Areas Related To Circles

Important Questions

Here are some important questions for Class 10 Mathematics Chapter 11, Areas Related to Circles, carefully selected to help students prepare effectively for the CBSE Class 10 Mathematics Examination in 2024-25. By practicing these varied problems, students can better understand the concepts of Areas Related to Circles and improve their problem-solving skills. These questions aim to clarify doubts and enhance performance in the chapter.

Table of Contents

In Chapter 11 of Class 10 Mathematics, Areas Related To Circles, we explore concepts related to the Area of Sectors and Segments of a circle. The problems in this chapter focus on calculating the areas and perimeters (or circumferences) of these plane figures. When calculating the area of a segment of a circle, the problems are restricted to central angles of 60°, 90°, and 120°. This chapter delves into understanding how to find the areas of different portions of a circle, as well as solving problems that involve both areas and perimeters/circumferences.

Ans: The area of a sector of a circle is the region enclosed by an arc and two radii of the circle. It is calculated using the formula,
A = (θ/360) × πr²,\\
where ‘θ’ is the central angle and ‘r’ is the radius.

(b) 45°

(c) 62.5°

(d) 25°

**Ans.**(b) 45°

**Explanation:**

Given, ∠AOD = 135°

∴ ∠BOC = 180° – 135° = 45°

Since, angles subtended at the center by a pair of opposite sides are supplementary.

**Ans.** (b) 1:64

**Explanation:**

The shaded region consists of a chord and the arcs of the circle. The region in the white has the major arc, so it is the major segment and the shaded region consists of the minor arc, so it is the minor segment.

\bigg (\text{Use π =} \dfrac{22}{7} \bigg)

**Ans.** \text{Area = 5.48 cm}^2

**Explanation:
**

Given, AB = BC = CD = DA = 8 cm,

AM = AN = PB = QB = 1.4 cm and radius of circle with center O = 4.2 cm

Thus, area of the remaining portion

= Area of ABCD

- [Area of circle + 2 Area of quadrants] \\= 8 × 8 - \bigg \{\dfrac{22}{7}(4.2)^2 + 2 × \dfrac{1}{4}× \dfrac{22}{7} (1.4)^2 \bigg\}\\ \text{= 64 - (55.44 + 3.08) cm}^2\\ \text{= 5.48 cm}^2

**Ans.** Area swept by the minute hand in covering 5 minutes = \text{51.33 cm}^2

**Explanation:**

Given, length of the minute hand = r = 14 cm

Total number of divisions in the clock = 12

Hence the angle subtended at the center between two digits covering five minutes = \dfrac{360^o}{12}= 30^o\\
Hence, area swept by the minute hand in covering 5 minutes \\ = \dfrac{\theta}{360^o} \pi r^2 \\[4.5 bp]
= \dfrac{30^o}{360^o} × \dfrac{22}{7}× (14)^2\\[4.5 bp]
=\dfrac{154}{3} \\[4.5 bp]
= 51.33 cm^3.

**Ans.** Area of the remaining part of the metal sheet \text{= 6.125 cm}^2 \\
**Explanation:**

Given, AB =BC =3.5 cm and DE = 2 cm

As BC and EC are the radii of the circle, so

BC = EC

Thus, DC = (3.5+2) cm

= 5.5 cm

Hence , area of remaining part of the metal sheet

= Area of the trapezium - Area of the quadrant

= \dfrac{1}{2}× \text{(sum of parallel sides)}\\[4.5 bp] × \text{(distance between parallel sides )-} \dfrac{1}{4}\pi r^2 \\[4.5 bp]
= \dfrac{1}{2} (3.5 + 5.5)× 3.5 - \dfrac{1}{4} × \dfrac{22}{7} × (3.5)^2 \\[4.5 bp]= \dfrac{1}{2}× 9 × 3.5 - \dfrac{11×0.5×3.5}{2}\\[4.5 bp]= \dfrac{31.5}{2}-\dfrac{19.25}{2}\\[4.5 bp]
= \dfrac{12.25}{2}\\[4.5 bp]
= 6.125 \text{cm}^2

Chapter No. | Chapter Name |
---|---|

Chapter 1 | Real Numbers |

Chapter 2 | Polynomials |

Chapter 3 | Pair of Linear Equations in Two Variable |

Chapter 4 | Quadratic Equations |

Chapter 5 | Arithmetic Progressions |

Chapter 6 | Triangles |

Chapter 7 | Coordinate Geometry |

Chapter 8 | Introduction to Trigonometry |

Chapter 9 | Some Applications of Trigonometry |

Chapter 10 | Circles |

Chapter 11 | Areas Related to Circle |

Chapter 12 | Surface Areas and Volumes |

Chapter 13 | Statistics |

Chapter 14 | Probability |

To enhance your understanding of the Areas Related to Circles chapter, consider exploring oswal.io. This website provides a variety of practice questions designed to facilitate effective learning. By working on these questions, you can improve your grasp of circle-related concepts and enhance your problem-solving skills in this crucial math topic.

Ans: A segment of a circle is the region bounded by an arc and a chord. It is like a “slice” of the circle. There are minor and major segments.

Ans: The area of a circle is used in various real-life scenarios, such as calculating the area of circular fields, pools, circular objects, and designing circular structures.

Ans: The perimeter (circumference) of a sector consists of the arc’s length and two radii. The area of the sector includes the region enclosed by the arc and radii.

Ans: For a major segment, subtract the area of a triangle from the area of a sector, while for a minor segment, add the area of a triangle to the area of a sector.

Ans: The area of a sector is directly proportional to the measure of its central angle. If two sectors have the same central angle, their areas will be proportional to the squares of their radii.

Chapter Wise Important Questions for CBSE Board Class 10 Maths |
---|

Real Numbers |

Polynomials |

Pair of Linear Equations in Two Variables |

Quadratic Equations |

Arithmetic Progressions |

Triangles |

Coordinate Geometry |

Introduction to Trigonometry |

Some Applications of Trigonometry |

Circles |

Areas Related to Circles |

Surface Areas and Volumes |

Statistics |

Probability |

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