12. Surface Areas and Volumes

Introduction

Chapter 12: Surface Areas and Volumes

Page 1 of 5: Introduction to Surface Area of Combination of Solids

Welcome to the fascinating world of three-dimensional shapes! In previous classes, you've mastered the basics of solids like cubes, cylinders, cones, and spheres. You know how to calculate the space they occupy (volume) and the total area of their surfaces (surface area).

But look around you. The world isn't made of perfect, isolated shapes. Think of an ice cream cone, a sharpened pencil, or a water tanker. These are combination solids—objects created by joining two or more basic shapes. To find the surface area of such an object, we can't simply add up the total surface areas of the individual parts. Why? Because when we join them, some surfaces get hidden! This lesson is all about mastering the art of "seeing" the new, exposed surface and calculating its area accurately.

{{FORMULA: expr=TSA_new = Σ(Exposed Surface Areas) | symbols=TSA:Total Surface Area, Σ:Summation}}

Definitions & Formulas

Before we combine solids, let's refresh our memory of the surface area formulas for the basic building blocks. These are your essential tools.

Note: For a cone, the slant height l is crucial and can be found using the Pythagorean theorem: l = √(r² + h²).

The Logic of Combining Solids

Finding the surface area of a combined solid is a game of visualization. It’s not about memorizing a new formula, but about applying a logical process. When we join two solids, the faces that meet are no longer part of the surface.

{{VISUAL: diagram: On the left, a separate cylinder and cone with all surfaces (top, bottom, curved) shown. On the right, the cone is placed on top of the cylinder, and the circular base where they join is highlighted and marked as 'hidden surface, not included in TSA'.}}

Here is the step-by-step thinking process:

1. Deconstruct the Object: Identify the basic solid shapes that make up the new object. For example, a toy top is a cone combined with a hemisphere.

2. Visualize the Join: Imagine putting the pieces together. The surfaces that press against each other disappear from the final object's "skin." For the toy top, the flat circular base of the cone and the flat circular base of the hemisphere are joined and hidden.

3. Identify Exposed Surfaces: Determine which surfaces remain visible on the outside. For the toy top, you can only see the curved surface of the cone and the curved surface of the hemisphere.

4. Formulate the Equation: Write an expression for the total surface area of the new solid by adding the areas of only the exposed surfaces.

   • TSA of Toy Top = (CSA of Cone) + (CSA of Hemisphere)

5. Handle Surface-Mounted Objects: If one object is placed on the face of another (like a hemisphere on a cube), the logic is slightly different. The area of the cube's face covered by the hemisphere is lost.

   • TSA of new block = (TSA of Cube) – (Base Area of Hemisphere) + (CSA of Hemisphere)

{{KEY: type=concept | title=The Golden Rule of Combined Surfaces | text=The Total Surface Area of a new solid formed by combining basic solids is the SUM of the Curved Surface Areas of the individual parts. Never simply add their Total Surface Areas.}}

Solved Examples

Let's apply this logic to some problems, starting from easy and moving to more challenging ones.

Example 1: The Simple Toy Top (Easy)

Given: A toy is shaped like a cone surmounted by a hemisphere. The radius of the hemisphere is 3.5 cm, and the total height of the toy is 15.5 cm. (Take π = 22/7)

To Find: The total surface area of the toy.

Solution:

1. The toy consists of a hemisphere and a cone. The radius r for both is the same.

   • r = 3.5 cm
   • The total surface area will be the sum of their curved surface areas.

2. Calculate the height of the conical part. The height of the hemisphere is equal to its radius.

   • Height of cone h = Total height – Height of hemisphere
   • h = 15.5 cm – 3.5 cm = 12 cm

3. Calculate the slant height l of the cone using l = √(r² + h²).

   • l = √(3.5² + 12²) = √(12.25 + 144) = √156.25

   l = 12.5 cm
   

4. Calculate the Curved Surface Area (CSA) of the cone.

   • CSA_cone = πrl

   CSA_cone = (22/7) × 3.5 × 12.5 = 137.5 cm²
   

5. Calculate the Curved Surface Area (CSA) of the hemisphere.

   • CSA_hemisphere = 2πr²

   CSA_hemisphere = 2 × (22/7) × 3.5 × 3.5 = 77 cm²
   

6. Add the two CSAs to get the total surface area of the toy.

   • TSA_toy = CSA_cone + CSA_hemisphere

   TSA_toy = 137.5 + 77 = 214.5 cm²
   

Final Answer: The total surface area of the toy is 214.5 cm².

Example 2: The Medicine Capsule (Medium)

Given: 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 is 5 mm.

To Find: The surface area of the capsule.

{{VISUAL: diagram: A labeled medicine capsule. The central part is marked as a cylinder with height 'h'. The two ends are marked as hemispheres with radius 'r'. The total length is labeled as 14 mm and the diameter as 5 mm.}}

Solution:

1. The capsule combines one cylinder and two hemispheres. The radius r is the same for all parts.

   • Diameter = 5 mm, so r = 5/2 = 2.5 mm

2. The total surface area will be the CSA of the cylinder plus the CSA of the two hemispheres. Two hemispherical ends make one full sphere.

   • TSA_capsule = (CSA of Cylinder) + 2 × (CSA of Hemisphere) = 2πrh + 2(2πr²) = 2πrh + 4πr²

3. Calculate the height h of the cylindrical part. The total length includes two radii from the two hemispherical ends.

   • h = Total length – r – r = 14 – 2.5 – 2.5

   h = 9 mm
   

4. Now, substitute the values of r and h into the surface area formula.

   • TSA_capsule = 2πr(h + 2r)

   TSA_capsule = 2 × (22/7) × 2.5 × (9 + 2 × 2.5)
   

5. Simplify the calculation.

   TSA_capsule = 2 × (22/7) × 2.5 × (9 + 5) = 2 × (22/7) × 2.5 × 14
   

   • The 7 in the denominator cancels with 14.

   TSA_capsule = 2 × 22 × 2.5 × 2 = 220 mm²
   

Final Answer: The surface area of the capsule is 220 mm².

Example 3: The Scooped-Out Cube (Hard)

Given: 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 a of the cube. The edge of the cube is 7 cm.

To Find: The surface area of the remaining solid.

{{VISUAL: diagram: A cube with edge 'a' = 7 cm. From the top face, a hemisphere is shown being scooped out. The diameter of the scoop is also marked as 7 cm. Arrows indicate the surfaces to be included: 5 full faces of the cube, the top face with a circular hole, and the inner curved surface of the hemisphere.}}

Solution:

1. This problem is tricky. When we scoop out the hemisphere, we remove a circular area from the top face of the cube, but we add the curved surface area of the hemisphere inside.

   • The total surface area will be: (TSA of Cube) – (Area of circular base of hemisphere) + (CSA of hemisphere).

2. Identify the given values.

   • Edge of cube a = 7 cm
   • Diameter of hemisphere d = 7 cm, so radius r = 3.5 cm.

3. Calculate the TSA of the original cube.

   • TSA_cube = 6a²

   TSA_cube = 6 × 7² = 6 × 49 = 294 cm²
   

4. The area of the removed circle is πr² and the added area is the CSA of the hemisphere, 2πr².

   • Net change = –πr² + 2πr² = +πr²
   • So, TSA_solid = TSA_cube + πr².

5. Calculate πr².

   Area = (22/7) × 3.5² = (22/7) × 12.25 = 38.5 cm²
   

6. Calculate the total surface area of the new solid.

   • TSA_solid = 294 + 38.5

   TSA_solid = 332.5 cm²
   

Final Answer: The surface area of the remaining solid is 332.5 cm².

Example 4: The Multi-Colored Rocket (Tricky)

Given: A toy rocket is in the shape of a cone mounted on a cylinder. The total height is 26 cm and the height of the conical part is 6 cm. The base diameter of the cone is 5 cm, and the base diameter of the cylinder is 3 cm. The conical part is painted orange and the cylindrical part is painted yellow. (Take π = 3.14)

To Find: The area painted orange and the area painted yellow.

Solution:

1. This problem requires us to calculate two separate areas for the two different colors.

2. Part 1: Area painted Orange (Conical Part)

   • The orange area consists of the CSA of the cone PLUS the ring-shaped base of the cone that is visible around the top of the cylinder.
   • Radius of cone R = 5/2 = 2.5 cm. Height of cone h_cone = 6 cm.
   • Radius of cylinder r = 3/2 = 1.5 cm.

3. First, find the slant height L of the cone.

   • L = √(R² + h_cone²) = √(2.5² + 6²) = √(6.25 + 36) = √42.25

   L = 6.5 cm
   

4. Calculate the CSA of the cone.

   • CSA_cone = πRL

   CSA_cone = 3.14 × 2.5 × 6.5 = 51.025 cm²
   

5. Calculate the area of the cone's base that is exposed (the ring).

   • Area_ring = Area_base_cone – Area_base_cylinder = πR² – πr²

   Area_ring = 3.14 × (2.5² – 1.5²) = 3.14 × (6.25 – 2.25) = 3.14 × 4 = 12.56 cm²
   

6. Total orange area = CSA_cone + Area_ring.

   Area_Orange = 51.025 + 12.56 = 63.585 cm²
   

7. Part 2: Area painted Yellow (Cylindrical Part)

   • The yellow area consists of the CSA of the cylinder PLUS the area of the bottom circular base of the cylinder.
   • Height of cylinder h_cyl = Total height – height of cone = 26 – 6 = 20 cm.
   • Radius of cylinder r = 1.5 cm.

8. Calculate the CSA of the cylinder.

   • CSA_cyl = 2πrh_cyl

   CSA_cyl = 2 × 3.14 × 1.5 × 20 = 188.4 cm²
   

9. Calculate the area of the bottom base of the cylinder.

   • Area_base = πr²

   Area_base = 3.14 × 1.5² = 3.14 × 2.25 = 7.065 cm²
   

10. Total yellow area = CSA_cyl + Area_base.

    Area_Yellow = 188.4 + 7.065 = 195.465 cm²
    

Final Answer: Area painted orange is 63.585 cm². Area painted yellow is 195.465 cm².

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. A solid cylinder of height 10 cm and radius 3.5 cm has a conical cavity of the same height and same radius hollowed out. What is the total surface area of the remaining solid?

   💡 Answer: The new surface area is the CSA of the cylinder + Area of the cylinder's base + the CSA of the inner cone. l = √(3.5² + 10²) = √112.25 ≈ 10.6 cm. Area = 2πrh + πr² + πrl = πr(2h + r + l) = (22/7)×3.5×(20 + 3.5 + 10.6) = 11 × 34.1 = 375.1 cm².

2. From a solid cube of side 14 cm, a sphere of the largest possible diameter is carved out. What percentage of the cube's original surface area is removed? (The removed surface is the sphere itself).

   💡 Answer: This is a trick question about wording. The "removed surface" is the surface of the sphere that is created. The question asks for the surface area of the sphere as a percentage of the cube's TSA. Largest diameter = cube side = 14 cm, so r = 7 cm. TSA_cube = 6a² = 6 × 14² = 1176 cm². SA_sphere = 4πr² = 4 × (22/7) × 7² = 616 cm². Percentage = (616 / 1176) × 100 ≈ 52.38%.

3. A bird-bath is shaped like a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath.

   💡 Answer: Convert units: h = 1.45 m = 145 cm, r = 30 cm. The total surface area is the outer CSA of the cylinder + the area of the bottom circular base + the inner CSA of the hemispherical depression. Area = (2πrh) + (πr²) + (2πr²) = 2πrh + 3πr² = πr(2h + 3r). Area = (22/7) × 30 × (2×145 + 3×30) = (22/7) × 30 × (290 + 90) = (22/7) × 30 × 380 ≈ 35828.57 cm² or 3.58 m².

Mini Cheatsheet

Surface Area of a Combination of Solids — Part 1

Page 2: Surface Area of a Combination of Solids — Part 1

Welcome back! In the previous section, we refreshed our memory on the surface areas of individual solid shapes. But the world around us is rarely that simple. From medicine capsules to architectural domes and even toys, objects are often a combination of these basic shapes.

This section teaches you the fundamental principle of calculating the surface area of such composite solids. The key is to think like a painter: if you can see a surface and touch it, you need to include its area. We will break down complex objects into familiar parts—cylinders, cones, and hemispheres—and learn how to piece their areas together correctly.

Definitions & Formulas

Before we combine solids, let's have the formulas for their individual surfaces at our fingertips. These will be our building blocks.

Important Note: The slant height l of a cone is found using the Pythagorean theorem: l = √(r² + h²).

The Logic of Combining Solids

Calculating the surface area of a combined solid is not as simple as adding the total surface areas of the individual shapes. Why? Because when you join two solids, some surfaces get hidden!

The core principle is to find the sum of the areas of all visible or exposed surfaces.

Here's the step-by-step logic:

1. Identify the Components: First, look at the combined solid and identify the basic shapes it's made of. For example, a toy top might be a cone joined to a hemisphere.

2. Visualize the Join: Imagine the two solids before they are joined. A cone has a circular base, and a hemisphere has a circular base. When you place the cone on the hemisphere, these two circular bases press against each other.

{{VISUAL: diagram: A cone and a hemisphere shown separately with their circular bases highlighted. An arrow shows them coming together, and a final image shows the combined solid where the circular bases have vanished from the exterior view.}}

3. Identify Hidden Surfaces: The surfaces that are joined together are now internal to the new object. They are no longer part of the surface area you can touch. In our toy top example, the circular base of the cone and the circular base of the hemisphere are hidden.

4. Sum the Visible Areas: The total surface area of the new, combined solid is the sum of the areas of the parts that are still visible. For the toy top, this would be the Curved Surface Area (CSA) of the cone and the Curved Surface Area (CSA) of the hemisphere.

Therefore, the general formula is: TSA of New Solid = (Sum of CSAs of all components) + (Area of any other exposed surfaces)

Let's see this in action with some examples.

Solved Examples

We will now apply this logic to problems ranging from simple to complex. Pay close attention to how we identify the visible surfaces in each case.

Example 1: The Classic Toy Top (Easy)

Given: Rasheed's playing top is a cone surmounted by a hemisphere. The total height is 5 cm and the diameter is 3.5 cm. Use π = 22/7.

To Find: The total area Rasheed has to colour.

Solution:

1. The area to be coloured is the sum of the curved surface of the hemisphere and the curved surface of the cone.

2. First, let's find the radius. The diameter is 3.5 cm.

   r = Diameter / 2 = 3.5 / 2 = 1.75 cm
   

3. Now, calculate the height of the conical part. The total height is 5 cm, and the height of the hemispherical part is just its radius.

   Height of cone (h) = Total height - Radius of hemisphere
   h = 5 cm - 1.75 cm = 3.25 cm
   

4. We need the slant height l for the cone's CSA. We use the formula l = √(r² + h²).

   l = √((1.75)² + (3.25)²)
   l = √(3.0625 + 10.5625) = √13.625 ≈ 3.7 cm
   

5. Calculate the CSA of the hemisphere (2πr²).

   CSA_hemisphere = 2 × (22/7) × (1.75)²
   CSA_hemisphere = 2 × (22/7) × 1.75 × 1.75 = 19.25 cm²
   

6. Calculate the CSA of the cone (πrl).

   CSA_cone = (22/7) × 1.75 × 3.7 = 20.35 cm²
   

7. The total area to be coloured is the sum of these two areas.

   Total Surface Area = CSA_hemisphere + CSA_cone
   Total Surface Area = 19.25 + 20.35 = 39.6 cm²
   

Final Answer:

The area Rasheed has to colour is approximately 39.6 cm².

Example 2: The Block and Hemisphere (Medium)

Given: A decorative block is a cube with an edge of 5 cm. A hemisphere with a diameter of 4.2 cm is fixed on top. Use π = 22/7.

To Find: The total surface area of the block.

{{VISUAL: diagram: A cube with side 5cm. A hemisphere with diameter 4.2cm is sitting on the top face of the cube. The circular base of the hemisphere is shown, indicating the area that is covered on the cube's top face.}}

Solution:

1. This problem is a bit different. The hemisphere covers a part of the cube's top surface. So, we must subtract this covered area.

2. The total surface area of the new block will be: (TSA of the cube) - (Base area of the hemisphere) + (CSA of the hemisphere).

3. First, calculate the TSA of the cube (6a²).

   TSA_cube = 6 × (5)² = 6 × 25 = 150 cm²
   

4. Next, find the radius of the hemisphere. Diameter = 4.2 cm.

   r = 4.2 / 2 = 2.1 cm
   

5. Calculate the base area of the hemisphere (πr²), which is the part of the cube that gets covered.

   Area_base = (22/7) × (2.1)² = (22/7) × 2.1 × 2.1 = 13.86 cm²
   

6. Calculate the CSA of the hemisphere (2πr²), which is the new curved surface added to the block.

   CSA_hemisphere = 2 × (22/7) × (2.1)² = 2 × 13.86 = 27.72 cm²
   

7. Now, combine these values as planned in Step 2.

   TSA_block = 150 - 13.86 + 27.72
   TSA_block = 163.86 cm²
   

   Alternatively, notice that TSA_cube - πr² + 2πr² simplifies to TSA_cube + πr². 150 + 13.86 = 163.86 cm².

Final Answer:

The total surface area of the block is 163.86 cm².

{{KEY: type=concept | title=Subtracting the Overlap | text=When one solid is placed on another, the area of the base of the top solid is hidden. This area must be subtracted from the surface area of the bottom solid before adding the curved surface area of the top solid.}}

Example 3: The Toy Rocket (Hard)

Given: A toy rocket has a conical top and a cylindrical body. Total height = 26 cm, conical part height h_cone = 6 cm. Diameter of cone base = 5 cm. Diameter of cylinder base = 3 cm. The cone is orange, the cylinder is yellow. Use π = 3.14.

To Find: The area painted orange and the area painted yellow.

{{VISUAL: diagram: A toy rocket. The top part is a cone labeled with height 6cm and base diameter 5cm. The bottom part is a cylinder labeled with base diameter 3cm. The total height from the tip of the cone to the base of the cylinder is 26cm. An exposed ring on the cone's base is highlighted.}}

Solution:

This is tricky because the cone's base is wider than the cylinder's base, leaving an exposed ring that needs to be painted.

Part 1: Area to be painted Orange

1. The orange area consists of the CSA of the cone plus the area of the exposed ring on its base. Area_orange = (CSA of cone) + (Area of cone's base) - (Area of cylinder's base).

2. Let's find the cone's dimensions. Diameter = 5 cm, so radius r_cone = 2.5 cm. Height h_cone = 6 cm.

3. Calculate the slant height l of the cone.

   l = √((r_cone)² + (h_cone)²) = √(2.5² + 6²)
   l = √(6.25 + 36) = √42.25 = 6.5 cm
   

4. Calculate the three area components for the orange part. CSA_cone = π × r_cone × l = 3.14 × 2.5 × 6.5 = 51.025 cm² Area_cone_base = π × (r_cone)² = 3.14 × (2.5)² = 19.625 cm² Area_cylinder_base = π × (r_cylinder)². Cylinder diameter is 3 cm, so r_cylinder = 1.5 cm. Area_cylinder_base = 3.14 × (1.5)² = 7.065 cm²

5. Combine them to get the total orange area.

   Area_orange = 51.025 + 19.625 - 7.065 = 63.585 cm²
   

Part 2: Area to be painted Yellow

1. The yellow area consists of the CSA of the cylinder plus the area of its bottom circular base.

2. Let's find the cylinder's height h_cyl. h_cyl = Total height - cone height = 26 - 6 = 20 cm. The cylinder's radius r_cyl is 1.5 cm.

3. Calculate the CSA of the cylinder (2πrh).

   CSA_cylinder = 2 × 3.14 × 1.5 × 20 = 188.4 cm²
   

4. The area of the bottom base is the Area_cylinder_base we already calculated: 7.065 cm².

5. Add them to get the total yellow area.

   Area_yellow = CSA_cylinder + Area_bottom_base
   Area_yellow = 188.4 + 7.065 = 195.465 cm²
   

Final Answer:

Area to be painted orange = 63.585 cm².
Area to be painted yellow = 195.465 cm².

Example 4: The Bird Bath (Tricky)

Given: A bird-bath is a cylinder with a hemispherical depression at one end. Cylinder height h = 1.45 m, radius r = 30 cm. Use π = 22/7.

To Find: The total surface area of the bird-bath.

Solution:

1. This problem is tricky because a part is removed, but this increases the surface area. Scooping out the hemisphere creates a new inner surface that needs to be coated.

2. The total surface area will be the CSA of the cylinder plus the CSA of the hemispherical depression. Note that the top circular face of the cylinder is gone, replaced by the hemisphere's inner surface. The bottom circular base of the cylinder is assumed to be on the ground and not part of the 'bird-bath' surface area, or sometimes problems will specify to include it. In this context, the bird-bath surface is what holds water and is exposed, so we don't count the flat bottom. TSA = (CSA of Cylinder) + (CSA of Hemisphere).

3. First, ensure units are consistent. Let's convert everything to cm. Height h = 1.45 m = 145 cm. Radius r = 30 cm.

4. Calculate the CSA of the cylinder (2πrh).

   CSA_cylinder = 2 × (22/7) × 30 × 145 = 27342.86 cm² (approx)
   

5. Calculate the CSA of the hemisphere (2πr²).

   CSA_hemisphere = 2 × (22/7) × (30)² = 2 × (22/7) × 900 = 5657.14 cm² (approx)
   

6. Add them together.

   TSA_bird-bath = 27342.86 + 5657.14 = 33000 cm²
   

   A simpler calculation can be done by factoring out 2πr: TSA = 2πr(h + r) TSA = 2 × (22/7) × 30 × (145 + 30) = 2 × (22/7) × 30 × 175 Since 175 is divisible by 7 (175/7 = 25), the calculation becomes: TSA = 2 × 22 × 30 × 25 = 33000 cm².

7. Convert the final answer to square meters if required (1 m² = 10000 cm²).

   33000 cm² = 3.3 m²
   

Final Answer:

The total surface area of the bird-bath is 33000 cm² or 3.3 m².

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Test your understanding with these slightly more challenging problems.

1. A solid wooden cube of side 14 cm has a conical cavity hollowed out from its center. The cavity has a height of 14 cm and a base radius of 7 cm. What is the total surface area of the remaining solid?

   💡 Answer: The new surface area is TSA_cube - Area_base_cone + CSA_cone. TSA_cube = 6 × 14² = 1176 cm². Area_base_cone = πr² = (22/7) × 7² = 154 cm². Slant height l = √(7² + 14²) = √(49 + 196) = √245 ≈ 15.65 cm. CSA_cone = πrl = (22/7) × 7 × 15.65 = 344.3 cm². Total Area = 1176 - 154 + 344.3 = 1366.3 cm².

2. Two solid hemispheres of the same radius r are joined together along their bases. What is the surface area of the new shape formed?

   💡 Answer: The new shape is a sphere of radius r. When joined, the two circular bases disappear. The remaining surfaces are the two curved surfaces of the hemispheres. Total Area = CSA_hemisphere + CSA_hemisphere = 2πr² + 2πr² = 4πr². This is the formula for the surface area of a sphere.

3. An ice cream cone is composed of a cone (height 10 cm, radius 3 cm) surmounted by a hemisphere of the same radius. What is the total outer surface area of the ice cream, assuming the cone is completely filled and the top is a perfect hemisphere?

   💡 Answer: The surface area is the CSA of the cone plus the CSA of the hemisphere. First, find the slant height of the cone: l = √(r² + h²) = √(3² + 10²) = √(9 + 100) = √109 ≈ 10.44 cm. CSA_cone = πrl = π × 3 × 10.44 = 31.32π cm². CSA_hemisphere = 2πr² = 2π × 3² = 18π cm². Total Surface Area = 31.32π + 18π = 49.32π cm².

Mini Cheatsheet

Screenshot this table for a quick revision of the core concepts from this page.

Surface Area of a Combination of Solids — Part 2

Surface Area of a Combination of Solids — Part 2

Welcome back! In the previous section, we learned the fundamental principle of finding the surface area of combined solids: we only count the surfaces we can see and touch. Now, we'll apply this logic to more complex and interesting shapes, including objects where one solid is carved out of another.

The key remains the same: break down the complex object into its simple components (cubes, cylinders, cones, spheres) and carefully consider which faces are exposed and which are hidden. Let's explore how to handle situations involving overlapping bases and internal surfaces.

{{FORMULA: expr=TSA_block = 6a² - πr² + 2πr² | symbols=a:edge of cube, r:radius of hemisphere}}

The Core Logic: Addition and Subtraction

When we combine solids, we often need to perform a careful "surgery" on their surface area formulas. The logic follows a simple real-world principle. Imagine you have a wooden cube and you glue a hemisphere on top of it.

1. You start with the total surface area of the cube (6a²).
2. But the base of the hemisphere covers a circular part of the cube's top face. This area is no longer exposed, so we must subtract it. The area of this covered circle is πr².
3. Finally, you have the new curved surface of the hemisphere that has been added to the object. We must add this new area. The curved surface area (CSA) of the hemisphere is 2πr².

This process of starting with a total area, subtracting the covered part, and adding the new visible part is the central theme of this lesson.

{{KEY: type=concept | title=The "Subtract and Add" Principle | text=To find the surface area of a combined solid, start with the total surface area of the base shape, subtract the area of the base covered by the attached shape, and then add the curved surface area of the attached shape.}}

Recap: Key Formulas

Before we dive into examples, let's refresh the essential formulas for the Curved Surface Area (CSA) and Base Area of our building blocks.

Remember: The slant height of a cone l is found using the Pythagorean theorem: l = √(r² + h²).

Solved Examples

Let's work through some problems, increasing in complexity, to master this concept.

Example 1: The Canvas Tent (Easy)

Given: A tent is in the shape of a cylinder surmounted by a conical top. The height of the cylindrical part is 2.1 m and the diameter is 4 m. The slant height of the conical top is 2.8 m.

To Find: The area of the canvas used for making the tent. Also, find the cost of the canvas at the rate of ₹500 per m². (Note: The base of the tent is not covered with canvas).

Solution:

1. First, identify the shapes and their dimensions. We have a cylinder and a cone.

   • Cylinder: Diameter = 4 m, so radius r = 2 m. Height h = 2.1 m.
   • Cone: It's surmounted on the cylinder, so its radius r is also 2 m. Slant height l = 2.8 m.

2. The total area of the canvas is the sum of the curved surface areas of the cone and the cylinder, as the base is open to the ground.

   Area of canvas = CSA of cone + CSA of cylinder

3. Calculate the CSA of the conical part.

   CSA_cone = πrl = (22/7) × 2 × 2.8
   

   CSA_cone = 22 × 2 × 0.4 = 17.6 m²
   

4. Calculate the CSA of the cylindrical part.

   CSA_cylinder = 2πrh = 2 × (22/7) × 2 × 2.1
   

   CSA_cylinder = 2 × 22 × 2 × 0.3 = 26.4 m²
   

5. Add the two areas to get the total canvas area.

   Total Area = 17.6 + 26.4 = 44 m²
   

6. Now, calculate the total cost.

   Cost = Total Area × Rate = 44 × 500
   

   Cost = ₹22,000
   

Final Answer:

Area of canvas = 44 m²
Total cost = ₹22,000

Example 2: The Scooped-Out Block (Medium)

Given: A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. The height of the cylinder is 10 cm, and its base radius is 3.5 cm.

To Find: The total surface area of the article.

{{VISUAL: diagram: A solid cylinder of height 'h' and radius 'r'. From the top circular face and the bottom circular face, two identical hemispheres of radius 'r' are shown scooped out, creating two bowl-like depressions.}}

Solution:

1. Visualize the final object. The outer curved surface of the cylinder is visible. The flat top and bottom are gone. In their place, we now have the inner curved surfaces of the two hemispherical depressions.

2. The total surface area will be the sum of the CSA of the cylinder and the CSAs of the two hemispheres.

   TSA of article = CSA of cylinder + 2 × (CSA of hemisphere)

3. Identify the dimensions.

   • Cylinder: Height h = 10 cm, Radius r = 3.5 cm.
   • Hemisphere: Radius r = 3.5 cm.

4. Calculate the CSA of the cylindrical part.

   CSA_cylinder = 2πrh = 2 × (22/7) × 3.5 × 10
   

   CSA_cylinder = 2 × 22 × 0.5 × 10 = 220 cm²
   

5. Calculate the CSA of one hemispherical part.

   CSA_hemisphere = 2πr² = 2 × (22/7) × 3.5 × 3.5
   

   CSA_hemisphere = 2 × 22 × 0.5 × 3.5 = 77 cm²
   

6. Calculate the total surface area of the article.

   TSA = CSA_cylinder + 2 × CSA_hemisphere
   

   TSA = 220 + 2 × 77 = 220 + 154 = 374 cm²
   

Final Answer:

Total surface area = 374 cm²

Example 3: The Decorative Block (Hard)

Given: A decorative block is made of a cube and a hemisphere. The base is a cube with an edge of 7 cm, and a hemisphere is fixed on the top with a diameter of 4.2 cm.

To Find: The total surface area of the block.

{{VISUAL: diagram: A cube with side length 'a' = 7 cm. On the center of its top face sits a hemisphere with diameter 4.2 cm. The circular base of the hemisphere is shown covering a part of the cube's top face.}}

Solution:

1. This problem is a perfect example of the "Subtract and Add" principle. We can see the entire surface of the cube except for the circular area covered by the hemisphere's base. We can also see the curved surface of the hemisphere on top.

2. Therefore, the formula is: TSA of block = TSA of cube - Base area of hemisphere + CSA of hemisphere

3. List the dimensions.

   • Cube: Edge a = 7 cm.
   • Hemisphere: Diameter = 4.2 cm, so radius r = 2.1 cm.

4. Calculate the Total Surface Area (TSA) of the cube.

   TSA_cube = 6a² = 6 × 7² = 6 × 49
   

   TSA_cube = 294 cm²
   

5. Calculate the base area of the hemisphere (the part to be subtracted).

   Area_base = πr² = (22/7) × 2.1 × 2.1
   

   Area_base = 22 × 0.3 × 2.1 = 13.86 cm²
   

6. Calculate the Curved Surface Area (CSA) of the hemisphere (the part to be added).

   CSA_hemisphere = 2πr² = 2 × ( (22/7) × 2.1 × 2.1 )
   

   CSA_hemisphere = 2 × 13.86 = 27.72 cm²
   

7. Combine these values to find the final surface area.

   TSA_block = 294 - 13.86 + 27.72
   

   TSA_block = 294 + 13.86 = 307.86 cm²
   

   Notice that (-πr² + 2πr²) simplifies to (+πr²). So an alternate calculation is TSA_cube + πr² = 294 + 13.86 = 307.86 cm².

Final Answer:

Total surface area = 307.86 cm²

Example 4: The Rocket Toy (Tricky)

Given: A toy rocket is in the shape of a cone mounted on a cylinder. The entire rocket is 26 cm tall. The conical part is 6 cm high. The base of the cone has a diameter of 5 cm, while the base of the cylinder has a diameter of 3 cm. The conical part is painted orange and the cylindrical part is painted yellow.

To Find: The area painted orange and the area painted yellow. (Use π = 3.14)

{{VISUAL: diagram: A rocket shape. The top is a cone with height 6 cm and base diameter 5 cm. This cone is mounted on a cylinder with base diameter 3 cm. The total height is labeled 26 cm. An exposed ring-shaped area between the cone's base and the cylinder's top is visible.}}

Solution:

This is tricky because the radii are different, creating an exposed ring (annulus) that needs to be painted.

Part 1: Area to be Painted Orange

1. The orange area consists of the CSA of the cone plus the area of its base that is visible. The visible base is a ring formed by the cone's base minus the cylinder's top.

   Area_orange = CSA_cone + Area_cone_base - Area_cylinder_base

2. Dimensions for the cone:

   • Height h = 6 cm.
   • Diameter = 5 cm, so radius r = 2.5 cm.
   • We need the slant height l. l = √(r² + h²) = √(2.5² + 6²) = √(6.25 + 36) = √42.25 = 6.5 cm.

3. Dimensions for the cylinder:

   • Diameter = 3 cm, so radius r' = 1.5 cm.

4. Calculate the areas for the orange part.

   • CSA_cone = πrl = 3.14 × 2.5 × 6.5 = 51.025 cm²
   • Area_cone_base = πr² = 3.14 × 2.5² = 19.625 cm²
   • Area_cylinder_base = π(r')² = 3.14 × 1.5² = 7.065 cm²

5. Combine them to find the total orange area.

   Area_orange = 51.025 + 19.625 - 7.065
   

   Area_orange = 70.65 - 7.065 = 63.585 cm²
   

Part 2: Area to be Painted Yellow

1. The yellow area consists of the CSA of the cylinder plus the area of its bottom base.

   Area_yellow = CSA_cylinder + Area_cylinder_base

2. Dimensions for the cylinder:

   • Radius r' = 1.5 cm.
   • Height h' = Total height - Cone height = 26 - 6 = 20 cm.

3. Calculate the areas for the yellow part.

   • CSA_cylinder = 2πr'h' = 2 × 3.14 × 1.5 × 20 = 188.4 cm²
   • Area_cylinder_base = π(r')² = 7.065 cm² (calculated before)

4. Combine them to find the total yellow area.

   Area_yellow = 188.4 + 7.065 = 195.465 cm²
   

Final Answer:

Area painted orange = 63.585 cm²
Area painted yellow = 195.465 cm²

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

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 surface area of the solid in terms of π.

   💡 Answer: The surface area is CSA of cone + CSA of hemisphere. r = 1 cm, h = 1 cm. l = √(r² + h²) = √(1² + 1²) = √2 cm. TSA = πrl + 2πr² = π(1)(√2) + 2π(1)² = (√2 + 2)π cm².

2. From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the total surface area of the remaining solid.

   💡 Answer: The new surface area is TSA of cube - Area of top circle + CSA of inner cone. TSA_cube = 6 × 7² = 294 cm². Area of circle = πr² = (22/7) × 3² = 198/7 cm². l_cone = √(r² + h²) = √(3² + 7²) = √(9 + 49) = √58 ≈ 7.62 cm. CSA_cone = πrl = (22/7) × 3 × √58 ≈ 71.79 cm². TSA_remaining ≈ 294 - (198/7) + 71.79 ≈ 294 - 28.28 + 71.79 = 337.51 cm².

3. A medicine capsule is a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter is 5 mm. What is its surface area?

   💡 Answer: The surface area is CSA of cylinder + 2 × CSA of hemisphere. Diameter = 5 mm, so r = 2.5 mm. The length of the cylindrical part h = Total length - 2 × (radius of hemisphere) = 14 - 2 × 2.5 = 14 - 5 = 9 mm. CSA_cylinder = 2πrh = 2π(2.5)(9) = 45π mm². CSA_2_hemispheres = 2 × (2πr²) = 4πr² = 4π(2.5)² = 25π mm². (This is just the surface area of a sphere). Total Area = 45π + 25π = 70π = 70 × (22/7) = 220 mm².

Mini Cheatsheet

Volume of a Combination of Solids

Volume of a Combination of Solids

Welcome back! In our last session, we mastered calculating the surface area of combined solids. We learned that when we join two shapes, some of their surface areas disappear. Now, we shift our focus to volume, and you'll find the logic here is much more straightforward.

Imagine an ice cream cone. It's a combination of a cone and a hemisphere of ice cream scooped on top. To find out how much ice cream you have, you wouldn't do any complex subtractions. You would simply calculate the volume of the cone, calculate the volume of the hemispherical scoop, and add them together. This is the beautiful simplicity of calculating the volume of combined solids: the total volume is just the sum of the individual volumes. It's an additive process. If a shape is carved out of another, we simply subtract its volume.

Let's dive into the formulas and methods to master this concept.

{{KEY: type=concept | title=The Additive Principle of Volume | text=When solids are combined, the volume of the resulting solid is the sum of the volumes of the individual solids. Unlike surface area, no part of the volume is lost in the process of joining them.}}

Definitions & Formulas

Before we combine solids, let's refresh the volume formulas for our basic 3D shapes.

The Logic of Calculating Combined Volume

Calculating the volume of a composite solid is a logical, step-by-step process.

1. Deconstruct the Solid: First, look at the complex shape and identify the basic solids it's made of. For instance, a shed might be a cuboid with a half-cylinder on top. A toy might be a cone mounted on a hemisphere.

2. List the Dimensions: For each basic solid you identified, carefully list its dimensions (radius r, height h, length l, etc.). Pay close attention to shared dimensions. For example, the radius of a cone and the hemisphere it stands on will be the same.

3. Calculate Individual Volumes: Apply the correct volume formula for each basic solid using the dimensions you listed. Calculate the volume of the cuboid, the cylinder, the cone, and so on, separately.

4. Combine the Volumes:

   • If the solid is formed by joining shapes, add their individual volumes. V_total = V₁ + V₂.
   • If the solid is formed by scooping out or removing a shape from a larger one, subtract the volume of the removed shape from the larger one. V_total = V_larger - V_removed.

Solved Examples

Let's work through some examples, starting from simple ones and moving to more complex problems.

Example 1: The Classic Toy (Easy)

Given: A solid toy is in the form of a cone standing on a hemisphere. Both their radii are 1 cm. The height of the cone is equal to its radius.

To Find: The volume of the solid in terms of π.

Solution:

1. First, let's identify the components and their dimensions. The toy consists of a hemisphere and a cone.

   • Hemisphere radius, r = 1 cm
   • Cone radius, r = 1 cm
   • Cone height, h = radius = 1 cm

2. The total volume of the toy is the sum of the volume of the hemisphere and the volume of the cone.

   V_toy = V_hemisphere + V_cone
   

3. Now, substitute the formulas for each shape.

   V_toy = (⅔πr³) + (⅓πr²h)
   

4. Substitute the given values for r and h.

   V_toy = (⅔ × π × 1³) + (⅓ × π × 1² × 1)
   

5. Simplify the expression.

   V_toy = ⅔π + ⅓π
   

6. Add the terms to get the final volume.

   V_toy = (2+1)/3 × π = 3/3 × π = π cm³
   

Final Answer: The volume of the solid is π cm³.

Example 2: The Industrial Shed (Medium)

{{VISUAL: diagram: A shed shaped like a cuboid with dimensions 15m x 7m x 8m, surmounted by a half-cylinder. The diameter of the half-cylinder is 7m and its length is 15m.}}

Given: A shed is shaped like a cuboid (base 7 m × 15 m, height 8 m) surmounted by a half-cylinder. Machinery occupies 300 m³ and 20 workers each occupy 0.08 m³.

To Find: The volume of air in the shed after accounting for machinery and workers. (Use π = 22/7)

Solution:

1. This is a multi-step problem. First, we find the total volume of the shed. The shed is a combination of a cuboid and a half-cylinder.

   • Cuboid dimensions: l = 15 m, b = 7 m, h_cuboid = 8 m.
   • Half-cylinder dimensions: The diameter is 7 m, so radius r = 7/2 m. The height (or length) of the cylinder h_cylinder is 15 m.

2. Calculate the volume of the cuboidal part.

   V_cuboid = l × b × h = 15 × 7 × 8 = 840 m³
   

3. Calculate the volume of the half-cylinder part.

   V_half-cylinder = ½ × (Volume of Cylinder) = ½ × πr²h
   

   V_half-cylinder = ½ × (22/7) × (7/2)² × 15
   

   V_half-cylinder = ½ × (22/7) × (49/4) × 15 = (11 × 7 × 15) / 4 = 1155 / 4 = 288.75 m³
   

4. Calculate the total volume of the shed by adding the two parts.

   V_shed = V_cuboid + V_half-cylinder = 840 + 288.75 = 1128.75 m³
   

5. Now, calculate the total volume occupied by machinery and workers.

   • Volume of machinery = 300 m³
   • Volume of workers = 20 × 0.08 = 1.6 m³

6. Calculate the total occupied space.

   V_occupied = 300 + 1.6 = 301.6 m³
   

7. Finally, find the volume of air by subtracting the occupied space from the total shed volume.

   V_air = V_shed - V_occupied = 1128.75 - 301.6 = 827.15 m³
   

Final Answer: The volume of air in the shed is 827.15 m³.

Example 3: The Wooden Pen Stand (Hard)

{{VISUAL: diagram: A wooden cuboid pen stand with dimensions 15cm x 10cm x 3.5cm. Four identical conical depressions are carved into it to hold pens, with labels for the radius (0.5cm) and depth (1.4cm) of each cone.}}

Given: A cuboid pen stand (15 cm × 10 cm × 3.5 cm) has four conical depressions. Each depression has a radius of 0.5 cm and a depth of 1.4 cm.

To Find: The volume of wood in the entire stand.

Solution:

1. This problem involves subtraction. The volume of the wood is the volume of the original cuboid minus the volume of the four conical depressions that were carved out.

2. First, calculate the volume of the cuboid before the depressions were made.

   • Cuboid dimensions: l = 15 cm, b = 10 cm, h = 3.5 cm.

   V_cuboid = l × b × h = 15 × 10 × 3.5 = 525 cm³
   

3. Next, calculate the volume of a single conical depression.

   • Cone dimensions: radius r = 0.5 cm, depth (height) h = 1.4 cm.

   V_cone = ⅓πr²h
   

   V_cone = ⅓ × (22/7) × (0.5)² × 1.4
   

   V_cone = ⅓ × (22/7) × 0.25 × 1.4 = ⅓ × 22 × 0.25 × 0.2 = 1.1/3 cm³
   

4. Now, calculate the total volume of the four conical depressions.

   V_removed = 4 × V_cone = 4 × (1.1/3) = 4.4/3 cm³
   

   V_removed ≈ 1.47 cm³
   

5. Finally, subtract the volume of the removed cones from the volume of the cuboid to find the volume of the wood.

   V_wood = V_cuboid - V_removed = 525 - 4.4/3
   

   V_wood = (1575 - 4.4) / 3 = 1570.6 / 3 ≈ 523.53 cm³
   

Final Answer: The volume of wood in the entire stand is approximately 523.53 cm³.

Example 4: The Displaced Water (Tricky)

{{VISUAL: diagram: A large right circular cylinder filled with water. A solid object (a cone of height 120cm on a hemisphere of radius 60cm) is placed upright inside it, touching the bottom. Labels show cylinder height=180cm, cylinder radius=60cm.}}

Given: A solid (cone of height 120 cm on a hemisphere of radius 60 cm) is placed upright in a cylinder full of water. The cylinder has a radius of 60 cm and a height of 180 cm.

To Find: The volume of water left in the cylinder.

Solution:

1. The key insight here is that the volume of water left in the cylinder is the total volume of the cylinder minus the volume of the solid submerged in it.

2. Identify the dimensions for all three shapes.

   • Cylinder: R = 60 cm, H = 180 cm.
   • Hemisphere: r = 60 cm.
   • Cone: r = 60 cm, h = 120 cm.
   • Note that the radius is the same for all three shapes.

3. First, calculate the volume of the water that the cylinder can hold, which is the volume of the cylinder itself.

   V_cylinder = πR²H = π × (60)² × 180 = π × 3600 × 180 = 648000π cm³
   

4. Next, calculate the volume of the solid object that is placed inside.

   V_solid = V_hemisphere + V_cone
   

   V_solid = (⅔πr³) + (⅓πr²h)
   

5. Substitute the values for the solid's components.

   V_solid = (⅔ × π × 60³) + (⅓ × π × 60² × 120)
   

   We can factor out ⅓π(60)² to simplify the calculation.

   V_solid = ⅓π(60)² [ (2 × 60) + 120 ]
   

   V_solid = ⅓π(3600) [ 120 + 120 ] = 1200π [ 240 ] = 288000π cm³
   

6. Finally, subtract the volume of the solid from the volume of the cylinder to find the volume of the remaining water.

   V_water_left = V_cylinder - V_solid
   

   V_water_left = 648000π - 288000π = 360000π cm³
   

Final Answer: The volume of water left in the cylinder is 360000π cm³.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. A toy rocket is in the shape of a cylinder with a cone attached to one end and a hemisphere to the other. The cylinder is 20 cm long with a diameter of 6 cm. The cone has a height of 4 cm, and the hemisphere has the same diameter as the cylinder. Find the total volume of the rocket. (Use π = 3.14)

   💡 Answer: Radius r = 6/2 = 3 cm. V_cylinder = πr²h = 3.14 × 3² × 20 = 565.2 cm³. V_cone = ⅓πr²h = ⅓ × 3.14 × 3² × 4 = 37.68 cm³. V_hemisphere = ⅔πr³ = ⅔ × 3.14 × 3³ = 56.52 cm³. Total Volume = 565.2 + 37.68 + 56.52 = 659.4 cm³.

2. A solid is formed by a cylinder of height 10 cm and a hemisphere of radius 7 cm attached to its base. If the total volume of the solid is 2254 cm³, what is the radius of the cylinder? (Hint: The radii are not the same).

   💡 Answer: Let the cylinder's radius be R. The hemisphere's radius r = 7 cm. V_hemisphere = ⅔ × (22/7) × 7³ = ⅔ × 22 × 49 ≈ 718.67 cm³. V_total = V_cylinder + V_hemisphere. 2254 = (πR²h) + 718.67 2254 = ((22/7) × R² × 10) + 718.67 1535.33 = (220/7) × R² R² = (1535.33 × 7) / 220 ≈ 48.8 R ≈ √49 = 7 cm. The radius of the cylinder is 7 cm.

3. A solid metal toy, in the shape of a cone of height 10 cm and base radius 5 cm mounted on a hemisphere of the same radius, is melted and recast into a solid sphere. What is the radius of this new sphere?

   💡 Answer: The volume of the metal remains conserved. So, V_toy = V_sphere. V_toy = V_cone + V_hemisphere = (⅓πr²h) + (⅔πr³) V_toy = (⅓ × π × 5² × 10) + (⅔ × π × 5³) = (250/3)π + (250/3)π = (500/3)π cm³. Let the new sphere's radius be R. V_sphere = ⁴⁄₃πR³ (500/3)π = (4/3)πR³ 500 = 4R³ R³ = 125 R = 5 cm. The radius of the new sphere is 5 cm.

Mini Cheatsheet

Summary & Quick Revision

Chapter 12: Surface Areas and Volumes

Page 5 of 5: Summary & Quick Revision

Concept Introduction

Welcome to the final revision page for Surface Areas and Volumes! So far, we've explored the surfaces of 3D shapes. Now, we dive into the space they occupy: their volume. Imagine an ice cream scoop. The total surface area is the cone plus the curved part of the hemisphere. But what about the total amount of ice cream? That's its volume.

Unlike surface area, where joining two shapes hides some area at the interface, volume is much simpler. When you join two solid shapes, their total volume is simply the sum of their individual volumes. If you scoop a piece out of a solid, the remaining volume is the original volume minus the volume of the piece you removed. This additive and subtractive nature is the key to mastering the volume of combined solids. Let's solidify this understanding.

{{KEY: type=concept | title=Volume vs. Surface Area Logic | text=When combining solids, their volumes simply ADD up (V = V₁ + V₂). This is different from surface area, where the joining area is SUBTRACTED from the total (A = A₁ + A₂ - A_base).}}

Definitions & Formulas

Before we combine shapes, let's refresh the volume formulas for the basic building blocks.

The Logic of Combining Volumes

Calculating the volume of a composite solid is a straightforward process of addition or subtraction. Let's break down the thinking process with an example of a toy shaped like a cone placed on top of a hemisphere.

{{VISUAL: diagram: A composite toy shape. A cone with height 'h' and radius 'r' is perfectly placed on top of a hemisphere of the same radius 'r'. The total height of the toy is labeled as H = h + r.}}

1. Deconstruct the Shape: First, identify the basic solids that form the object. Our toy is made of one cone and one hemisphere.

2. List Dimensions: Note down the dimensions for each part.

   • For the cone: We need its radius (r) and height (h).
   • For the hemisphere: We need its radius (r).
   • Crucially, notice that the radius r is the same for both parts.

3. Determine the Operation: Are the solids being joined together or is one being removed from another? Here, the cone is placed on the hemisphere, so we are combining them. This means we must add their volumes.

4. Formulate the Expression: Write the total volume as the sum of the individual volumes.

   • Volume of Toy = Volume of Cone + Volume of Hemisphere

   V_toy = (⅓πr²h) + (⅔πr³)
   

5. Simplify and Substitute: Before plugging in numbers, look for common factors to make calculation easier. Here, we can factor out ⅓πr².

   V_toy = ⅓πr² (h + 2r)
   

6. Calculate the Final Answer: Substitute the known values for r and h into the simplified expression and compute the final volume. Remember to include the correct units (like cm³ or m³).

Solved Examples

Let's apply this logic to a few problems, ranging from easy to tricky.

Example 1: The Classic Toy (Easy)

Given: A solid toy is a hemisphere surmounted by a cone. The height of the cone is 2 cm and the diameter of the base is 4 cm.

To Find: The volume of the toy. (Use π = 3.14)

Solution:

1. Identify the shapes and dimensions. The toy consists of a hemisphere and a cone.

   • Diameter = 4 cm, so the radius for both shapes is r = 4/2 = 2 cm.
   • Height of the cone, h = 2 cm.

2. The total volume is the sum of the volume of the cone and the volume of the hemisphere.

   V_toy = Volume of Cone + Volume of Hemisphere
   

3. Substitute the formulas for each shape.

   V_toy = (⅓πr²h) + (⅔πr³)
   

4. Substitute the given values for r, h, and π.

   V_toy = (⅓ × 3.14 × 2² × 2) + (⅔ × 3.14 × 2³)
   

5. Calculate each part.

   V_toy = (⅓ × 3.14 × 8) + (⅔ × 3.14 × 8)
   

   V_toy = (25.12 / 3) + (50.24 / 3)
   

6. Sum the values to get the final volume.

   V_toy = 75.36 / 3 = 25.12 cm³
   

Final Answer: The volume of the toy is 25.12 cm³.

Example 2: The Pen Stand (Medium)

Given: A wooden pen stand is a cuboid with dimensions 15 cm × 10 cm × 3.5 cm. Four conical depressions are drilled to hold pens. The radius of each depression is 0.5 cm and the depth is 1.4 cm.

To Find: The volume of the wood in the entire stand. (Use π = 22/7)

Solution:

1. This is a subtraction problem. The volume of wood is the volume of the cuboid minus the volume of the four cones that were removed.

{{VISUAL: diagram: A cuboid representing the pen stand. Four identical conical depressions are shown on the top surface, indicating where wood has been removed.}}

2. Calculate the volume of the cuboid.

   • Dimensions: l = 15 cm, b = 10 cm, h = 3.5 cm.

   V_cuboid = l × b × h = 15 × 10 × 3.5 = 525 cm³
   

3. Calculate the volume of one conical depression.

   • Radius r = 0.5 cm, Depth (height) h = 1.4 cm.

   V_cone = ⅓πr²h
   

   V_cone = ⅓ × (22/7) × (0.5)² × 1.4
   

   V_cone = ⅓ × (22/7) × 0.25 × 1.4 = ⅓ × 22 × 0.25 × 0.2 = 1.1 / 3 cm³
   

4. Calculate the total volume of the four conical depressions.

   V_4cones = 4 × (1.1 / 3) = 4.4 / 3 ≈ 1.47 cm³
   

5. Subtract the volume of the depressions from the volume of the cuboid.

   V_wood = V_cuboid - V_4cones
   

   V_wood = 525 - (4.4 / 3) = 525 - 1.47 = 523.53 cm³
   

Final Answer: The volume of the wood in the stand is 523.53 cm³.

Example 3: The Gulab Jamun (Hard)

Given: A gulab jamun is shaped like a cylinder with two hemispherical ends. Its total length is 5 cm and diameter is 2.8 cm. It contains sugar syrup up to 30% of its volume.

To Find: The approximate volume of syrup in 45 such gulab jamuns. (Use π = 22/7)

Solution:

{{VISUAL: diagram: A capsule shape representing a gulab jamun. It's broken down into a central cylinder and two hemispheres at the ends. The total length (5 cm) and diameter (2.8 cm) are labeled.}}

1. First, let's find the volume of a single gulab jamun. It's composed of one cylinder and two hemispheres.

   • Diameter = 2.8 cm, so radius r = 1.4 cm.
   • The two hemispheres form a single sphere of radius 1.4 cm.
   • The length of the cylindrical part is the total length minus the two radii of the hemispherical ends: h_cyl = 5 cm - (1.4 cm + 1.4 cm) = 2.2 cm.

2. Calculate the volume of the cylindrical part.

   V_cyl = πr²h = (22/7) × (1.4)² × 2.2 = (22/7) × 1.96 × 2.2 = 13.552 cm³
   

3. Calculate the volume of the two hemispherical ends (which is one sphere).

   V_2hemi = V_sphere = ⁴⁄₃πr³ = ⁴⁄₃ × (22/7) × (1.4)³ = ⁴⁄₃ × (22/7) × 2.744 = 11.498 cm³
   

4. Find the total volume of one gulab jamun.

   V_one_gj = V_cyl + V_2hemi = 13.552 + 11.498 = 25.05 cm³
   

5. Find the volume of syrup in one gulab jamun (30% of its volume).

   V_syrup_one = 30% of 25.05 = 0.30 × 25.05 = 7.515 cm³
   

6. Calculate the total volume of syrup in 45 gulab jamuns.

   V_syrup_total = 45 × V_syrup_one = 45 × 7.515 = 338.185 cm³
   

Final Answer: Approximately 338 cm³ of syrup would be found in 45 gulab jamuns.

Example 4: Water Displacement (Tricky)

Given: A solid toy (a 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. The cylinder has a radius of 60 cm and a height of 180 cm.

To Find: The volume of water left in the cylinder. (Use π = 22/7)

Solution:

1. The volume of water left is the volume of the cylinder minus the volume of the solid toy that displaced the water.

2. First, calculate the volume of the cylinder.

   • Radius r_cyl = 60 cm, Height h_cyl = 180 cm.

   V_cyl = πr²h = (22/7) × 60² × 180 = (22/7) × 3600 × 180 = 14256000 / 7 cm³
   

3. Next, calculate the volume of the solid toy. The toy consists of a cone and a hemisphere.

   • Radius for both r_toy = 60 cm.
   • Height of cone h_cone = 120 cm.

4. Calculate the volume of the conical part.

   V_cone = ⅓πr²h = ⅓ × (22/7) × 60² × 120 = ⅓ × (22/7) × 3600 × 120 = 5280000 / 7 cm³
   

5. Calculate the volume of the hemispherical part.

   V_hemi = ⅔πr³ = ⅔ × (22/7) × 60³ = ⅔ × (22/7) × 216000 = 3168000 / 7 cm³
   

6. Find the total volume of the toy.

   V_toy = V_cone + V_hemi = (5280000 / 7) + (3168000 / 7) = 8448000 / 7 cm³
   

7. Finally, subtract the toy's volume from the cylinder's volume.

   V_left = V_cyl - V_toy = (14256000 / 7) - (8448000 / 7)
   

   V_left = 5808000 / 7 ≈ 829714.28 cm³
   

   To express this in m³, divide by 1,000,000 (since 1m = 100cm, 1m³ = 100³ cm³).

   V_left ≈ 0.83 m³
   

Final Answer: The volume of water left in the cylinder is approximately 829714.28 cm³ (or 0.83 m³).

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Brain-Teaser Questions

1. A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed.

   💡 Answer: The volume of the material remains constant. Volume of Sphere = n × Volume of one Cone (⁴⁄₃)πR³ = n × (⅓)πr²h 4R³ = n × r²h 4 × (10.5)³ = n × (3.5)² × 3 4 × 1157.625 = n × 12.25 × 3 4630.5 = n × 36.75 n = 4630.5 / 36.75 = 126. So, 126 cones can be formed.

2. 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. What is the volume of the remaining solid in cm³?

   💡 Answer: The volume of the remaining solid is the volume of the cylinder minus the volume of the cone. Radius r = 1.4/2 = 0.7 cm. Height h = 2.4 cm. V_remaining = V_cylinder - V_cone V_remaining = πr²h - ⅓πr²h = ⅔πr²h V_remaining = ⅔ × (22/7) × (0.7)² × 2.4 V_remaining = ⅔ × (22/7) × 0.49 × 2.4 = 2.464 cm³.

3. A sphere, a cylinder, and a cone have the same radius r. The height of the cylinder and cone is equal to the diameter of the sphere. What is the ratio of their volumes?

   💡 Answer: Let radius be r. Height of cylinder and cone h = diameter of sphere = 2r. V_cone = ⅓πr²h = ⅓πr²(2r) = (⅔)πr³ V_sphere = (⁴⁄₃)πr³ V_cylinder = πr²h = πr²(2r) = 2πr³ Ratio V_cone : V_sphere : V_cylinder is (⅔)πr³ : (⁴⁄₃)πr³ : 2πr³. Divide by (⅔)πr³: 1 : 2 : 3. The ratio is 1:2:3.

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