RD Sharma Solutions for Class 10 Maths Chapter 15 Areas Related to Circles
RD Sharma Solutions for Class 10 Maths Chapter 15 – Areas Related to Circles provide a comprehensive resource to help students strengthen their understanding and boost their performance in board exams. This chapter holds significant importance as circles and circular shapes are commonly encountered in various real-life situations, making the concepts both practical and essential.
Chapter 15 focuses on calculating the areas of sectors and segments—two special parts of a circular region. The RD Sharma Solutions offer detailed, step-by-step explanations that simplify complex problems, allowing students to grasp core mathematical concepts effectively and apply them confidently in exams.
Designed to reinforce conceptual clarity, these solutions guide students through multiple problem-solving techniques, improving both accuracy and speed. With clear and descriptive answers, RD Sharma Class 10 Maths Solutions serve as an ideal study tool for mastering the chapter on Areas Related to Circles and preparing efficiently for the Class 10 board examination.
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Q. Find the circumference of a circle whose area is 301.84 cm2.
Solution:
Given,
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Area of the circle = 301.84 cm2
We know that
Area of a Circle = πr2 = 301.84 cm2
(22/7) × r2 = 301.84
r2 = 13.72 x 7 = 96.04
r = √96.04 = 9.8
So, the radius is = 9.8 cm.
Now, the circumference of a circle = 2πr
= 2 × (22/7) × 9.8 = 61.6 cm
Hence, the circumference of the circle is 61.6 cm.
Q. The circumference of a circle exceeds the diameter by 16.8 cm. Find the circumference of the circle.
Solution:
Let the radius of the circle be r cm.
So, the diameter (d) = 2r [As radius is half the diameter]
We know that,
Circumference of a circle (C) = 2πr
From the question,
The circumference of the circle exceeds its diameter by 16.8 cm.
C = d + 16.8
2πr = 2r + 16.8 [d = 2r]
2πr – 2r = 16.8
2r (π – 1) = 16.8
2r (3.14 – 1) = 16.8
r = 3.92 cm
Thus, radius = 3.92 cm
Now, the circumference of the circle (C) = 2πr
C = 2 × 3.14 × 3.92
= 24.64 cm
Q. Find the circumference and area of a circle of radius of 4.2 cm.
Solution: Given,
Radius (r) = 4.2 cm
We know that
Circumference of a circle = 2πr
= 2 × (22/7) × 4.2 = 26.4 cm2
Area of a circle = πr2
= (22/7) x 4.22
= 22 x 0.6 x 4.2 = 55.44 cm2
Q. The sum of the radii of two circles is 140 cm, and the difference of their circumference is 88 cm. Find the diameters of the circles.
Solution:
Let the radii of the two circles be r1 and r2.
And the circumferences of the two circles be C1 and C2.
We know that the circumference of the circle (C) = 2πr
Given,
Sum of radii of two circle, i.e., r1 + r2 = 140 cm … (i)
Difference of their circumference,
C1 – C2 = 88 cm
2πr1 – 2πr2 = 88 cm
2(22/7)(r1 – r2) = 88 cm
(r1 – r2) = 14 cm
r1 = r2 + 14…..(ii)
Substituting the value of r1 in equation (i), we have
r2 + r2 + 14 = 140
2r2 = 140 – 14
2r2 = 126
r2 = 63 cm
Substituting the value of r2 in equation (ii), we have
r1 = 63 + 14 = 77 cm
Therefore,
Diameter of circle 1 = 2r1 = 2 x 77 = 154 cm
Diameter of circle 2 = 2r2 = 2 × 63 = 126 cm
Q.The radii of the two circles are 8 cm and 6 cm, respectively. Find the radius of the circle having its area equal to the sum of the areas of two circles.
Solution: Given,
The Radii of the two circles are 6 cm and 8 cm.
The area of circle with radius 8 cm = π (8)2 = 64π cm2
The area of circle with radius 6cm = π (6)2 = 36π cm2
The sum of areas = 64π + 36π = 100π cm2
Let the radius of the circle be x cm.
Area of the circle = 100π cm2 (from above)
πx2 = 100π
x = √100 = 10 cm
Q. Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 15cm and 18cm.
Solution:
Given,
Radius of circle 1 = r1 = 15 cm
Radius of circle 2 = r2 = 18 cm
We know that the circumference of a circle (C) = 2πr
So, C1 = 2πr1 and C2 = 2πr2
Let the radius be r of the circle which is to be found and its circumference (C).
Now, from the question,
C = C1 + C2
2πr = 2πr1 + 2πr2
r = r1 + r2 [After dividing by 2π both sides]
r = 15 + 18
r = 33 cm
Q. The radii of the two circles are 19 cm and 9 cm, respectively. Find the radius and area of the circle which has circumferences equal to the sum of the circumference of two circles.
Solution: Given,
Radius of circle 1 = r1 = 19 cm
Radius of circle 2 = r2 = 9 cm
We know that the circumference of a circle (C) = 2πr
So, C1 = 2πr1 and C2 = 2πr2
Let the radius be r of the circle which is to be found and its circumference (C).
Now, from the question,
C = C1 + C2
2πr = 2πr1 + 2πr2
r = r1 + r2 [After dividing by 2π both sides]
r = 19 + 9
r = 28 cm
Thus, the radius of the circle = 28 cm
So, the area of required circle = πr2 = (22/7) × 28 × 28 = 2464 cm2
RD Sharma Solutions for Class 10 Maths Chapter 15 FAQs
What is covered in RD Sharma Solutions for Class 10 Chapter 15?
RD Sharma Solutions for Class 10 Chapter 15 – Areas Related to Circles cover detailed explanations and step-by-step solutions for finding the area of sectors and segments of a circle. The chapter focuses on real-life applications of circle-based geometry and enhances conceptual clarity through solved examples and exercises.
How do RD Sharma Solutions help in understanding Areas Related to Circles?
The solutions offer a clear and structured approach to solving problems related to sectors and segments. By breaking down formulas and problem-solving techniques, RD Sharma Solutions help Class 10 students build a strong foundation and improve their performance in exams.
Are RD Sharma Class 10 Chapter 15 solutions useful for board exam preparation?
Yes, RD Sharma Chapter 15 solutions are highly beneficial for Class 10 board exam preparation. They provide comprehensive coverage of important topics, including formula derivations, conceptual insights, and a variety of practice problems aligned with the CBSE syllabus.
Where can I download RD Sharma Solutions for Class 10 Maths Chapter 15 PDF?
You can access or download RD Sharma Solutions for Class 10 Maths Chapter 15 – Areas Related to Circles PDF from various educational platforms like Infinity Learn that provide reliable and free study materials for students.