# Area Related To Circle – Chapter 12.2 (NCERT Solutions) Mathematics (Class 10th)

SUBJECT  MATHS
CLASS  X (10th)
QUESTION FROM – Area Related To Circle – Chapter 12 (12.2)
ALL EXTRA & IMPORTANT / EXERCISE QUESTION ANSWER WITH VIDEOS

Hi Friends, Welcome To – Study24Hours.Com, Here We Are Giving You Some Very Important Question & Answers From The Chapter Area Related To Circle – Chapter 12.2, Area Related To Circle – Is The 12th Chapter Of Your Mathematics NCERT Book Of Class X. So Check Out All Of Them & Please Revise All Of These Questions Before Your Exam. So Please Remember To Visit –Daily For All Study Materials And More.

#### Exercise 12.2 – Area Related To Circle – Chapter 12 (NCERT Solutions) Mathematics (Class 10th)

Unless stated otherwise, Use π = 22/7

Q1.  Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution –

Given, radius of circle (r) = 6 cm
Angle of the sector (θ) = 60°
Now, area of sector = (θ/360°) * πr2
= (60°/360°) * (22/7) * 62
= (1/6) * (22/7) * 6 * 6

= (22/7) * 6
= 132/7 cm2 Q2. Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution

Let radius of circle be r cm.
Given, circumference of the circle = 22
=> 2πr = 22
=> 2 * (22/7) * r = 22
=> 2r/7 = 1
=> r = 7/2 cm
So, area of the quadrant (1/4)th of the circle = (θ/360) * πr2
= (90°/360°) * (22/7) * (7/2)2
= (1/4) * (22/7) * (7/2) * (7/2)

= (1/4) * 11 * (7/2)
= 77/8 cm2 Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution –  Length of minute hand = radius of the circle,
=> r = 14 cm
Angle swept by the minute hand in 60 minutes = 360°
Angle swept by the minute hand in 5 minutes = (360°/60) * 5 = 6 * 5 = 30°

Now, area of the sector with r = 14 and θ = 30° is given as
= (θ/360°) * πr2
= (30°/360°) * (22/7) * 142
= (1/12) * (22/7) * 14 * 14

= (11 * 14)/3
= 154/3 cm2
Thus, the required area swept by the minute hand by 5 minutes = 154/3 cm
2. Q4.  A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14).
Solution – Length of the radius (r) = 10 cm
Sector angle θ = 90°
Area of the sector with θ = 90° and r = 10 cm is given as
= (θ/360°) * πr2
= (90°/360°) * 3.14 * 102
= (1/4) * 3.14 * 10 * 10

= 3.14 * 5 * 5
= 78.5 cm2

(i) Area of the minor segment = Area of minor sector – Area of ΔAOB
= 78.5 – (1/2) * 10 * 10
= 78.5 – 50
= 28.5 cm2

(ii) Area of major segment = Area of the circle] –
Area of the minor segment
= πr2 – 78.5
= 3.14 * 10 * 10 – 78.5
= 314 – 78.5
= 235.5 cm2 Q5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc (ii) area of the sector formed by the arc,
(iii) area of the segment formed by the corresponding chord,
Solution –

Given, radius = 21 cm and θ = 60°

(i) Circumference of the circle = 2πr

= 2 * (22/7) * 21
= 2 * 22 * 3 = 132 cm

Now, length of the arc APB = (60°/360°) * 132 = 132/6 = 22 cm

(ii) Area of the sector with angle 60° = (θ/360°) * πr2

= (60°/360°) * (22/7) * 212
= (1/6) * (22/7) * 21 * 21

= 11 * 21
= 231 cm2

(iii) Area of the segment APQ = Area of the sector AOB – Area of ΔAOB   ………….1
In ΔAOB, OA = OB = 21 cm
So, ∠A = ∠B = 60°       [∠O = 60°]

=> AOB is an equilateral triangle,
Hence, AB = 21 cm
Draw OM ⊥ AB such that OM/OA = sin 60°
=> OM/OA = √3/2
=> OM = (√3/2) * 21

Now, area of triangle OAB = (1/2) * AB * OM
= (1/2) * 21 * (√3/2) * 21
= 441√3/4 cm2      ………….2
From equation 1 and 2, we get
Area of segment = (231 – 441√3/4) cm2. Q6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and 3 = 1.73)
Solution – Here, radius (r) = 15 cm, Sector angle θ = 60°

Area of the sector with angle 60° = (θ/360°) * πr2
= (60°/360°) * 3.14 * 152
= (1/6) * 3.14 * 15 * 15

= 1.57 * 5 * 15 = 117.75 cm2

Since ∠O = 60° and OA = OB = 15 cm
∠AOB is an equilateral triangle.
=> AB = 15 cm and ∠A = 60°

Draw OM Ʇ AB

So, OM/OA = sin 60° = √3/2
=> OM = OA * √3/2
=> OM = 15 * √3/2

Now, AR (ΔAOB) = 1/2 * AB * OM

= 1/2 * 15 * 15 * √3/2
= 225√3/4
= (225 * 3.14)/4 = 20.4375 cm2

Now area of the minor segment = (Area of minor sector) – (AR ΔAOB)
= (117.75 – 97.3125) cm2 = 20.4375 cm2

Area of the major segment = Area of the circle – Area of the minor segment
= πr2 – 20.4375
= 3.14 * 15 * 15 – 20.4375
= 706.5 – 20.4375 cm = 686.0625 cm2. Q7.  A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and 3 = 1.73)
Solution – Given, θ = 120° and r = 12 cm
Area of the sector = (θ/360°) * πr2
= (120°/360°) * 3.14 * 122
= (1/3) * 3.14 * 12 * 12

= 3.14 * 12 * 4
= 150.72 cm2     …………..1

Now, area of ΔAOB = 1/2 * AB * OM    ………..2
In ΔOAB, ∠O = 120°

=> ∠A + ∠B = 180° – 120°
=> ∠A + ∠B = 60°
Since, OB = OA = 12 cm
So, ∠A = ∠B = 30°

Again, OM/OA = sin 30° = 1/2
=> OM = OA * 1/2
=> OM = 12/2 = 6 cm

In right angle triangle AMO, AM2 = 122 – 62

=> AM2 = 144 – 36
=> AM2 = 108
=> AM = √108
=> AM = 6√3
=> 2AM = 6√3
=> AM = 12√3
From equation 2, we get

Area of ΔAOB = 1/2 * AB * OM

= 1/2 * 12√3 * 6
= 36√3
= 36 * 1.73
= 62.28 cm2    ……….3

From equation 1 and 3, we get
Area of the minor segment = Area of minor segment – Area of ΔAOB
= 150.72 – 62.28 cm  = 88.44 cm2. Q8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 12.11). Find
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)
Solution –

Here, Length of the rope = 5 m
Radius of the circular region grazed by the horse = 5 m

(i) Area of the circular portion grazed = (θ/360°) * πr2
= (90°/360°) * 3.14 * 52 = (1/4) * 3.14 * 5 * 5   = 78.5/4 = 19.625 m2

(ii) When length of the rope is increased to 10 m,
So, r = 10 m

Area of the circular region where θ = 90°.
= (θ/360°) * πr2
= (90°/360°) * 3.14 * 102 = (1/4) * 3.14 * 10 * 10   = 314/4   = 78.5 m2

So, increase in the grazing area = 78.5 – 19.625 = 58.875 m2 Q9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 12.12. Find :
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Solution – Diameter of the circle = 35 mm, Radius (r) = 35/2 mm
(i) Circumference = 2πr
= 2 * (22/7) * (35/2) = 22 * 5 = 110 mm,
Length of 1 piece of wire used to make diameter to divide the circle into 10 equal sectors = 35 mm

So, length of 5 pieces = 5 * 35 = 175 mm
Hence, total length of the silver wire = 110 + 175 mm = 285 mm

(ii) Since the circle is divided into 10 equal sectors,
Hence, Sector angle θ = 360°/10 = 36°
Now, area of each sector = (θ/360°) * πr2

= (36°/360°) * (22/7) * (35/2) * (35/2)
= (11 * 35)/4  = 385/4 mm2 Q10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
Solution –

Here, radius (r) = 45 cm
Since circle is divided in 8 equal parts,
Now, sector angle corresponding to each part θ = 360°/8 = 45°
Now, area of a sector(part) = (θ/360°) * πr2
= (45°/360°) * (22/7) * 45 * 45

= (1/8) * (22/7) * 45 * 45
= (1/4) * (11/7) * 45 * 45
= (11 * 45 * 45)/(4 * 7) = 22275/28 mm2
Hence, the required area between two pins = 22275/28 mm
2. Q11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Solution – Given Here, radius (r) = 25 cm
Sector angle (θ) = 115°
Area cleaned by each sweep of the blades = (θ/360°) * πr2 * 2
[Since each sweep will have to and from movement]

= (115°/360°) * (22/7) * 25 * 25 * 2
= (23 * 11 * 25 * 25)/(18 * 7)

Q12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)
Solution – Here, Radius (r) = 16.5 km
Sector angle (θ) = 80°
Now, Area of the sea surface over which the ships are warned
= (θ/360°) * πr2
= (80°/360°) * 3.14 * 16.5 * 16.5
= (80°/360°) * 314/100 * 165/10 * 165/10
= (157 * 11 * 11)/100
= 18997/100
= 189.97 km2 Q13. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)
Solution – Given, r = 28 cm
Since, the circle is divided into six equal sectors.
So, sector angle θ = 360°/6 = 60°
Area of the sector with angle 60° and radius 28 cm = (θ/360°) * πr2

= (60°/360°) * (22/7) * 28 * 28
= (1/6) * 22/7 * 28 * 4 = (44 * 28)/3
= 1232/3 = 410.67 cm2    ……………1

Now, area of 1 design = Area of segment APB

= Area of sector – Area of ΔAOB      …………..2
In ΔAOB, ∠AOB = 60°, OA = PN = 28 cm

So, ∠OAB = 60° and ∠OBA = 60°

=> ΔAOB is an equilateral triangle.
=> AB = AO = BO
=> AB = 28 cm
=> Draw OM ⊥AB

Now, in right ΔAOM, we have

OM/OA = sin 60° = √3/2

=> OM = OA * √3/2
=> OM = 28 * √3/2
=> OM = 14√3 cm

Area of ΔAOB = 1/2 * AB * OM

= 1/2 * 28 * 14√3 = 14 * 14 * 1.7

= 333.3 cm2   ………..3

Now, from (1), (2) and (3), we have:
Area of segment APQ = 410.67 – 333.2 = 77.47 cm2
=> Area of 1 design = 77.47 cm2
So, area of the 6 equal designs = 6 * 77.47 = 464.82 cm2
Cost of making the design at the rate of Rs. 0.35 per cm2 = Rs. 0.35 * 464.82

= Rs. 162.68 Q14. Tick the correct answer in the following :
Area of a sector of angle p (in degrees) of a circle with radius R is :-
(a) P/180 * 2 Pie R
(b) P/180 * Pie R2
(c) P/360 * 2 Pie R
(d) P/360 * 2 Pie R2
Solution – Given, radius (r) = R
Angle of sector (θ) = p
Area of the sector = (θ/360°) * πr2 = (p/360°) * πR2
= (2/2) * (p/360°) * πR2 = (p/720°) * 2πR2
Hence, option (D) is the right answer. area related to circle class 10 exercise 12.3 area related to circle class 10 exercise 12.2 area related to circle class 10 formulas area related to circle class 10 exercise 12.1 area related to circle class 10 extra questions area related to circle class 10 teachoo area related to circle class 10 ncert area related to circle class 10 ncert solution area related to circle class 10 all formulas area related to circle class 10 answers area related to circle class 10 assignment area related to circle class 10 rs aggarwal area related to the circle class 10 all formulas of area related to circle class 10 pdf area related to circle class 10 project area related to circle class 10 board questions area related to circle class 10 by sameer kohli area related to circle class 10 byju’s area related to circles class 10 cbse important questions formulas of area related to circle class 10 cbse formula of chapter area related to circle class 10 class 10 chapter 12 area related to circle class 10 maths ch area related to circle ch area related to circle class 10 area related to circle class 10 difficult questions area related to circle class 10 ppt download area related to circle class 10 exercise area related to circle class 10 example 4 area related to circle class 10 exercise 12.3 question 6 area related to circle class 10 explanation area related to circle for class 10 ncert solution for class 10 area related to circle area related to circle class 10 gseb area related to circle class 10 hots questions with solutions area related to circle class 10 in hindi area related to circle class 10 introduction area related to circle class 10 jsunil area related to circle class 10 formulas list area related to circle class 10 mcq studyrankers class 10 maths area related to circle area related to circle class 10 ncert solutions area related to circle class 10 notes area related to circle class 10 notes pdf area related to circle class 10 ncert solutions pdf area related to circles class 10 ncert solutions study rankers area related to circle of class 10 formula of area related to circle class 10 solution of area related to circle class 10 answers of area related to circle class 10 question of area related to circle class 10 ppt on area related to circle class 10 important questions of area related to circle class 10 extra questions of area related to circle class 10 area related to circle class 10 pdf area related to circle class 10 pdf ncert area related to circle class 10 sample paper area related to circle class 10 solution pdf area related to circle class 10 test paper area related to circle class 10 extra questions pdf area related to circle class 10 questions area related to circle class 10 questions with solutions area related to circle class 10 questions with solutions study ranker area related to circle class 10 solutions area related to circle class 10 solutions 12.3 area related to circle class 10 subject teacher area related to circle class 10 solutions gseb area related to circle class 10 ncert solution pdf area related to circle class 10 rd solution area related to circle class 10 exercise 12.3 solutions all the formulas of area related to circle class 10 area related to circle class 10 videos class 10 area related to circles worksheet area related to circle class 10 youtube area related to circle class 10 12.3 area related to circle class 10 12.1 area related to circle class 10 exercise 12.4

# Area Related To Circle – Chapter 12.1 (NCERT Solutions) Mathematics (Class 10th)

SUBJECT – MATHS
CLASS  X (10th)
QUESTION FROM – Area Related To Circle – Chapter 12 (12.1)
ALL EXTRA & IMPORTANT / EXERCISE QUESTION ANSWER WITH VIDEOS

Hi Friends, Welcome To – Study24Hours.Com, Here We Are Giving You Some Very Important Question & Answers From The Chapter Area Related To Circle – Chapter 12. Area Related To Circle – Is The 12th Chapter Of Your Mathematics NCERT Book Of Class X. So Check Out All Of Them & Please Revise All Of These Questions Before Your Exam. So Please Remember To Visit – www.Study24Hours.Com Daily For All Study Materials And More.

#### Exercise 12.1 – Area Related To Circle – Chapter 12 (NCERT Solutions) Mathematics (Class 10th)

Unless stated otherwise, use π =22/7

Q1. The radii of the two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.

SolutionAccording to the question, We got here

Radius of circle R1 = 19 cm
Radius of circle-II R2 = 9 cm

Circumference of circle-I = 2πR1 = 2π * 19 cm
Circumference of circle-II = 2πR2 = 2π * 9 cm

Therefore, the Sum of the circumferences of circle-I and circle-II = 2π * 19 + 2π * 9

= 2π(19 + 9)
= (2π * 28) cm

Let R be the radius of the circle-III.
and, Circumference of circle-III is = 2πR

According to the condition, 2πR = 2π * 28

=> R = (2π * 28)/2π
=> R = 28

Thus, the radius of the new circle = 28 cm

Q2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Solution – According to the question, We got here

Radius of circle-I, r1 = 8 cm
So, Area of circle-I = πr12 = π(8)2 cm2
Area of circle-II = πr22 = π(6) 2 cm2

So, Let the radius of the circle-III be R
∴ Area of circle-III = πR2

Now, according to the condition,
πr12 + πr12 = πR2

=> π(8)2 + π(6)2 = πR2
=> π(82 + 62) = πR2
=> 64 + 36 = R2
=> 100 = R2
=> 102 = R2
=> R = 10

Thus, the radius of the new circle = 10 cm.

Q3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions. ? Solution – According to the question, We got here

Diameter of the innermost region = 21 cm
Radius of the innermost (Gold Scoring) region = 21/2 = 10.5 cm
So, area of Gold region = π(10.5)2

= (22/7) * 10.5 * 10.5
= 22 * 1.5 * 10.5
= 346.50 cm2

Area of Red region = π(10.5 + 10.5)2 – π(10.5)2

= π(21)2 – π(10.5)2
= (22/7) * 21 * 21 – 346.50
= 22 * 21 * 3 – 346.50
= 1386 – 346.50
= 1039.50 cm2

Area of Blue region = π(21 + 10.5)2 – π(21)2

= π(31.5)2 – π(21)2
= π{(31.5)2 – (21)2}
= π(31.5 – 21)(31.5 + 21)
= (22/7) * 10.5 * 52.5
= 22 * 1.5 * 52.5
= 1732.50 cm2

Area of Black region = π(31.5 + 10.5)2 – π(31.5)2

= π(42)2 – π(31.5)2
= π{(42)2 – (31.5)2}
= π(42 – 31.5)(42 + 31.5)
= (22/7) * 10.5 * 73.5
= 22 * 1.5 * 73.5
= 2425.50 cm2

Area of White region = π(42 + 10.5)2 – π(42)2

= π(52.5)2 – π(42)2
= π{(52.5)2 – (42)2}
= π(52.5 – 42)(52.5 + 42)
= (22/7) * 10.5 * 94.5
= 22 * 1.5 * 94.5
= 3118.50 cm2

Q4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Answer – According to the question, We got here

Given diameter of the wheel = 80
So, the radius of the wheel r = 80/2 = 40
Now circumference of the wheel = 2πr

=2π * 40
= 80π

Now Speed of the car = 66 km/hour = (66*1000 *100)/60

(1 km = 1000m and 1 m = 100 cm, 1 hour = 60 minutes)

= (66*1000 *10)/6 (100 and 60 is divided by 10)
= 11*10000 (66 and 6 is divided by 6)
= 110000 cm/minute

Distance travelled by car in 10 minutes = 110000*10 = 1100000 cm
Let the number of revolution of the wheel is n.
So, n * distance travelled in one minute = Distance travelled in 10 minutes

=> n * circumference of the wheel = Distance travelled in 10 minutes
=> n * 80π = 1100000
=> n = 1100000/80π
=> n = (110000)/8π (1100000 and 80 is divided by 10)
=> n = (110000*7)/(22*8) (π = 22/7)
=> n = (10000*7)/(2*8) (110000 and 22 is divided by 11)
=> n = (5000*7)/8 (10000 and 2 is divided by 2)
=> n = 625*7 (5000 and 8 is divided by 8)
=> n = 4375

So, the number of revolution in 10 minutes at speed of 66 km/hour is 4375

Q5. Tick the correct answer in the following and justify your choice: If the perimeter and the
area of a circle are numerically equal, then the radius of the circle is
(A) 2 units (B) π units (C) 4 units (D) 7 units
Answer – Numerical area of the circle = Numerical circumference of the circle

=> πr2 = 2πr
=> πr2 – 2πr = 0
=> π(r2 – 2r) = 0
=> r2 – 2r = 0 [Since π ≠ 0]
=> r(r – 2) = 0
=> r = 0 or r = 2

But r cannot be zero. So, r = 2 units.
Thus, the option (A) is the correct answer.

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# Class 10th – Maths Important Questions For 2019 Exams (CBSE & All State Boards)

SUBJECT
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CLASS X (10th)
QUESTION FROM – Whole Syllabus (CH 1 To 15)
ALL IMPORTANT QUESTION ANSWER FOR BOARD EXAMS 2019

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#### Class 10th – Maths Important Questions For 2019 Exams (CBSE & All State Boards)

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##### Download Exam Test Papers For 2019 Board Exams (Class 10) | CBSE & All State Boards

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# Ex 4.1 – Quadratic Equation (Question No. 2) MATHS – NCERT | Explained In Hindi

SUBJECT – MATHS
CLASS – X (10th)
QUESTION FROM – Quadratic Equation (Exercise 4.1)
ALL EXTRA & IMPORTANT QUESTION ANSWER FOR 2019 EXAMS

###### Q2. Represent the following situation in the form of quadratic equation:

(i) The area of a rectangular plot is 528 m2. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find the Rohan’s age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Answer – Solutions For Question Number 2 Is In The Video Below (Explained In Hindi) :-

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# Ex 4.1 – Quadratic Equation (Question No. 1) MATHS – NCERT | Explained In Hindi

SUBJECT – MATHS
CLASS – X (10th)
QUESTION FROM – Quadratic Equation (Exercise 4.1)
ALL EXTRA & IMPORTANT QUESTION ANSWER FOR 2019 EXAMS

###### 1. Check whether the following are quadratic equations :-

(i) (x+1)2=2(x–3)
(ii) x2–2x=(-2)(3–x)
(iii) (x–2)(x+1)=(x–1)(x+3)
(iv) (x–3)(2x+1)=x(x+5)
(v) (2x–1)(x–3)=(x+5)(x–1)
(vi) x2+3x+1=(x–2)2
(vii) (x+2)3=2x(x2–1)
(viii) x3–4×2–x+1=(x–2)3

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# Ex 4.2 – Quadratic Equation (Full Exercise Solution + Video) MATHS – NCERT | Class 10

SUBJECT – MATHS
CLASS – X (10th)
QUESTION FROM – Quadratic Equation (Exercise 4.2)
ALL EXTRA & IMPORTANT QUESTION ANSWER FOR 2019 EXAMS

##### Ex 4.2 – Quadratic Equation (Full Exercise Solution + Video) MATHS – NCERT | Class 10

###### Q1. Find the roots of the following quadratic equations by factorisation:

(i) x2 -3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2x2 + 7x + 5√2 = 0
(iv) 2x2 – x + $\frac { 1 }{ 8 }$ = 0 8
(v) 100 x2 – 20 X + 1 = 0

Ex 4.2 Quadratic Equation (Q.1) Class 10 Maths NCERT Solution In Hindi

###### Question 2 :: Solve the following situations mathematically:

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹750. We would like to find out the number of toys produced on that day.

Question 3 :: Find two numbers whose sum is 27 and product is 182.

Ex 4.2 Quadratic Equation (Q.3) class 10 Maths NCERT Solution In Hindi

###### Question 5 :: The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Question 6 :: A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90, find the number of articles produced and the cost of each article.

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For More Questions And Answers, Please Use Our Search Option To Search Those Answers. Please Keep Visiting – www.Study24Hours.Com For More Study Materials. Thank You For Visiting This Site !! # Ex 4.1 – Quadratic Equation (Full Exercise Solution + Video) MATHS – NCERT | Class 10

SUBJECT – MATHS
CLASS – X (10th)
QUESTION FROM – Quadratic Equation (Exercise 4.1)
ALL EXTRA & IMPORTANT QUESTION ANSWER FOR 2019 EXAMS

##### Ex 4.1 – Quadratic Equation (Full Exercise Solution + Video) NCERT | Explained In Hindi

###### Q1. Check whether the following are quadratic equations :-

(i) (x+1)2=2(x–3)
(ii) x2–2x=(-2)(3–x)
(iii) (x–2)(x+1)=(x–1)(x+3)
(iv) (x–3)(2x+1)=x(x+5)
(v) (2x–1)(x–3)=(x+5)(x–1)
(vi) x2+3x+1=(x–2)2
(vii) (x+2)3=2x(x2–1)
(viii) x3–4×2–x+1=(x–2)3

Solutions For Question Number 1 Is In The Video Below (Explained In Hindi) :-

###### Q2. Represent the following situation in the form of quadratic equation:

(i) The area of a rectangular plot is 528 m2. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find the Rohan’s age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Answer – Solutions For Question Number 2 Is In The Video Below (Explained In Hindi) :-

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For More Questions And Answers, Please Use Our Search Option To Search Those Answers. Please Keep Visiting – www.Study24Hours.Com For More Study Materials. Thank You For Visiting This Site !!