Tiêu đề 08-Circumference and Area of Circles.pdf ✅

8 CHAPTER CIRCUMFERENCE AND AREA OF CIRCLES CONTENTS  Circle  Circumference of a circle  Area of a circle  Area between two concentric circles CIRCLE A circle is a geometrical figure consisting of all those points in a plane which are at a fixed distance from a point, called centre of the circle.  Radius The constant distance from centre to any point on circle is called the radius of the circle and is denoted by ‘r’. The radius is also used to mean the line segment joining the centre with any point on the circle. Radius Centre O Circle A B There are infinite number of radii for a given circle and all are equal in length.  Chord of a Circle A line segment with its end points lying on a circle is called the chord.  Diameter The diameter of a circle is any line segment passing through the centre of the circle and having its end points on the circle. Diameter = 2 × radius O Diameter Circle B A The longest chord is chord diameter.  Concentric Circles Two or more circles with the same centre are called concentric circles. Following are concentric circles : (i) (ii) The following circles are not concentric circles, because they have different centres. A B
CIRCUMFERENCE OF A CIRCLE Circumference means, ‘the perimeter of a circle’. The word has been derived from the Latin word circumferre means to carry around. The distance around a circular region is also known as its circumference. Note : (1) The ratio of circumference to diameter is approximately the same around 3.142. i.e. The circumference of a circle is slightly more than 3 times its diameter. Thus, we have Diameter of thecircle Circumfere nce of a circle =Constant or , d C   The constant ratio of circumference to diameter, i.e., 3.142 is denoted by Greek letter , read as pi (). (2) For calculation purposes, the value of  is taken as 7 22 or 3.14 approx.  C =  × d  C =  × 2r  C = 2r, where r is the radius of the circle. i.e., Circumference of the Circle = 2 × radius of the circle ×  or Circumference of the Circle = diameter of the circle ×  (3) Circumference of a semi-circle = r 2 2 r    and the perimeter of a semi-circular shape = ( + 2) r units. EXAMPLES  Ex.1 Find the circumference of a circle of diameter 7 cm (Take  = 7 22 ). Sol. We know that the formula for a circumference of a circle =  × d  Circumference = 7 22 × 7 = 22 cm. Ex.2 The radius of a hoop is 60 cm. Find its circumference (Take  = 3.142). Sol. We know that, Circumference = 2 ×  × r = 2 × 3.142 × 60 = 377.04 cm. Ex.3 The radius of a wheel is 35 cm. Find its circumference (Take  = 7 22 ). Sol. Radius = 35 cm, Therefore, circumference = 2 ×  × r = 2 × 7 22 × 35 = 220 cm Ex.4 A circular flower bed has a diameter of 1.5 m. A metal edging is to be placed around it. Find the length of edging needed and the cost of the edging if it is sold by the metre and costs j 60 a metre. (You can only buy a whole number of metres) Sol. First find the circumference of the circle, how many metres you need.  C =  × d = 3.14 × 1.5 = 4.71 m. As the required length is 4.71 m, therefore we have to buy 5 m of edging. So, the cost for buying 5 m = 5 × j 60 = j 300. Ex.5 The diameter of a wheel of a car is 77 cm. How many revolutions will it take to travel 10 km ? Sol. We know that, Circumference of the wheel =  × d = 7 22 × 77 = 242 cm This means that in one revolution, the car will travel a distance = 242 cm In other words, distance travelled by a car in 1 revolution = 242 cm Given 10 km = 10,000 m = 10,000 × 100 cm = 10,00,000 cm Now, 242 cm = 1 revolution  1 cm = 242 1 revolution So, in 10,00,000 cm = 242 10,00,000 revolutions = 4132.23 (approx) revolutions
Ex.6 There is a circular pond and a footpath runs along its boundary. A man walks around it, exactly once, keeping close to the edge. If his step is 66 cm long and he takes exactly 400 steps to go around the pond. What is the diameter of the pond ? Pond Footpath Sol. We know that perimeter of the circle =  d Length of one step of man = 66 cm Length of 400 steps of man = 400 × 66 cm = 26400 cm This means circumference of the pond is 26400 cm   d = 26400 cm Or d = cm 26400  = 7cm 22 26400  = 1200 × 7 cm = 8400 cm = 84 m. Ex.7 The circumference of a bicycle is 198 cm. What is the approximate length of a spoke? Sol. Circumference of bicycle = 198 cm Here, the spoke of bicycle is the radius of the wheel of bicycle. If r is the radius (length of a spoke), and C is the circumference, then C = 2r or 198 = 2r or r = 2 63 2 22 198 7 7 22 2 198 2 198        = 31.5 cm Hence, the length of a spoke is 31.5 cm. Ex.8 A running track has two semi-circular ends of radius 63 metres at the two straight lengths. The perimeter of the track is 1000 metres. Find the length of each straight length (correct to the nearest metres). F E D C A x B 63 Sol. If we draw the sketch, we find that track is like as figure. Then circumference of track = AB + BCD + DE + EFA Let AB = DE = x and BCD (Semi-circular arc) + EFA (Semi-circular arc) makes a complete circle. Therefore, circumference of track = 2x + circumference of a circle of radius 63 or 1000 m = 2x + 2 ×  × 63 or 2x = 1000 – 2 × 7 22 × 63 or 2x = 1000 – 44 × 9 = 1000 – 396 or 2x = 604 m or x = 302 m. Ex.9 A circular table cloth has a circumference of 220 cm. (a) Is the cloth large enough to fit on a round table which is 50 cm in diameter ? (b) If so, what length of the table cloth would hang down on each side ? Table cloth on round table Sol. (a) The diameter of the table is 50 cm. Therefore, circumference of the table =  × d =  × 50 cm ......(i) and circumference of cloth = 220 cm
 2r = 220  r = 2 22 220 7 7 22 2 220     = 35 cm  Circumference of cloth = 220 cm = 2 ×  × 35 or 70  .....(ii) Clearly from (i) and (ii), we have The cloth is large enough to fit on a round table. (b) Radius of the table = 2 50 cm = 25 cm and radius of circular cloth =     2 70 2 Circumfere nce = 35 cm  Hanging length = 35 cm – 25 cm = 10 cm Clearly, radius of cloth is much longer than the radius of table. Therefore, the cloth will hang down 10 cm on each side. Ex.10 Some cotton thread is wound on a reel with a radius of 35 cm. (a) What length of cotton round on one turn of the reel? (b) How many turns of the reel are needed to wind 44 m of cotton on the reel? Reel Sol. (a) Clearly, to calculate the length of cotton fits round on one turn of the reel, we have to calculate the circumference of reel. Now, circumference of reel = 2r, where r is the radius of the reel. = 2 × 7 22 × 35 cm = 220 cm Thus, 220 cm is the required length of cotton round on one turn of the reel. (b) Since, we know 1 m = 100 cm  44 m = 4400 cm  220 cm is the required length to complete one turn. Now, the number of turns to complete 1 cm length = 220 1 turns So, the required number of turns to complete the length 4400 cm = 4400 × 220 1 turns = 20 turns. Ex.11 Find the circumference of each of these shapes in figure (i) and (ii). Sol. (i) Clearly, in the given figure (i), there are two straight lengths of 3.5 m each and two quarter circles. R S P Q A B 3.5m 6.5m Now, to calculate the circumference of the given figure, we first have to calculate the radius of quarter circles. Since AB = 6.5 m, PQ = RS = 3.5 m and AP = QB  AB = AP + PQ + QB or AB = PQ + 2QB [ QB = AP] or 6.5 = 3.5 + 2QB or 2QB = 6.5 – 3.5 = 3 m or QB = 1.5 m Therefore, the perimeter of given figure = 6.5 m + 3.5 m + Circumference of two quarter circular arcs = 10 m + Perimeter of a semi-circular arc = 10 m + 2 Perimeter of a circle = 10 m + 2 21.5m = 10 m + 1.5  m = 10 m + 1.5 × 3.14 m = 14.71 m