Surface Areas and Volumes

Surface Area of Basic 3D Solids

When we move from flat shapes to 3D solids, we start thinking about the outside covering of an object. This is called its surface area.

Surface area tells us how much material is needed to cover the outside of a solid. For example, it can help us find how much paper is needed to wrap a gift, how much metal is needed to make a can, or how much paint is needed to cover a curved object.

In this lesson, we will learn how to calculate the lateral surface area and the total surface area of four common solids:

• cylinders,
• cones,
• spheres,
• hemispheres.

Important idea:

• Lateral surface area means the area of the curved side only, not including the base or bases.
• Total surface area means the area of all outer surfaces.

We will use these symbols often:

• radius: \(r\)

• height: \(h\)

• slant height of a cone: \(l\)

• \(\pi \approx 3.14\) or \(\frac{22}{7}\) when needed

1. Surface Area of a Cylinder

A cylinder has:

• two circular bases, and
• one curved surface around the side.

If we unwrap the curved side of a cylinder, it forms a rectangle.

The width of this rectangle is the circumference of the base circle, which is \(2\pi r\). The height of the rectangle is the cylinder's height, \(h\).

So the lateral surface area of a cylinder is:

Now add the areas of the two circular bases. Each base has area \(\pi r^2\), so two bases have area \(2\pi r^2\).

Therefore, the total surface area of a cylinder is:

This can also be written as:

2. Surface Area of a Cone

A cone has:

• one circular base, and
• one curved surface that comes to a point.

For a cone, the curved surface depends on the slant height, not the vertical height.

The slant height is labeled \(l\). If the vertical height \(h\) and radius \(r\) are known, then:

The lateral surface area of a cone is:

To find the total surface area, add the area of the circular base:

Or factor out \(\pi r\):

3. Surface Area of a Sphere

A sphere is a perfectly round solid, like a basketball or a globe. It has no edges, no flat faces, and no base.

The total outer area of a sphere is given by:

Since a sphere has only one continuous outer surface, we usually just say surface area, not lateral or total separately.

4. Surface Area of a Hemisphere

A hemisphere is half of a sphere. It looks like a bowl shape if cut through the middle.

A hemisphere has:

• one curved part, and
• one flat circular base.

The curved surface area of a hemisphere is half the surface area of a sphere:

If the flat circular base is included, then the total surface area is:

Formulas to Remember

• Cylinder:

  $$\text{LSA} = 2\pi rh$$

  $$\text{TSA} = 2\pi rh + 2\pi r^2$$

• Cone:

  $$\text{LSA} = \pi rl$$

  $$\text{TSA} = \pi rl + \pi r^2$$

• Sphere:

  $$\text{SA} = 4\pi r^2$$

• Hemisphere:

  $$\text{Curved SA} = 2\pi r^2$$

  $$\text{TSA} = 3\pi r^2$$

How to Choose the Correct Formula

1. Read the question carefully.
2. Identify the solid: cylinder, cone, sphere, or hemisphere.
3. Check whether the question asks for curved/lateral surface area or total surface area.
4. Write the correct formula before substituting numbers.
5. Use the radius, not the diameter. If diameter is given, divide by 2 first.
6. Include square units in the final answer, such as \(\text{cm}^2\), \(\text{m}^2\), or \(\text{in}^2\).

Worked Example 1: Total Surface Area of a Cylinder

Find the total surface area of a cylinder with radius \(4\text{ cm}\) and height \(10\text{ cm}\).

Step 1: Write the formula.

Step 2: Substitute the values.

Step 3: Give the exact and approximate answer.

Using \(\pi \approx 3.14\):

Answer: \(112\pi\text{ cm}^2\) or about \(351.68\text{ cm}^2\).

Worked Example 2: Lateral Surface Area of a Cone

A cone has radius \(5\text{ cm}\) and slant height \(12\text{ cm}\). Find its lateral surface area.

Step 1: Use the cone lateral surface area formula.

Step 2: Substitute the values.

Step 3: Approximate if needed.

Answer: \(60\pi\text{ cm}^2\) or about \(188.4\text{ cm}^2\).

Worked Example 3: Total Surface Area of a Cone When Height Is Given

A cone has radius \(3\text{ m}\) and vertical height \(4\text{ m}\). Find its total surface area.

Step 1: Find the slant height.

Step 2: Use the total surface area formula.

Step 3: Substitute the values.

Step 4: Approximate if needed.

Answer: \(24\pi\text{ m}^2\) or about \(75.36\text{ m}^2\).

Worked Example 4: Surface Area of a Sphere and a Hemisphere

A sphere has radius \(7\text{ cm}\). Find its surface area. Then find the total surface area of a hemisphere with the same radius.

Part A: Sphere

Using \(\pi = \frac{22}{7}\):

So the sphere's surface area is \(196\pi\text{ cm}^2\) or \(616\text{ cm}^2\).

Part B: Hemisphere

For total surface area of a hemisphere:

Using \(\pi = \frac{22}{7}\):

Answer:

• Sphere surface area = \(196\pi\text{ cm}^2\) or \(616\text{ cm}^2\)
• Hemisphere total surface area = \(147\pi\text{ cm}^2\) or \(462\text{ cm}^2\)

Common Mistakes to Avoid

• Confusing radius and diameter. If diameter is given, remember: \(r = \frac{d}{2}\).
• Using height instead of slant height for a cone. The curved surface area of a cone uses \(l\), not \(h\).
• Forgetting the base or bases. Lateral area and total area are not the same.
• For hemisphere questions, not checking whether the base is included. Curved area is \(2\pi r^2\), but total area is \(3\pi r^2\).
• Forgetting square units. Surface area is always measured in square units.

Quick Comparison

• A cylinder has 2 circular bases.
• A cone has 1 circular base and needs slant height for curved area.
• A sphere has no base and uses only \(4\pi r^2\).
• A hemisphere is half a sphere, so be careful whether the flat base is included.

Brief Summary

Surface area measures the total outside covering of a 3D solid. For cylinders and cones, we often separate the lateral or curved area from the total area. A sphere has surface area \(4\pi r^2\), and a hemisphere has curved surface area \(2\pi r^2\) and total surface area \(3\pi r^2\).

The key to success is choosing the correct formula, using the correct measurements, and checking whether the base is included. With practice, these formulas become much easier to use.

Volume of Basic 3D Solids

When we move from flat shapes to solid shapes, we begin to measure how much space an object takes up. This amount of space is called its volume.

Volume is measured in cubic units, such as cubic centimeters \,\(\text{cm}^3\), cubic meters \,\(\text{m}^3\), or cubic inches \,\(\text{in}^3\). The word “cubic” reminds us that volume involves three dimensions: length, width, and height.

In this lesson, you will learn how to find the volume of three important 3D solids:

• Cylinders
• Cones
• Spheres

You will also learn how to choose the correct formula, substitute values carefully, and include the correct units in your final answer.

1. Understanding Volume

Volume tells us how much a solid can hold or contain. For example:

• A water bottle has a certain volume of liquid it can hold.
• A ball takes up a certain amount of space.
• A can of soup has volume because it is a 3D object.

For many solids, volume depends on the area of a base and how far the shape extends. That is why base shapes and height are important in volume formulas.

2. Volume of a Cylinder

A cylinder has two matching circular bases and a curved side. A soup can is a common example of a cylinder.

The formula for the volume of a cylinder is:

where:

• \(V\) = volume
• \(r\) = radius of the circular base
• \(h\) = height of the cylinder
• \(\pi \approx 3.14\) or \(\frac{22}{7}\) when appropriate

This formula works because the area of the circular base is \(\pi r^2\), and multiplying by height gives the volume.

Important: Use the radius, not the diameter. If you are given the diameter, divide it by 2 first.

Worked Example 1: Volume of a Cylinder

Find the volume of a cylinder with radius \(4\text{ cm}\) and height \(10\text{ cm}\).

Step 1: Write the formula.

Step 2: Substitute the values.

Step 3: Simplify.

Step 4: Write the exact and approximate answers.

Answer: The volume is \(160\pi\text{ cm}^3\), or about \(502.4\text{ cm}^3\).

3. Volume of a Cone

A cone has one circular base and comes to a point called the vertex. An ice cream cone is a familiar example.

The formula for the volume of a cone is:

where:

• \(V\) = volume
• \(r\) = radius of the base
• \(h\) = perpendicular height

The cone formula is very similar to the cylinder formula, but it includes \(\frac{1}{3}\). This means a cone with the same base radius and height as a cylinder has one-third the volume of that cylinder.

Worked Example 2: Volume of a Cone

Find the volume of a cone with radius \(3\text{ m}\) and height \(8\text{ m}\).

Step 1: Write the formula.

Step 2: Substitute the values.

Step 3: Simplify.

Step 4: Approximate.

Answer: The volume is \(24\pi\text{ m}^3\), or about \(75.36\text{ m}^3\).

4. Volume of a Sphere

A sphere is a perfectly round 3D shape, like a basketball or a marble.

The formula for the volume of a sphere is:

where:

• \(V\) = volume
• \(r\) = radius of the sphere

Notice that the radius is cubed in this formula. That means you multiply the radius by itself three times.

Worked Example 3: Volume of a Sphere

Find the volume of a sphere with radius \(5\text{ cm}\).

Step 1: Write the formula.

Step 2: Substitute the value.

Step 3: Simplify.

Step 4: Approximate.

Answer: The volume is \(\frac{500}{3}\pi\text{ cm}^3\), or about \(523.33\text{ cm}^3\).

5. Comparing the Formulas

It helps to study the formulas together:

• Cylinder: $$V = \pi r^2 h$$
• Cone: $$V = \frac{1}{3}\pi r^2 h$$
• Sphere: $$V = \frac{4}{3}\pi r^3$$

Notice these patterns:

• Cylinder and cone both use \(r^2h\) because they depend on a circular base and a height.
• The cone has one-third the volume of a matching cylinder.
• The sphere uses \(r^3\) and does not have a height in the formula.

6. Be Careful with Radius and Diameter

Many mistakes happen because students use the diameter instead of the radius.

Remember:

If the diameter of a sphere is \(12\text{ cm}\), then the radius is:

Always check whether the problem gives you radius or diameter before using a formula.

Worked Example 4: Using Diameter First

A sphere has a diameter of \(14\text{ in}\). Find its volume.

Step 1: Find the radius.

Step 2: Use the sphere formula.

Step 3: Substitute.

Step 4: Approximate.

Answer: The volume is \(\frac{1372}{3}\pi\text{ in}^3\), or about \(1435.01\text{ in}^3\).

7. Units of Volume

Since volume measures space inside a 3D object, the units must always be cubic units.

Examples:

• \(\text{cm}^3\)
• \(\text{m}^3\)
• \(\text{ft}^3\)

If your measurements are in centimeters, your final answer should be in cubic centimeters. If your measurements are in meters, your answer should be in cubic meters.

8. Common Mistakes to Avoid

• Using diameter instead of radius without dividing by 2.
• Forgetting the \(\frac{1}{3}\) in the cone formula.
• Forgetting to cube the radius in the sphere formula.
• Leaving off units or writing square units instead of cubic units.
• Entering values incorrectly into a calculator. Use parentheses when needed.

9. Problem-Solving Tips

1. Identify the solid: cylinder, cone, or sphere.
2. Write the correct formula.
3. Check whether you have radius or diameter.
4. Substitute carefully.
5. Simplify step by step.
6. Include cubic units in your final answer.

10. Quick Practice Check

Try these on your own:

• A cylinder with radius \(2\text{ cm}\) and height \(9\text{ cm}\)
• A cone with radius \(6\text{ m}\) and height \(4\text{ m}\)
• A sphere with radius \(3\text{ in}\)

Answers:

• Cylinder: $$V = \pi (2)^2(9) = 36\pi\text{ cm}^3$$
• Cone: $$V = \frac{1}{3}\pi (6)^2(4) = 48\pi\text{ m}^3$$
• Sphere: $$V = \frac{4}{3}\pi (3)^3 = 36\pi\text{ in}^3$$

Summary

Volume measures how much space a 3D solid takes up, and it is written in cubic units. For a cylinder, use $$V = \pi r^2 h$$. For a cone, use $$V = \frac{1}{3}\pi r^2 h$$. For a sphere, use $$V = \frac{4}{3}\pi r^3$$.

To solve volume problems correctly, first identify the shape, then make sure you are using the radius. Substitute carefully, simplify step by step, and always include the correct cubic units in your answer.

Cavalieri's Principle is a powerful idea in geometry that helps us compare the volumes of solids.

It says that if two solids have the same height, and if the areas of their cross-sections are equal at every level, then the two solids have the same volume.

In simpler words: imagine slicing two different 3D shapes horizontally. If every slice in one shape has the same area as the slice at the same height in the other shape, then both shapes take up the same amount of space.

This idea is called Cavalieri's Principle, named after the mathematician Bonaventura Cavalieri.

Why is this useful? Sometimes two solids look different, but their slices match perfectly in area. Instead of finding volume in a complicated way, we can use Cavalieri's Principle to show the volumes are equal.

The principle:

If two solids have:

• the same height, and
• equal cross-sectional areas at every height,

then their volumes are equal.

We can write this idea as:

if for every height level,

$$A_1 = A_2,$$

and both solids have the same height, then

$$V_1 = V_2.$$

What is a cross-section?

A cross-section is the 2D shape made when a solid is sliced by a plane.

For Cavalieri's Principle, we usually imagine slices that are:

• parallel to the base,
• very thin, and
• taken at the same height in both solids.

For example, if you slice a stack of papers straight across, each paper acts like a thin cross-section.

Key idea to remember: The solids do not need to have the same shape. They only need:

• equal height, and
• equal slice area at every matching height.

This is why a slanted solid and a straight solid can sometimes have the same volume.

A common example: prisms

Think about a right prism and an oblique prism.

A right prism stands straight up. An oblique prism leans to one side. They may look different, but if they have:

• the same base area, and
• the same height,

then every horizontal slice has the same area.

So by Cavalieri's Principle, their volumes are equal.

This supports the prism volume formula:

$$V = Bh,$$

where:

• \(B\) is the area of the base

• \(h\) is the height

Another common example: cylinders

A right cylinder and a slanted cylinder can also have the same volume if they have the same base area and the same height.

Even though one leans, each horizontal slice is still a circle of the same area at the same height.

So their volumes are equal too.

How to use Cavalieri's Principle

1. Check that the two solids have the same height.

2. Compare the cross-sectional area at any height.

3. If those areas are equal for every height, then the volumes are equal.

Worked Example 1: Right prism and oblique prism

A right prism and an oblique prism each have base area \(20 \text{ cm}^2\) and height \(8 \text{ cm}\). Do they have the same volume? What is the volume?

Step 1: Compare heights

Both have height \(8\) cm.

Step 2: Compare cross-sections

At every horizontal level, each slice has the same area as the base, which is \(20 \text{ cm}^2\).

Step 3: Apply Cavalieri's Principle

Their cross-sectional areas match at every height, so the two prisms have the same volume.

Step 4: Calculate the volume

$$V = Bh = 20 \times 8 = 160 \text{ cm}^3$$

Answer: Yes, they have the same volume, and each volume is \(160 \text{ cm}^3\).

Worked Example 2: Two cylinders

Cylinder A stands straight. Cylinder B is slanted. Both have radius \(3\) cm and height \(10\) cm. Show that they have equal volume.

Step 1: Find the area of a horizontal slice

Each horizontal slice is a circle with radius \(3\) cm.

So the slice area is:

$$A = \pi r^2 = \pi(3)^2 = 9\pi \text{ cm}^2$$

Step 2: Compare heights

Both cylinders have height \(10\) cm.

Step 3: Apply Cavalieri's Principle

At every height, both cylinders have equal cross-sectional area \(9\pi \text{ cm}^2\). Since the heights are equal, the volumes are equal.

Step 4: Calculate volume

$$V = Bh = 9\pi \times 10 = 90\pi \text{ cm}^3$$

Answer: The cylinders have equal volume, and each has volume \(90\pi \text{ cm}^3\).

Worked Example 3: Matching slices

Two solids each have height \(12\) cm. At every height \(h\), the cross-sectional area of Solid A is \(15 \text{ cm}^2\), and the cross-sectional area of Solid B is also \(15 \text{ cm}^2\). Compare their volumes.

Step 1: Compare heights

Both heights are \(12\) cm.

Step 2: Compare slice areas

At every height, the cross-sectional areas are equal:

$$A_A = 15 \text{ cm}^2 \quad \text{and} \quad A_B = 15 \text{ cm}^2$$

Step 3: Use Cavalieri's Principle

Since the heights are equal and all matching cross-sections have equal area, the volumes are equal.

Answer: The two solids have the same volume.

Notice that in this example we do not even need to know the exact shape of either solid.

Worked Example 4: When Cavalieri's Principle does not apply

Solid X and Solid Y each have height \(9\) cm. At some heights, their cross-sectional areas are equal, but at other heights they are different. Can we conclude that their volumes are equal?

Step 1: Check the height

Yes, the heights are equal.

Step 2: Check the slice areas

The areas are only equal at some heights, not at every height.

Conclusion

No, we cannot use Cavalieri's Principle.

Answer: We cannot conclude that the volumes are equal, because the cross-sectional areas must match at every level.

Important reminders

• The solids can look different and still have equal volume.

• The slices must be compared at the same height.

• The cross-sectional areas must be equal for all heights, not just one or two.

• The solids must have the same total height.

Common mistakes

• Thinking two solids need to have the same shape. They do not.

• Checking only the base areas. Equal bases alone are not enough unless the slice areas match all the way up.

• Ignoring height. Even if slice areas match, different heights can give different volumes.

• Comparing slices taken in different directions. The slices should be parallel and measured at matching heights.

Why this matters in volume formulas

Cavalieri's Principle helps explain why many volume formulas work even when solids are slanted.

For example, a slanted prism has the same volume as a straight prism with the same base area and height. A slanted cylinder has the same volume as a straight cylinder with the same base area and height.

So the formula

$$V = Bh$$

still works for these solids.

Quick check questions

1. If two solids have equal height and equal cross-sectional area at every level, what can you say about their volumes?

2. Can two solids have the same volume even if they look different?

3. If two solids have the same base area and same height, does Cavalieri's Principle always apply automatically?

Answers:

1. Their volumes are equal.

2. Yes, as long as the conditions of Cavalieri's Principle are met.

3. No. We need equal cross-sectional areas at every height, not just equal base area.

Summary

Cavalieri's Principle says that if two solids have the same height and equal cross-sectional areas at every height, then they have equal volume.

This principle helps us compare different-looking solids and explains why slanted and straight versions of some shapes can have the same volume.

When using it, always check both conditions: same height and equal slice area at every level.

Surface Area of Composite Solids is about finding the total area of all the outside faces that are exposed on a 3D object made by joining two or more simple solids.

In Grade 10, this often means combining shapes such as prisms, cylinders, cones, hemispheres, and pyramids. The key idea is simple: only count the surfaces you can see from the outside. Any face where two solids are joined together is inside the object, so it is not included in the surface area.

This lesson will show you how to recognize visible surfaces, choose the correct surface area formulas, and avoid the common mistake of counting hidden bases.

1. What is a composite solid?

A composite solid is a 3D shape made by putting together two or more basic solids.

Examples include:

• a cylinder with a hemisphere on top
• a cone mounted on a hemisphere
• two rectangular prisms joined together
• a cube with a square pyramid on top

To find the surface area of a composite solid, you do not always add the total surface areas of the separate solids directly. You must first think about which parts are hidden where the solids touch.

2. Main idea: Add exposed surfaces only

When two solids are attached, the touching faces are inside the shape. Since they are not exposed to the outside, they should not be counted.

A good strategy is:

1. Draw or imagine the composite solid clearly.
2. Identify each surface on the outside.
3. Mark any surfaces that are hidden because of joining.
4. Use the correct formulas for only the visible parts.
5. Add those visible areas together.

3. Useful surface area formulas

Here are the formulas you will most often use.

Rectangular prism

If the length is \(l\), width is \(w\), and height is \(h\), then

$$SA = 2(lw + lh + wh)$$

Cube

If each edge is \(a\), then

$$SA = 6a^2$$

Cylinder

Total surface area:

$$SA = 2\pi r^2 + 2\pi rh$$

The two circular bases give \(2\pi r^2\), and the curved surface gives \(2\pi rh\).

Cone

Total surface area:

$$SA = \pi r^2 + \pi rl$$

where \(r\) is the radius and \(l\) is the slant height.

The curved surface area only is:

$$\pi rl$$

Sphere

$$SA = 4\pi r^2$$

Hemisphere

A hemisphere is half a sphere.

Its curved surface area is:

$$2\pi r^2$$

If you include the flat circular base, the total becomes:

$$3\pi r^2$$

In composite solids, the flat base is often attached to another solid, so usually you count only the curved part.

Square pyramid

If the base side length is \(a\) and slant height is \(l\), then the lateral surface area is

$$2al$$

If you include the base, total surface area is

$$a^2 + 2al$$

4. Important idea: Total surface area vs curved/lateral surface area

For composite solids, sometimes you need the whole surface area formula, and sometimes you need only the curved or lateral part.

For example:

• If a cone sits on a hemisphere, the cone’s base circle and the hemisphere’s flat circular face are touching, so both are hidden.
• That means you use the cone’s curved area \(\pi rl\), not its full formula \(\pi r^2 + \pi rl\).
• You also use the hemisphere’s curved area \(2\pi r^2\), not \(3\pi r^2\).

This is one of the most common places students lose marks.

5. A step-by-step method

Whenever you see a composite solid, use this method:

1. Break the object into basic solids.
2. List the outside surfaces.
3. Remove hidden surfaces. If two shapes are joined, the touching faces are not exposed.
4. Write the formula for each visible part.
5. Substitute values carefully.
6. Add the visible areas.
7. Write the units. Surface area is always in square units such as \(\text{cm}^2\), \(\text{m}^2\), or \(\text{in}^2\).

Worked Example 1: Two rectangular prisms joined together

A solid is made by joining two identical cubes of side \(4\text{ cm}\) face to face. Find the exposed surface area.

Step 1: Find the surface area of one cube.

$$SA = 6a^2 = 6(4^2) = 6(16) = 96\text{ cm}^2$$

So two separate cubes would have

$$2 \times 96 = 192\text{ cm}^2$$

Step 2: Subtract the hidden faces.

When the cubes are joined, one square face from each cube is hidden.

Area of one square face:

$$4^2 = 16\text{ cm}^2$$

Two hidden faces:

$$2 \times 16 = 32\text{ cm}^2$$

Step 3: Subtract.

$$192 - 32 = 160\text{ cm}^2$$

Answer: The exposed surface area is \(160\text{ cm}^2\).

Why this works: We started with all faces from both solids, then removed the parts that became internal when the solids were joined.

Worked Example 2: Cylinder with a hemisphere on top

A cylinder of radius \(3\text{ cm}\) and height \(8\text{ cm}\) has a hemisphere of the same radius attached on top. Find the exposed surface area.

Step 1: Identify visible parts.

• curved surface of the cylinder
• bottom circular base of the cylinder
• curved surface of the hemisphere

The top circle of the cylinder and the flat circular base of the hemisphere are touching, so they are hidden.

Step 2: Write formulas.

Cylinder curved surface area:

$$2\pi rh = 2\pi(3)(8) = 48\pi$$

Bottom base of cylinder:

$$\pi r^2 = \pi(3^2) = 9\pi$$

Hemisphere curved surface area:

$$2\pi r^2 = 2\pi(3^2) = 18\pi$$

Step 3: Add visible areas.

$$48\pi + 9\pi + 18\pi = 75\pi$$

Answer:

$$75\pi\text{ cm}^2$$

or approximately

$$75\pi \approx 235.6\text{ cm}^2$$

Worked Example 3: Cone mounted on a hemisphere

A cone is mounted on a hemisphere. Both have radius \(5\text{ cm}\). The cone has slant height \(12\text{ cm}\). Find the exposed surface area.

Step 1: Identify visible surfaces.

• the curved surface of the cone
• the curved surface of the hemisphere

The cone’s circular base and the hemisphere’s flat base are joined together, so neither is exposed.

Step 2: Find the cone’s curved surface area.

$$\pi rl = \pi(5)(12) = 60\pi$$

Step 3: Find the hemisphere’s curved surface area.

$$2\pi r^2 = 2\pi(5^2) = 2\pi(25) = 50\pi$$

Step 4: Add.

$$60\pi + 50\pi = 110\pi$$

Answer:

$$110\pi\text{ cm}^2$$

or approximately

$$110\pi \approx 345.6\text{ cm}^2$$

Important note: A common mistake is to add \(\pi r^2\) for the cone’s base or another \(\pi r^2\) for the hemisphere’s flat face. Those are inside the solid, so they must not be counted.

Worked Example 4: Cube with a square pyramid on top

A cube has side length \(6\text{ cm}\). A square pyramid with base side length \(6\text{ cm}\) and slant height \(5\text{ cm}\) is placed on top of the cube. Find the exposed surface area.

Step 1: Identify hidden and visible parts.

The square base of the pyramid sits exactly on the top face of the cube.

So:

• the top face of the cube is hidden
• the base of the pyramid is hidden
• the other 5 faces of the cube are exposed
• the 4 triangular faces of the pyramid are exposed

Step 2: Find exposed area of the cube.

Each face of the cube has area

$$6^2 = 36\text{ cm}^2$$

Five exposed faces:

$$5 \times 36 = 180\text{ cm}^2$$

Step 3: Find lateral area of the pyramid.

For a square pyramid, lateral area is

$$2al$$

Substitute \(a=6\) and \(l=5\):

$$2(6)(5) = 60\text{ cm}^2$$

Step 4: Add.

$$180 + 60 = 240\text{ cm}^2$$

Answer: The exposed surface area is \(240\text{ cm}^2\).

6. Common mistakes to avoid

• Counting hidden faces. If two solids touch, that shared face is inside and should not be included.
• Using total surface area when only curved area is needed. For cones, cylinders, and hemispheres, check whether bases are exposed.
• Forgetting the bottom. In shapes like a cylinder with something on top, the bottom may still be visible.
• Mixing up radius and diameter. If the diameter is given, divide by 2 to get the radius.
• Wrong units. Surface area must be written in square units.

7. Quick checklist before finishing a problem

• Did I identify all the solids correctly?
• Did I include only the outer, visible surfaces?
• Did I exclude all joined faces?
• Did I use the correct formulas?
• Did I write the answer in square units?

8. How to think about these problems

Imagine wrapping the solid in paper or painting its outside. The amount of paper or paint needed matches the exposed surface area.

If a face is glued to another solid, paper or paint cannot reach it. That means it should not be counted.

This way of thinking can help you decide which surfaces belong in the final answer.

Summary

To find the surface area of a composite solid, break the shape into simpler solids and add the areas of only the visible outside surfaces. Do not count faces or bases that are hidden where solids are joined. Always choose the correct formula for each exposed part, then add carefully and include square units in your final answer.

Lesson: Volume of Composite Solids

In 2D geometry, you find the area of shapes. In 3D geometry, you find the volume of solids. Volume tells us how much space a solid takes up.

A composite solid is a 3D object made by joining two or more simple solids, such as prisms, cylinders, cones, pyramids, or hemispheres. To find the volume of a composite solid, you usually break it into simpler parts, find the volume of each part, and then add the volumes.

This lesson focuses on cases where the total volume is found by simply summing the volumes of the pieces.

1. Key Idea

If a solid is made from several smaller solids joined together, then:

This works when the parts fit together without overlapping in a way that would count the same space twice.

2. Common Volume Formulas

Before working with composite solids, you need to know the volume formulas for basic solids.

• Rectangular prism: \(V = lwh\)
• Cube: \(V = s^3\)
• Cylinder: \(V = \pi r^2 h\)
• Triangular prism: \(V = (\text{area of triangular base}) \times (\text{length})\)
• Sphere: \(V = \frac{4}{3}\pi r^3\)
• Hemisphere: \(V = \frac{1}{2}\left(\frac{4}{3}\pi r^3\right) = \frac{2}{3}\pi r^3\)
• Cone: \(V = \frac{1}{3}\pi r^2 h\)
• Pyramid: \(V = \frac{1}{3}(\text{area of base})h\)

3. Steps for Finding the Volume of a Composite Solid

1. Identify the simple solids that make up the composite figure.
2. Write the dimensions for each part clearly.
3. Choose the correct volume formula for each part.
4. Calculate each volume carefully.
5. Add the volumes.
6. Write the answer in cubic units, such as \(\text{cm}^3\), \(\text{m}^3\), or \(\text{in}^3\).

4. Important Reminders

• All measurements must be in the same unit before calculating volume.
• Volume is always measured in cubic units.
• If the figure looks complicated, draw lines to split it into familiar solids.
• Check that you are adding because the solids are combined. In some other problems, you may need subtraction, but this lesson focuses on addition only.

Worked Example 1: Two Rectangular Prisms

A solid is made by joining two rectangular prisms.

• Prism A: length \(8\text{ cm}\), width \(4\text{ cm}\), height \(3\text{ cm}\)
• Prism B: length \(5\text{ cm}\), width \(4\text{ cm}\), height \(2\text{ cm}\)

Find the total volume.

Step 1: Find the volume of Prism A

Step 2: Find the volume of Prism B

Step 3: Add the volumes

Answer: The total volume is \(136\text{ cm}^3\).

Worked Example 2: Rectangular Prism and Cylinder

A toy is made from a rectangular prism with a cylinder attached on top.

• Rectangular prism: length \(10\text{ cm}\), width \(6\text{ cm}\), height \(4\text{ cm}\)
• Cylinder: radius \(2\text{ cm}\), height \(7\text{ cm}\)

Find the total volume. Use \(\pi \approx 3.14\).

Step 1: Volume of the rectangular prism

Step 2: Volume of the cylinder

Step 3: Add the volumes

Answer: The total volume is \(327.92\text{ cm}^3\).

Worked Example 3: Triangular Prism and Rectangular Prism

A solid is made from a rectangular prism with a triangular prism on top, like a simple house shape.

• Rectangular prism: length \(12\text{ m}\), width \(5\text{ m}\), height \(4\text{ m}\)
• Triangular prism on top: triangular base has base \(5\text{ m}\) and height \(3\text{ m}\), and the prism length is \(12\text{ m}\)

Find the total volume.

Step 1: Volume of the rectangular prism

Step 2: Find the area of the triangular base

Step 3: Volume of the triangular prism

Step 4: Add the volumes

Answer: The total volume is \(330\text{ m}^3\).

Worked Example 4: Cylinder and Hemisphere

A solid is made of a cylinder with a hemisphere on top.

• Cylinder: radius \(3\text{ cm}\), height \(10\text{ cm}\)
• Hemisphere: radius \(3\text{ cm}\)

Find the total volume. Leave your answer in terms of \(\pi\), then give a decimal approximation.

Step 1: Volume of the cylinder

Step 2: Volume of the hemisphere

Step 3: Add the volumes

Step 4: Decimal approximation

Answer: The total volume is \(108\pi\text{ cm}^3\), or about \(339.12\text{ cm}^3\).

5. How to Recognize Composite Solids in Problems

Many real-life objects are composite solids. For example:

• a building made of prism-shaped sections
• a water tank with a cylindrical body and a rounded top
• a toy block made by stacking shapes
• a tent shape made from a prism and a triangular prism

When you see a shape like this, ask yourself:

• What simple solids do I recognize?
• What dimensions belong to each part?
• Do I need to add all the volumes?

6. Common Mistakes to Avoid

• Using the wrong formula: Make sure you know whether the part is a prism, cylinder, cone, or another solid.
• Forgetting part of the shape: Check that every section is included.
• Mixing units: Do not combine centimeters and meters without converting first.
• Using square units instead of cubic units: Volume should be written as units cubed.
• Confusing radius and diameter: If given the diameter, divide by 2 to get the radius.

7. Quick Check

Try these on your own:

1. A composite solid is made of two cubes. One has side length \(3\text{ cm}\), and the other has side length \(2\text{ cm}\). What is the total volume?
2. A rectangular prism has dimensions \(9\text{ m} \times 4\text{ m} \times 2\text{ m}\). A cylinder with radius \(1\text{ m}\) and height \(5\text{ m}\) is attached. What is the total volume in terms of \(\pi\)?

Answers:

1. \(3^3 + 2^3 = 27 + 8 = 35\text{ cm}^3\)
2. Rectangular prism: \(9 \times 4 \times 2 = 72\text{ m}^3\); cylinder: \(\pi(1^2)(5) = 5\pi\text{ m}^3\); total: \(72 + 5\pi\text{ m}^3\)

Summary

The volume of a composite solid is found by splitting the solid into simple 3D shapes, finding the volume of each one, and adding them together.

The most important skills are recognizing the parts, choosing the correct formula for each part, and keeping units consistent. If you work step by step, even complicated solids become much easier to handle.

Concept of Cavities and Material Removed

In many 3D geometry problems, a solid object does not stay completely full. Sometimes a part is cut out, drilled out, or hollowed out. The empty part inside the solid is called a cavity.

To solve these questions, we usually need to find one or both of these:

• Volume of material left after removing a part
• New surface area after the cavity is made

This idea is very useful in real life. For example, a metal block may be drilled to make a pipe opening, or a wooden cube may have a hole cut through it. In such cases, the amount of material used changes, and the visible surface also changes.

In this lesson, you will learn how to:

• Find the volume of a solid after a cavity is removed
• Find the surface area after drilling or cutting out a shape
• Decide which surfaces are lost and which new surfaces are created

1. Main Idea: Volume After Material Is Removed

When a shape is cut out from a solid, the remaining volume is:

This is the most important rule for cavity questions.

For example, if a cylindrical hole is drilled through a cuboid, then:

• First find the volume of the cuboid
• Then find the volume of the cylinder removed
• Subtract the cylinder from the cuboid

Useful volume formulas

• Cuboid: \(V = lbh\)
• Cube: \(V = a^3\)
• Cylinder: \(V = \pi r^2 h\)
• Cone: \(V = \frac{1}{3}\pi r^2 h\)
• Sphere: \(V = \frac{4}{3}\pi r^3\)

2. Surface Area After Material Is Removed

Surface area questions are a little more careful than volume questions. When material is removed, two things may happen:

• Some old surface may disappear
• Some new inner surface may be exposed

So we do not always just subtract areas. We must think about the new shape carefully.

General method for surface area

1. Start with the original surface area.
2. Subtract the area of any surface parts that are cut away.
3. Add the area of any new surfaces created inside the cavity.

For a cylinder drilled straight through a block:

• The circular openings remove two circles from the outer faces.
• The curved wall of the cylindrical hole becomes a new exposed surface.

The curved surface area of a cylinder is:

The area of one circular opening is:

3. How to Think About a Drilled Hole

Suppose a cylindrical hole of radius \(r\) is drilled completely through a block of thickness \(h\).

Then:

• Volume removed = volume of the cylinder = \(\pi r^2 h\)
• New inside area added = curved surface area of cylinder = \(2\pi rh\)
• Area removed from outer faces = area of the circular openings

If the hole goes through both ends, then two circles are removed, so total area removed from the original outside faces is:

So if a cylindrical hole goes right through a solid, the new surface area often becomes:

This formula works only when the hole goes fully through and creates two circular openings.

4. Blind Cavities and Through Holes

There are two common kinds of cavities:

• Through hole: the hole passes completely through the solid
• Blind hole: the hole goes in only part of the way

For a blind cylindrical hole:

• Only one circle is removed from the outer face
• The curved surface inside is added
• The circular base at the bottom of the hole is also a new exposed surface

So for a blind cylindrical cavity of depth \(h\):

• Volume removed = \(\pi r^2 h\)
• New surfaces added = curved area \(2\pi rh\) and bottom circle \(\pi r^2\)
• Area removed from outside = one top circle \(\pi r^2\)

Notice something interesting: the one removed circle and the one bottom circle are equal. So in some cases they cancel each other, and only the curved surface changes the total area. But you should still think through the surfaces instead of memorizing blindly.

5. Worked Example 1: Volume Left After a Cylindrical Hole Is Drilled Through a Cuboid

A cuboid has length \(12\) cm, breadth \(8\) cm, and height \(10\) cm. A cylindrical hole of radius \(2\) cm is drilled completely through the height of the cuboid. Find the volume of material left.

Step 1: Find the volume of the cuboid

Step 2: Find the volume of the cylindrical hole

The height of the cylinder is the same as the height of the cuboid, so \(h = 10\) cm.

Using \(\pi \approx 3.14\):

Step 3: Subtract

Approximately,

Answer: The volume of material left is \(960 - 40\pi\text{ cm}^3\), or about \(834.4\text{ cm}^3\).

6. Worked Example 2: New Surface Area of the Same Cuboid

Using the same cuboid and cylindrical hole as above, find the new total surface area.

Step 1: Find the original surface area of the cuboid

Step 2: Subtract the two circular parts removed

The hole passes completely through, so two circular openings are formed.

Step 3: Add the curved surface area of the cylindrical hole

Step 4: Find the new surface area

Using \(\pi \approx 3.14\):

Answer: The new surface area is \(592 + 32\pi\text{ cm}^2\), or about \(692.48\text{ cm}^2\).

7. Worked Example 3: Blind Cylindrical Cavity in a Cube

A cube has side \(10\) cm. A cylindrical cavity of radius \(3\) cm and depth \(6\) cm is drilled into one face of the cube. Find:

• the volume of material left
• the new surface area

Step 1: Volume of the cube

Step 2: Volume removed

Step 3: Volume left

Approximately,

Now the surface area

Step 4: Original surface area of cube

Step 5: Surface changes

• One circular top piece is removed: \(\pi r^2 = 9\pi\)
• Curved inner surface is added: \(2\pi rh = 2\pi \times 3 \times 6 = 36\pi\)
• Bottom circular base of cavity is added: \(\pi r^2 = 9\pi\)

Step 6: New surface area

Approximately,

Answer:

• Volume left = \(1000 - 54\pi\text{ cm}^3\) \(\approx 830.44\text{ cm}^3\)
• New surface area = \(600 + 36\pi\text{ cm}^2\) \(\approx 713.04\text{ cm}^2\)

8. Worked Example 4: Conical Cavity in a Cylinder

A solid cylinder has radius \(5\) cm and height \(12\) cm. A conical cavity with the same radius \(5\) cm and height \(12\) cm is carved out from the top. Find the volume of material left.

Step 1: Volume of cylinder

Step 2: Volume of cone removed

Step 3: Volume left

Approximately,

If a surface area question is asked here, remember that the top circular face of the cylinder is removed and the curved inner surface of the cone becomes exposed.

9. Common Mistakes to Avoid

• Forgetting to subtract the removed volume. Always start with original volume, then subtract the cavity volume.
• Using the wrong height. In a drilled hole, the height of the cylinder is the depth or thickness through which it is drilled.
• Ignoring new inner surfaces. A hole creates inside walls that count in surface area.
• Subtracting surface area only. Surface area must be adjusted by both removing old exposed parts and adding new exposed parts.
• Mixing radius and diameter. If diameter is given, remember \(r = \frac{d}{2}\).

10. Problem-Solving Steps

Whenever you see a cavities question, use this simple plan:

1. Identify the original solid.
2. Identify the shape removed.
3. Write the needed formulas.
4. For volume, do:
   \(\text{Original volume} - \text{Removed volume}\)
5. For surface area, carefully list:
   
   • surfaces lost
   • surfaces added
6. Check units: volume in \(\text{cm}^3\), surface area in \(\text{cm}^2\).

11. Quick Recap of Important Formulas

• Cuboid volume: \(lbh\)
• Cube volume: \(a^3\)
• Cylinder volume: \(\pi r^2 h\)
• Cone volume: \(\frac{1}{3}\pi r^2 h\)
• Cuboid surface area: \(2(lb + bh + lh)\)
• Cube surface area: \(6a^2\)
• Curved surface area of cylinder: \(2\pi rh\)
• Area of circle: \(\pi r^2\)

Summary

A cavity means some material is removed from a solid. To find the remaining volume, subtract the volume of the removed part from the original volume.

For surface area, think carefully about what outer surface is cut away and what new inner surface is exposed. This is the key to solving drilling and hollowing problems correctly.

With practice, these questions become easier if you always separate the problem into original solid, removed shape, volume change, and surface area change.

Lesson: Conversion of Solids via Melting and Recasting

In many real-life situations, a solid object is melted and then recast into a new shape. For example, a metal cube may be melted and made into small spheres, or a cylindrical candle may be melted and shaped into another cylinder.

The most important idea in these questions is this: when a solid is melted and recast, its volume stays the same, as long as there is no loss of material.

This means we do not compare surface areas. We compare only volumes.

Main Principle

If one solid is melted and turned into another solid, then

$$\text{Volume of original solid} = \text{Volume of new solid}$$

This single idea helps us solve all recasting problems.

Why volume stays the same

Volume tells us how much space a solid occupies. When a material like metal, wax, or clay is melted and reshaped, the amount of material does not change. Only the shape changes.

So if a large object is converted into many smaller objects, the total volume of all the smaller objects will be equal to the volume of the large object.

Important note

If the question says there is some loss of material, then the volumes will not be exactly equal. But in most standard school questions, we assume no loss unless told otherwise.

Common solids and their volume formulas

• Cube: \(V = a^3\), where \(a\) is the side
• Cuboid: \(V = l \times b \times h\)
• Cylinder: \(V = \pi r^2 h\)
• Cone: \(V = \frac{1}{3}\pi r^2 h\)
• Sphere: \(V = \frac{4}{3}\pi r^3\)
• Hemisphere: \(V = \frac{2}{3}\pi r^3\)

How to solve melting and recasting questions

1. Identify the original solid and write its volume formula.
2. Identify the new solid and write its volume formula.
3. If one solid is made into many identical solids, multiply the volume of one new solid by the number of solids.
4. Set the volumes equal.
5. Solve for the unknown quantity.

In short:

$$\text{Original volume} = \text{Total recast volume}$$

Case 1: One solid is recast into one new solid

If a solid is melted and made into one other solid, then we simply equate the two volumes.

For example, if a cube is melted into a sphere:

$$a^3 = \frac{4}{3}\pi r^3$$

Case 2: One solid is recast into many identical solids

If one large solid is melted to make many smaller identical solids, then:

$$\text{Volume of large solid} = n \times \text{Volume of one small solid}$$

where \(n\) is the number of smaller solids.

For example, if a sphere is melted to make \(8\) identical smaller spheres:

$$\frac{4}{3}\pi R^3 = 8 \times \frac{4}{3}\pi r^3$$

Worked Example 1: Cylinder recast into smaller cylinders

A solid cylinder of radius \(6\) cm and height \(16\) cm is melted and recast into \(4\) identical cylinders of radius \(3\) cm each. Find the height of each small cylinder.

Step 1: Volume of original cylinder

$$V = \pi r^2 h = \pi \times 6^2 \times 16 = \pi \times 36 \times 16 = 576\pi$$

Step 2: Volume of one small cylinder

If the height of each small cylinder is \(h\) cm, then

$$V = \pi \times 3^2 \times h = 9\pi h$$

Step 3: Total volume of 4 small cylinders

$$4 \times 9\pi h = 36\pi h$$

Step 4: Equate volumes

$$576\pi = 36\pi h$$

Cancel \(\pi\):

$$576 = 36h$$

$$h = \frac{576}{36} = 16$$

Answer: The height of each small cylinder is 16 cm.

Observation: The radius became half, but since 4 cylinders were made, the height remained the same in this case.

Worked Example 2: Cube melted and recast into spheres

A metallic cube of side \(6\) cm is melted and recast into \(9\) identical solid spheres. Find the radius of each sphere.

Step 1: Volume of the cube

$$V = a^3 = 6^3 = 216 \text{ cm}^3$$

Step 2: Volume of 9 spheres

If the radius of each sphere is \(r\) cm, then volume of one sphere is

$$\frac{4}{3}\pi r^3$$

So volume of 9 spheres is

$$9 \times \frac{4}{3}\pi r^3 = 12\pi r^3$$

Step 3: Equate volumes

$$216 = 12\pi r^3$$

$$r^3 = \frac{216}{12\pi} = \frac{18}{\pi}$$

Using \(\pi = \frac{22}{7}\):

$$r^3 = \frac{18 \times 7}{22} = \frac{63}{11}$$

So

$$r = \sqrt[3]{\frac{63}{11}}$$

This is approximately

$$r \approx 1.79 \text{ cm}$$

Answer: The radius of each sphere is approximately 1.79 cm.

Worked Example 3: Sphere recast into smaller spheres

A solid sphere of radius \(8\) cm is melted and recast into small spheres of radius \(2\) cm each. How many small spheres are formed?

Step 1: Volume of the large sphere

$$V = \frac{4}{3}\pi (8)^3 = \frac{4}{3}\pi \times 512$$

Step 2: Volume of one small sphere

$$V = \frac{4}{3}\pi (2)^3 = \frac{4}{3}\pi \times 8$$

Step 3: Number of small spheres

$$n = \frac{\text{Volume of large sphere}}{\text{Volume of one small sphere}}$$

$$n = \frac{\frac{4}{3}\pi \times 512}{\frac{4}{3}\pi \times 8}$$

Cancel common factors:

$$n = \frac{512}{8} = 64$$

Answer: 64 small spheres are formed.

Shortcut idea: Since both shapes are spheres, the number formed is the cube of the ratio of radii:

$$\left(\frac{8}{2}\right)^3 = 4^3 = 64$$

Worked Example 4: Cone recast into a cylinder

A solid cone of radius \(6\) cm and height \(12\) cm is melted and recast into a cylinder of radius \(3\) cm. Find the height of the cylinder.

Step 1: Volume of cone

$$V = \frac{1}{3}\pi r^2 h$$

$$V = \frac{1}{3}\pi \times 6^2 \times 12$$

$$V = \frac{1}{3}\pi \times 36 \times 12 = 144\pi$$

Step 2: Volume of cylinder

If the height of the cylinder is \(h\), then

$$V = \pi r^2 h = \pi \times 3^2 \times h = 9\pi h$$

Step 3: Equate volumes

$$144\pi = 9\pi h$$

Cancel \(\pi\):

$$144 = 9h$$

$$h = 16$$

Answer: The height of the cylinder is 16 cm.

Things students often get wrong

• Using surface area instead of volume: Melting and recasting depends on volume, not surface area.
• Forgetting the number of new solids: If many objects are made, multiply the volume of one object by the number of objects.
• Using wrong formulas: Be careful to choose the correct volume formula for each solid.
• Not writing units: Volume is usually in cubic units such as \(\text{cm}^3\), and lengths are in cm or m.
• Not checking if material is lost: If the question says some material is wasted, then subtract that loss before equating volumes.

Quick strategy for solving any recasting problem

1. Write the volume formula of the first shape.
2. Write the volume formula of the second shape.
3. If there are many new shapes, multiply by the number.
4. Set both sides equal.
5. Simplify and solve.

Practice thinking

Ask yourself these questions when solving:

• What is the original solid?
• What is the new solid?
• Is it one object or many objects?
• Which quantity is unknown: radius, height, number of solids, or side length?
• Have I used volume and not surface area?

Brief Summary

In conversion of solids by melting and recasting, the key idea is that volume remains unchanged if no material is lost. So we equate the volume of the original solid to the volume of the new solid, or to the total volume of all new solids.

Once you know the volume formulas for common solids like cube, cuboid, cylinder, cone, sphere, and hemisphere, you can solve these questions by forming one equation and finding the unknown value.

Liquid Flow Rate and Capacity Problems connect ideas from geometry and measurement. In these questions, we use the size of a tank or container together with the rate at which liquid flows in or out to find time, volume, or flow rate.

This topic is important because many real-life problems involve filling and emptying containers, such as water tanks, swimming pools, fuel containers, and pipes. To solve these problems correctly, we must understand both capacity and rate.

In this lesson, you will learn how to:

• find the volume or capacity of a tank,
• use flow rate formulas,
• connect cross-sectional area, speed, and flow rate,
• calculate the time needed to fill a container,
• solve problems where liquid is flowing in or out.

1. Understanding capacity

Capacity is the amount of liquid a container can hold. In mathematics, capacity is usually measured using volume units such as cubic centimetres, cubic metres, or litres.

The most useful conversion is:

Also,

This means you must pay close attention to units. A very common mistake is mixing cubic units and litre units without converting.

2. Volume formulas you may need

To solve flow problems, you often first find the volume of the tank. Here are some common formulas:

• Rectangular prism: $$V = lwh$$
• Cylinder: $$V = \pi r^2 h$$
• Cuboid tank filled partly: $$V = (\text{base area})(\text{depth of liquid})$$

If a tank has a constant cross-section, then the volume of liquid inside is:

3. Flow rate

Flow rate tells us how much liquid moves each second, minute, or hour. It can be measured in units like:

• litres per second,
• litres per minute,
• cubic metres per hour,
• cubic centimetres per second.

The key relationship is:

From this, we can rearrange to get two more useful formulas:

These three formulas are the main tools for solving capacity and filling problems.

4. Flow rate from cross-sectional area and speed

Sometimes the question gives the speed of the liquid in a pipe instead of the volume per second. In that case, flow rate is found using:

This works because in 1 second, the liquid moves a distance equal to its speed. So the volume passing through the pipe in 1 second is:

and since distance per second is speed, we get:

where:

• \(Q\) = flow rate,
• \(A\) = cross-sectional area of the pipe,
• \(v\) = speed of the liquid.

If the pipe is circular, then its cross-sectional area is:

5. A step-by-step method for solving problems

1. Read the question carefully.
2. Identify what shape the tank or pipe is.
3. Find the volume or capacity if needed.
4. Make sure all units match.
5. Use the correct formula: \(Q=\frac{V}{t}\), \(V=Qt\), \(t=\frac{V}{Q}\), or \(Q=Av\).
6. Check if the answer should be in seconds, minutes, litres, or cubic metres.

Worked Example 1: Filling a rectangular tank

A rectangular water tank is \(2\text{ m}\) long, \(1.5\text{ m}\) wide, and \(1.2\text{ m}\) high. Water flows into it at \(0.3\text{ m}^3\) per minute. How long will it take to fill the tank?

Step 1: Find the volume of the tank.

Step 2: Use the time formula.

So, the tank will take 12 minutes to fill.

Worked Example 2: Converting litres and cubic centimetres

A container holds \(18\) litres of water. A tap pours water at \(600\text{ cm}^3\) per second. How many seconds will it take to fill the container from empty?

Step 1: Convert litres to cubic centimetres.

Since \(1\text{ litre} = 1000\text{ cm}^3\),

Step 2: Use the time formula.

So, it takes 30 seconds to fill the container.

Worked Example 3: Using area and speed to find flow rate

Water flows through a circular pipe of radius \(4\text{ cm}\) at a speed of \(5\text{ cm/s}\). Find the flow rate.

Step 1: Find the cross-sectional area of the pipe.

Step 2: Use \(Q=Av\).

Step 3: Give an approximate value if needed.

So, the flow rate is \(80\pi\text{ cm}^3/\text{s}\), or about \(251.3\text{ cm}^3/\text{s}\).

Worked Example 4: Pipe filling a cylindrical tank

A cylindrical tank has radius \(35\text{ cm}\) and height \(80\text{ cm}\). Water enters through a pipe of radius \(2\text{ cm}\) at a speed of \(10\text{ cm/s}\). How long will it take to fill the tank?

Step 1: Find the volume of the tank.

Step 2: Find the flow rate of the pipe.

The cross-sectional area of the pipe is:

Now use \(Q=Av\):

Step 3: Find the filling time.

So, the tank takes 2450 seconds to fill.

Step 4: Convert to minutes and seconds.

So, the time is 40 minutes 50 seconds.

6. Problems with liquid already in the tank

Sometimes a tank is not empty at the start. Then you only need to fill the remaining volume.

For example, if a tank holds \(500\) litres and already contains \(120\) litres, then the amount still needed is:

You then use the remaining volume in the formula:

7. Problems with inflow and outflow

Some questions involve one pipe filling a tank while another pipe empties it. In that case, use the net flow rate.

If water enters at \(15\) L/min and leaves at \(4\) L/min, then the actual increase in water is:

Then use:

If the outflow is greater than the inflow, the tank will empty instead of fill.

8. Common mistakes to avoid

• Forgetting unit conversions: litres and cubic centimetres are not the same number.
• Using diameter instead of radius: in formulas like \(\pi r^2\), make sure \(r\) is half the diameter.
• Not matching time units: if flow is in litres per minute, time must be in minutes unless you convert.
• Using total tank volume when only part needs filling.
• Ignoring outflow when liquid is also leaving the tank.

9. Quick check questions

Try these on your own:

1. A tank holds \(2400\text{ cm}^3\). Water flows in at \(80\text{ cm}^3/s\). How long does it take to fill?
2. A pipe delivers \(12\) litres per minute. How much water flows in \(5\) minutes?
3. A circular pipe has radius \(3\text{ cm}\) and water speed \(4\text{ cm/s}\). Find the flow rate.

Answers

1. $$t=\frac{2400}{80}=30\text{ s}$$
2. $$V=Qt=12\times 5=60\text{ litres}$$
3. $$A=\pi(3)^2=9\pi$$ $$Q=Av=9\pi\times 4=36\pi\text{ cm}^3/\text{s}$$

Summary

Liquid flow rate and capacity problems combine volume, area, speed, and time. First find the volume of the tank if needed, then use the correct formula for flow rate or time. Always keep units consistent, especially when converting between litres, cubic centimetres, and cubic metres.

If liquid moves through a pipe and the speed is given, use:

If the flow rate is already known, use:

With careful unit conversion and clear steps, these problems become much easier to solve.

Deriving Surface Area of a Frustum of a Cone

In this lesson, we will learn what a frustum of a cone is and how to find its curved surface area and total surface area.

A frustum of a cone is formed when the top part of a cone is cut off by a plane parallel to its base. The shape left behind has:

• a larger circular base of radius \(R\),
• a smaller circular top of radius \(r\), and
• a slant height \(l\).

This shape appears in many real-life objects, such as buckets, lampshades, cups, and flower pots.

1. Understanding the shape

Imagine a full cone. If we slice it horizontally near the top and remove the small cone above the cut, the remaining solid is a frustum.

So, a frustum can be thought of as:

$$\text{Frustum} = \text{Large cone} - \text{Small cone}$$

This idea helps us derive its surface area formula.

2. Curved surface area of a frustum

The curved surface area of the frustum is the curved surface area of the large cone minus the curved surface area of the small cone.

For a cone, the curved surface area is:

$$\pi r l$$

Suppose:

• the large cone has radius \(R\) and slant height \(L\),
• the small cone has radius \(r\) and slant height \(l_1\).

Then the slant height of the frustum is:

$$l = L - l_1$$

So its curved surface area is:

$$\pi R L - \pi r l_1$$

At first, this does not yet look like the formula we want. To simplify it, we use the fact that the two cones are similar.

Since the cones are similar, their corresponding radii and slant heights are in the same ratio:

$$\frac{R}{L} = \frac{r}{l_1}$$

Cross-multiplying gives:

$$Rl_1 = rL$$

Now look at the curved surface area again:

$$\pi R L - \pi r l_1 = \pi(RL - rl_1)$$

Using \(Rl_1 = rL\), we can rewrite:

$$RL - rl_1 = R(L-l_1) + l_1(R-r)$$

But there is a simpler standard result from similarity and subtraction, which gives:

$$\text{Curved Surface Area of Frustum} = \pi (R+r)l$$

So the formula is:

$$\boxed{\text{CSA} = \pi (R+r)l}$$

where:

• \(R\) = radius of the larger base,
• \(r\) = radius of the smaller base,
• \(l\) = slant height of the frustum.

3. Why this formula makes sense

For a full cone, the curved surface area is \(\pi r l\).

For a frustum, there are two radii, not just one. The formula uses their sum, \(R+r\), because the curved surface lies between the two circular edges.

So instead of:

$$\pi r l$$

we get:

$$\pi(R+r)l$$

4. Total surface area of a frustum

The total surface area includes:

• the curved surface area,
• the area of the top circular face,
• the area of the bottom circular base.

The top circle has area \(\pi r^2\), and the bottom circle has area \(\pi R^2\).

So,

$$\text{Total Surface Area} = \pi(R+r)l + \pi R^2 + \pi r^2$$

Therefore,

$$\boxed{\text{TSA} = \pi(R+r)l + \pi R^2 + \pi r^2}$$

5. Finding the slant height

Sometimes the slant height \(l\) is not given directly. Instead, you may know the vertical height \(h\) of the frustum.

If so, use the right triangle formed by:

• height \(h\),
• difference of radii \((R-r)\),
• slant height \(l\).

By Pythagoras' Theorem,

$$l = \sqrt{h^2 + (R-r)^2}$$

This is very important when solving questions.

6. Steps for solving frustum surface area questions

1. Identify the larger radius \(R\) and smaller radius \(r\).
2. Find the slant height \(l\), if needed.
3. Use $$\text{CSA} = \pi(R+r)l$$ for curved surface area.
4. Use $$\text{TSA} = \pi(R+r)l + \pi R^2 + \pi r^2$$ for total surface area.
5. Simplify the answer, and include units such as \(\text{cm}^2\), \(\text{m}^2\), etc.

Worked Example 1: Finding curved surface area directly

A frustum has larger radius \(R=7\text{ cm}\), smaller radius \(r=4\text{ cm}\), and slant height \(l=5\text{ cm}\). Find its curved surface area.

Step 1: Write the formula.

$$\text{CSA} = \pi(R+r)l$$

Step 2: Substitute the values.

$$\text{CSA} = \pi(7+4)(5)$$

$$= \pi(11)(5)$$

$$= 55\pi$$

Answer:

$$\boxed{55\pi\text{ cm}^2}$$

Using \(\pi \approx 3.14\),

$$55\pi \approx 172.7\text{ cm}^2$$

Worked Example 2: Finding total surface area

A frustum has larger radius \(R=6\text{ cm}\), smaller radius \(r=3\text{ cm}\), and slant height \(l=8\text{ cm}\). Find its total surface area.

Step 1: Find the curved surface area.

$$\text{CSA} = \pi(R+r)l = \pi(6+3)(8) = 72\pi$$

Step 2: Find the areas of the two circular ends.

$$\pi R^2 = \pi(6^2) = 36\pi$$

$$\pi r^2 = \pi(3^2) = 9\pi$$

Step 3: Add them.

$$\text{TSA} = 72\pi + 36\pi + 9\pi$$

$$= 117\pi$$

Answer:

$$\boxed{117\pi\text{ cm}^2}$$

Using \(\pi \approx 3.14\),

$$117\pi \approx 367.38\text{ cm}^2$$

Worked Example 3: Finding slant height first

A frustum has vertical height \(12\text{ cm}\), larger radius \(10\text{ cm}\), and smaller radius \(6\text{ cm}\). Find its curved surface area.

Step 1: Find the difference in radii.

$$R-r = 10-6 = 4$$

Step 2: Use Pythagoras' Theorem to find the slant height.

$$l = \sqrt{h^2 + (R-r)^2}$$

$$l = \sqrt{12^2 + 4^2}$$

$$l = \sqrt{144 + 16}$$

$$l = \sqrt{160}$$

$$l = 4\sqrt{10}$$

Step 3: Use the curved surface area formula.

$$\text{CSA} = \pi(R+r)l$$

$$= \pi(10+6)(4\sqrt{10})$$

$$= 64\pi\sqrt{10}$$

Answer:

$$\boxed{64\pi\sqrt{10}\text{ cm}^2}$$

Approximate value:

$$64\pi\sqrt{10} \approx 635.6\text{ cm}^2$$

Worked Example 4: A real-life style problem

A metal bucket without a lid is shaped like a frustum. Its larger radius is \(14\text{ cm}\), its smaller radius is \(8\text{ cm}\), and its slant height is \(10\text{ cm}\). Find the area of metal used to make the bucket.

Because the bucket has no lid, we need:

• the curved surface area, and
• only the bottom circular base.

Step 1: Find curved surface area.

$$\text{CSA} = \pi(R+r)l$$

$$= \pi(14+8)(10)$$

$$= 220\pi$$

Step 2: Find the bottom base area.

$$\pi R^2 = \pi(14^2) = 196\pi$$

Step 3: Add them.

$$\text{Area of metal} = 220\pi + 196\pi = 416\pi$$

Answer:

$$\boxed{416\pi\text{ cm}^2}$$

Using \(\pi \approx 3.14\),

$$416\pi \approx 1306.24\text{ cm}^2$$

7. Common mistakes to avoid

• Mixing up radius and diameter: If a question gives diameter, divide by 2 to get radius.
• Using height instead of slant height: In surface area formulas, use \(l\), not the vertical height \(h\).
• Forgetting one or both circular ends: Total surface area includes both circles unless the question says otherwise.
• Adding the wrong base in real-life questions: A bucket without a lid has one circular base, not two.
• Using \(R-r\) instead of \(R+r\) in the CSA formula: The correct formula is \(\pi(R+r)l\).

8. Key formulas to remember

For a frustum of a cone:

• Curved surface area: $$\boxed{\text{CSA} = \pi(R+r)l}$$
• Total surface area: $$\boxed{\text{TSA} = \pi(R+r)l + \pi R^2 + \pi r^2}$$
• Slant height from vertical height: $$\boxed{l = \sqrt{h^2 + (R-r)^2}}$$

9. Brief summary

A frustum is the part of a cone left after cutting off the top parallel to the base. Its curved surface area is found using:

$$\pi(R+r)l$$

Its total surface area is the curved area plus the areas of the two circular ends:

$$\pi(R+r)l + \pi R^2 + \pi r^2$$

Always check whether the question gives slant height or vertical height, and be careful about which surfaces are included.

Deriving the Volume of a Frustum of a Cone

In real life, many objects are not full cones. Buckets, flower pots, paper cups, and some drinking glasses are shaped like a frustum of a cone.

A frustum of a cone is formed when the top part of a cone is cut off by a plane parallel to its base. What remains is a solid with two circular ends of different sizes.

In this lesson, we will learn how to derive the volume formula for a frustum of a cone, not just memorize it. Once you understand where the formula comes from, it becomes much easier to use correctly.

1. What does a frustum look like?

A frustum has:

• a larger circular base of radius \(R\)
• a smaller circular top of radius \(r\)
• a vertical height \(h\)

Its volume is the amount of space inside it.

2. Review: Volume of a cone

Before deriving the frustum formula, let us recall the volume of a cone:

This means the volume of a cone depends on:

• the radius of its circular base
• its perpendicular height

3. Main idea behind the derivation

A frustum can be thought of as:

• a large cone
• with a smaller cone removed from the top

So,

This is the key idea.

4. Setting up the cones

Suppose the original large cone has:

• radius \(R\)
• height \(H\)

The small cone that is cut away has:

• radius \(r\)
• height \(x\)

The height of the frustum is the difference between these two heights:

So we can also write:

5. Using similarity of cones

The small cone and the original large cone are similar because the cut is parallel to the base. That means corresponding lengths are in the same ratio.

So,

Since \(H = h + x\), we could work with this relation. But a cleaner way is to express both cone volumes and simplify using similarity.

6. Write the cone volumes

Volume of the large cone:

Volume of the small cone:

Therefore, volume of the frustum is:

Now we want to rewrite this only in terms of \(R\), \(r\), and \(h\).

7. Express heights using similarity

From similar cones,

This gives:

Also, remember:

Using the same ratio idea, the heights are proportional to the radii. This leads to:

From this, the total height \(H\) and small height \(x\) can be written in terms of \(h\):

Now substitute these into the frustum volume expression:

Now use the algebra identity:

So,

The \((R-r)\) terms cancel:

Final formula:

This is the volume of a frustum of a cone.

8. Meaning of the formula

In the formula

• \(R\) is the radius of the larger circular base
• \(r\) is the radius of the smaller circular top
• \(h\) is the perpendicular height of the frustum

Notice that the formula includes:

• \(R^2\)
• \(r^2\)
• the middle term \(Rr\)

That middle term is important. A common mistake is to forget it.

9. Step-by-step method for solving questions

1. Identify the larger radius \(R\), smaller radius \(r\), and height \(h\).
2. Use the formula $$V=\frac{1}{3}\pi h(R^2+Rr+r^2).$$
3. Substitute carefully.
4. Simplify the bracket first.
5. Multiply by \(\frac{1}{3}\pi h\).
6. Write the answer in cubic units, such as \(\text{cm}^3\) or \(\text{m}^3\).

10. Worked Examples

Example 1: Direct use of the formula

A frustum has larger radius \(R=6\) cm, smaller radius \(r=3\) cm, and height \(h=8\) cm. Find its volume.

Step 1: Write the formula

Step 2: Substitute the values

Answer:

Approximate value:

Example 2: A bucket-shaped object

A bucket is shaped like a frustum of a cone. Its top radius is \(10\) cm, bottom radius is \(6\) cm, and height is \(15\) cm. Find its capacity.

Here, the larger radius is:

Use the formula:

Answer:

Approximate value:

So the bucket holds about \(3078.8\text{ cm}^3\), which is about \(3.08\) litres because \(1000\text{ cm}^3 = 1\) litre.

Example 3: Finding volume when diameters are given

A frustum has top diameter \(8\) cm, bottom diameter \(14\) cm, and height \(12\) cm. Find the volume.

Step 1: Convert diameters to radii

Step 2: Use the formula

Answer:

Approximate value:

Example 4: Using the derivation idea

A frustum is formed by cutting a small cone from a larger cone. The larger cone has radius \(9\) cm and height \(18\) cm. The removed small cone has radius \(3\) cm and height \(6\) cm. Find the volume of the frustum.

Since frustum volume = large cone volume \(-\) small cone volume:

Answer:

Let us check using the frustum formula.

The frustum has:

• larger radius \(R=9\)
• smaller radius \(r=3\)
• height \(h=18-6=12\)

The answers match, which confirms the formula.

11. Common mistakes to avoid

• Using diameters instead of radii: Always check whether the question gives radius or diameter.
• Forgetting the middle term \(Rr\): The formula is not just \(R^2+r^2\).
• Using slant height instead of vertical height: In the volume formula, \(h\) must be the perpendicular height.
• Wrong units: Volume must be written in cubic units.

12. Why this formula makes sense

If \(r=0\), then the frustum becomes a full cone. Let us see what happens:

This is exactly the cone formula.

So the frustum formula fits perfectly with what we already know.

13. Quick practice questions

1. A frustum has \(R=5\) cm, \(r=2\) cm, and \(h=9\) cm. Find its volume.
2. A cup is shaped like a frustum with top diameter \(12\) cm, bottom diameter \(6\) cm, and height \(10\) cm. Find its capacity.
3. A frustum has volume \(\frac{1}{3}\pi(7)(49+21+9)\). Identify \(R\), \(r\), and \(h\).

14. Brief summary

A frustum of a cone is made by cutting the top off a cone parallel to the base.

Its volume is found by subtracting the volume of the small cone from the volume of the large cone. Using similarity and algebra, we get the formula:

When solving problems, make sure you use the radii, the perpendicular height, and write the answer in cubic units.