25 CHAP

# 25 CHAP - 25.VOLUME AND SURFACE AREA IMPORTANT FORMULAE I....

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25.VOLUME AND SURFACE AREA IMPORTANT FORMULAE I. CUBOID Let length = 1, breadth = b and height = h units. Then, 1. Volume = (1 x b x h) cubic units. 2. Surface area= 2(lb + bh + lh) sq.units. 3. Diagonal .= l 2 +b 2 +h 2 units II. CUBE Let each edge of a cube be of length a. Then, 1. Volume = a 3 cubic units. 2. Surface area = 6a 2 sq. units. 3. Diagonal = 3 a units. III. CYLINDER Let radius of base = r and Height (or length) = h. Then, 1. Volume = ( Π r 2 h) cubic units. 2. Curved surface area = (2 Π rh). units. 3. Total surface area =2 Π r (h+r) sq. units IV. CONE Let radius of base = r and Height = h. Then, 1. Slant height, l = h 2 +r 2 2. Volume = (1/3) Π r 2 h cubic units. 3. Curved surface area = ( Π rl) sq. units. 4. Total surface area = ( Π rl + Π r 2 ) sq. units. V. SPHERE Let the radius of the sphere be r. Then, 1. Volume = (4/3) Π r3 cubic units. 2. Surface area = (4 Π r 2 ) sq. units. VI. HEMISPHERE Let the radius of a hemisphere be r. Then, 1. Volume = (2/3) Π r 3 cubic units. 2. Curved surface area = (2 Π r 2 ) sq. units. 3. Total surface area = (3 Π r 2 ) units. Remember : 1 litre = 1000 cm3. SOLVED EXAMPLES .

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Ex. 1. Find the volume and surface area of a cuboid 16 m long, 14 m broad and 7 m high. Sol . Volume = (16 x 14 x 7) m3 = 1568 m3. 1 Surface area = [2 (16 x 14 + 14 x 7 + 16 x 7)] cm 2 = (2 x 434) cm 2 = 868 cm 2 . Ex. 2. Find the length of the longest pole that can be placed in a room 12 m long 8m broad and 9m high. i Sol . Length of longest pole = Length of the diagonal of the room = (12 2 +8 2 +9 2 = . (289)= 17 m. Ex. 3. Tbe volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 12.8 cu. metres. Find the breadth of the wall. Sol. Let the breadth of the wall be x metres. Then, Height = 5x metres and Length = 40x metres. x * 5x * 40x = 12.8 x 3 =12.8/200 = 128/2000 = 64/1000 So, x = (4/10) m =((4/10)*100)cm = 40 cm Ex. 4. Find the number of bricks, each measuring 24 cm x 12 cm x 8 cm, required to construct a wall 24 m long, 8m high and 60 cm thick, if 10% of the wall is filled with mortar? Sol . Volume of the wall = (2400 x 800 x 60) cu. cm. Volume of bricks = 90% of the volume of the wall =((90/100)*2400 *800 * 60)cu.cm. Volume of 1 brick = (24 x 12 x 8) cu. cm. Number of bricks=(90/100)*(2400*800*60)/(24*12*8)=45000. Ex. 5. Water flows into a tank 200 m x 160 m througb a rectangular pipe of 1.5m x 1.25 m @ 20 kmph . In what time (in minutes) will the water rise by 2 metres? Sol . Volume required in the tank = (200 x 150 x 2) m 3 = 60000 m 3 . . . Length of water column flown in1 min =(20*1000)/60 m =1000/3 m Volume flown per minute = 1.5 * 1.25 * (1000/3) m 3 = 625 m 3 .
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## This note was uploaded on 08/13/2011 for the course FSD 011 taught by Professor Vinh during the Spring '11 term at Beacon FL.

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25 CHAP - 25.VOLUME AND SURFACE AREA IMPORTANT FORMULAE I....

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