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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online