1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Explanation:
cube’s volume =
= 64
Side = 4 cm
cuboid, length = 4 + 4 = 8 cm, breadth = 4 cm and height = 4 cm
Resulting cuboid’s surface area =
=
= 2 (32 + 16 + 32)
=
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.
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.
Explanation:
hollow hemisphere’s Diameter = 14 cm
hollow hemisphere’s Radius = = 7 cm
vessel’s Total height = 13 cm
hollow cylinder’s Height = 13 – 7 = 6 cm
vessel’s Inner surface area
= hollow hemisphere’s inner surface area + hollow cylinder’s inner surface area
=
=
= =
3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Explanation:
Cone’s Radius = 3.5 cm
hemisphere’s Radius = 3.5 cm
toy’s total height = 15.5 cm
cone’s Height = 15.5 – 3.5 = 12 cm
cone’s Slant height =
=
= = 12.5 cm
toy’s TSA = hemisphere’s CSA + cone’s CSA
=
=
= =
= =
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.
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.
Explanation:
Hemisphere’s Greatest diameter = cubical block’s side = 7 cm
Therefore solid’s TSA = cubical block’s External surface area+ hemisphere’s CSA
=
=
= =
= = 294 + 38.5 =
5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Explanation:
Therefore hemisphere’s Diameter = hemisphere’s radius =
cube’s edge length =
remaining solid’s Surface area
=
=
=
=
6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
Explanation:
Hemisphere’s Radius = mm
Consider radius = = 2.5 mm
Cylindrical height = height’s total – sphere’s Diameter = = 14 – (2.5 + 2.5) = 9 mm
capsule’s Surface area = cylinder’s CSA + hemisphere’s Surface area
=
=
=
= =
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 Rs. 500 per. (Note that the base of the tent will not be covered with canvas.)
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 Rs. 500 per. (Note that the base of the tent will not be covered with canvas.)
Explanation:
cylindrical part’s Diamter = 4 cm
cylindrical part’s Radius = 2 cm
tent’s TSA = cylindrical part’s CSA + conical cap’s CSA
=
=
=
=
=
Cost of the tent's canvas at the current rate of Rs. 500 per
= = Rs. 22000.
8. From a solid cylinder whose height is 2.4 cm and diameter is 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.
8. From a solid cylinder whose height is 2.4 cm and diameter is 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.
Explanation:
solid cylinder’s Diameter = 1.4 cm
Solid cylinder’s Radius = 0.7 cm
The base of the conical cavity’s Radius = 0.7 cm
solid cylinder’s Height = 2.4 cm
conical cavity’s Height = 2.4 cm
conical cavity’s Slant height =
=
= = 2.5 cm
remaining solid’s TSA
=
=
=
= = 17.6
=
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder as shown in the figure. 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.
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder as shown in the figure. 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.
Explanation:
Article’s TSA =
=
=
=
=
=
10. 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 being equal to its radius. Find the volume of the solid in terms of
10. 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 being equal to its radius. Find the volume of the solid in terms of
Explanation:
Hemisphere’s Radius = 1 cm
Volume of hemisphere=
=
= cm3
cone’s base radius = 1 cm
Cone’s Height = 1 cm
Cone’s Volume =
=
= cm3
solid’s Volume = hemisphere’s volume + Cone’s Volume
= + =
11. 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 aluminum 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 that Rachel made. (Assume the outer and inner dimensions of the model are nearly the same.)
11. 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 aluminum 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 that Rachel made. (Assume the outer and inner dimensions of the model are nearly the same.)
Explanation:
upper conical portion, base’s Radius = 1.5 cm
Cone’s Height = 2 cm
Cone’s Volume=
=
=
lower conical portion, Volume =
central cylindrical portion
base’s Radius = 1.5 cm
Height = 12 – (2 + 2) = 8 cm
Volume = = =
Model’s Volume= + +
=
= =
12. 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 a length of 5 cm and a diameter of 2.8 cm (see figure).
12. 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 a length of 5 cm and a diameter of 2.8 cm (see figure).
Explanation:
Gulab Jamun's Volume =
=
=
=
= =
45 Gulab Jamuns’s Volume
=
=
=
Syrup’s Volume =
= = (approx.)
13. 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 id 1.4 cm. Find the volume of wood in the entire stand (see figure).
13. 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 id 1.4 cm. Find the volume of wood in the entire stand (see figure).
Explanation:
cuboid’s Volume =
=
=
conical depression’s Volume =
=
=
Four conical depressions’ Volume = =
Entire stand having Wood’s Volume = 525 – 1.47 =
14. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of the 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.
14. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of the 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.
Explanation:
cone’s top Radius = 5 cm and Cone’s height = 8 cm
Cone’s Volume =
=
= cm3
spherical lead shot’s Radius (R) = 0.5 cm
spherical lead shot’s Volume =
=
=
water’s Volume that emanates out = cone’s Volume
= =
Consider the number of lead shots dropped in the vessel be
= 100.
15. 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 iron has approximately 8 g mass.
15. 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 iron has approximately 8 g mass.
Explanation:
lower cylinder’s Base radius = = 12 cm
Lower Cylinder’s Height = 220 cm
Lower Cylinder’s Volume =
=
=
Upper cylinder’s Base Radius (R) = 8 cm
Upper Cylinder's Height (H) = 60 cm
Upper Cylinder‘sVolume =
=
=
Solid Iron pole’s Volume
=lower cylinder’s Volume +upper cylinder’s Volume
= + =
= = 111532.8
16. 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.
16. 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.
Explanation:
Right circular cone, base’s Radius = 60 cm
Cone’s Height = 120 cm
Cone’s Volume =
=
=
Hemisphere, base’s Radius = 60 cm
Hemisphere’s Volume =
=
=
Right circular cylinder, Base’s Radius = 60 cm
Cylinder’s Height = 180 cm
Cylinder’s Volume =
=
=
The water’s Volume remains in the cylinder
= right circular cylinder’s Volume – (right circular cone’s Volume + hemisphere’s Volume)
= – ( + )
=
=
= = (approx.)
17. A spherical glass vessel has a cylindrical neck 8 cm long, and 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. Check whether she is correct, taking the above as the inside measurements and = 3.14.
17. A spherical glass vessel has a cylindrical neck 8 cm long, and 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. Check whether she is correct, taking the above as the inside measurements and = 3.14.
Explanation:
It holds this much water=
=
=
= 321.39 + 25.12
=
Thus, she is false. The true volume is.
18. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
18. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Explanation:
sphere’s Radius = 4.2 cm
Sphere’s Volume = =
Cylinder’s Radius (R) = 6 cm
Consider the Cylinder’s Height be H cm.
Cylinder’s Volume = =
By question, sphere’s Volume = cylinder’s Volume
=
H =
H = 2.74 cm.
19. 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.
19. 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.
Explanation:
Consider the resulting sphere’s Volume be cm.
By question,
= 216 + 512 + 1000
= 1728
= 12 cm
20. A 20 m deep well with a diameter of 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.
20. A 20 m deep well with a diameter of 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.
Explanation:
Well’s Diameter = 7 m
well’s Radius = m
Earth dug’s Depth = 20 m
platform’s Length = 22 m, platform’s Breadth = 14 m
Consider the platform’s Height to be m
By question,
Earth dug’s Volume = Platform’s Volume
= 2.5 m
21. 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.
21. 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.
Explanation:
well’s Diamter = 3 m
well’s Radius = m and earth dug’s Depth = 14 m
Embankment’s Width = 4 m
well ‘s Radius including embankment = m
Consider the embankment’s Height be m
By question,
embankment’s Volume = earth dug’s Volume
=
=
= 1.125 m
22. A container shaped like a right circular cylinder having a diameter of 12 cm and a height of 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.
22. A container shaped like a right circular cylinder having a diameter of 12 cm and a height of 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.
Explanation:
The right circular cylinder’s Diameter = 12 cm
Cylinder’s Radius = = 6 cm and Cylinder’s height = 15 cm
Cone’s Diameter = 6 cm
cone’s Radius = = 3 cm and cone’s height = 12 cm
Consider cones are ice cream-filled.
cones’ Volume = Right circular cylinder’s Volume
=
= 15
23. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions?
23. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions?
Explanation:
silver coin’s Diameter = 1.75 cm
silver coin’s Radius = cm and silver coin’s Thickness = 2 mm = cm
Cuboid’s Length = 5.5 cm, Cuboid’s Breadth = 10 cm and Cuboid’s Height = 3.5 cm
Consider coins will be melted.
coins’s Volume = cuboid’s Volume
=
= 400
24. A cylindrical bucket, 32 cm high and with a radius of base of 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.
24. A cylindrical bucket, 32 cm high and with a radius of base of 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.
Explanation:
Cylindrical bucket having the base Radius = 18 cm and a Cylindrical bucket height = 32 cm
Cylindrical bucket’s Volume = =
=
conical heap’s Height = 24 cm
Consider the radius be cm.
conical heap’s Volume =
= =
bucket’s Volume = conical heap’s Volume
=
= 1296
= 36 cm
Consider Slant height =
= =
= = cm
25. Water in a canal 6 m wide and 1.5 m deep is flowing at a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
25. Water in a canal 6 m wide and 1.5 m deep is flowing at a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Explanation:
Canal’s Width = 6 m and Canal’s Depth
= 1.5 m = m
The rate of water flow = 10 km/h
= = 10000 m/h
= m/min = m/min
the rate of water flow in 30 minutes
= m/min
Water’s Volume that flows in 30 minutes
= =
The area that will be irrigated =
=
= hectares = 56.25 hectares.
26. 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?
26. 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?
Explanation:
Cylindrical tank’s Diameter = 10 m
Cylindrical tank’s Radius = 5 m and the Cylindrical tank’s Depth = 2 m
Cylindrical tank’s Volume = = =
the water's flow rate = 3 km/h = 3000 m/h = m/min = 50 m/min
Pipe’s Internal diameter = 20 cm, Pipe’s radius = 10 cm = 0.1 m
water’s Volume that flows per minute =
= =
Needed time = = 100 minutes.
27. 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.
27. 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.
Explanation:
= 2 m,
= 1 m and = 14 m
Glass’s Capacity =
=
=
=
28. 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.
28. 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.
Explanation:
Consider cm and cm be the end’s Radius of the frustum of the cone.
= 4 cm
= 18 cm
= 9 cm
= 6 cm
= 3 cm
frustum’s CSA =
=
=
29. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, the radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
29. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see figure). If its radius on the open side is 10 cm, the radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
Explanation:
= 10 cm,
= 4 cm and = 15 cm
Surface area =
=
= =
30. A container, opened from the top and 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 total cost of milk which can completely fill the container at the rate of Rs. 20 per liter. Also find the cost of the metal sheet used to make the container, if it costs Rs. 8 per 100 cm2.
30. A container, opened from the top and 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 total cost of milk which can completely fill the container at the rate of Rs. 20 per liter. Also find the cost of the metal sheet used to make the container, if it costs Rs. 8 per 100 cm2.
Explanation:
= 20 cm,
= 8 cm and = 16 cm
Container’s Volume =
=
=
=
= = 10.44992 liters
milk’s Price =
= Rs. 208.8894 = Rs. 209
surface area = +
= +
= +
=
= 1158.4 + 200.96
=
Metal sheet’s Area utilized =
metal sheet’s Price =
= 156.7488 = Rs. 156.75.
31. A metallic right circular cone 20 cm high and whose vertical angle 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 cm, find the length of the wire.
31. A metallic right circular cone 20 cm high and whose vertical angle 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 cm, find the length of the wire.
Explanation:
cm
cm
= 10 cm
Cone’s Volume =
=
=
= =
Wire’s Diameter = cm
Wire’s Radius = cm
Consider the wire’s Length be cm.
Wire’s Volume = = =
=
796444.44 cm = 7964.4 m
32. 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 be 8.88 g per.
32. 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 be 8.88 g per.
Explanation:
This many rounds must be covered 12 cm, i.e. 120 mm = = 40
Diameter = 10 cm, Radius cm
the wire's length when playing a round
= cm
length of the wire after 40 rounds have been played
= cm
Copper wire’s Radius = mm
= cm
wire’s Volume =
=
wire’s Mass =
= 787.98 g
33. 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 the value as found appropriate)
33. 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 the value as found appropriate)
Explanation:
Hypotenuse = = 5 cm
ADB CAB [By AA similarity]
AD = cm
DB = cm
CD = BC – DB = cm
Double cone’s Volume
=
= =
Double cone’s Surface Area
=
= =
=
34. A cistern, internally measuring has 129600 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 overflowing the water, each brick being?
34. A cistern, internally measuring has 129600 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 overflowing the water, each brick being?
Explanation:
cistern’s Volume = =
water’s Volume =
cistern’s Volume to be filled
= 1980000 – 129600 =
Brick’s Volume =
=
Consider bricks be required
bricks absorbed water =
= 1792 (approx.)
35. 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, 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.
35. 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, 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.
Explanation:
Rainfall’s Volume =
=
Three rivers Volume =
=
As a result, the two are not roughly equal.
36. An oil funnel made of a tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, the 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 (see figure).
36. An oil funnel made of a tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, the 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 (see figure).
Explanation:
Frustum of the cone’s Slant height
= = 13 cm
The Tin sheet’s Area required
= Cylinder’s CSA + frustum’s CSA
=
=
= = =
37. Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
37. Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Explanation:
The difference between the two cones, OAB and OCD, is the frustum (in the figure).
frustum’s height = frustum’s slant height = and frustum’s radii of the bases = and
OP = OA = OB =
Cone’s Height =
OQD OPB [By AA similarity]
……….(i)
Cone’s Height OCD =
= = ……….(ii)
frustum’s Volume
= Cone’s Volume OAB – Cone’s Volume OCD
=
=
[Considering eq. (i) & (ii)]
=
=
38. Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
38. Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Explanation:
When two circular bases' surface areas are, then
A1 = and A2 =
Frustum’s Volume
=
=
DEB,
OQD OPB [By AA similarity]
……….(iii)
= ……….(iv)
frustum of the cone’s CSA =
=
[Considering eq. (i) and (ii)]
= = ,
where
Frustum of the cone’s TSA
=
Also Read: Surface Area and Volume Class 10 Extra Questions
CHAPTER-13, SURFACE AREA AND VOLUMES