CBSE Notes For Class 10 Maths Chapter 11 Areas Related To Circles

1.0Introduction to Circles 

A circle is a closed curve in which all the points are equidistant from a fixed point, which is also called the centre. The distance from the centre to any point on the circle is called the radius. The area covered by a circle can be calculated using the given formula. 

$\text{ Area of a circle }=\pi r^2$

Where, r = radius of the circle. 

Perimeter/circumference of the circle = $2\pi r$

Diameter = 2r 

2.0CBSE Class 10 Maths Chapter 11 Area Related to Circles - Revision Notes

What are the sectors, segments, and lengths of an arc of a circle? 

Sector of a circle 

The sector is the region between two radii and the arc of a circle. Sectors are of two types:

• Minor Sector: It is the sector corresponding to an angle less than 180 degrees. 
• Major Sector: This sector corresponds to an angle more than or equal to 180 degrees. 

Area related to the sector:

If the angle of a sector is given by $\theta$ then;

$\text{ Area of the sector }=\frac{\theta }{360}\times \pi r^2$

Segment of a circle

A segment of a circle is the region enclosed by a chord and the arc intercepted by that chord. Essentially it's an area between the chord and the arc, part of the circle lying between the chord and the curved part of it. In maths, Segment is also of two types: 

• Minor Segment: The segment is made of a chord and a minor arc.
• Major Segment: The arc greater than the semicircle created by a chord.

Area related to segment: 

The area of a segment of a circle is the area of the sector minus the area of a triangle formed by two radii and the corresponding chord. 

Area of segment = Area of the sector - Area of a triangle

That is, 

$\text{ Area of segment }=\frac{\theta }{360}\pi r^2-\text{ Area of triangle }$

Length of Arc of the Circle

The length of an arc is defined as the distance along the curved part of the circle between two points on the circumference as given in the figure. 

The length of an arc depends upon the radius of the circle and the central angle subtended by the arc. It is denoted by l.

Area related to the length of an arc of the circle: 

If the angle of a sector is given by then;

$\text{ Length of an }\mathrm{arc}(l)=\frac{\theta }{360}2\pi r$

3.0Solved Problems 

Question 1: The area of a sector of a circle of radius 36 cm is $54\pi {\mathrm{cm}}^2$. Find the length of the corresponding arc of the sector. 

Solution:

$\text{ Area of the sector }=\frac{\theta }{360}\times \pi r^2$

$54\pi =\frac{\theta }{360}\times \pi \times 36^2$

$\theta =\frac{54\pi \times 360}{\pi \times 36\times 36}=3\times 5=15$

$\text{ Length of an }\mathrm{arc}(l)=\frac{\theta }{360}\times 2\pi r$

$l=\frac{15}{360}\times 2\times \pi \times 36=3\pi$

Question 2: In the given figure, find the area of the shaded region given that the radius of the circle that is inscribed in a square is 7.5cm. $\text{ (Use }\pi =3.14)$

Solution:

$\text{ Area of a circle }=\pi r^2=3.14\times 7.5^2$$=176.625c{m}^2$

The Diameter of the circle = side of the square = 15cm

Area of the shaded region =Area of square-Area of circle

$\text{ Area of the shaded region }=4a^2-176.625$

$=4\times 15\times 15-176.625$

$=48.375{\mathrm{cm}}^2$

Question 3: A chord of a circle of radius 20 cm subtends an angle of 90° at the center. Find the area of the corresponding major segment of the circle. (Use π = 3.14).

Solution: Area of segment = 360r2- Area of triangle

Area of minor segment = 903603.14202-122020 = 314-200=114cm2

Area of a circle = r2= 3.1420 × 20=1256cm2

Area of major segment= Area of circle-Area of minor segment

= 1256-114=1142cm2

Question 4: A calf is tied with a rope of length 6 m at the corner of a square grassy lawn of side 20 m. If the length of the rope is increased by 5.5m, find the increase in the area of the grassy lawn in which the calf can graze.

Solution:  The increase in area = Difference between the two sectors of central angle 90° each and radii 11.5 m (6 m + 5.5 m) and 6 m, 

So, the required increase in area =

$=9036011.52-9036062$

$=4(11.5+6)(11.5-6)=\mathrm{227417.55.5}=75.625c{m}^2$

4.0Key features of CBSE Maths Notes for Class 10 Chapter 11

• The notes are aligned with the latest CBSE curriculum. 
• A step-by-step guide is also provided, along with solved problems to help you better understand the chapter. 
• Visual aids and diagrams are provided for a better understanding of concepts. 
• The language used is easy to understand, making notes ideal for self-learning.