ch_12_surface_areas_and_volumes - Surface Areas and Volumes

Sep 18, 2026

Study material for CBSE Class 10 Mathematics Chapter 12: Surface Areas and Volumes, compiling key formulas for solids including cubes, cuboids, cylinders, cones, spheres, hemispheres, and frustums. The document provides expressions for total, lateral/curved, and volume calculations, then applies them through 1-mark, 2-mark, and 3-mark problem sets with step-by-step solutions. Exercises cover joining solids to form new cuboids, melting/recasting shapes, computing slant height, inner surface area of combined vessels, volume left after carving a sphere from a cube, and ratios of volumes and surface areas.

LanguageEnglish

CategoryExam

FormatPDF

Pages6

Reading Time15 min

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Ch12. Surface Areas and Volumes

Formulae

1 Marks Questions

2 Marks Questions

3 Marks Questions

Mathematics Kolabari Tutorial CBSE X Comprehensive Study material Ch12. SURFACE AREAS AND VOLUMES Formulae 1) Total Surface area of cube = 6 2) Lateral Surface area of cube = 4 3) Volume of cube = 4) Total Surface area of cuboid = 2(lb+bh+hl) 5) Lateral Surface area of cuboid = 2(l+b)h 6) Volume of cuboid = l×b×h 7) Curved Surface area of Cylinder = 2πrh 8) Total Surface area of Cylinder = + 2π = 2πr r+h 9) Volume of Cylinder = π h 10) Total Surface area of Cone = πr l+r 11) Lateral Surface area of Cone = πrl 12) Volume of Cone = ~~ ~~ π h 13) Surface area of Sphere = 4π 14) Volume of Sphere = ~~ ~~ π 15) Total Surface area of Hemisphere = 3 16) Curved Surface area of Hemisphere = 2π 17) Volume of Hemisphere = ~~ ~~ π 18) Total Surface area of Frustum = + + π +π = π R+r l +π + ), where l = + ,R & r are the radii of circular bases ofthe frustum. 19) Curved Surface area of Frustum = π R+r l 20) Volume of a Frustum = ~~ ~~ πh + +Rr) 1 MARKS QUESTIONS Q1. Two cubes each with 12cm. edge are joined end to end. Find the surface area of the resulting cuboid. Solution: Length of resulting cuboid = 12+12 =24cm. Breadth= 12cm. Height = 12cm S.A. of resulting cuboid = 2(lb+bh+hl) =2(288+144+288) =2 x 720 =1440 cm.2 Q2. Three cubes of metal whose edges are 3cm., 4cm. and 5cm. are melted and formed into a single cube. Find its edge. Solution: Volume of new cube= Volume of 3 cubes= 33+ 43 +53=27+64+125=216 cm.3, then edge of new cube = 2 =6 cm. Q3 . Find the curved surface area of a right circular cylinder of height 13.5 cm. and radius of whose base is 7 cm. Solution: height of cylinder(h)=13.5 cm. Radius of base(r)= 7 cm. C.S.A of cylinder = 2πrh =2 x (22/7) x 7 13.5 = 704 cm. Q4.The radii of end of frustum of a cone 40 cm high are 20 cm and 11 cm. Find its slant height. Solution: l= + = 4 + 2 cm = cm = 41 cm Q5. A vessel is in the form of hollow hemisphere mounted by a hollow cylinder .The diameter of hemisphere is 14 cm and total height of vessel is 13 cm. Find the inner surface area of vessel. Solution: Here the radius of hemisphere = radius of cylinder = r cm =7 cm and height of cylinder, h= 13-7=6 cm Now, inner S.A of vessel = curved surface area of cylindrical part + curved surface area of hemispherical part = (2 + 2 2 ) = 2 + = 2 x ~~ ~~ x 7 x (6 +7) =2 x 22 x 13 =572cm2 Q6. The largest possible sphere is curved out of a wooden solid cube of side 7 cm. Find the volume of wood left. Sol. Diameter of sphere carved out = side of cube = 7 cm Thus, R= 3.5 cm Mathematics Kolabari Tutorial CBSE X Comprehensive Study material => Volume of sphere carved out = (4/3) 3 = 179.66 cm3 Also, Volume of cube = a3=73 =343 cm3 => Volume of wood left = 343-179.66 =163 .34 cm3 Q7. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of wire. Solution: The volume ofthe rod = π(1/2)2 x 8 cm3 = 2 cm3 The length of new wire of the same volume = 18 m =1800 cm If r is the radius of cross-section of wire its volume = r2 x 1800 Then r2 x 1800=2 => r =1/30 => The thickness of wire of cross section is 1/15 cm. = 0.67 mm 2 MARKS QUESTIONS Q8. A solid sphere of radius 12cm. is melted and recast into spherical balls of radius 1 cm. Find the number of balls made? Solution: Vol. of sphere = (4/3) π x 23 ; Vol. of spherical ball = (4/3) π x 3 Let n is the no. of balls then, n x (4/3) π x 3 = (4/3) π x 23 => n = 1728 balls Q9. The material of a cone is converted into the shape of a cylinder of equal radius. If the height of a cylinder is 5cm. , find the height of a cone. Solution: As per given condition,(1/3) πr2h = π r2 H => h = 3H = 3 x 5 =15cm. Q10. A cone, a hemisphere and a cylinder stand on equal bases and have the same heights. Show that their volumes are in the ratio 1:2:3 . Solution: Ratio of volumes = (1/3) π x 2 x h : (2/3) π x r3 : π x r2 x h = (1/3) h : (2/3) r : h = (1/3) h : (2/3) h :

FAQs

What type of question appears in the 1-mark and 2-mark sections?

The 1-mark section includes direct calculation problems (e.g., resulting cuboid surface area after joining cubes, cylinder curved surface area). The 2-mark section extends to conversion/melting scenarios and relationships involving dimensions and derived lengths.

Sep 18, 2026

Uploaded by Aria Callaghan