Chapter 12 of NCERT Maths Class 10 delves into the concepts of surface area and volume for various 3D shapes, such as cubes, cuboids, spheres, cylinders, and cones.
Formulas for common 3D shapes:
- Cube:
- Surface area = 6a² (where a is the side length)
- Volume = a³
- Cuboid:
- Surface area = 2(lb + bh + lh) (where l, b, and h are the length, breadth, and height)
- Volume = l * b * h
- Sphere:
- Surface area = 4πr² (where r is the radius)
- Volume = (4/3)πr³
- Cylinder:
- Total surface area = 2πr(r + h) (where r is the radius and h is the height)
- Lateral surface area = 2πrh
- Volume = πr²h
- Cone:
- Total surface area = πr(r + l) (where r is the radius and l is the slant height)
- Lateral surface area = πrl
- Volume = (1/3)πr²h
Applications:
- Real-world problems involving the calculation of surface area and volume
- Understanding the relationship between the dimensions of a shape and its surface area and volume
- Applications in fields like architecture, engineering, and science
In essence, this chapter provides a comprehensive understanding of surface area and volume for various 3D shapes, equipping students with the necessary tools to solve a wide range of geometric problems.
Exercise 12.1
1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Ans :
Step 1: Find the side length of each cube
- Volume of a cube = side³
- 64 cm³ = side³
- side = 4 cm
Step 2: Determine the dimensions of the resulting cuboid
- When two cubes of side length 4 cm are joined end to end, the resulting cuboid has:
- Length = 4 cm + 4 cm = 8 cm
- Breadth = 4 cm
- Height = 4 cm
Step 3: Calculate the surface area of the cuboid
- Surface area of a cuboid = 2(lb + bh + lh)
- Substituting the values: Surface area = 2(84 + 44 + 8*4)
- Surface area = 2(32 + 16 + 32)
- Surface area = 2 * 80
- Surface area = 160 cm²
Therefore, the surface area of the resulting cuboid is 160 square centimeters.
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.
Ans :
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.
Ans :
1. Finding the height of the cone:
- The radius of the hemisphere (and the cone) is 3.5 cm.
- So, the height of the cone is 15.5 cm – 3.5 cm = 12 cm.
2. Finding the slant height of the cone:
- The slant height (l) of the cone can be found using the Pythagorean theorem: l = √(r² + h²)
- Substituting the values: l = √(3.5² + 12²) = √156.25 = 12.5 cm
3. Finding the surface area of the hemisphere:
- The surface area of a hemisphere is (2/3)πr²
- Substituting the value of r: (2/3)π(3.5)² = 77/3 cm²
4. Finding the surface area of the cone:
- The surface area of a cone is πrl
- Substituting the values of r and l: π(3.5)(12.5) = 137.5π/7 cm²
5. Finding the total surface area of the toy:
- Total surface area = Surface area of hemisphere + Surface area of cone
- Total surface area = (77/3)π + (137.5π/7)
- Total surface area = (539 + 408.75)π/21
- Total surface area = 947.75π/21 cm²
- Total surface area ≈ 142.17 cm²
Therefore, the total surface area of the toy is approximately 214.5 cm².
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.
Ans :
5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter d of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Ans :
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.
Ans :
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 m2. (Note that the base of the tent will not be covered with canvas.)
Ans :
Finding the Area of Canvas:
- We need to find the curved surface areas of the cylindrical and conical parts.
- The base of the tent is not covered, so we don’t need to consider the base area of the cylinder.
Curved Surface Area of Cylinder (CSA_cylinder):
- CSA_cylinder = 2πrh
- Substituting the values: CSA_cylinder = 2 * π * 2 * 2.1
- CSA_cylinder = 8.4π m²
Curved Surface Area of Cone (CSA_cone):
- CSA_cone = πrl
- Substituting the values: CSA_cone = π * 2 * 2.8
- CSA_cone = 5.6π m²
Total Surface Area (TSA):
- TSA = CSA_cylinder + CSA_cone
- TSA = 8.4π m² + 5.6π m²
- TSA = 14π m²
Cost of Canvas:
- Cost = TSA * Rate per m²
- Cost = 14π m² * ₹500/m²
- Cost ≈ ₹22000 (using π ≈ 22/7)
Therefore, the area of the canvas used for making the tent is approximately 14π square meters, and the cost of the canvas is approximately ₹22,000.
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.
Ans :
9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in 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.
Ans :
Exercise 12.2
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 π.
Ans :
Volume of the cone:
- The formula for the volume of a cone is (1/3)πr²h, where r is the radius and h is the height.
- In this case, r = 1 cm and h = 1 cm.
- So, the volume of the cone is (1/3)π(1)³ = π/3 cm³.
Volume of the hemisphere:
- Formula : (2/3)πr³.
- In this case, r = 1 cm.
- So, the volume of the hemisphere is (2/3)π(1)³ = 2π/3 cm³.
Total volume of the solid:
- Total volume = Volume of cone + Volume of hemisphere
- Total volume = π/3 + 2π/3
- Total volume = 3π/3
- Total volume = π cm³
Therefore, the total volume of the solid is π cubic centimeters.
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 that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Ans :
Volume of the cylinder:
- Formula : πr²h,
- In this case, r = 3/2 cm (radius of the cylinder) and h = 12 – 2 – 2 = 8 cm (height of the cylinder).
- So, the volume of the cylinder is π(3/2)²(8) = 18π cm³.
Volume of each cone:
- Formula : (1/3)πr²h,
- In this case, r = 3/2 cm (radius of the cylinder) and h = 2 cm (height of each cone).
- So, the volume of each cone is (1/3)π(3/2)²(2) = 3π/2 cm³.
Total volume of the model:
- Total volume = Volume of cylinder + 2 * Volume of cone
- Total volume = 18π + 2 * (3π/2)
- Total volume = 18π + 3π
- Total volume = 21π cm³
Therefore, the volume of air contained in the model Rachel made is 21π cubic centimeters.
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.
Ans :
Step 1: Find the volume of a single gulab jamun
- Radius of the cylinder and hemispheres = 2.8 cm / 2 = 1.4 cm
- Height of the cylindrical part = 5 cm – (1.4 cm + 1.4 cm) = 2.2 cm
- Volume of one gulab jamun = Volume of cylinder + Volume of two hemispheres
- Volume of cylinder = πr²h = π(1.4)²(2.2)
- Volume of one hemisphere = (2/3)πr³ = (2/3)π(1.4)³
- Total volume = π(1.4)²(2.2) + (4/3)π(1.4)³
- Total volume ≈ 25.05 cm³
Step 2: Find the volume of sugar syrup in one gulab jamun
- Volume of syrup = 30% of the total volume
- Volume of syrup ≈ (30/100) * 25.05 cm³
- Volume of syrup ≈ 7.515 cm³
Step 3: Find the total volume of sugar syrup in 45 gulab jamuns
- Total volume of syrup = 45 * 7.515 cm³
- Total volume of syrup ≈ 338.18 cm³
Therefore, approximately 338.18 cubic centimeters of sugar syrup would be found in 45 gulab jamuns.
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.
Ans :
Volume of the cuboid:
- Volume of a cuboid = length * breadth * height
- Volume of cuboid = 15 cm * 10 cm * 3.5 cm
- Volume of cuboid = 525 cm³
Volume of one conical depression:
- Volume of a cone = (1/3)πr²h
- Radius (r) = 0.5 cm
- Height (h) = 1.4 cm
- Volume of one depression = (1/3) * π * (0.5)² * 1.4
- Volume of one depression ≈ 0.3665 cm³
Volume of four conical depressions:
- Volume of four depressions = 4 * 0.3665 cm³
- Volume of four depressions ≈ 1.466 cm³
Volume of wood in the stand:
- Volume of wood = Volume of cuboid – Volume of four depressions
- Volume of wood = 525 cm³ – 1.466 cm³
- Volume of wood ≈ 523.53 cm³
Therefore, the volume of wood in the entire pen stand is approximately 523.53 cubic centimeters.
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.
Ans :
Step 1: Calculate the volume of the cone (vessel)
- Volume of a cone = (1/3)πr²h
- Radius (r) = 5 cm
- Height (h) = 8 cm
- Volume of cone = (1/3)π(5)²(8) = (200/3)π cm³
Step 2: Calculate the volume of water that flows out
- One-fourth of the water flows out.
- Volume of water flowed out = (1/4) * (200/3)π cm³
- Volume of water flowed out = (50/3)π cm³
Step 3: Calculate the volume of a single lead shot
- Volume of a sphere = (4/3)πr³
- Radius of lead shot = 0.5 cm
- Volume of lead shot = (4/3)π(0.5)³ = (π/6) cm³
Step 4:
- Number of lead shots = Volume of water flowed out / Volume of one lead shot
- Number of lead shots = ((50/3)π cm³) / ((π/6) cm³)
- Number of lead shots = 50 * 2
- Number of lead shots = 100
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 8g mass.
Ans :
Understanding the Pole:
- The pole consists of two cylindrical sections.
- The lower cylinder has a height of 220 cm and a diameter of 24 cm.
- The upper cylinder has a height of 60 cm and a diameter of 16 cm (radius of 8 cm).
Finding the Volumes:
- Radius of the lower cylinder: 24 cm / 2 = 12 cm
- Volume of the lower cylinder: πr²h = π * 12² * 220 = 31680π cm³
- Volume of the upper cylinder: πr²h = π * 8² * 60 = 3840π cm³
- Total volume of the pole: 31680π + 3840π = 35520π cm³
Finding the Mass:
- Given that 1 cm³ of iron has approximately 8 g mass.
- Mass of the pole = Volume of the pole * Density of iron
- Mass of the pole ≈ 35520π cm³ * 8 g/cm³
- Mass of the pole ≈ 890525.35 g ≈ 890.53 kg
Therefore, the mass of the pole is approximately 890.53 kilograms.
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.
Ans :
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.
Ans :
