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## VOLUME AND SURFACE AREA -> IMPORTANT FACTS AND FORMULAE

 I. CUBIOD Let length = l, breadth = b and height = h units. Then, 1. Volume = (l x b x h) cubic units. 2. Surface area = 2 (lb + bh + lh) II. CUBE Let each edge of a cube be of length a. Then, 1. Volume = a³ cubic units. 2. Surface area = 6a² sq. units. 3. Diagonal = √3 a units. III. CYLINDER Let radius of base = r and Height (or length) = h Then, 1. Volume = (∏r²h) cubic units. 2. Curved surface area = (2∏rh) sq. units. 3. Total surface area = (2∏rh + 2∏r² sq. units) = 2∏r (h + r) sq. units. IV. CONE Let radius of base = r and Height = h. Then, 1. Slant height, l = √h² + r ² units. 2. Volume = [1/3 ∏r²h] cubic units. 3. Total surface area = (∏rl + ∏r²) sq.units. V. SPHERE Let the radius of the sphere be r. Then, 1. Volume = [4/3 ∏r3] cubic units. 2. Surface area = (4∏r²) sq. units. VI. HEMISPHERE Let the radius of a hemisphere be r. Then, 1. Volume = [2/3 ∏r3] cubic units. 2. Curved surface area = (3∏r²) sq. units. 3. Total surface area = (3∏r²) sq. units. Remember : 1 litre = 1000 cm³.

## VOLUME AND SURFACE AREA -> SOLVED EXAMPLES

1. The diagonal of a cube is 6√3 cm. Find its volume and surface area. Sol. Let the edge of the cube be a. ∴ √3a = 6 √3 ⇒ a = 6. So, Volume = a³ = (6 * 6 * 6) cm³ = 216 cn³ Surface area = 6a² = (6 * 6 * 6) cm² = 216 cm² Find the volume and surface area of a cuboid 16m long, 14m broad and 7m high. Sol. Volume = (16 * 14 * 7) m³ = 1568 m³ Surface area = [2 (16 * 14 + 14 * 7 + 16 * 7)] cm² = (2 * 434) cm² = 868 cm². If the capacity of a cylindrical tank is 1848 m³ and the diameter of its base is 14m, then find the depth of the tank. Sol.Let the depth of the tank be h metres. then, ∏ * (7)² * h = 1848 ⇔ h = [1848 * 7/22 * 1/7*7] = 12 m. How many iron rods, each of length 7m and diameter 2cm can be made out of 0.88 cubic metre of iron? Sol. Volume of 1 rod = [22/7 * 1/100 * 1/100 * 7] cu. m = 11/5000 cu.m. Volume of iron = 0.88 cu. m. Number of rods = [0.88 * 5000/11] = 400. A cube of edge 15 cm is imersed completely in a rectangular vessel containing water. If the dimensions of the base of vessel are 20 cm * 15 cm, find the rise in waer level. Sol. Increase in volume = Volume of the cube = (15 * 15 * 15) cm³. ∴ Rise in water level = [Volume/Area] = [15 * 15 * 15/20 * 15] cm = 11.25 cm. How many spherical bullets can be made out of a lead cylinder 28 cm high and with base radius 6 cm, each bullet being 1.5 cm ub duaneter? Sol. Volume of cylinder = (∏ * 6 * 6 * 28) cm³ = (36 * 28) ∏ cm³. Bolume of each bullet = [4/3 ∏ * 3/4 * 3/4 * 3/4] cm³ = 9∏/16 cm³. Number of bullets = Volume of cylinder / Volume of each bullet = [(36 * 28)∏ * 16/9∏] = 1792.

## VOLUME AND SURFACE AREA -> EXERCISE

1. Three cubes of iron whose edges are 6cm, 8cm and 10cm respectively are melted and formed into a single cube. The edge of the new cube formed is

• A. 10 cm
• B. 12 cm
• C. 16 cm
• D. 18 cm
 Ans: B. Sol. Volume of the new cube = (63 + 83 + 103) cm3 = 1728cm3. Let the edge of the new cube be a cm. ∴ a3 = 1728 ⇒ a = 12.

2. A metallic sheet is of rectangular shape with dimensions 48 m x 36 m. From each of its corner, a square is cut off so as to make an open box. If the length of the square is 8 m, the volume of the box (in m3) is

• A. 6420
• B. 8960
• C. 5120
• D. 4830
 Ans: C. Sol. Clearly, l = (48 - 16) m = 32 m, b = (36 - 16) m = 20 m, h= 8 m. ∴ Volume of the box = (32 x 20 x 8) m3 = 5120 m3

3. A rectangular box measures internally 1.6 m long, 1 m broad and 60 cm deep. The number of cubical blocks each of edge 20 cm that can be packed inside the box is

• A. 30
• B. 60
• C. 120
• D. 150
Ans: C.
Sol. Number of blocks =
 160x100x60
20x20x20] = 120.