Areas Related to Circles, Class 10, Maths Detailed Chapter Notes PDF Download

INTRODUCTION
In earlier classes, we have studied methods of finding perimeters and areas of simple plane figures such as rectangles, squares, parallelograms, triangles and circles. In our daily life, we come across many objects which are related to circular shape in some form or the other. For example, cycle wheels, wheel arrow, drain cover, bangles, flower beds, circlular paths etc. That is why the problem of finding perimeters and areas related to circular figures is of great practical importance. In this chapter, we shall discuss problems on finding the areas of some combinations of plane figures involving circles or parts of circles. Let us first recall the concepts related to their perimeter and area of a circle.

HISTORICAL FACTS
Mensuration is that branch of mathematics which studies the method of measurements. Measurment is a very important human activity. We measure the length of a cloth for stitching. The area of a wall for painting, the perimeter of a plot for fencing. We do many other measurements of similar nature in our daily life. All these measurements, we shall study in this capter called Mensuration.
π(pi) occupies the most significant place in measurement of surface area as well as volume of various solid and plane figures. The value of π is not exactly known. The story of the accuracy by which the value of ? was estimated is an interesting one.

According to S. Ramanujan, the value of π

According to Archimedes ; the vlaue of π is given below :

Note : π (pi) is an irrational number. It cannot be expressed as the ratio of whole numbers. However, the ratio 22 : 7 is ofter used as approximation for it.

RECALL
(A) Circle : Circle is the locus of a point which moves in such a manner that its distance from a fixed point O remains constant (the same). The fixed point is called the centre O and the constant distance OA is called its radius.

(B) Chord : A line segment joining two points on a circle is called a chord of the circle. In fig, AB and CD are
two chords of the circle.

(C) Diameter : A chord passing through the centre of the circle is called the diameter. In fig, AOB and COD are diameter of the circle i.e., the diameter is the largest chord of the circle.

Length of diameter = Twice the radius = 2 × r = AOB = COD

(D) Circumference : The perimeter of the circle or the length of boundary of the circle is called its circumference i.e. the distance covered by travelling once around a circle is called the perimeter or circumference. The circumference of a circle is given by 2πr. It is well-known fact that the ratio of the circumference of a circle to its diameter bears a constant ratio.

 where r is the radius of the circle.

(E) Arc : Any part of a circle is called an arc of the circle. Two points A and B on a circle divides it into two
arcs. In general one arc is greater than other. The smaller arc is called minor arc and greater arc is called
major arc.
In the given fig, AB is an arc of a circle with centre O, denoted by AB. The remaining part of the circle shown
by the dotted lines is represented by BA.

(F) Central Angle : Angle subtended by an arc at the centre of a circle is called its central angle. In fig, the
centre of the circle is O.
Central angle made by AB at the centre O = AOB = θ
If θ° < 180° then the arc AB is called the minor arc and the arc BA is called major arc.

(G) Semi-circle : A diameter divides a circle into two congruent arcs. Each of these two arc is called a semicircle.
In the given fig. of a circle with centre O, APB and BQA are semicircles. Semicircle is the half of the circle.

(H) Major arc : An arc whose length is more than the length of the semi-circle is called a major arc.

(I) Minor arc : An arc whose length is less than the length of semi-circle is called a minor arc.

(J) Segment : A segment of a circle is the region bounded by an arc and its chord, including the arc and the chord.

The shaded segment containing the minor arc is called a minor segment, while the unshaded segment containing the major arc is called the major segment.

(K) Sector of a circle : A sector of a circle is a region enclosed by an arc and its two bounding radii. In the
fig OACBO is a sector of the circle with centre O.

If arc AB is a minor arc then OACBO is a called the minor segment of the circle. The remaining part OADBO
of the circle is called the major sector of the circle.

FORMULAE
I. For a circle of radius = r units, we have

(a) Circumference of the circle = (2πr) units = (π d) units,
where d is the diameter. 
(b) Area of the circle = (πr²) sq. units.

II. For a semi-circle of radius = r units, we have

(a) Area of the semi-circle
(b) Perimeter of the semi-circle = (πr + 2r) units.

III. Area of a Circular Ring :

If R and r be the outer and inner radii of a ring, then 
Area of the ring = π(R² – r²) sq. units.

IV. Results on Sectors and Segments :

Suppose an arc ACB makes an angle θ at the centre O of a circle of radius = r units. Then :

(c) Perimeter of sector OACBO = length of arc ACB + OA + OB 

(d) Area of segment ACBA = (Area of sector OACBO) – (Area of ΔOAB) =

(e) Perimeter of segment ACBA = (arc ACB + chord AB) units.

(f) Area of Major segment BDAB = (Area of circle) – (Area of segment ACBA).

V. Rotations Made By a Wheel :

(a) Distance moved by a wheel in 1 revolution = Circumference of the wheel

(b) Number of rotations made by a wheel in unit time  

VI. Facts About Clocks :
(a) Angle described by minute hand in 60 minutes = 360°

(b) Angle described by minute hand in 5 minutes = 

(c) Angle described by hour hand in 12 hours = 360°.
Angle described by hour hand in 1 hour = 30°.

VII. In an equilateral triangle of side a units, we have :

Ex.1 Calculate the circumference and area of a circle of radius 5·6 cm.

Sol. We have :
Circumference of the circle  
Area of the circle

Ex.2 The circumference of a circle is 123·2 cm. Calculate :
(i) the radius of the circle in cm,
(ii) the area of the circle, correct to nearest cm².

Sol. (i) Let the radius of the circle be r cm.
Then, its circumference = (2πr) cm.

Ex.3 The area of a circle is 301·84 cm². Calculate:

(i) the radius of the circle in cm.

(ii) the circumference of the circle, correct to nearest cm.

Sol. (i) Let the radius of the circle be r cm.
Then, its area = πr² cm² = 301.84
 

= 61.6 cm.

Circumference of the circle, correct to nearest cm = 62 cm.

Ex.4 The perimeter of a semi-circular protractor is 32·4 cm. Calculate :

(i) the radius of the protractor in cm,

(ii) the area of the protractor in cm².

Sol. (i) Let the radius of the protractor be r cm.

Radius of the protractor = 6.3 cm.

(ii) Area of the protractor

Area of the protractor = 62·37 cm².

Ex.5 The area enclosed by the circumferences of two concentric circles is 346.5 cm². If the circumference of the inner circle is 88 cm, calculate the radius of the outer circle.

Sol. Let the radius of inner circle be r cm.
Then, its circumference = (2πr) cm.

Hence, the radius of the outer circle is 17·5 cm.

Ex.6 Two circles touch externally. The sum of their areas is 130π sq. cm and the distance between their centres is 14 cm. Determine the radii of the circles.

Sol. Let the radii of the given circles be R cm and r cm respectively. As the circles touch externally, distance between their centres = (R + r) cm.

Ex.7 Two circles touch internally. The sum of their areas is 116π sq. cm and the distance between their centres is 6 cm. Find the radii of the given circles.

Sol. Let the radii of the given circles be R cm and r cm respectively. As the circles touch internally, distance between their centres = (R – r) cm.

Ex.8 The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 cm, find its speed in km/hr. Give your answer, correct to nearest km.

Sol. Radius of the wheel = 42 cm.

Circumference of the wheel =  
Distance moved by the wheel in 1 revolution = 264 cm.
Distance moved by the wheel in 5 revolutions = (264 × 5) cm = 1320 cm.
Distance moved by the wheel in 1 second = 1320 cm.
Distance moved by the wheel in 1 hour = (1320 × 60 × 60) cm.

Hence, the speed of the cart, correct to nearest km/hr is 48 km/hr.

Ex.9 The diameter of the driving wheel of a bus is 140 cm. How many revolutions must the wheel make in order to keep a speed of 66 km/hr?

Sol. Distance to be covered in 1 min. = 

Hence, the wheel must make 250 revolutions per minute.

Ex.10 A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77 cm. Given that the bucket ascends in 1 min. 28 seconds with a uniform speed of 1·1 m / sec, calculate the number of complete revolutions the wheel makes in raising the bucket.

Sol. Time taken by bucket to ascend = 1 min. 28 sec. = 88 sec. Speed = 1.1 m/sec.

Length of the rope = Distance covered by bucket to ascend
= (1·1 × 88) m = (1·1 × 88 × 100) cm = 9680 cm.

Hence, the wheel makes 40 revolutions to raise the bucket.