ncert solutions for class 9 maths

1. https://www.entrancei.com/ https://www.entrancei.com/ncert-solutions-class-10-maths

2. https://www.entrancei.com/ AREAS RELATED TO CIRCLES S.No . Nomenclatur e Name Figure Perimeter Area r = radius 2r = d = diameter 2r 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 2R + 2r 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

3. 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

4. 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

5. 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

6. https://www.entrancei.com/ https://www.entrancei.com/ncert-solutions-class-10-maths