top of cube + Inner curved Surface area of hemisphere
= 6 (side)² – πR² + 2πR²
= 6(a)² + πR²
= 6(a)² + π \(\frac{a}{2}\)²
= 6a² + π \(\frac{a^{2}}{4}\)
= a² 6 + \(\frac{\pi}{4}\) cm²

Question 6.
A medicine capsule is ¡n the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:
Diameter of capsule = Diameter of hemisphere = Diameter of cylinder = 5 mm
∴ 2R = 5 mm
R = \(\frac{5}{2}\) mm

Length of entire capsule = 14 mm
Height of cylinderical part = (14 – \(\frac{5}{2}\) – \(\frac{5}{2}\)) mm
= (14 – 5) mm
H = 9 mm
Surface area of capsule = Surface area of cylinder + 2 Surface area of hemisphere
= 2πRH + 2 (2πR²)
= 2πRH + 4πR²
= 2πR [H + 2R]
= 2 × \(\frac{22}{7} \times \frac{5}{2}\left[9+2\left(\frac{5}{2}\right)\right]\)
= 2 × \(\frac{22}{7}\) × \(\frac{5}{2}\) [9 + 5]
= \(\frac{22}{7}\) × 5 × 14
= 220 mm²
Hence, Surface area of capsule = 220 mm².

Question 7.
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of 500 per (Note that the base of the tent will not be
covered with canvas.)
Solution:
Diameter of cone = Diameter of cylinder
2R = 4
R = 2 m

Radius of cone = Radius of cylinder
Height of cylinder (H) = 2.1 in
Slant height of cone (L) = 2.8 m
Curved surface area of tent = Curved surface of cylinder + Curved surface of conical part
= 2πRH + πRL
= πR [2H + L]
= \(\frac{22}{7}\) × 2[2(2.1) + 2.8]
= \(\frac{22}{7}\) × 2[4.2 + 2.8]
= \(\frac{22}{7}\) × 2 × 7
= 44 m²
∴ Curved surface area of tent = 44 m²
Cost of 1m² canvas = ₹ 500
Cost of 44 m² canvas = 44 × 500 = ₹ 22000
Hence, Total cost of canvas = ₹ 22000.

Question 8.
From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Solution:
Diameter of cylinder (D) = 1.4 cm = Diameter of cone
∴ Radius of cylinder = Radius of cone (R) = 0.7 cm
Height of cylinder (H) = 2.4 cm

As we know, L² = R² + H² + (2.4)²
L = \(\sqrt{(0.7)^{2}+(2.4)^{2}}\)
= \(\sqrt{0.49+5.76}\) = \(\sqrt{6.25}\)
L = 2.5 cm
Total surface area of remaining solid = curved surface area of cylinder + Area of base of cylinder + Surface area of cone
= 2πRH + πR² + πRL
= πR [2R +R + L]
= \(\frac{22}{7}\) × 0.7 [2(2.4) + 0.7 + 2.5]

=\(\frac{22}{7}\) × \(\frac{7}{10}\) [4.8 + 3.2]

= \(\frac{22}{10}\) [8]

= \(\frac{176}{10}\) = 17.6 cm²

Hence, Total surface area remaining solid to nearest cm² = 18 cm².

Question 9.
A wooden article was made by scooping out a hemisphere form each end of a solid cylinder, as shown in Fig.

If the height of the cylinder is 10 cm, and its base ¡s of radius 3.5 cm, find the total surface area of the article.
Solution:
Height of cylinder (H) = 10 cm
Radius of cylinder = Radius of hemisphere (R) = 3.5 cm

Surface area of article = curved surface area of cylinder + 2 curved surface area of hemisphere
= 2πRH + 2 (2πR²)
= 2πR [H + 2R]
= 2 × \(\frac{22}{7}\) × 3.5 [10 + 2(3.5)]
= \(\frac{44}{7}\) × \(\frac{35}{10}\) [10 + 7]
= 44 × \(\frac{5}{10}\) × 17
= 44 × \(\frac{1}{2}\) × 17
= 22 × 17 = 374 cm²
Hence, total surface area of article = 374 cm².

Punjab State Board PSEB 7th Class English Book Solutions English Grammar Tense Exercise Questions and Answers, Notes.

PSEB 7th Class English Grammar Tense

Study the Chart

Exercise 1

I. Fill in the blanks with the right tense form of the verb given in the brackets:

1. Our school ……………………… (begin) with a prayer everyday.
2. Ahmed ……………………… (keep awake) till midnight these days.
3. What…………….(make) you do so ?
4. This box ………………….. (contain) a gift for him.
5. This road is closed. They …………………….. (repair) it.
6. The clerk ……………. (type) the letter still, he never …………… (finish) the work in time.
7. Don’t make a noise. An important meeting ……………………. (go on) here. 8. She
8. ………… (say) a prayer to God regularly before going to bed.
9. The peon ……………….. (ring) the bell now.
10. We ……………. (feel) uneasy on a very hot day.
Hints:
1. begins
2. keeps awake
3. makes
4. contains
5. are repairing
6. is typing, finishes
7. is going on
8. says
9. is ringing
10. feel.

Exercise 2

Fill in the blanks with the right tense form of the verb given in the brackets: (Use Present Perfect Tense)

1. The water level …………………… (go up) because of rains.
2. The doctor ……………………. (examine) the patient. He is improving now.
3. I ……………………… (finish) my work. I am going home now.
4. They ………………….. (leave) the place.
5. Sheela …………………. (learn) her lesson.
6. Father ………………………. (not come) home for lunch yet.
7. ………………… you ………………. (finish) your work ? Can you come with me now?
8. I ……………………. (study) the problem. It is easy to solve.
9. ………….. he ………. (take charge) of his new assignment ?
10. The train …………………………… (leave) the station. The platform looks deserted.
Hints:
1. has gone up
2. has examined
3. have finished
4. have left
5. has learnt
6. has not come
7. Have, finished
8. have studied
9. Has, taken charge
10. has left.

Exercise 3

Fill in the blanks with the right tense form (Past Continuous or Simple Past) of the verbs given in brackets:

1. She ………………… (look) for a book shop, when I ………………….. (meet) her.
2. The policeman …………………… (arrest) the thief, when I …………………… (see) him.
3. I……………………………… (go) to the cattle-shed, when I …………………….. (hear) someone quarrelling.
4. The sarpanch ………… (take) the horse out of the stable, when I ……. (call) him.
5. Yesterday as I …………. (walk) along the street. I ……………………… (meet) my friend.
6. In January 1948 Gandhiji ……………………… (stay) in Delhi. He was shot while he ……………. (come out) of the prayer meeting.
7. While I …………………. (watch) the T.V., the lights ………….. (go off).
Hints:
1. was looking, met
2. was arresting, saw
3. was going, heard
4. was taking, called
5. was walking, met
6. was staying, was coming out
7. was watching, went off.

Exercise 4

Fill in the blanks using the verbs in the brackets according to the tense form indicated:

1. A group of officials ……….. (go) to Delhi tomorrow. (Simple Future)
2. We …………………… (visit) Kashmir next January. (Future Continuous)
3. What …………………….. you ……………(do) tomorrow evening ? (Future Continuous)
4. I …….. (go) with my brother. I am sure I ……………… (have) a very nice time on this (Simple Future)
5. I want to give this book to Jaspreet …………………… you ……………………… (go) to give this book to her ? (Simple Future)
Hints:
1. will go
2. shall be visiting
3. will, be doing
4. shall go, will have
5. will, go.

Exercise 5

Fill in the blanks to express Future Perfect aspects of the verbs given in the brackets.

1. The police …………………… (arrest) the thief by tomorrow.
2. Davinder. ………………… (tell) me all before you talk to him.
3. He ………………….. (return) before you arrive.
4. Gobind ……………………. (write) the story before you return.
5. We …………………… (enjoy) our holidays for about a month before he arrives.
Hints:
1. will have arrested
2. will have told
3. will have returned
4. will have written
5. shall have enjoyed.

Exercise 6
(Miscellaneous)

I. Underline the verb and write in the space given whether the sentence is in the Present, Past or Future Perfect tense. One has been done for you.

1. The rain had stopped before I arrived. (Past Perfect Tense)
2. I have lived in Bhatinda since childhood. (Present Perfect Tense)
3. We shall have reached home before they arrive. (Future Perfect Tense)
4. The children shall have eaten something by the time I reach home, (Future Perfect Tense)
5. She has not finished writing the book. (Present Perfect Tense)
6. The watchman had run away before the owner reached. (Past Perfect Tense)
7. The children have learnt the song. (Present Perfect Tense)
8. Naaz has posted the letter. (Present Perfect Tense)
9. My uncle has given me a room in the new house. (Present Perfect Tense)
10. They will have left for Patiala by night. (Future Perfect Tense)

II. Complete the following letter by using the verb in the bracket in present perfect/ past tense. One has been done for you:

Dear sir
I wrote (write) to you some time ago-asking about conditions of entry to your competition. You replied (reply) enclosing an entry form which I filled up (fill up) and sent (send) without delay. I have heard (hear) nothing from you and am beginning to wonder if my application has gone (go) astray. Please check if you have received (receive) it or not. In case you have not got (not get) it. Kindly inform me.

Thanking you
Yours faithfully
Amarjit Singh

III. You have planned to undertake a railway journey in your summer holidays. In about ten sentences describe your forth coming trip using simple future tense.

I will go to Amritsar during summer holidays. I will go by train. I will catch the train from Ludhiana. I will go to the railway station. I will buy a ticket and go to the platform to board the train. The train will reach Amritsar within three hours. There, I will stay in a hotel. I will visit Sri Harmandar Sahib. I will have a holy dip in the tank there. I will also visit the Durgiana Mandir. I will not miss to see the Jallianwala Bagh before I come back.

IV. Your house is flooded due to heavy rains. You saved yourself by sitting on the roof top for almost three days and nights. Using simple and continuous past tense, write your experience.

I found water all around me. It was rising continually. I, with my family, took shelter on the roof of my house. We took with us all the eatables available in home. There was a large carpet at the roof. At night we slept on it. But the rain did not stop for three days. All our eatables were consumed. We grew weaker and weaker. We prayed to God for help. Then one day, it stopped raining. We came down, took buckets and threw the water out of the house. We thanked God for saving our lives.

V. Supply for the blanks the future perfect tense of the verb given in the brackets:

1. Our maid will have broken all the cups (break).
2. He …………..by that time. (return)
3. The sun …………… when we reach home. (set)
4. We …………. all the cakes by evening. (eat)
5. She ……………for the family by night. (cook)
6. He …………. his pledge. (keep)
7. Each child …………… a new pull over. (bought)
8. The shoeshine …………… my shoes. (polish)
9. The commander ………….. the army to march. (order)
10. I……………. the job by sunset. (complete)
11. She …………. to speak English by the year end. (learn)
12. You ………….. tea before we reach there. (drink)
13. Meeta ………….. the beauty contest. (win)
14. Mini …………… to India by early September. (fly)
Hints:
2. will have returned
3. will have set
4. shall have eaten
5. will have cooked
6. will have kept
7. will have bought
8. will have polished
9. will have ordered
10. shall have completed
11. will have learnt
12. will have drunk
13. will have won
14. will have flown.

VI. Given below is a complaint letter. Fill in the blanks with the correct form of the verb given in the brackets. One has been done.

The Secretary
DIV/4901
Vasant Kunj
New Delhi.
Dear Sir

I regret to bring (bring/brought) to your notice that Bihari Lal, the sweeper is not doing (is not doing / has not done) his duty well. He sweeps (sweeps / sweep) the road only once a day. He leaves (leave /leaves) the garbage on the road or throws (threw / throws) them all around. As a result the area is filthy. I have requested (am requesting /have requested) him many times but he refused (refuses / refused) to obey. It seems (seem / seems) he does not care (do not care/does not care).

Yours sincerely
Anil Sharma

VII. Rewrite the following sentences after changing the verbs into the present or past continuous tense:

1. Sudha lies on the bed.
2. Raja plays with his brother.
3. The servant rang the bell.
4. The children scream.
5. The sun sets in the west.
6. They go out for a picnic.
7. She likes the game.
8. I eat my food.
Answer:
1. Sudha is lying on the bed.
2. Raja is playing with his brother.
3. The servant was ringing the bell.
4. The children are screaming.
5. The sun is setting in the west.
6. They are going out for a picnic.
7. She is liking the game.
8. I am eating my food.

VIII. Fill in the blanks with the past perfect tense of the verb given in brackets:

1. He …………. a tiger before I reached the forest. (kill)
2. She ………….. a sweater before I bought a new one. (knit)
3. I ……………. money from my friend before I received my salary. (borrow)
4. The river ……………. its bank before the dam was built. (overflow)
5. She ……………. my book before I could check her. (steal)
6. The train …………….. before I could reach the station. (arrive)
7. I ………….. a funny story before the sad one. (hear)
8. The thief ………….. from the jail before the police arrived. (escape)
Answer:
1. He had killed a tiger before I reached the forest. (kill)
2. She had knitted a sweater before I bought a new one. (knit)
3. I had borrowed money from my friend before I received my salary. (borrow)
4. The river had overflown its bank before the dam was built. (overflow)
5. She had stolen my book before I could check her. (steal)
6. The train had arrived before I could reach the station. (arrive)
7. I had heard a funny story before the sad one. (hear)
8. The thief had escaped from the jail before the police arrived. (escape)

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 12 Areas Related to Circles Ex 12.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3

Question 1.
Find the area of the shaded region in Fig., If PQ = 24 cm, PR =7 cm and O is the centre of the circle.

Solution:
PQ = 24 cm
PR = 7 cm
RQ is diameter of circle
∠RPQ = 90° Angle in semi circle
In ∆PQR,
QR² = RP² + PQ²
QR = \(\sqrt{(7)^{2}+(24)^{2}}=\sqrt{49+576}\)
= \(\sqrt{625}\)
QR = 25 cm
∴ Diameter of circle (QR) = 25 cm
Radius of circle (R) = \(\frac{25}{2}\) cm
Area of shaded region = Area of the semicircle – Area of ∆RPQ
= \(\frac{1}{2} \pi \mathrm{R}^{2}-\frac{1}{2} \mathrm{RP} \times \mathrm{PQ}\)

= \(\left[\frac{1}{2} \times \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2}-\frac{1}{2} \times 7 \times 24\right]\) cm²

= \(\left[\frac{6875}{28}-84\right]\)
= 245.53 – 84 = 161.53 cm²
∴ Area of shaded region = 161.53 cm².

Question 2.
Find the area of the shaded region in Fig., if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Solution:
Radius of smaller circle (r) = 7 cm
Radius of bigger circle (R) = 14 cm
Central angle ∠AOC (θ) = 40°
Area of shaded region = Area of bigger sector OAC – Area of smaller sector OBD
= \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}-\frac{\pi r^{2} \theta}{360^{\circ}}\)

= \(\frac{\pi \theta}{360^{\circ}}\) [R² – r²]

= \(\frac{22}{7} \times \frac{40}{360}\) × [14² – 7²]

= \(\frac{22}{63}\) [196 – 49]

= \(\frac{22}{63}\) × 147 = 51.33 cm²
∴ Shaded Region = 51.33 cm².

Question 3.
Find the area of the shaded region in fig., if ABCD is a square of side 14 cm and APD and BPC are semi circles.

Solution:
Side of square = 14 cm
Diameter of semicircle (AB = BC) = 14 cm
Radius of semi circle (R) = 7 cm
Area of square = (Side)²
= 14 × 14 = 196cm²
Area of a semi circles = \(\frac{1}{2}\) πR²
= \(\frac{1}{2} \times \frac{22}{7}\) × 7 × 7
= 77 cm²
Area of two semi circle = 2(77) = 154 cm²
Area of shaded region = Area of square ABCD – Area of two semi circles
= (196 – 154) = 42 cm²
Area of shaded region = 42 cm².

Question 4.
Find the area of the shaded region in fig., where a circular arc of radius 6 cm has been drawn iith vertex O of an equilateral triangle OAB of side 12 cm as centre.

[Each angle of equilateral triangle is 60°]
Area of major sector of circle = Area of circle – Area of sector
= πR² – \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7}\) × 6 × 6 – \(\frac{22}{7}\) × 6 × 6 × \(\frac{60}{360}\)

= \(\frac{22}{7}\) × 6 × 6 1 – \(\frac{60}{360}\)

= \(\frac{22}{7}\) × 36 1 – \(\frac{1}{6}\)

= \(\frac{22}{7}\) × 36 × \(\frac{5}{6}\)
= 94.28 cm²
∴ Area of major sector of circle = 94.28 cm²
Area of equilateral triangle OAB = \(\frac{\sqrt{3}}{4}\) (side)²
= \(\frac{1.73}{4}\) × 12 × 12
= 1.73 × 36 = 62.28 cm²
Shaded Area = Area of equilateral triangle OAB + Area of major sector of circle
= 62.28 + 94.28 = 62.28 cm²
Shaded Area = Area of equilateral triangle OAR + Area of major sectoç of circle
= 62.28 + 94.28 = 156.56
Shaded Area = 156.56 cm².

Question 5.
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in fig. Find the area of the remaining portion of the square.
Solution:
Side of square = 4 cm
Radius of each semi circle cut out (r) = 1 cm

Diameter of circle (R) = 2 cm
.. Radius of circle (R) = 1 cm
Area of square = (Side)²
= (4)² = 16 cm²
Area of 4 quadrants = 4\(\left[\frac{\pi^{2} \theta}{360^{\circ}}\right]\)

= \(\frac{4 \times 90}{360} \times \frac{22}{7} 1 \times 1\)

= 1 × \(\frac{22}{7}\) × 1 × 1 = 3.14 cm²
Area of circle = πR²
= \(\frac{22}{7}\) × 1 × 1
Area of circle = 3.14 cm²
Required area = Area of square – Area of 4 quadrants – Area of circle
= (16 – 3.14 – 3.14) cm² = 9.72 cm²
Required Area = 9.72 cm².

Question 6.
In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as
shown in fig. Find the area of the design (shaded region).

Solution:
Radius of table cover (R) = 32 cm
OA = OB = OC = 32 cm

∆ABC is equilateral triangle with AB = AC = BC
∠AOB = ∠BOC = ∠COA = 120°
Now, in ∆BOC,
From O draw, angle bisector of ∠BOC as well as perpendicular bisector 0M of BC.
∴ BM = MC = \(\frac{1}{2}\) BC
Also, OB = OC [radii of the circle]
∴ ∠B = ∠C
∴ ∠O + ∠B + ∠C = 180°
120° + 2∠B = 180°
∠B = 30°
and ∠B = ∠C = 30°
Also, ∠BOM = ∠COM = 60°
∆OMB ≅ ∆OMC [RHS Cong.]
∴ In ∠OMB,
∠OBM = 30° [∠O = 60° and ∠M = 90°]
∴ \(\frac{\mathrm{BM}}{\mathrm{OB}}\) = cos 30°

\(\frac{\mathrm{BM}}{32}=\frac{\sqrt{3}}{2}\)

BM = 16√3 cm.

∴ BC = 2 MB = 32√3 cm
Area of circle = πR² = \(\frac{22}{7}\) × (32)²
= \(\frac{22}{7}\) × 32 × 32 = 3218.28 cm²

Area of ∆ABC = 4 (side)²
= \(\frac{1.73}{4}\) × 32√3 × 32√3 = 1328.64 cm²

∴ Required Area = Area of circle – Area of ∆ABC
= 3218.28 – 1328.64 = 1889.64 cm²

Question 7.
In fig., ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Solution:
Side of square ABCD = 14 cm
Radius of circle (R) = 7 cm
Sector angle (θ) = 90° [Each angle of square 90°]
Area of square = (side)²
= 14 × 14 = 196 cm²
Area of four quadrants = 4 \(\left[\frac{\pi R^{2} \theta}{360}\right]\)
= 4 × \(\frac{22}{7} \times \frac{7 \times 7 \times 90}{360}\)
= 22 × 7 = 154 cm²
∴ Required shaded area = Area of square – Area of 4 quadrants
= 196 – 154 = 42 cm².

 

Question 8.
Fig. depicts a racing track whose left and right ends are semicircular. The distahce between the two inner parallel line segments is 60 m and they are each 106 m long. 1f the track is 10 m wide, find
(i) the distance around the track along its inner edge.
(ii) the area of the track.

Solution:

(i) Here AB = DC = 106 m
AF = BE = CG = HD = 10m
Diameter of inner semicircle (APD and BRC) =60 m
∴ Radius of inner semicircle (APD (r) = 30 m
Radius of outer semicircle (R) = r + 10 = 30 + 10 = 40 m
Distance around the track along inner edge = AB + circumference of semi circle BRC + CD + circumference of semi circle DPA
= 2 AB + 2 [circumference of semi circle BRC]
= 2 (106) + 2(\(\left(\frac{2 \pi r}{2}\right)\))
= 212 + 2πr
= 212 + 2 × \(\frac{22}{7}\) × 30
= 212 + \(\frac{60 \times 22}{7}\)
= 212 + 188.57 = 400.57 m.
∴ Distance around the track along its inner edge = 400.57 m

(ii) Area of track = Area of rectangle ABEF + Area of region BEMGCRB + Area of rectangle CGHD + area of region.
= 2 Area of rectangle ABCD + 2 Area of region (II)
= 2 (AB × AF) + 2
[Area of semi circle with Radius 60 cm – Area of semi circle with radius 30 cm]
= 2 [106 × 10] + 2 [latex]\frac{\pi \mathrm{R}^{2}}{2}-\frac{\pi r^{2}}{2}[/latex]
= 2 × 1060 + \(\frac{2 \pi}{2}\) [R² – r²]
= 2120 + \(\frac{22}{7}\) (40² – 30²)
= 2120 + \(\frac{22}{7}\) [1600 – 900]
= 2120 + \(\frac{22}{7}\) [700]
= 2120 + 2200 = 4320 m²
Area of track = 4320 m²

Question 9.
In Fig., AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Solution:
Diameter of circle = 14 cm
Radius of circle = 7 cm
Diameter of smaller circle = 7 cm
∴ Radius of smaller circle = \(\frac{7}{2}\) cm
Since AB and CD are to perpendicular the diameters of a circle,
∴ AO ⊥ CD
Area of bigger circle = πR² × 7 × 7 = 154 cm²
Area of bigger semicircle = \(\frac{154}{2}\) = 77 cm²
Area of smaller circle = πr²
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)
= 38.50 cm²

Area of ∆ABC = \(\frac{1}{2}\) Base × Altitude
= \(\frac{1}{2}\) × 14 × 7 = 49 cm²
∴ Shaded Area = Area of bigger semi circle + Area of smaller circle – Area of triangle
= (77 – 49 + 38.5) cm² = 66.5 cm²

Question 10.
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half of the length of the side of the triangle (see Fig.). Find the area of the shaded region.
(Use n = 3.14 and ,√3 = 1.73205)

Solution:
Area of equilateral triangle ABC = 17320.5 cm²

\(\frac{\sqrt{3}}{4}\) (side)² = 17320.5

(side)² = \(\frac{17320.5 \times 4}{1.73205}\)

(side)² = \(\frac{173205}{10} \times \frac{100000 \times 4}{173205}\)

side = \(\sqrt{4 \times 100 \times 100}\)
side = 2 × 100 = 200 cm
AB = BC = AC
Radius of circle (R) = \(\frac{A B}{2}=\frac{200}{2}\) = 100 cm
Sector angle, θ = 60°
Area of sector APN = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{3.14 \times 100 \times 100 \times 60}{360}\)

= \(\frac{15700}{3}\)

Area of three sector = 3 × \(\frac{15700}{3}\) cm²
∴ Required shaded Area = Area of triangle – Area of three sectors
= 17320.5 – 15700 = 1620.5 cm²
∴ Hence, Required shaded Area = 1620.5 cm²

Question 11.
On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig). Find the area of the remaining portion of the handkerchief.

Solution:
Radius of circle (R) = 7 cm
Diameter of circle = 2 × R = 2 × 7 = 14 cm
Since there are three circles along a side of square
∴ side of squrae = 3 [14] = 42 cm
Total area of handkerchief = Area of square = (side)²
= (42)² = 1764 cm².
Area of 9 circular designs = 9πR²
= 9 × \(\frac{22}{7}\) × (7)2
= 9 × \(\frac{22}{7}\) × 7 × 7
= 9 × 154 = 1386 cm²
∴ Required area of remaining portion = Area of square – Area of 9 circular designs
= 1764 – 1386 = 378 cm²
∴ Required area of remaining portion = 378 cm².

Question 12.
In Fig., OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the
(i) quadrant OACB,
(ii) shaded region.

Solution:

Radius of quadrant (R) = 3.5 cm
Angle of sector (θ) = 90°
OD = 2 cm.

(i) Area of quadrant OACB = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{22}{7} \times \frac{3.5 \times 3.5 \times 90}{360}\) = 9.625 cm².

(ii) Area of ODB = \(\frac{1}{2}\) Base × Altitude
= \(\frac{1}{2}\) × 3.5 × 2 = 3.5 cm²

∴ Shaded Area = Area of quadrant OACB – Area of ∆ODB
= 9.625 – 3.5 = 6.125 cm²
∴ Hence, Shaded Area = 6.125 cm².

Question 13.
In fig., a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.

Solution:
Side of square ABCO = 20 cm
∠AOC = 90°
AB = OA

OB² = OA² + AB²

OB = \(\sqrt{(20)^{2}+(20)^{2}}\)

= \(\sqrt{400+400}\)

= \(\sqrt{800}=\sqrt{400 \times 2}\)
OB = 20√2 cm
Area of square OABC = (side)² = (20)²
∴ Area of square = 400 cm²
Radius of quadrant (R) = 20√2 cm
Sector angle (θ) = 90°
∴ Area of sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)
= \(\frac{3.14 \times 20 \sqrt{2} \times 20 \sqrt{2} \times 90}{360}\)
= 2 × 314 cm² = 628 cm²
Required shaded Area = Area of sector – Area of square
= (628 – 400) cm² = 228 cm²

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If ZAOB = 30°, find the area of the shaded region.

Solution:
Radius of sector OBA (R) =21 cm
Radius of sector ODC (r) 7 cm
Sector angle (θ) = 30°
Area of bigger sector (OAB) = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{21 \times 21 \times 30}{360}\) = 115.5 cm²

Area of smaller sector (ODC) = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{7 \times 7 \times 30}{360}\) = 12.83 cm²

Area of smaller sector (ODC) = 12.83 cm²
Now, Shaded Area = Area of bigger sector OAB – Area of smaller sector OCD
= 115.5 – 12.83 = 102.66
Hence, Shaded Area = 102.66 cm².

Question 15.
In fig., ABC is a quadrant of a circle of radius 14 cm and a semi circle is drawn with BC as diameter. Find the area of the shaded region.

Solution:
Radius of quadrant ACPB (r) = 14 cm
Sector angle (θ) = 90°
AB = AC = 14 cm

Area of triangle = \(\frac{1}{2}\) AB × AC
= \(\frac{1}{2}\) × 14 × 14
= 98 cm²

Area of sector ACPB = \(\frac{\pi \mathrm{R}^{2} \theta}{360^{\circ}}\)

= \(\frac{22}{7} \times \frac{14 \times 14 \times 30}{360}\) = 154 cm²

∴ Area of BOCPB = Area of sector ABPC – Area of \ABC
= 154 cm² – 98 cm² = 56 cm²
In ∆BAC, AB² + AC² = BC²
(14)² + (14)² = BC²
BC = \(\sqrt{196+196}=\sqrt{2(196)}\) = 14√2

∴ Radius of semi circle BOCR = \(\frac{14 \sqrt{2}}{2}\) = 7√2

Area of semi circle = \(\frac{\pi \mathrm{R}^{2}}{2}\)

= \(\frac{22}{7} \times \frac{7 \sqrt{2} \times 7 \sqrt{2}}{2}\)

= \(\frac{22}{7} \times \frac{7 \times 7 \times 2}{2}\)
= 154 cm²

Required Area = Area of semi circle – [Area of sector – Area of ∆BAC]
= 154 – [154 – 98]
= (154 – 56) cm² = 98 cm²
Hence, Shaded Area = 98 cm².

Question 16.
Calculate the area of the designed region in fig. common between the two quadrants of circles of radius 8 cm each.

Solution:
Side of square = 8 cm
Area of square = (8)² = 64 cm²
Line BD divides square ABCD into the equal parts
Area of ∆ABD = ar of ∆BDC
Sector angle θ = 90°
Area of sector = \(\frac{\pi \mathrm{R}^{2} \theta}{360}\)

= \(\frac{22}{7} \times \frac{8 \times 8 \times 90}{360}\)

Area of sector = 50.28 cm²
Area of ∆ABD = \(\frac{1}{2}\) × AB × AD
= \(\frac{1}{2}\) × 8 × 8
= 32 cm²

∴ Area of segment DMBPD = Area of sector ∆BPD – Area of ∆ABD
= 50.28 – 32 = 18.28 cm²
Hence, Shaded area = 2 area of segment DMBPD = 2 (18.28) = 36.56 cm²