Surface Areas and Volumes - Solutions

Exercise 12.1
Exercise 12.2
Key Formulas

Exercise 12.1 - Surface Areas

1. 2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Solution:

Volume of each cube = 64 cm³

Side of cube = ∛64 = 4 cm

When two cubes are joined end to end, the resulting cuboid has:

Length = 4 + 4 = 8 cm

Breadth = 4 cm

Height = 4 cm

Surface area of cuboid = 2(lb + bh + hl)

= 2(8×4 + 4×4 + 4×8)

= 2(32 + 16 + 32) = 2×80 = 160 cm²

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 = 14 cm

Radius (r) = 7 cm

Total height = 13 cm

Height of cylinder (h) = 13 - 7 = 6 cm

Inner surface area = CSA of cylinder + CSA of hemisphere

= 2πrh + 2πr² = 2πr(h + r)

= 2 × (22/7) × 7 × (6 + 7)

= 44 × 13 = 572 cm²

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 (r) = 3.5 cm

Total height = 15.5 cm

Height of cone (h) = 15.5 - 3.5 = 12 cm

Slant height of cone (l) = √(r² + h²) = √(3.5² + 12²) = √(12.25 + 144) = √156.25 = 12.5 cm

Total surface area = CSA of cone + CSA of hemisphere

= πrl + 2πr² = πr(l + 2r)

= (22/7) × 3.5 × (12.5 + 7)

= 11 × 19.5 = 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.

Solution:

Side of cube = 7 cm

Greatest diameter of hemisphere = side of cube = 7 cm

Radius (r) = 3.5 cm

Surface area of solid = Surface area of cube - area of base of hemisphere + CSA of hemisphere

= 6a² - πr² + 2πr² = 6a² + πr²

= 6×7² + (22/7)×3.5²

= 294 + 38.5 = 332.5 cm²

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 edge of cube = a

Diameter of hemisphere = a

Radius (r) = a/2

Surface area of remaining solid = Surface area of cube - area of base of hemisphere + CSA of hemisphere

= 6a² - πr² + 2πr² = 6a² + πr²

= 6a² + π(a/2)² = 6a² + (πa²)/4

= a²(6 + π/4)

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:

Total length = 14 mm

Diameter = 5 mm, Radius (r) = 2.5 mm

Length of cylindrical part = 14 - 2.5 - 2.5 = 9 mm

Surface area = CSA of cylinder + 2 × CSA of hemisphere

= 2πrh + 2 × 2πr² = 2πr(h + 2r)

= 2 × (22/7) × 2.5 × (9 + 5)

= (110/7) × 14 = 110 × 2 = 220 mm²

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.

Solution:

Diameter of cylinder = 4 m, Radius (r) = 2 m

Height of cylinder (h) = 2.1 m

Slant height of cone (l) = 2.8 m

Canvas area = CSA of cylinder + CSA of cone

= 2πrh + πrl = πr(2h + l)

= (22/7) × 2 × (4.2 + 2.8)

= (44/7) × 7 = 44 m²

Cost of canvas = 44 × 500 = ₹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.

Solution:

Height (h) = 2.4 cm

Diameter = 1.4 cm, Radius (r) = 0.7 cm

Slant height of cone (l) = √(r² + h²) = √(0.49 + 5.76) = √6.25 = 2.5 cm

Total surface area = CSA of cylinder + CSA of cone + Area of base

= 2πrh + πrl + πr² = πr(2h + l + r)

= (22/7) × 0.7 × (4.8 + 2.5 + 0.7)

= 2.2 × 8 = 17.6 cm² ≈ 18 cm² (to nearest cm²)

9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. 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 (r) = 3.5 cm

Total surface area = CSA of cylinder + 2 × CSA of hemisphere

= 2πrh + 2 × 2πr² = 2πr(h + 2r)

= 2 × (22/7) × 3.5 × (10 + 7)

= 22 × 17 = 374 cm²

Exercise 12.2 - Volumes

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.

Solution:

Radius (r) = 1 cm

Height of cone (h) = 1 cm

Volume of solid = Volume of cone + Volume of hemisphere

= (1/3)πr²h + (2/3)πr³

= (1/3)π(1)²(1) + (2/3)π(1)³

= (1/3)π + (2/3)π = π cm³

2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends. 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.

Solution:

Diameter = 3 cm, Radius (r) = 1.5 cm

Total length = 12 cm

Height of each cone (h) = 2 cm

Height of cylinder = 12 - 2 - 2 = 8 cm

Volume of model = Volume of cylinder + 2 × Volume of cone

= πr²H + 2 × (1/3)πr²h

= π(1.5)²[8 + (2/3)×2]

= 2.25π[8 + 4/3] = 2.25π × (28/3) = 21π cm³

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:

Diameter = 2.8 cm, Radius (r) = 1.4 cm

Total length = 5 cm

Length of cylinder = 5 - 1.4 - 1.4 = 2.2 cm

Volume of one gulab jamun = Volume of cylinder + 2 × Volume of hemisphere

= πr²h + 2 × (2/3)πr³ = πr²[h + (4/3)r]

= (22/7) × (1.4)² × [2.2 + (4/3)×1.4]

= (22/7) × 1.96 × [2.2 + 1.867]

= 6.16 × 4.067 ≈ 25.05 cm³

Volume of syrup in one gulab jamun = 30% of 25.05 = 7.515 cm³

Volume of syrup in 45 gulab jamuns = 45 × 7.515 ≈ 338.175 cm³

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:

Volume of cuboid = l × b × h = 15 × 10 × 3.5 = 525 cm³

Volume of one conical depression = (1/3)πr²h = (1/3) × (22/7) × (0.5)² × 1.4

= (1/3) × (22/7) × 0.25 × 1.4 = (1/3) × 1.1 = 0.367 cm³

Volume of 4 depressions = 4 × 0.367 = 1.467 cm³

Volume of wood = Volume of cuboid - Volume of 4 depressions

= 525 - 1.467 = 523.533 cm³

5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top 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:

Volume of cone = (1/3)πr²h = (1/3) × (22/7) × 5² × 8

= (1/3) × (22/7) × 25 × 8 = (4400/21) cm³

Volume of water that flows out = (1/4) × (4400/21) = 1100/21 cm³

Volume of one lead shot = (4/3)πr³ = (4/3) × (22/7) × (0.5)³

= (4/3) × (22/7) × 0.125 = (11/21) cm³

Number of lead shots = (Volume of water that flowed out) / (Volume of one lead shot)

= (1100/21) ÷ (11/21) = 1100/11 = 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.

Solution:

Volume of larger cylinder = πR²H = 3.14 × (12)² × 220

= 3.14 × 144 × 220 = 99475.2 cm³

Volume of smaller cylinder = πr²h = 3.14 × (8)² × 60

= 3.14 × 64 × 60 = 12057.6 cm³

Total volume = 99475.2 + 12057.6 = 111532.8 cm³

Mass = Volume × Density = 111532.8 × 8 = 892262.4 g ≈ 892.26 kg

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:

Volume of cylinder = πr²h = π × (60)² × 180 = 648000π cm³

Volume of solid = Volume of cone + Volume of hemisphere

= (1/3)πr²h + (2/3)πr³ = (1/3)π(60)²(120) + (2/3)π(60)³

= (1/3)π × 3600 × 120 + (2/3)π × 216000

= 144000π + 144000π = 288000π cm³

Volume of water left = Volume of cylinder - Volume of solid

= 648000π - 288000π = 360000π cm³

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.

Solution:

Radius of spherical part (R) = 8.5/2 = 4.25 cm

Volume of spherical part = (4/3)πR³ = (4/3) × 3.14 × (4.25)³

= (4/3) × 3.14 × 76.765625 ≈ 321.39 cm³

Radius of cylindrical neck (r) = 2/2 = 1 cm

Volume of cylindrical neck = πr²h = 3.14 × (1)² × 8 = 25.12 cm³

Total volume = 321.39 + 25.12 = 346.51 cm³

The child's measurement of 345 cm³ is approximately correct.

Key Formulas

Surface Areas

Cuboid: TSA = 2(lb + bh + hl), CSA = 2h(l + b)

Cube: TSA = 6a², CSA = 4a²

Cylinder: TSA = 2πr(h + r), CSA = 2πrh

Cone: TSA = πr(l + r), CSA = πrl, where l = √(r² + h²)

Sphere: TSA = 4πr²

Hemisphere: TSA = 3πr², CSA = 2πr²

Volumes

Cuboid: V = l × b × h

Cube: V = a³

Cylinder: V = πr²h

Cone: V = (1/3)πr²h

Sphere: V = (4/3)πr³

Hemisphere: V = (2/3)πr³

Important Notes

1. When combining solids, the total surface area is not simply the sum of individual surface areas.

2. For volume calculations of combined solids, we can add the volumes of individual solids.

3. When a solid is surmounted by another solid, the common surface area is not included in the total surface area.

4. For hollow objects, we consider only the surfaces that are exposed.