Page 1 AREAS RELA TED TO CIRCLES 223 12 12.1 Introduction You are already familiar with some methods of finding perimeters and areas of simple plane figures such as rectangles, squares, parallelograms, triangles and circles from your earlier classes. Many objects that we come across in our daily life are related to the circular shape in some form or the other. Cycle wheels, wheel barrow (thela), dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1). So, the problem of finding perimeters and areas related to circular figures is of great practical importance. In this chapter, we shall begin our discussion with a review of the concepts of perimeter (circumference) and area of a circle and apply this knowledge in finding the areas of two special â€˜partsâ€™ of a circular region (or briefly of a circle) known as sector and segment. We shall also see how to find the areas of some combinations of plane figures involving circles or their parts. Fig. 12.1 AREAS RELATED TO CIRCLES Page 2 AREAS RELA TED TO CIRCLES 223 12 12.1 Introduction You are already familiar with some methods of finding perimeters and areas of simple plane figures such as rectangles, squares, parallelograms, triangles and circles from your earlier classes. Many objects that we come across in our daily life are related to the circular shape in some form or the other. Cycle wheels, wheel barrow (thela), dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1). So, the problem of finding perimeters and areas related to circular figures is of great practical importance. In this chapter, we shall begin our discussion with a review of the concepts of perimeter (circumference) and area of a circle and apply this knowledge in finding the areas of two special â€˜partsâ€™ of a circular region (or briefly of a circle) known as sector and segment. We shall also see how to find the areas of some combinations of plane figures involving circles or their parts. Fig. 12.1 AREAS RELATED TO CIRCLES 224 MATHEMA TICS 12.2 Perimeter and Area of a Circle â€” A Review Recall that the distance covered by travelling once around a circle is its perimeter, usually called its circumference. You also know from your earlier classes, that circumference of a circle bears a constant ratio with its diameter. This constant ratio is denoted by the Greek letter p (read as â€˜piâ€™). In other words, circumference diameter = p or, circumference = p × diameter = p × 2r (where r is the radius of the circle) =2pr The great Indian mathematician Aryabhatta (A.D. 476 â€“ 550) gave an approximate value of p. He stated that p = 62832 , 20000 which is nearly equal to 3.1416. It is also interesting to note that using an identity of the great mathematical genius Srinivas Ramanujan (1887â€“1920) of India, mathematicians have been able to calculate the value of p correct to million places of decimals. As you know from Chapter 1 of Class IX, p is an irrational number and its decimal expansion is non-terminating and non-recurring (non-repeating). However, for practical purposes, we generally take the value of p as 22 7 or 3.14, approximately. You may also recall that area of a circle is pr 2 , where r is the radius of the circle. Recall that you have verified it in Class VII, by cutting a circle into a number of sectors and rearranging them as shown in Fig. 12.2. Fig 12.2 Page 3 AREAS RELA TED TO CIRCLES 223 12 12.1 Introduction You are already familiar with some methods of finding perimeters and areas of simple plane figures such as rectangles, squares, parallelograms, triangles and circles from your earlier classes. Many objects that we come across in our daily life are related to the circular shape in some form or the other. Cycle wheels, wheel barrow (thela), dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1). So, the problem of finding perimeters and areas related to circular figures is of great practical importance. In this chapter, we shall begin our discussion with a review of the concepts of perimeter (circumference) and area of a circle and apply this knowledge in finding the areas of two special â€˜partsâ€™ of a circular region (or briefly of a circle) known as sector and segment. We shall also see how to find the areas of some combinations of plane figures involving circles or their parts. Fig. 12.1 AREAS RELATED TO CIRCLES 224 MATHEMA TICS 12.2 Perimeter and Area of a Circle â€” A Review Recall that the distance covered by travelling once around a circle is its perimeter, usually called its circumference. You also know from your earlier classes, that circumference of a circle bears a constant ratio with its diameter. This constant ratio is denoted by the Greek letter p (read as â€˜piâ€™). In other words, circumference diameter = p or, circumference = p × diameter = p × 2r (where r is the radius of the circle) =2pr The great Indian mathematician Aryabhatta (A.D. 476 â€“ 550) gave an approximate value of p. He stated that p = 62832 , 20000 which is nearly equal to 3.1416. It is also interesting to note that using an identity of the great mathematical genius Srinivas Ramanujan (1887â€“1920) of India, mathematicians have been able to calculate the value of p correct to million places of decimals. As you know from Chapter 1 of Class IX, p is an irrational number and its decimal expansion is non-terminating and non-recurring (non-repeating). However, for practical purposes, we generally take the value of p as 22 7 or 3.14, approximately. You may also recall that area of a circle is pr 2 , where r is the radius of the circle. Recall that you have verified it in Class VII, by cutting a circle into a number of sectors and rearranging them as shown in Fig. 12.2. Fig 12.2 AREAS RELA TED TO CIRCLES 225 Y ou can see that the shape in Fig. 12.2 (ii) is nearly a rectangle of length 1 2 2 r ×p and breadth r. This suggests that the area of the circle = 1 2 × 2pr × r = pr 2 . Let us recall the concepts learnt in earlier classes, through an example. Example 1 : The cost of fencing a circular field at the rate of Rs 24 per metre is Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m 2 . Find the cost of ploughing the field (Take p = 22 7 ). Solution : Length of the fence (in metres) = Total cost Rate = 5280 = 220 24 So, circumference of the field = 220 m Therefore, if r metres is the radius of the field, then 2pr = 220 or, 2 × 22 7 × r = 220 or, r = 220 × 7 2× 22 = 35 i.e., radius of the field is 35 m. Therefore, area of the field = pr 2 = 2 7 2 × 35 × 35 m 2 = 22 × 5 × 35 m 2 Now, cost of ploughing 1 m 2 of the field = Rs 0.50 So, total cost of ploughing the field = Rs 22 × 5 × 35 × 0.50 = Rs 1925 EXERCISE 12.1 Unless stated otherwise, use p = 22 7 . 1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles. 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. 3. Fig. 12.3 depicts an archery target marked with its five scoring areas 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. Fig. 12.3 Page 4 AREAS RELA TED TO CIRCLES 223 12 12.1 Introduction You are already familiar with some methods of finding perimeters and areas of simple plane figures such as rectangles, squares, parallelograms, triangles and circles from your earlier classes. Many objects that we come across in our daily life are related to the circular shape in some form or the other. Cycle wheels, wheel barrow (thela), dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1). So, the problem of finding perimeters and areas related to circular figures is of great practical importance. In this chapter, we shall begin our discussion with a review of the concepts of perimeter (circumference) and area of a circle and apply this knowledge in finding the areas of two special â€˜partsâ€™ of a circular region (or briefly of a circle) known as sector and segment. We shall also see how to find the areas of some combinations of plane figures involving circles or their parts. Fig. 12.1 AREAS RELATED TO CIRCLES 224 MATHEMA TICS 12.2 Perimeter and Area of a Circle â€” A Review Recall that the distance covered by travelling once around a circle is its perimeter, usually called its circumference. You also know from your earlier classes, that circumference of a circle bears a constant ratio with its diameter. This constant ratio is denoted by the Greek letter p (read as â€˜piâ€™). In other words, circumference diameter = p or, circumference = p × diameter = p × 2r (where r is the radius of the circle) =2pr The great Indian mathematician Aryabhatta (A.D. 476 â€“ 550) gave an approximate value of p. He stated that p = 62832 , 20000 which is nearly equal to 3.1416. It is also interesting to note that using an identity of the great mathematical genius Srinivas Ramanujan (1887â€“1920) of India, mathematicians have been able to calculate the value of p correct to million places of decimals. As you know from Chapter 1 of Class IX, p is an irrational number and its decimal expansion is non-terminating and non-recurring (non-repeating). However, for practical purposes, we generally take the value of p as 22 7 or 3.14, approximately. You may also recall that area of a circle is pr 2 , where r is the radius of the circle. Recall that you have verified it in Class VII, by cutting a circle into a number of sectors and rearranging them as shown in Fig. 12.2. Fig 12.2 AREAS RELA TED TO CIRCLES 225 Y ou can see that the shape in Fig. 12.2 (ii) is nearly a rectangle of length 1 2 2 r ×p and breadth r. This suggests that the area of the circle = 1 2 × 2pr × r = pr 2 . Let us recall the concepts learnt in earlier classes, through an example. Example 1 : The cost of fencing a circular field at the rate of Rs 24 per metre is Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m 2 . Find the cost of ploughing the field (Take p = 22 7 ). Solution : Length of the fence (in metres) = Total cost Rate = 5280 = 220 24 So, circumference of the field = 220 m Therefore, if r metres is the radius of the field, then 2pr = 220 or, 2 × 22 7 × r = 220 or, r = 220 × 7 2× 22 = 35 i.e., radius of the field is 35 m. Therefore, area of the field = pr 2 = 2 7 2 × 35 × 35 m 2 = 22 × 5 × 35 m 2 Now, cost of ploughing 1 m 2 of the field = Rs 0.50 So, total cost of ploughing the field = Rs 22 × 5 × 35 × 0.50 = Rs 1925 EXERCISE 12.1 Unless stated otherwise, use p = 22 7 . 1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles. 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. 3. Fig. 12.3 depicts an archery target marked with its five scoring areas 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. Fig. 12.3 226 MATHEMA TICS Fig. 12.5 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 travelling at a speed of 66 km per hour? 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) p units (C) 4 units (D) 7 units 12.3 Areas of Sector and Segment of a Circle You have already come across the terms sector and segment of a circle in your earlier classes. Recall that the portion (or part) of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle and the portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle. Thus, in Fig. 12.4, shaded region OAPB is a sector of the circle with centre O. ? AOB is called the angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of the circle. For obvious reasons, OAPB is called the minor sector and OAQB is called the major sector. You can also see that angle of the major sector is 360° â€“ ? AOB. Now, look at Fig. 12.5 in which AB is a chord of the circle with centre O. So, shaded region APB is a segment of the circle. You can also note that unshaded region AQB is another segment of the circle formed by the chord AB. For obvious reasons, APB is called the minor segment and AQB is called the major segment. Remark : When we write â€˜segmentâ€™ and â€˜sectorâ€™ we will mean the â€˜minor segmentâ€™ and the â€˜minor sectorâ€™ respectively, unless stated otherwise. Now with this knowledge, let us try to find some relations (or formulae) to calculate their areas. Let OAPB be a sector of a circle with centre O and radius r (see Fig. 12.6). Let the degree measure of ? AOB be ?. You know that area of a circle (in fact of a circular region or disc) is pr 2 . Fig. 12.4 Fig. 12.6 Page 5 AREAS RELA TED TO CIRCLES 223 12 12.1 Introduction You are already familiar with some methods of finding perimeters and areas of simple plane figures such as rectangles, squares, parallelograms, triangles and circles from your earlier classes. Many objects that we come across in our daily life are related to the circular shape in some form or the other. Cycle wheels, wheel barrow (thela), dartboard, round cake, papad, drain cover, various designs, bangles, brooches, circular paths, washers, flower beds, etc. are some examples of such objects (see Fig. 12.1). So, the problem of finding perimeters and areas related to circular figures is of great practical importance. In this chapter, we shall begin our discussion with a review of the concepts of perimeter (circumference) and area of a circle and apply this knowledge in finding the areas of two special â€˜partsâ€™ of a circular region (or briefly of a circle) known as sector and segment. We shall also see how to find the areas of some combinations of plane figures involving circles or their parts. Fig. 12.1 AREAS RELATED TO CIRCLES 224 MATHEMA TICS 12.2 Perimeter and Area of a Circle â€” A Review Recall that the distance covered by travelling once around a circle is its perimeter, usually called its circumference. You also know from your earlier classes, that circumference of a circle bears a constant ratio with its diameter. This constant ratio is denoted by the Greek letter p (read as â€˜piâ€™). In other words, circumference diameter = p or, circumference = p × diameter = p × 2r (where r is the radius of the circle) =2pr The great Indian mathematician Aryabhatta (A.D. 476 â€“ 550) gave an approximate value of p. He stated that p = 62832 , 20000 which is nearly equal to 3.1416. It is also interesting to note that using an identity of the great mathematical genius Srinivas Ramanujan (1887â€“1920) of India, mathematicians have been able to calculate the value of p correct to million places of decimals. As you know from Chapter 1 of Class IX, p is an irrational number and its decimal expansion is non-terminating and non-recurring (non-repeating). However, for practical purposes, we generally take the value of p as 22 7 or 3.14, approximately. You may also recall that area of a circle is pr 2 , where r is the radius of the circle. Recall that you have verified it in Class VII, by cutting a circle into a number of sectors and rearranging them as shown in Fig. 12.2. Fig 12.2 AREAS RELA TED TO CIRCLES 225 Y ou can see that the shape in Fig. 12.2 (ii) is nearly a rectangle of length 1 2 2 r ×p and breadth r. This suggests that the area of the circle = 1 2 × 2pr × r = pr 2 . Let us recall the concepts learnt in earlier classes, through an example. Example 1 : The cost of fencing a circular field at the rate of Rs 24 per metre is Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m 2 . Find the cost of ploughing the field (Take p = 22 7 ). Solution : Length of the fence (in metres) = Total cost Rate = 5280 = 220 24 So, circumference of the field = 220 m Therefore, if r metres is the radius of the field, then 2pr = 220 or, 2 × 22 7 × r = 220 or, r = 220 × 7 2× 22 = 35 i.e., radius of the field is 35 m. Therefore, area of the field = pr 2 = 2 7 2 × 35 × 35 m 2 = 22 × 5 × 35 m 2 Now, cost of ploughing 1 m 2 of the field = Rs 0.50 So, total cost of ploughing the field = Rs 22 × 5 × 35 × 0.50 = Rs 1925 EXERCISE 12.1 Unless stated otherwise, use p = 22 7 . 1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles. 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. 3. Fig. 12.3 depicts an archery target marked with its five scoring areas 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. Fig. 12.3 226 MATHEMA TICS Fig. 12.5 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 travelling at a speed of 66 km per hour? 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) p units (C) 4 units (D) 7 units 12.3 Areas of Sector and Segment of a Circle You have already come across the terms sector and segment of a circle in your earlier classes. Recall that the portion (or part) of the circular region enclosed by two radii and the corresponding arc is called a sector of the circle and the portion (or part) of the circular region enclosed between a chord and the corresponding arc is called a segment of the circle. Thus, in Fig. 12.4, shaded region OAPB is a sector of the circle with centre O. ? AOB is called the angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of the circle. For obvious reasons, OAPB is called the minor sector and OAQB is called the major sector. You can also see that angle of the major sector is 360° â€“ ? AOB. Now, look at Fig. 12.5 in which AB is a chord of the circle with centre O. So, shaded region APB is a segment of the circle. You can also note that unshaded region AQB is another segment of the circle formed by the chord AB. For obvious reasons, APB is called the minor segment and AQB is called the major segment. Remark : When we write â€˜segmentâ€™ and â€˜sectorâ€™ we will mean the â€˜minor segmentâ€™ and the â€˜minor sectorâ€™ respectively, unless stated otherwise. Now with this knowledge, let us try to find some relations (or formulae) to calculate their areas. Let OAPB be a sector of a circle with centre O and radius r (see Fig. 12.6). Let the degree measure of ? AOB be ?. You know that area of a circle (in fact of a circular region or disc) is pr 2 . Fig. 12.4 Fig. 12.6 AREAS RELA TED TO CIRCLES 227 In a way, we can consider this circular region to be a sector forming an angle of 360° (i.e., of degree measure 360) at the centre O. Now by applying the Unitary Method, we can arrive at the area of the sector OAPB as follows: When degree measure of the angle at the centre is 360, area of the sector = pr 2 So, when the degree measure of the angle at the centre is 1, area of the sector = 2 360 r p · Therefore, when the degree measure of the angle at the centre is ?, area of the sector = 2 360 r p ×? = 2 360 r ? ×p . Thus, we obtain the following relation (or formula) for area of a sector of a circle: Area of the sector of angle ? ? ? ? ? = × 2 360 ? pr , where r is the radius of the circle and ? the angle of the sector in degrees. Now, a natural question arises : Can we find the length of the arc APB corresponding to this sector? Yes. Again, by applying the Unitary Method and taking the whole length of the circle (of angle 360°) as 2pr, we can obtain the required length of the arc APB as 2 360 r ? ×p . So, length of an arc of a sector of angle ? ? ? ? ? = × ? p 2 360 r . Now let us take the case of the area of the segment APB of a circle with centre O and radius r (see Fig. 12.7). You can see that : Area of the segment APB = Area of the sector OAPB â€“ Area of ? OAB = 2 â€“ area of OAB 360 r ? ×p ? Note : From Fig. 12.6 and Fig. 12.7 respectively, you can observe that : Area of the major sector OAQB = pr 2 â€“ Area of the minor sector OAPB and Area of major segment AQB = pr 2 â€“ Area of the minor segment APB Fig. 12.7Read More

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### Ex 12.2 NCERT Solutions- Areas Related to Circles

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### Area of a Circle, Major and Minor Segments(Hindi)

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### Ex 12.3 NCERT Solutions- Areas Related to Circles

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### Example: Area Related to Circles- 3

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### Example: Area Related to Circles- 4

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### Example: Area Related to Circles- 5

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- Test: Area Related To Circles- 4
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