JKBOSE 10th Class Maths Solutions chapter – 13 Surface Areas and Volumes
JKBOSE 10th Class Maths Solutions chapter – 13 Surface Areas and Volumes
J&K 10th Class Maths Solutions chapter – 13 Surface Areas and Volumes
Jammu & Kashmir State Board JKBOSE 10th Class Maths Solutions
INTRODUCTION
Cuboid
Volume of cuboid = LBH
Total surface area = 2[LB + BH + HL]
Lateral surface area = 2[L + B]H
Cube
Side of each side of cube = x
Volume of cube = (side)³
Lateral surface area = 6 (side)²
Total surface area of cube = 4 (side)²
Right circular cylinder
Volume of cylinder = πR2H
Curved surface area of cylinder = 2лRH
Total surface area of cylinder = 2πR [R+H]
Hollow Cylinder
Volume of material = π [R2 – r2] H
Curved surface area = 2лH [R + r]
Area of Base = π [R2 + r2]
Right Circular cone
Area of base = πR2
Curved surface area = лRL
Total surface area = πR2 + πRL
= πR [R+L]
Sphere
Hollow sphere.
Hemi-sphere
Curved surface area of hemi-sphere = 2πr²
Total surface of a hemi-sphere = 3πr².
[Surface Area of a Combination Solids]
In our daily life, we saw various interesting designs, where two or more standard solid figures are combined to form a new solid. In such types of problems, firstly, we divide it into different known solids. So, we find surface area of each solid separately and then adding all these areas, we find area of required combined solid.
TEXT BOOK EXERCISE 13.1
Unless stated otherwise, use π = 22/7
Q. 1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Solution. Let side of cube = x cm
Volume of cube = 64 cm³
[volume of cube = (side)³]
When cubes are joined end to end and cuboid is formed
whose Length = 2x cm = 2(4) = 8 cm
Width = x cm = 4 cm
Height = x cm = 4 cm
Surface area of cuboid = 2[LB + BH + HL]
= 2 (8 × 4 + 4 × 4 + 4 × 8] cm²
= 2 [32 + 16 + 32] cm²
= 2 [80] cm² = 160 cm²
∴ Surface area of cuboid 160 cm² Ans.
Q. 2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution.
Diameter of hemisphere = Diameter of cylinder
= 14 cm
2R = 14 cm
Radius of hemisphere (R) = 7 cm
Total height of vessel = 13 cm
∴ Height of cylinder = (13-7) cm = 6 cm
Inner surface area of vessel
= Inner surface area of cylinder
+ Inner surface area of hemisphere
= 2лRH+ 2лR²
= 2πR [H + R]
= 2 × 22/7 (6 + 7 ) cm²
= 44 × 13 = 572 cm²
Hence, Inner surface area of vessel = 572 cm² Ans.
Q. 3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution. Radius of cone = Radius of hemisphere (R) = 3.5 cm
Total height of toy = 15.5 cm
∴ Height of cone (H) = (15.5 – 3.5) = 12 cm
Slant height cone (l) = 12.5 cm
Total surface area of toy
= Surface area of cone + Surface area of hemisphere
∴ Total surface area of toy = 214.5 cm² Ans.
Q. 4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have ? Find the surface area of the solid.
Solution. Side of cubical box = 7 cm
Diameter of hemisphere = Side of cubical box = 7 cm
Surface area of solid
= (surface area of the cube)
– (area of base of hemisphere)
+ (curved surface area of hemisphere)
= 332.5 cm² Ans.
Q.5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution. Let side of cube = a
∴ Diameter of hemisphere = Side of cube
Surface area of remaining solid
= Total surface area of cuboid
– Area of the top of cube
+ Inner curved Surface area of hemisphere
Q. 6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Solution. Diameter of capsule
= Diameter of hemisphere
= Diameter of cylinder = 5 mm
∴ 2R = 5 mm
= (14 – 5) mm = 9 mm
Surface area of capsule
= Surface area of cylinder + 2 Surface area of hemisphere
Hence, Surface area of capsule = 220 mm² Ans.
Q. 7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of 500 per m² (Note that the base of the tent will not be covered with canvas.)
Solution. Diameter of cone = Diameter of cylinder
or 2R = 4
R = 2 m
Radius of cone = Radius of cylinder
Height of cylinder (H) = 2.1 m
Slant height of cone (L) = 2.8 m
Curved surface area of tent
= Curved surface of cylinder
+ Curved surface of conical part
= 2лRH + лRL
= πR [2H + L]
∴ Curved surface area of tent = 44 m²
Cost of 1m² canvas = ₹ 500
Cost of 44 m² canvas = ₹ 44 × 500
= ₹ 22000
Hence, Total cost of canvas = ₹ 22000 Ans.
Q. 8. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
Solution. Diameter of cylinder (D) = 1.4 cm
= Diameter of cone
∴ Radius of cylinder Radius of cone (R) = 0.7 cm
Height of cylinder (H) = 2.4 cm
Total surface area of remaining solid
= curved surface area of cylinder
+ Area of base of cylinder + Surface area of cone
Hence, Total surface area remaining solid to nearest cm² = 18 cm² Ans.
Q. 9. A wooden article was made by scooping out a hemisphere form each end of a solid cylinder, as shown in Fig. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
Solution. Height of cylinder (H) = 10 cm
Radius of cylinder = Radius of hemisphere (R) = 3.5 cm
Surface area of remaining
= curved surface area of cylinder
+ 2 curved surface area of hemisphere
= 2лRH + 2 (2лR²)
= 2лR [H + 2R]
Hence, total surface area of article = 374 cm² Ans.
TEXT BOOK EXERCISE 13.2
Unless stated otherwise, use π = 22/7
Q. 1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of л.
Solution. Radius of cone = Radius of hemisphere = 1 cm
R = 1 cm
Height of cone (H) = 1 cm
Volume of solid = volume of cone + volume of hemisphere
Hence, Volume of solid = π cm³ Ans.
Q. 2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model the Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution.
Radius of cone = Radius of cylinder (R) = 3/2 CM
∴ R = 1.5 cm
Height of each cone (h) = 2 cm
∴ Height of cylinder = (12-2-2) cm = 8 cm
Volume of air in cylinder
= volume of cylinder + 2 (volume of cone)
Hence, Volume of air in cylinder = 66 cm³ Ans.
Q. 3. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.
Solution. Gulab Jamun is in the shape of cylinder
Diameter of cylinder = Diameter of hemisphere = 2.8 cm
Radius of cylinder = Radius of hemisphere (R)
R = 1.4 cm
Height of cylindrical part = (5-1.4-1.4) cm
= (5-2.8) cm = 2.2 cm.
Volume of one gulab Jamun
= Volume of cylinder + 2 [Volume of hemisphere]
Volume of one gulab Jamun = 25.05 cm³
Now volume of 45 gulab Jamuns = 45 × 25.05 cm³ = 1127.28 cm³
Volume of sugar syrup
= 30% volume of 45 gulab Jamuns
Hence, approximately sugar syrup = 338 cm³ Ans.
Q. 4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.
Solution. Length of cuboid (L) = 15 cm
Width of cuboid (B) = 10 cm
Height of cuboid (H) = 3.5 cm
Radius of conical cavity (r) = 0.5 cm
Height of conical cavity (h) = 1.4 cm
Volume of wood in Pen stand = volume of cuboid-4 [volume of cone]
= 525 – 1.466 cm³ = 523.533 cm³.
Hence, volume of wood in Pen stand = 523.53 cm³ Ans.
Q. 5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution.
Radius of cone (R) = 5 cm
Height of cone (H) = 8 cm
Radius of each spherical lead shot (r) = 0.5 cm
Let number of shot put into the cone = N
According to Question,
N [Volume of one lead shot] = 1/4 Volume of water is cone
Hence, Number of lead shots = 100 Ans.
Q. 6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass. (Use π = 3.14)
Solution. Diameter of lower cylinder = 24 cm
Radius of lower cylinder (R) = 12 cm
Height of lower cylinder (H) = 220 cm
Radius of upper cylinder (r) = 8 cm
Height of upper cylinder (h) = 60 cm
Volume of pole = Volume of lower cylinder
+ volume of upper cylinder
= πR² H + πr²h
= 3.14 × 12 × 12 × 220 + 3.14 × 8 × 8 × 60
= 99475.2 + 12057.6
Volume of pole = 111532.8 cm³
Mass of 1 cm³ = 8 gm
Mass of 111532.8 cm³ = 8 × 111532.8
= 892262.4 gm
= 892262.4/1000 kg
= 892.2624 kg
Hence, Mass of Pole=892.2624 kg Ans.
Q. 7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution. Radius of cone
= Radius of hemisphere
= Radius of cylinder
R = 60 cm
Height of cone (h)=120 cm
Height of cylinder (H) = 180 cm
Volume of cylindrical vessel = лR2H .
= 22/7 × 60 × 60 V 180 cm³
= 2036571.4 cm³
Volume of solid inserted in cylinder
= Volume of hemisphere + Volume of cone
Volume of water flows out = 905142.86 cm³
Hence, Volume of water left in cylinder = 1.131 m³ Ans.
Q. 8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution. Diameter of neck, (cylindrical portion) = 2 cm
∴ Radius of neck (r) = 1 cm
Height of cylindrical portion (H) = 8 cm
Diameter of spherical portion = 8.5 cm
= (321.39 + 25.12) cm³
= 346.51 cm³
Hence, Volume of water in vessel = 346.51 cm³ and
She is wrong. Ans.
TEXT BOOK EXERCISE 13.3
Unless stated otherwise, use π = 22/7
Q. 1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of cylinder of radius 6 cm. Find the height of the cylinder.
Solution.
Radius of sphere (r) = 4.2 cm
Radius of cylinder (R) = 6 cm
Let height of cylinder be H cm
On recasting volume remain same
Volume of sphere = Volume of cylinder
Hence, Height of cylinder (H) = 2.744 cm Ans.
Q. 2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution.
Radius of first sphere (r1) = 6 cm
Radius of second sphere (r2) = 8 cm
Radius of third sphere (r3) = 10 cm
Let radius of new sphere formed be R cm
On recasting volume remain same
Volume of all three spheres = Volume of big sphere
Hence, Radius of sphere = 12 cm Ans.
Q. 3. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
Solution. Diameter of well = 7 m
Radius of well (cylinder) R = 7/2 m
Height of well (H) = 20 m
Length of Platform (L) = 22 m
Width of Platform (B) = 14 m
Let height of Platform be H m
Volume of earth dug out from well = Volume of platform formed
Hence, Height of Platform H = 2.5 cm Ans.
Q. 4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Solution. Depth of well (h) = 14 m
Embankment is in the shape of hollow cylinder whose inner radius is same as radius of well and width of embankment 4 m
Inner radius of embankment = Radius of well
Volume of earth dug out = Volume of embankment (so formed)
πr²h = volume of outer cylinder – volume of inner cylinder
лr²h = πR²H – лr²H
= πH [R² – r²]
Hence, Height of embankment H = 1.125 m Ans.
Q. 5. A container shaped like a right circular cylinder having diameter 12 cm and hight 15 cm is full of ice-cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.
Solution.
Diameter of cylinder (D) = 12 cm
∴ Radius of cylinder (R) = 6 cm
Height of cylinder (H) = 15 cm
Diameter of cone = 6 cm
Radius of cone (r) = 3 cm
Radius of hemisphere (r) = 3 cm
Height of cone (h) = 12 cm
Let us suppose number of cones used to fill the ice-cream = n
Volume of ice cream in container
= n [Volume of ice cream in one cone]
πR²H =n [Volume of cone + Volume of hemisphere]
n = 10
Hence, Number of cones formed = 10 Ans.
Q. 6. How many silver coins 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm ?
Solution. Silver coin is in the form of cylinder
Diameter of silver coin = 1.75 cm
Let number of coins formed on recasting the cuboid on recasting volume remain same
Volume of cuboid = n [volume of one silver coin]
Hence, Number of coins so formed = 400 Ans.
Q.7. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution.
Radius of cylindrical bucket (R) = 18 cm
Height of cylindrical bucket (H) = 32 cm
Height of conical heap (h) = 24 cm
Let ‘r’ cm and ‘l’ cm be the radius and slant height of cone
Volume of sand in bucket = Volume of sand in cone
∴ Radius of cone (r) 36 cm. Ans.
As we know,
(Slant height)2 = (Radius)2 + (Height)2
Q. 8. Water in a canal 6 m wide and 1.5 m deep is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Solution. Width of canal = 6 m
Depth of water in canal = 1.5 m
Velocity at which water is flowing = 10 km/hr
Volume of water discharge in one hour
= (Area of cross section) velocity
= (6 × 1.5 m²) × 10 km
= 6 x 1.5 × 10 × 10 × 1000 m³
Let us suppose area to be irrigate = (x) m²
According to question, 8 cm standing water is required in field
∴ Volume of water discharge by canal in ½ hours = Volume of water in field
450000 m³ = (Area of field) × Height of water
Hence, Area of field = 56.25 hectares Ans.
Q. 9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled ?
Solution. Velocity of water = 3 km/hr
Diameter of pipe = 20 cm
Diameter of tank = 10 m
Radius of tank (R) = 5 m
Depth of tank (H) = 2 m
Let us suppose pipe filled a tank in n minutes
Volume of water tank
= Volume of water through the pipe in n minutes
πR²H = n[Area of cross section x Velocity of water]
πR²H = n[(πr²) × 3 km/h]
⇒ n = 100 minutes
Hence, Time taken to fill the tank = 100 minutes Ans.
TEXT BOOK EXERCISE 13.4
Unless stated otherwise, use π = 22/7
Q. 1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution.
Radius of upper end (R) = 2 cm
Radius of lower end (r) = 1 cm
Height of glass (H) = 14 cm
Glass is in the shape of frustum
Volume of frustum
Hence, Volume of glass = 102.67 cm³ Ans.
Q. 2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Solution. Slant height of frustum = 4 cm
Let radius of upper end and lower ends are R and r
Circumference of upper end = 18 cm
Hence, Curved surface area of frustum = 48 cm² Ans.
Q. 3. A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
Solution.
Radius of lower end of frustum (R) = 10 cm
Radius of upper end of frustum (r) = 4 cm
Slant height of frustum (1) = 15 cm
Curved surface area of frustum = πL [R + r]
Hence, Total material used = 710.28 cm² Ans.
Q. 4. A container opened from the top is made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate, of ₹ 20 per litre. Also find the cost of metal sheet used to make the container, if its costs ₹ 8 per 100 cm². (Take π = 3.14.)
Solution.
Radius of upper end of container (R) = 20 cm
Radius of lower end of container (r) = 8 cm
Height of container (H) = 16 cm
∴ Milk in the container = 10.45 litres
Cost of 1 lt. milk = ₹ 20
∴ Cost of 10.45 litres = ₹ 20 × 10.45
Total cost of milk = ₹ 209
Curved surface area of frustum = πL [R + r]
= 3.14 × 20 [20 + 8] cm²
= 3.14 × 20 × 28 cm²
= 1758.4 cm²
Area of base of container = πr²
= 3.14 × (8)2 cm²
= 3.14 × 64 cm²
= 200.96 cm²
Total metal used to make container
= curved surface area of frustum + area of base
= (1758. 4 + 200.96) cm²
= : 1959.36 cm²
Cost of 100 cm² metal sheet used = * 8
Cost of 1 cm² metal sheet used = ₹ 8
Hence, Total cost of sheet used is ₹ 156.75 and Total cost of milk is ₹ 209 Ans.
Q. 5. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.
Solution. Vertical angle of cone = = 60°
Altitude of cone divide vertical angle
∠EOF = 30°
In ΔODB,
Frustum is made into wire, which is in shape of cylinder having diameter 1/16 cm
∴ Radius of cylindrical wire (r1)
Hence, Length of cylindrical wire (H) = 7964.44 m Ans.
TEXT BOOK EXERCISE 13.5
(OPTIONAL)
Q. 1. A copper wire 3 mm in diameter is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to 8.88 g per cm³.
Solution. Diameter of wire (d) = 3 mm
Diameter of cylinder = 10 cm
Radius of cylinder (R) = 5 cm
Height of cylinder (H) = 12 cm
Circumference of cylinder
= length of wire used in one turn
2πR = length of wire used in one turn
Mass of 1 cm³ = 8.88 g
Mass of 88.89 cm³ = 8.88 × 88.89 = 789.41 gm
Hence, length of wire is 1257.14 cm
and volume of wire is 789.41 gm Ans.
Q. 2. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.)
Solution. Let ΔABC be the right triangle right angled triangle at A whose sides AB and AC measure 3 cm and 4 cm respectively.
The length of the side BC (hypotenuse)
Here, AO (or A’O) is the radius of the common base of the double cone formed revolving the right triangle about BC.
Height of the cone BAA’ is BO and slant height is 3 cm.
Height of the cone CAA’ is CO and slant height is 4 cm.
Now, ΔAOB ∼ ΔCAB (AA similarity)
Q. 3. A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm³ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without the water overflowing, each brick being 22.5 cm × 7.5 cm x 6.5 cm ?
Solution. Volume of bricks
= 22.5 × 7.5 × 6.5 cm³
= 1096.87 cm³
Volume of cistern: = 150 × 120 × 110 cm³
= 1980000
Let number of bricks used=n
Total volume of n bricks
= n (volume of one brick)
= n [1096.87] cm³
Volume of water = 129600 cm³
Volume of water available for bricks
= 1980000 – 129600
= 1850400 cm³
Hence, Number of bricks used = 1792 Ans.
Q. 4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 97280 km², show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.
Solution. Area of valley = 97280 km²
Rainfall in valley = 10 cm
∴ Volume of total rainfall
= 9.728 km³ Ans.
Q. 5. An oil funnel made of tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.
Solution. Diameter of top of funnel = 18 cm
∴ Radius of top of funnel (R) : = 18/2 cm = 9 cm
Diameter of bottom of funnel = 8 cm
Radius of bottom of funnel (r) = 4 cm
Height of cylindrical portion (h) = 10 cm
Height of frustum (H) =(22-10) = 12 cm
Slant height of frustum (l)
Slant height of frustum (l) = 13 cm
Area of metal sheet required = curved surface area of cylindrical base + curved surface of frustum
= 782.56 cm²
Hence, Total area of metal sheet used = 782.56 cm² Ans.
Q. 6. Derive the formula for the curved surface area and total surface area of a frustum of a cone, given to you in Section 13.5, using the symbols as explamed.
Solution. A frustum of a right circular cone has two unequal flat circular bases and a curved surface. Let ACDB be the frustum of the cone obtained by removing the portion VCD. The linesegment OP joining the centres of two bases is I called the height of the frustum. Each of the linesegments AC and BD of frustum ACDB is called its slant height.
Let R and r be the radii the circular end faces (R > r) of the frustum ACDB of the cone (V, AB). We complete the conical part VCD. Let h and l be the vertical height and slant height respectively. Then OP = h and AC = BD = l.
The frustum of the right circular cone can be viewed as the difference of the two right circular VAB and VCD.
Let the height of the cone VAB be h1 and its slant height l1 i.e. VP = h1 and VA = VB = l1.
Now from right ΔDEB, we have
Curved surface area of the frustum of cone
= πRl1 – πr(l1 – l)
[Curved surface area of a cone = π × r × l]
∴ Curved surface area of the frustum of right circular cone
and total surface area of frustum of right circular cone
= Curved surface area + Area of base + Area of top
= πl (R + r) + πR² + πl²
∴ π[R² + r² + l1 (R+ r)] sq. units.
Q. 7. Derive the formula for the volume of the frustum of a cone, given to you in section 13.5, using the symbols as explained.
Solution. A frustum of a right circular cone has two unequal flat circular bases and a curved surface. Let ACDB be the frustum of the cone obtained by removing the portion VCD. The linesegment OP joining the centres of two bases is called the height of the frustum. Each of the linesegments AC and BD of frustum ACDB is called its slant height.
Ler R and r be the radii of the circular end faces (R > r) of the frustum ACDB of the cone (V, AB). We complete the conical part VCD. Let h and l be the vertical height and slant height respectively. Then OP = h and AC = BD = l.
The frustum of the right circular cone can be viewed as the difference of the two right circular VAB and VCD.
Let the height of the cone VAB be h1 and its slant height l1 i.e. VP = h1 and VA = VB = l1.
∴ The height of the cone VCD = VP – OP = h1– h
Since right Δs VOD and VPB are similar
Volume of the frustum ACDB of the cone (V, AB)
= Volume of the cone (V, AB) – Volume of the cone (V, CD)
Hence, volume of the frustum of cone is
are the surface areas of the two circular bases, then
A1 = πR² and A2 = πr²
Now volume of the frustum of cone
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