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10th class is the foundation of a student's academic career and their future will be based on the marks they will get in class 10, so It is really mandatory to get the best study material for Scoring good marks in the most challenging subject Maths. You can see that so many websites are providing good NCERT Solutions for Class 10 Maths and students can download them free of cost very easily from: https://www.entrancei.com/ncert-solutions-class-10
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https://www.entrancei.com/ AREAS RELATED TO CIRCLES S.No . Nomenclatur e Name Figure Perimeter Area r = radius 2r = d = diameter 2r or d 2 Circle 1. r 1 Semicircl e 2. 2 r + r 2 r r = radius 2 A r 1 2 3. r + 2 r r = radius Quadrant r 4 2 O r B R = outer radius r = inner radius 2 2 R − or ( r ) Ring (shaded region) 4. r 2R + 2r R (R + r)(R–r) = angle of the sector r = radius of sector l = length of arc 5. r 2 r 1 Sector of a circle + = + l 2 r 2 r or lr 180 360 2 Area of the minor segment (ACB) = 2 r – 360 º area of AOB = − 360 Area of the major segment (ADB) = area of the circle – area of the minor segment = r – area of the minor segment C 6. A B 2 r 1 2 r sin r = radius = angle of the related sector Segment of a circle 2 180 2 r r r + 2 r sin O D 2 For Example: In a circle of radius 21 cm, an arc subtends an angle of 60º at the centre. Find (i) the length of the arc, https://www.entrancei.com/ncert-solutions-class-10-maths
https://www.entrancei.com/ (ii) the area of the sector, (iii) the area of the minor segment, and (iv) the area of the major segment. Let ACB be the given arc subtending an angle of 60º at the centre. Then, r = 21 cm and = 60º D r 2 O (i) Length of the arc ACB = cm 360 A B 22 60 = = 22 cm 2 21 cm C 7 360 2 r 2 (ii) Area of the sector OACBO = cm 360 22 60 2 = = 231 cm2 21 21 cm 7 360 (iii) Area of the minor segment ACBA = (area of the sector OACB) – (area of the OAB) 1 1 2 2 2 − = − = 231 r sin cm 231 21 21 sin 60 º cm 2 2 1 3 2 2 − = − 231 21 21 cm ( 231 190 . 953 ) cm = 2 2 2 = 40.047 cm (iv) Area of the major segment BDAB = (area of the circle) – (area of the minor segment) 22 2 − = 21 21 40 047 . cm 7 = (1386 – 40.047) cm2 = 1345.953 cm2 Distance covered by a wheel in 1 revolution = Circumference of the wheel. • • distance covered minute 1 in • Number of revolutions in 1 minute = . circumfere nce For example: https://www.entrancei.com/ncert-solutions-class-10-maths
https://www.entrancei.com/ The diameter of the wheels of a bus is 140 cm. How many revolutions per minute must a wheel make in order to move at a speed of 66 km per hour? Distance covered by a wheel in 1 minute 66 1000 100 = cm 110000 cm 60 Circumference of a wheel 22 = = 2 70 cm 440 cm 7 Number of revolutions in 1 min 110000 = = . 250 440 • • Angle subtended by the minute hand of a clock in 60 minutes = 360º. 360º= = • • 6 º Angle subtended by minute hand (of clock) in 1 minute = . 60 • • Angle described by the hour hand of a clock in 24 hours = 360º. For example: The minute hand of a clock is 12 cm long. Find the area of the face of the clock described by the minute hand in 35 minutes. Angle described by the minute hand in 60 minutes = 360º. A Angle described by the minute hand in 35 minutes º O 360 210º = = 35 210 º 60 B = 210º and r = 12 cm Area swept by the minute hand in 35 minutes 2 r 22 210 2 2 2 = = = cm 12 12 cm 264 cm 360 7 360 • Areas of combinations of plane figures For example: In the figure, two circular flowerbeds have been shown on two sides of a square lawn ABCD of side 56 m. If the centre of each circular flowerbed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flowerbeds. A B O 56m D C 2 Area of the square lawn ABCD = (side)2 = 56 × 56 m ... (i) https://www.entrancei.com/ncert-solutions-class-10-maths
https://www.entrancei.com/ Let OA = OB = x metres In AOB, AOB = 90 [Diagonals of square bisect at 90] 2 2 2 + x = So, [By Pythagoras theorem] x 56 2x2 = or, 56 56 x2 = or, ... (ii) 28 56 90 1 2 2 = Now, area of sector OAB = x x 360 4 1 22 = [From (ii)] ... (iii) 28 56 4 7 1 1 2 Also, area of OAB = = OA OB x 2 2 1 [AOB = 90º] ... (iv) = 28 56 2 area of flower bed AB = Area of sector OAB − area of OAB So, 1 22 1 2 − = 28 56 28 56 m 4 7 2 1 22 2 − = [From (iii) and (iv)] 28 56 2 m 4 7 1 8 2 28 56 m = ... (v) 4 7 Similarly, area of the other flower bed 1 8 2 28 56 m = ... (vi) 4 7 1 8 1 8 2 + + Therefore, totals area = 56 56 28 56 28 56 m 4 7 4 7 2 2 2 + + = [From (i), (v) and (vi)] 28 56 2 m 7 7 18 2 2 = = 28 56 m 4032 m 7 https://www.entrancei.com/ncert-solutions-class-10-maths
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