NCERT solutions for class 10

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

2. https://www.entrancei.com/ SURFACE AREAS AND VOLUMES Lateral/ curved surface area S. No. Total surface area Name Figure Volume Nomenclature l = length b = breadth h = height a = edge of cube  +  ( 2 l b b h + )  ( 2 l b h   l b h 1. Cuboid +  h l ) 2 3 2. Cube 6a a 2 4a r = radius h = height 2 Right circular cylinder 2rh + 2 r  2 rh r2 3.  or ( r h  + 2 r h ) r = radius of base h = height  = slant height = r + r = radius of the sphere 2 r  +  Right circular cone r 1 r  r2  h 4. or 3   + r ( r ) 2 2 h 4 3 r  2 2 5. Sphere 4 r  4 r  3 r = radius of the hemisphere 2 Hemi sphere 3 r  2 2 6. 2 r  3 r  3 R = external radius r = internal radius R and r = radii of the base and top respectively h = height  = slant height 4 Spherical shell 3 3 2 2  R −  R + ( r ) 2 2 7. 2 ( r )  +  3 R r 3  + ( R r ),  2 2  + +  +  2 ( R r ) R 2 r 2   + h [ 3 2 R Frustum of a cone 2 2 = where h   + 8. [  R r or + r Rr ] 2 + + ( R r )] + − ( R r ) 2 2 2 3 . a 10.Diagonal of cube = + + 9.Diagonal of cuboid = . l b h + )  + + = SA of open rectangular box = SA of open cylindrical box = Conversion of solid from one shape to another Whenever a solid of given shape is converted (melted or transformed) into a solid of different shape then the volume of the two solids of different shapes remains the same. • 2 h rh  ( l b r lb  2 + 2 r ( 2 h r ) For example: The diameter of a copper sphere is 6 cm. The sphere is melted and drawn into a long wire of uniform circular cross section. If the length of the wire is 36 cm, find its radius. Radius of the sphere = 3 cm. https://www.entrancei.com/ncert-solutions-class-10-maths

3. https://www.entrancei.com/     4 4 3 3 =  =     cm3. Volume of the sphere r cm 3 3 3         3 3 = (36) cm3. • Length (height) of wire = 36 cm. Let the radius of the wire be r cm. Volume of the wire = (r2h) cm3 = (r2x 36) cm3. But, volume of wire = volume of sphere  36r2 = 36  r2 = 1  r = 1 Hence, the radius of the wire is 1 cm. To find the number of objects If x cubic units of a liquid is filled in n cylindrical shaped containers of radius r and height h, then we have x 2   = = or n ( r h ) x n 2  r h If x cubic units of solid material is consumed for making n spherical balls of radius r, then we have    3 4  Volume  4 3 x 3   = = or n r x   n 3 r of bigger object or number of objects required = Volume of smaller object For example: A metallic sphere of radius 10.5 cm is melted and then recast into smaller cones, each of radius 3.5 cm and height 3 cm. How many cones are obtained? 21 = cm. Radius of the sphere 2   4 4 21 21 21 3  =     cm3. Volume of the sphere r      3 3  2 2 2  3087 = cm3.   7   2 = Radius of each cone cm and its height = 3 cm. 2 1   1 7 7 2h =  =     cm3. Volume of each cone r 3     3  3 2 2   49 cm3. =     4 volume of the sphere = Required number of cones volume  2 of each cone   3087 4  = = 126.      49 https://www.entrancei.com/ncert-solutions-class-10-maths

4. https://www.entrancei.com/ • Combination of solids In problems related to total surface area of the combination of solids, such as a circus tent consisting of a cylindrical base surmounted by a conical rod, a toy in the form of a cone mounted on a hemisphere etc, lateral/curved surface area of the each solid is calculated. For example: A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the total surface area of the toy, if the total height of the toy is 30 cm. The shape of the toy is as given: Radius of the hemisphere = 5 cm. Radius of the base of the cylinder = 5 cm. Radius of the base of the cone = 5 cm. Height of the hemisphere = its radius = 5 cm. Let the height of the cone be h cm. Then , (5+13+h) cm = 30 cm h = 12.  height of the cone = 12 cm. cm ) 12 ( 5 =  . 2+ 2 Slant height of the cone, = = 169 cm 13 cm Surface area of the toy = curved surface area of the hemisphere + curved surface area of the cylinder + curved surface area of the cone 2 2 ( rh r +  +  = 2 ) 5 ( 2 [(  +   =    7 . 2  r   sq units 13 +  ) 2    cm2 5 5 13 ]  22 =  cm2 = 770 cm2. 245   Volume of the earth dugout • Height of embankment= Area on which earth is to be spread For example: A well with 10 minsidediameter is dug 14 m deep. Earth taken out of it is spread all a round to a width of 5 m to form an embankment. Find the height of embankment. Here r = 5 m and R = r + width = 5 + 5 = 10 m = Volume of the earth dugout ( ) 2 3 r  22 h m 3 3 =    =  5 5 14 m 1100 m Volume of the earth dugout 7 2 2 Area of the embankment =  R − ( r ) 22 2 2 2 = 10 (  − =  m2 5 ) m 75 7 Volume of the earth dugout  = Height of the embankment Area 1100 of the 7 embankment 1100 =   = = . 4 66 m 22 22 75  75 7 For example: https://www.entrancei.com/ncert-solutions-class-10-maths

5. https://www.entrancei.com/ The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to 1 of the volume of the given cone, at what height above the base. If its volume be 27 the base is the section made? Height of the given cone = 30 cm. Let the radius of its base be R cm.   1 2 =   Volume of the given cone   R 30   3 = (10R2) cm3.   1 r2 =  Then, volume of smaller cone   h   3 cm3. 1 1 2 2 =  =   r h 10 ( R )[given] 30 3 27 2   R 9 h  =     … r 10 (i) Now, OAB ~ OCD. AB OA R 30  =  = … CD OC r h (ii) From (i) and (ii), we get 2    30 h 9 h 30 h 30 h 9 h =  =      10 10   30 30 10 3  h = = 1000 9 3= 3  h 10 ( ) .  =  Height of the section from the base = (30 – 10) cm = 20 cm. height of the smaller cone = 10 cm. h 10 https://www.entrancei.com/ncert-solutions-class-10-maths

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