Ncert Solution of Maths Class 10 Chapter 12
Complete solutions of Ex 12.1 Class 10 Maths Chapter 12 with additional questions and answers from new NCERT syllabus textbook Class 10 Maths.
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1.
- Areas Related to Circles Class 10 Ex 12.1
- Areas Related to Circles Class 10 Ex 12.2
- Areas Related to Circles Class 10 Ex 12.3
Board | CBSE |
Textbook | NCERT |
Class | Class 10 |
Subject | Maths |
Chapter | Chapter 12 |
Chapter Name | Areas Related to Circles |
Exercise | Ex 12.1 |
Number of Questions Solved | 5 |
Category | NCERT Solutions |
NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1
Ex 12.1 Class 10 Maths Question 1.
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.
Solution:
Radius of the bigger circle = 19 cm
∴ The Circumference of the bigger circle
= 2πr = 2π x 19 = 38π cm
Radius of the smallest circle = 9 cm
∴ The Circumference of the smaller circle
= 2πr = 2π x 9 = 18π cm
Sum of Circumference of these two circle
= 38π + 18π = 56π cm
Let r be the radius of the required circle.
Then, 2πr = 56π ⇒ r = .
Ex 12.1 Class 10 Maths Question 2.
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:
Radius of bigger circle = 8 cm
∴ Area of bigger circle = π x (8)2 = 64π cm2
Radius of smaller circle = 6 cm
∴ Area of smaller circle = π x (6)2 = 36π cm2
Sum of areas of two circles = 64π + 36π
= 100π cm2
Let r be the radius of the required circle.
Then πr2 = 100π
⇒ r2 = 100 ⇒ r = = 10 cm.
Ex 12.1 Class 10 Maths Question 3.
The given figure 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:
Now, radius of circular region scoring Gold and Red = radius of Gold scoring region + width of Red scoring region = 10.5 + 10.5 = 21 cm.
The combined area of Gold and Red scoring regions
= π(21)2 cm2
∴ Area of Red scoring region = combined area of Gold and Red scoring regions – Area of Gold scoring region
Radius of circular regions scoring Gold, Red, Blue and Black = 10.5 + 10.5 + 10.5 = 42 cm.
similarly, the area of a Black scoring region
Ex 12.1 Class 10 MathsQuestion 4.
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 traveling at a speed of 66 km per hour?
Solution:
Radius of one wheel of the car = 40 cm
= 0.4 m.
Distance covered by the wheel to complete one revolution – 2πr x 0.4 = 0.8 m.
Let the wheel of the car completest revolutions in 10 minutes at a speed of 66 km/h.
Then distance covered by the wheel in making r complete revolutions in 10 minutes
= (0.8π x n ) m
Also, distance travelled by car in 60 minutes
= (66 x 1000) m
∴ Distance travelled by car in 1 minute
m
∴ Dista nce travelled by car in 10 minutes
m = 11000 m.
According to the question, we have:
Ex 12.1 Class 10 Maths Question 5.
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) n units
(c) 4 units
(d) 7 units
Solution:
The correct answer is (a) 2 units
Justification:
Perimeter of a circle = 2πr
Area of a circle = πr2
When 2πr = πr2, then r = 2 units.
We hope the NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.1, drop a comment below and we will get back to you at the earliest.
Ncert Solution of Maths Class 10 Chapter 12
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