Spatial Geometry and Mensuration

Dimensional Analysis and Scale helps us answer two very important questions in geometry and measurement:

• How do we convert units correctly, especially when the units are more complicated than just metres or centimetres?
• What happens to length, area, and volume when a shape is enlarged or reduced?

This topic is especially useful in spatial geometry and mensuration, where we work with real objects, diagrams, models, maps, and solids.

In this lesson, you will learn how to use dimensional analysis to convert units and how to apply scale factors to 2D and 3D shapes.

1. What is dimensional analysis?

Dimensional analysis is a method of converting units by multiplying by conversion fractions that equal 1. The goal is to cancel unwanted units and end with the unit you want.

For example, since \(1\text{ m} = 100\text{ cm}\), we can write:

$$\frac{100\text{ cm}}{1\text{ m}} \quad \text{or} \quad \frac{1\text{ m}}{100\text{ cm}}$$

Both fractions are equal to 1, but we choose the one that makes the unwanted unit cancel.

Suppose we want to convert \(2.5\text{ m}\) to centimetres:

$$2.5\text{ m} \times \frac{100\text{ cm}}{1\text{ m}} = 250\text{ cm}$$

The unit \(\text{m}\) cancels, leaving \(\text{cm}\).

2. Converting squared and cubed units

When units involve area or volume, the conversion must also be squared or cubed.

For length:

$$1\text{ m} = 100\text{ cm}$$

For area:

$$1\text{ m}^2 = (100\text{ cm})^2 = 10{,}000\text{ cm}^2$$

For volume:

$$1\text{ m}^3 = (100\text{ cm})^3 = 1{,}000{,}000\text{ cm}^3$$

This is a very common place where mistakes happen. Students sometimes think:

$$1\text{ m}^2 = 100\text{ cm}^2$$

but that is incorrect. The conversion factor must match the dimension.

Key idea:

• Length units change by the scale factor itself.
• Area units change by the square of the scale factor.
• Volume units change by the cube of the scale factor.

3. Derived units

Derived units are units made from combining basic units. In geometry and measurement, common derived units include:

• Area: \(\text{cm}^2, \text{m}^2\)
• Volume: \(\text{cm}^3, \text{m}^3\)
• Rates such as density or speed may also appear in other topics, but here we mainly focus on geometric units.

To convert derived units, treat the units carefully and convert each part correctly.

For example, to convert \(0.4\text{ m}^3\) to \(\text{cm}^3\):

$$0.4\text{ m}^3 \times \frac{1{,}000{,}000\text{ cm}^3}{1\text{ m}^3} = 400{,}000\text{ cm}^3$$

4. Understanding scale

A scale factor tells us how much larger or smaller a figure becomes.

If the scale factor is:

• greater than 1, the figure is enlarged,
• between 0 and 1, the figure is reduced.

If a shape is enlarged by a scale factor of \(k\), then:

• every length is multiplied by \(k\),
• every area is multiplied by \(k^2\),
• every volume is multiplied by \(k^3\).

This is one of the most important rules in mensuration.

5. Scale in diagrams, maps, and models

Scale is often written as a ratio, such as \(1:50\) or \(1:1000\).

A scale of \(1:50\) means:

• 1 unit on the drawing represents 50 units in real life.

So if a line on a diagram is \(6\text{ cm}\) and the scale is \(1:50\), the real length is:

$$6 \times 50 = 300\text{ cm} = 3\text{ m}$$

Always make sure the units match before giving the final answer.

6. How resizing affects perimeter, area, and volume

When a 2D shape is resized, all lengths change by the scale factor. Since perimeter is made of lengths added together, perimeter also changes by the scale factor.

However, area covers surface, so it changes more quickly. If the scale factor is \(k\), then area changes by \(k^2\).

For 3D solids, volume changes even more quickly because it depends on three dimensions. If the scale factor is \(k\), then volume changes by \(k^3\).

Example of the pattern:

• If side lengths double \((k=2)\), perimeter doubles, area becomes 4 times as large, and volume becomes 8 times as large.
• If side lengths triple \((k=3)\), perimeter triples, area becomes 9 times as large, and volume becomes 27 times as large.

7. Worked Example 1: Converting area units

Question: Convert \(3.2\text{ m}^2\) to \(\text{cm}^2\).

Step 1: Use the length conversion:

$$1\text{ m} = 100\text{ cm}$$

Step 2: Square it for area:

$$1\text{ m}^2 = 10{,}000\text{ cm}^2$$

Step 3: Convert:

$$3.2\text{ m}^2 \times \frac{10{,}000\text{ cm}^2}{1\text{ m}^2} = 32{,}000\text{ cm}^2$$

Answer: \(3.2\text{ m}^2 = 32{,}000\text{ cm}^2\).

Worked Example 2: Using a map scale

Question: On a map with scale \(1:25{,}000\), the distance between two towns is \(8\text{ cm}\). What is the real distance in kilometres?

Step 1: Interpret the scale.

\(1\text{ cm}\) on the map represents \(25{,}000\text{ cm}\) in real life.

Step 2: Multiply by the map distance:

$$8 \times 25{,}000 = 200{,}000\text{ cm}$$

Step 3: Convert centimetres to metres, then kilometres.

$$200{,}000\text{ cm} = 2{,}000\text{ m} = 2\text{ km}$$

Answer: The real distance is \(2\text{ km}\).

Worked Example 3: Scale factor and area

Question: A rectangle has length \(5\text{ cm}\) and width \(3\text{ cm}\). It is enlarged by a scale factor of \(4\). Find the new area.

Method 1: Find new dimensions first.

New length:

$$5 \times 4 = 20\text{ cm}$$

New width:

$$3 \times 4 = 12\text{ cm}$$

New area:

$$20 \times 12 = 240\text{ cm}^2$$

Method 2: Use the area scale factor.

Original area:

$$5 \times 3 = 15\text{ cm}^2$$

Since the scale factor is \(4\), the area scale factor is:

$$4^2 = 16$$

So the new area is:

$$15 \times 16 = 240\text{ cm}^2$$

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

Worked Example 4: Scale factor and volume

Question: A cube has side length \(2\text{ cm}\). It is enlarged by a scale factor of \(3\). Find the new volume.

Step 1: Find the original volume.

$$V = s^3 = 2^3 = 8\text{ cm}^3$$

Step 2: Use the volume scale factor.

If the linear scale factor is \(3\), then the volume scale factor is:

$$3^3 = 27$$

Step 3: Multiply:

$$8 \times 27 = 216\text{ cm}^3$$

Answer: The new volume is \(216\text{ cm}^3\).

8. A useful problem-solving strategy

When solving dimensional analysis and scale problems, use this checklist:

1. Read the units carefully. Are you working with length, area, or volume?
2. Choose the correct conversion factor. If the unit is squared or cubed, the conversion must also be squared or cubed.
3. Identify the scale factor. Is it an enlargement or a reduction?
4. Apply the correct power of the scale factor.
   • Length or perimeter: \(k\)
   • Area: \(k^2\)
   • Volume: \(k^3\)
5. Check that your final unit makes sense.

9. Common mistakes to avoid

• Forgetting to square or cube the conversion factor.
  Example: \(1\text{ m}^2 \neq 100\text{ cm}^2\). It is \(10{,}000\text{ cm}^2\).
• Using the wrong scale rule.
  Lengths scale by \(k\), areas by \(k^2\), volumes by \(k^3\).
• Mixing units.
  Do not compare cm with m without converting first.
• Confusing map scale with scale factor of enlargement.
  A map scale like \(1:50{,}000\) describes drawing-to-real-life size, while a scale factor like \(2\) or \(\frac{1}{2}\) describes resizing.

10. Quick comparison table

• Length: multiply by \(k\)
• Perimeter: multiply by \(k\)
• Area: multiply by \(k^2\)
• Surface area: multiply by \(k^2\)
• Volume: multiply by \(k^3\)

11. Final summary

Dimensional analysis is a reliable way to convert units by using fractions that cancel unwanted units. It is especially important when working with area and volume, because squared and cubed units must be converted using squared and cubed factors.

Scale tells us how measurements change when figures are enlarged or reduced. If the linear scale factor is \(k\), then lengths and perimeters scale by \(k\), areas by \(k^2\), and volumes by \(k^3\).

If you remember to match the conversion and the scale rule to the dimension, you will solve these problems much more accurately.

Area of Complex Polygons

In real-life geometry, many shapes are not simple rectangles, triangles, or trapezoids. Instead, they are often complex polygons, which are shapes made from several straight sides and can be broken into smaller familiar shapes.

To find the area of a complex polygon, we usually decompose it into simpler polygons such as rectangles, triangles, parallelograms, and trapezoids. Then we find the area of each part and combine the results.

This skill is important in mensuration because floor plans, land plots, tiles, gardens, and composite diagrams often use irregular shapes. If you know how to split a shape carefully, you can still find its exact area.

What is a complex polygon?

A complex polygon in this lesson means a 2D shape with straight sides that is not one basic polygon formula by itself. It may look irregular, but it can still be analyzed by dividing it into simpler parts.

Main idea: find area by breaking the shape into parts or by finding a larger simple shape and subtracting the missing parts.

Two main strategies

• Additive method: Split the figure into smaller standard shapes and add their areas.
• Subtractive method: Enclose the figure in a larger known shape, then subtract the areas of the parts that are missing.

Useful area formulas

• Rectangle: \(A = lw\)
• Square: \(A = s^2\)
• Triangle: \(A = \frac{1}{2}bh\)
• Parallelogram: \(A = bh\)
• Trapezoid: \(A = \frac{1}{2}(a+b)h\)

Here, \(b\) is base, \(h\) is perpendicular height, \(l\) is length, \(w\) is width, and \(a\) and \(b\) in the trapezoid formula are the parallel sides.

Step-by-step method for finding the area of a complex polygon

1. Look at the shape carefully.
2. Decide how to split it into known polygons.
3. Label all needed lengths and heights.
4. Find the area of each simpler shape.
5. Add or subtract the areas as needed.
6. Write the final answer in square units.

Important reminders

• Area is always measured in square units, such as \(\text{cm}^2\), \(\text{m}^2\), or \(\text{units}^2\).
• Be careful not to use a slanted side as a height unless it is actually perpendicular.
• Draw extra lines lightly if needed to create rectangles or triangles.
• If a length is not given directly, try to find it by subtraction from the total length.

Worked Example 1: L-shaped polygon

Find the area of an L-shaped figure made from a vertical rectangle and a horizontal rectangle. Suppose the full outer dimensions are 10 m by 8 m, and a rectangle of 4 m by 3 m is missing from the top-right corner.

This is easiest with the subtractive method.

First find the area of the full outer rectangle:

$$A_{\text{outer}} = 10 \times 8 = 80$$

Now find the area of the missing rectangle:

$$A_{\text{missing}} = 4 \times 3 = 12$$

Subtract:

$$A_{\text{L-shape}} = 80 - 12 = 68$$

Answer: The area is \(68\text{ m}^2\).

Why this works: The L-shape is really a full rectangle with one piece cut out. So subtracting the missing part gives the exact area.

Worked Example 2: Polygon split into a rectangle and a triangle

A shape consists of a rectangle measuring 12 cm by 5 cm, with a triangle attached to one side. The triangle has base 5 cm and perpendicular height 4 cm. Find the total area.

Use the additive method.

Area of the rectangle:

$$A_{\text{rectangle}} = 12 \times 5 = 60$$

Area of the triangle:

$$A_{\text{triangle}} = \frac{1}{2} \times 5 \times 4 = 10$$

Add the two areas:

$$A_{\text{total}} = 60 + 10 = 70$$

Answer: The total area is \(70\text{ cm}^2\).

Worked Example 3: Irregular polygon using missing lengths

An irregular shape can be divided into two rectangles.

The bottom rectangle has length 15 m and height 6 m. On top of its left side is a smaller rectangle with width 9 m and height 4 m.

Find the total area.

Since the shape is made of two non-overlapping rectangles, we add their areas.

Area of the bottom rectangle:

$$A_1 = 15 \times 6 = 90$$

Area of the top rectangle:

$$A_2 = 9 \times 4 = 36$$

Total area:

$$A_{\text{total}} = 90 + 36 = 126$$

Answer: The area is \(126\text{ m}^2\).

Now suppose you also wanted the width of the uncovered top-right part of the bottom rectangle. Since the whole bottom length is 15 m and the top rectangle covers 9 m of it, the remaining length is:

$$15 - 9 = 6$$

This idea of finding unknown side lengths by subtraction is very useful in complex polygon problems.

Worked Example 4: Complex polygon with a trapezoid and a rectangle

A figure is made from a rectangle and a trapezoid.

• The rectangle has dimensions 8 cm by 6 cm.
• On top of it sits a trapezoid with height 4 cm and parallel sides 8 cm and 14 cm.

Find the total area.

First, find the area of the rectangle:

$$A_{\text{rectangle}} = 8 \times 6 = 48$$

Now find the area of the trapezoid:

$$A_{\text{trapezoid}} = \frac{1}{2}(8+14)(4)$$ $$A_{\text{trapezoid}} = \frac{1}{2}(22)(4) = 44$$

Add them:

$$A_{\text{total}} = 48 + 44 = 92$$

Answer: The total area is \(92\text{ cm}^2\).

How to choose the best method

When you see a complex polygon, ask yourself:

• Is it easier to split it into smaller known shapes?
• Or is it easier to place it inside a larger shape and subtract the unwanted parts?

There is often more than one correct method. Choose the one that makes the calculations simplest and avoids unnecessary steps.

Common mistakes to avoid

• Forgetting a piece: Make sure all parts of the figure are included.
• Overlapping pieces: If your split causes overlap, you may count area twice.
• Using the wrong height: For triangles and trapezoids, the height must be perpendicular to the base.
• Incorrect subtraction: In subtractive problems, subtract only the missing regions, not parts that still belong to the shape.
• Wrong units: Area must be written in square units.

Checking your answer

After solving, do a quick reasonableness check:

• Is your answer smaller than the enclosing rectangle if you used subtraction?
• Is your answer larger than each individual part if you used addition?
• Do the units make sense?
• Did you use all the measurements correctly?

Brief practice ideas

When practicing this topic, try these steps:

1. Redraw the shape neatly.
2. Add dashed lines to split it into familiar polygons.
3. Write the formula for each part before calculating.
4. Keep your work organized so you can see where each number comes from.

Summary

The area of a complex polygon can be found by breaking the figure into simpler shapes or by subtracting missing parts from a larger known shape. The key is to use standard area formulas correctly and identify any missing lengths carefully.

With practice, irregular figures become much easier to handle. If you stay organized and choose a good decomposition, you can find exact areas confidently.

Surface Area of Prisms and Cylinders

In spatial geometry, surface area tells us how much material is needed to cover the outside of a three-dimensional object. For example, if you wanted to wrap a box, paint a can, or cover a slanted prism with paper, you would need to know its surface area.

In this lesson, we will learn how to find the lateral surface area and the total surface area of right and oblique prisms and cylinders. A key idea is to use nets, which are flat patterns that show all the faces of a solid.

1. Key Vocabulary

• Prism: A solid with two congruent, parallel bases and flat side faces.
• Cylinder: A solid with two congruent, parallel circular bases and a curved side.
• Base: One of the parallel congruent faces.
• Lateral surface area (LSA): The area of the side surfaces only, not including the bases.
• Total surface area (TSA): The area of all outer surfaces, including the bases.
• Right solid: A prism or cylinder whose height is perpendicular to its bases.
• Oblique solid: A prism or cylinder that slants, so its height is not along the edge or side surface.
• Height: The perpendicular distance between the two bases.
• Slant length: The length of a lateral edge or side segment on an oblique solid.

2. Understanding Nets

A net helps us see how the surfaces of a solid fit together. When you unfold a prism or cylinder, each face becomes a flat region whose area can be calculated.

For a prism, the net usually includes:

• two identical base shapes, and
• rectangles or parallelograms forming the side faces.

For a cylinder, the net includes:

• two circles for the bases, and
• one rectangle for a right cylinder or a slanted parallelogram-shaped side region for an oblique cylinder.

The net makes the main idea very clear:

$$ \text{Total Surface Area} = \text{Lateral Surface Area} + 2(\text{Area of Base}) $$

3. Surface Area of a Right Prism

In a right prism, the side faces are rectangles. If the base has perimeter \(P\) and the prism has height \(h\), then the total area of all the side rectangles is:

$$ \text{LSA} = Ph $$

This works because each rectangle has one side equal to a side of the base and the other side equal to the height. Adding all these rectangles gives perimeter times height.

If the area of one base is \(B\), then the total surface area is:

$$ \text{TSA} = Ph + 2B $$

4. Surface Area of an Oblique Prism

In an oblique prism, the prism slants. The side faces are usually parallelograms, not rectangles. So we must be careful.

The total surface area still comes from adding the areas of all lateral faces and the two bases:

$$ \text{TSA} = \text{LSA} + 2B $$

However, for an oblique prism, you cannot always use \(Ph\) for the lateral area unless the dimensions given match the perpendicular height of each side face in the net. More often, you find the area of each parallelogram separately:

$$ \text{Area of a parallelogram} = \text{base} \times \text{perpendicular height} $$

So for an oblique prism:

• find the area of each lateral face from the net,
• add them to get the lateral surface area,
• then add the two base areas.

Important: The height of the prism is the perpendicular distance between the bases. The slant edge is usually longer than the height. Do not confuse them.

5. Surface Area of a Right Cylinder

A right cylinder has two circular bases and one curved surface. If we unwrap the curved surface, it becomes a rectangle.

The rectangle in the net has:

• length equal to the circumference of the base, which is \(2\pi r\),
• width equal to the height \(h\).

So the lateral surface area is:

$$ \text{LSA} = (2\pi r)(h) = 2\pi rh $$

Each circular base has area \(\pi r^2\), so the total surface area is:

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

This can also be written as:

$$ \text{TSA} = 2\pi r(h+r) $$

6. Surface Area of an Oblique Cylinder

An oblique cylinder is a slanted cylinder. The two bases are still congruent circles, but the side is not straight up and down.

Using a net, the lateral surface can be thought of as a slanted shape. In many school problems, the lateral surface area is found using the circumference of the base times the slant length \(s\):

$$ \text{LSA} = 2\pi r s $$

Then the total surface area is:

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

Here, \(s\) is the distance along the side surface in the net, not the perpendicular height between the bases.

7. Strategy for Solving Surface Area Problems

1. Identify the solid: prism or cylinder, right or oblique.
2. Find the base area \(B\).
3. Find the lateral surface area using the correct dimensions.
4. Add the two base areas if the question asks for total surface area.
5. Check units. Surface area is always in square units, such as \(\text{cm}^2\) or \(\text{m}^2\).

8. Worked Examples

Example 1: Right Rectangular Prism

A right rectangular prism has length \(8\text{ cm}\), width \(5\text{ cm}\), and height \(12\text{ cm}\). Find the lateral surface area and total surface area.

Step 1: Find the perimeter of the base.

The base is a rectangle with sides 8 cm and 5 cm, so:

$$ P = 2(8+5) = 26\text{ cm} $$

Step 2: Find the lateral surface area.

$$ \text{LSA} = Ph = 26(12) = 312\text{ cm}^2 $$

Step 3: Find the area of one base.

$$ B = 8 \times 5 = 40\text{ cm}^2 $$

Step 4: Find the total surface area.

$$ \text{TSA} = 312 + 2(40) = 312 + 80 = 392\text{ cm}^2 $$

Answer: Lateral surface area = \(312\text{ cm}^2\), total surface area = \(392\text{ cm}^2\).

Example 2: Right Triangular Prism

A right triangular prism has a triangular base with side lengths 6 cm, 8 cm, and 10 cm. The height of the prism is 15 cm. Find the total surface area.

Step 1: Find the perimeter of the base.

$$ P = 6 + 8 + 10 = 24\text{ cm} $$

Step 2: Find the lateral surface area.

$$ \text{LSA} = Ph = 24(15) = 360\text{ cm}^2 $$

Step 3: Find the area of the triangular base.

The sides 6, 8, and 10 form a right triangle, so:

$$ B = \frac{1}{2}(6)(8) = 24\text{ cm}^2 $$

Step 4: Find the total surface area.

$$ \text{TSA} = 360 + 2(24) = 360 + 48 = 408\text{ cm}^2 $$

Answer: \(408\text{ cm}^2\).

Example 3: Right Cylinder

A cylinder has radius \(4\text{ m}\) and height \(9\text{ m}\). Find the lateral and total surface area.

Step 1: Find the lateral surface area.

$$ \text{LSA} = 2\pi rh = 2\pi(4)(9) = 72\pi\text{ m}^2 $$

Step 2: Find the area of the two bases.

$$ 2\pi r^2 = 2\pi(4^2) = 2\pi(16) = 32\pi\text{ m}^2 $$

Step 3: Find the total surface area.

$$ \text{TSA} = 72\pi + 32\pi = 104\pi\text{ m}^2 $$

Answer: Lateral surface area = \(72\pi\text{ m}^2\), total surface area = \(104\pi\text{ m}^2\).

Example 4: Oblique Prism Using a Net

An oblique triangular prism has a triangular base with area \(12\text{ cm}^2\). The three side faces in the net are parallelograms with areas \(30\text{ cm}^2\), \(42\text{ cm}^2\), and \(36\text{ cm}^2\). Find the lateral and total surface area.

Step 1: Add the lateral face areas.

$$ \text{LSA} = 30 + 42 + 36 = 108\text{ cm}^2 $$

Step 2: Add the two bases.

$$ 2B = 2(12) = 24\text{ cm}^2 $$

Step 3: Find the total surface area.

$$ \text{TSA} = 108 + 24 = 132\text{ cm}^2 $$

Answer: Lateral surface area = \(108\text{ cm}^2\), total surface area = \(132\text{ cm}^2\).

9. Common Mistakes to Avoid

• Mixing up lateral and total surface area. Lateral area does not include the bases.
• Using the slant length when the formula needs perpendicular height. This is especially important for oblique solids.
• Forgetting both bases. Total surface area includes two bases, not one.
• Using the diameter instead of the radius in cylinder formulas.
• Forgetting square units. Surface area must be written in units like \(\text{cm}^2\).

10. Quick Formula Review

For a right prism:

$$ \text{LSA} = Ph $$ $$ \text{TSA} = Ph + 2B $$

For a right cylinder:

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

For oblique prisms and cylinders:

• use the net,
• find the lateral area from the actual side faces,
• then add \(2B\).

11. Final Summary

Surface area measures the outside covering of a solid. For prisms and cylinders, it is easiest to think in terms of a net: the side surfaces give the lateral surface area, and adding the two bases gives the total surface area.

For right prisms and right cylinders, there are direct formulas. For oblique solids, the safest method is often to use the net and add the areas of the lateral faces carefully. Always identify the correct dimensions and include square units in your answer.

Surface Area of Pyramids and Cones

When we talk about the surface area of a 3D object, we mean the total area of all the outside faces or curved surfaces that cover it.

For pyramids and cones, surface area often has two parts:

• Lateral surface area: the area of the sides only, not including the base.
• Total surface area: the lateral area plus the area of the base.

In this lesson, you will learn how to calculate both types of surface area and how to use the Pythagorean theorem to find a missing slant height.

1. Surface Area of a Pyramid

A pyramid has a polygon base and triangular faces that meet at one point called the apex.

The most common pyramid problems in 11th Grade involve a regular pyramid. In a regular pyramid:

• the base is a regular polygon, and
• all lateral faces are congruent triangles.

The key measurement for surface area is the slant height, written as \(l\). This is the height of each triangular face, measured from the midpoint of a base edge to the apex along the face.

For a regular pyramid, the lateral surface area formula is:

$$LA = \frac{1}{2}Pl$$

where:

• \(LA\) = lateral area
• \(P\) = perimeter of the base
• \(l\) = slant height

The total surface area is:

$$SA = \frac{1}{2}Pl + B$$

where \(B\) is the area of the base.

Why does this formula work?

Each side face is a triangle. The area of one triangle is \(\frac{1}{2}(\text{base})(\text{height})\). If you add all the triangular faces together, the sum of all their base lengths is the perimeter \(P\), and the common height of each face is the slant height \(l\). That gives:

$$LA = \frac{1}{2}Pl$$

2. Surface Area of a Cone

A cone has a circular base and one curved surface that meets at a point.

For a cone, the lateral surface area is:

$$LA = \pi rl$$

where:

• \(r\) = radius of the base
• \(l\) = slant height

The area of the circular base is:

$$B = \pi r^2$$

So the total surface area of a cone is:

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

This can also be factored as:

$$SA = \pi r(l+r)$$

3. Finding Slant Height Using the Pythagorean Theorem

Sometimes the slant height is given directly. Other times, you must find it first.

For both pyramids and cones, the slant height, the vertical height, and a horizontal distance in the base form a right triangle.

The Pythagorean theorem says:

$$a^2 + b^2 = c^2$$

In these problems, the slant height is usually the hypotenuse, so:

$$l^2 = h^2 + x^2$$

where:

• \(h\) = vertical height
• \(x\) = horizontal distance from the center of the base to the midpoint of a side, or to the edge depending on the shape
• \(l\) = slant height

For a cone, the horizontal distance is the radius \(r\), so:

$$l^2 = h^2 + r^2$$

For a square pyramid, the horizontal distance from the center of the square base to the midpoint of one side is half the side length. If the side length is \(s\), then:

$$l^2 = h^2 + \left(\frac{s}{2}\right)^2$$

Important reminder: Do not confuse the vertical height with the slant height. The vertical height goes straight down inside the solid. The slant height lies along the face of the pyramid or cone.

4. Step-by-Step Method

When solving surface area problems, use this order:

1. Identify the solid: pyramid or cone.
2. Decide whether the question asks for lateral area or total surface area.
3. Find the slant height if needed.
4. Use the correct formula.
5. Check whether to include the base.
6. Write the answer in square units.

Worked Example 1: Square Pyramid with Given Slant Height

A regular square pyramid has base side length \(10\text{ cm}\) and slant height \(13\text{ cm}\). Find:

• the lateral surface area
• the total surface area

Step 1: Find the perimeter of the base.

$$P = 4(10) = 40\text{ cm}$$

Step 2: Find the lateral area.

$$LA = \frac{1}{2}Pl$$ $$LA = \frac{1}{2}(40)(13)$$ $$LA = 20 \cdot 13 = 260\text{ cm}^2$$

Step 3: Find the base area.

The base is a square, so:

$$B = s^2 = 10^2 = 100\text{ cm}^2$$

Step 4: Find the total surface area.

$$SA = LA + B = 260 + 100 = 360\text{ cm}^2$$

Answer:

• Lateral surface area: \(260\text{ cm}^2\)
• Total surface area: \(360\text{ cm}^2\)

Worked Example 2: Cone with Given Radius and Slant Height

A cone has radius \(5\text{ m}\) and slant height \(12\text{ m}\). Find the total surface area.

Step 1: Use the cone surface area formula.

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

Step 2: Substitute the values.

$$SA = \pi(5)(12) + \pi(5^2)$$ $$SA = 60\pi + 25\pi$$ $$SA = 85\pi\text{ m}^2$$

Step 3: Give an approximate decimal if needed.

$$85\pi \approx 266.9\text{ m}^2$$

Answer: The total surface area is \(85\pi\text{ m}^2\), or about \(266.9\text{ m}^2\).

Worked Example 3: Cone with Slant Height Found by Pythagorean Theorem

A cone has radius \(6\text{ cm}\) and vertical height \(8\text{ cm}\). Find its lateral surface area.

Step 1: Find the slant height.

For a cone:

$$l^2 = h^2 + r^2$$ $$l^2 = 8^2 + 6^2$$ $$l^2 = 64 + 36 = 100$$ $$l = 10\text{ cm}$$

Step 2: Use the lateral area formula.

$$LA = \pi rl$$ $$LA = \pi(6)(10) = 60\pi\text{ cm}^2$$

Step 3: Approximate if needed.

$$60\pi \approx 188.5\text{ cm}^2$$

Answer: The lateral surface area is \(60\pi\text{ cm}^2\), or about \(188.5\text{ cm}^2\).

Worked Example 4: Square Pyramid with Slant Height Found First

A regular square pyramid has base side length \(14\text{ cm}\) and vertical height \(24\text{ cm}\). Find its total surface area.

Step 1: Find the slant height.

For a square pyramid, use half the side length:

$$\frac{s}{2} = \frac{14}{2} = 7\text{ cm}$$

Now apply the Pythagorean theorem:

$$l^2 = h^2 + \left(\frac{s}{2}\right)^2$$ $$l^2 = 24^2 + 7^2$$ $$l^2 = 576 + 49 = 625$$ $$l = 25\text{ cm}$$

Step 2: Find the perimeter of the base.

$$P = 4(14) = 56\text{ cm}$$

Step 3: Find the lateral area.

$$LA = \frac{1}{2}Pl$$ $$LA = \frac{1}{2}(56)(25)$$ $$LA = 28 \cdot 25 = 700\text{ cm}^2$$

Step 4: Find the base area.

$$B = 14^2 = 196\text{ cm}^2$$

Step 5: Find the total surface area.

$$SA = LA + B = 700 + 196 = 896\text{ cm}^2$$

Answer: The total surface area is \(896\text{ cm}^2\).

5. Common Mistakes to Avoid

• Using height instead of slant height. Surface area formulas for pyramids and cones use \(l\), not the vertical height \(h\).
• Forgetting the base. If the problem asks for total surface area, include the base area.
• Adding the wrong base area. A pyramid’s base depends on its shape. A cone’s base is always a circle.
• Using the full side length instead of half when finding the slant height of a square pyramid.
• Forgetting square units. Surface area must be written in units like \(\text{cm}^2\), \(\text{m}^2\), or \(\text{in}^2\).

6. Quick Formula List

• Regular pyramid lateral area: \(LA = \frac{1}{2}Pl\)
• Regular pyramid total surface area: \(SA = \frac{1}{2}Pl + B\)
• Cone lateral area: \(LA = \pi rl\)
• Cone total surface area: \(SA = \pi rl + \pi r^2\)
• Cone slant height: \(l^2 = h^2 + r^2\)
• Square pyramid slant height: \(l^2 = h^2 + \left(\frac{s}{2}\right)^2\)

7. Final Summary

To find the surface area of pyramids and cones, separate the problem into lateral area and base area. For pyramids, use \(\frac{1}{2}Pl\). For cones, use \(\pi rl\).

If the slant height is missing, use the Pythagorean theorem to find it. Then substitute carefully into the correct formula and check whether the question wants lateral surface area only or total surface area.

Volume of Polyhedra and Cylinders

In spatial geometry, volume measures how much space a three-dimensional object takes up. We usually measure volume in cubic units, such as \\(cm^3\\), \\(m^3\\), or \\(in^3\\).

In this lesson, you will learn how to calculate the volume of common solid shapes: prisms, pyramids, cylinders, and cones. You will also see an important pattern: pyramids and cones have one-third the volume of related prisms and cylinders when they have the same base area and height.

1. The Main Idea of Volume

Many volume formulas are built from the same simple idea:

$$ \text{Volume} = \text{base area} \times \text{height} $$

This works directly for solids with a constant cross-section, such as prisms and cylinders.

For solids that come to a point, such as pyramids and cones, the volume is one-third of the related prism or cylinder:

$$ \text{Volume} = \frac{1}{3}(\text{base area} \times \text{height}) $$

2. Volume of a Prism

A prism has two parallel, congruent bases, and the same cross-section all the way through. Examples include rectangular prisms and triangular prisms.

The formula for any prism is:

$$ V = Bh $$

where:

• \\(V\\) = volume
• \\(B\\) = area of the base
• \\(h\\) = perpendicular height of the prism

For a rectangular prism, the base area is length times width, so:

$$ V = lwh $$

3. Volume of a Cylinder

A cylinder is like a prism with a circular base. Since the base is a circle, its area is:

$$ B = \pi r^2 $$

So the volume of a cylinder is:

$$ V = \pi r^2 h $$

where:

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

4. Volume of a Pyramid

A pyramid has a polygon base and triangular faces that meet at a single point called the apex.

The volume formula is:

$$ V = \frac{1}{3}Bh $$

This means a pyramid with the same base area and height as a prism has only one-third of the prism’s volume.

5. Volume of a Cone

A cone is similar to a pyramid, but it has a circular base.

Its base area is \\(\pi r^2\\), so its volume formula is:

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

Like the pyramid, a cone has one-third the volume of a cylinder with the same base and height.

6. Understanding the One-Third Relationship

This relationship is very important:

• A pyramid and a prism with the same base area and height satisfy: \\(V_{\text{pyramid}} = \frac{1}{3}V_{\text{prism}}\\)
• A cone and a cylinder with the same base area and height satisfy: \\(V_{\text{cone}} = \frac{1}{3}V_{\text{cylinder}}\\)

You can also reverse this idea:

• The prism’s volume is 3 times the pyramid’s volume.
• The cylinder’s volume is 3 times the cone’s volume.

This helps a lot when comparing solids instead of calculating each one from the start.

7. Important Notes About Height

In every volume formula, the height must be the perpendicular distance from the base to the top or apex. It is not always the slanted side.

For example, in a cone, the slant height is used in surface area problems, but not in the volume formula. Volume uses the straight up-and-down height.

8. Worked Examples

Example 1: Volume of a Rectangular Prism

A rectangular prism has length 8 cm, width 5 cm, and height 3 cm. Find its volume.

Step 1: Use the prism formula.

$$ V = lwh $$

Step 2: Substitute the values.

$$ V = 8 \times 5 \times 3 $$ $$ V = 120 $$

Answer: The volume is \\(120\,cm^3\\).

Example 2: Volume of a Cylinder

A cylinder has radius 4 m and height 10 m. Find its volume.

Step 1: Use the cylinder formula.

$$ V = \pi r^2 h $$

Step 2: Substitute the values.

$$ V = \pi (4)^2(10) $$ $$ V = \pi (16)(10) = 160\pi $$

Step 3: Give an exact or approximate answer.

$$ V = 160\pi\,m^3 \approx 502.7\,m^3 $$

Answer: The volume is \\(160\pi\,m^3\\), or about \\(502.7\,m^3\\).

Example 3: Volume of a Pyramid

A pyramid has a square base with side length 6 cm and perpendicular height 9 cm. Find its volume.

Step 1: Find the base area.

$$ B = 6^2 = 36\,cm^2 $$

Step 2: Use the pyramid formula.

$$ V = \frac{1}{3}Bh $$

Step 3: Substitute.

$$ V = \frac{1}{3}(36)(9) $$ $$ V = 12 \times 9 = 108 $$

Answer: The volume is \\(108\,cm^3\\).

Example 4: Comparing a Cone and a Cylinder

A cone and a cylinder have the same radius, 3 cm, and the same height, 12 cm. Compare their volumes.

Step 1: Find the cylinder’s volume.

$$ V_{\text{cyl}} = \pi r^2 h = \pi (3)^2(12) $$ $$ V_{\text{cyl}} = \pi (9)(12) = 108\pi $$

Step 2: Use the one-third relationship for the cone.

$$ V_{\text{cone}} = \frac{1}{3}V_{\text{cyl}} = \frac{1}{3}(108\pi) = 36\pi $$

Step 3: State the comparison.

• Cylinder volume: \\(108\pi\,cm^3\\)
• Cone volume: \\(36\pi\,cm^3\\)
• The cylinder has 3 times the volume of the cone.

9. Common Mistakes to Avoid

• Using the wrong height: Use the perpendicular height, not a slanted edge.
• Forgetting the base area: In \\(V = Bh\\), you must calculate the full area of the base first.
• Forgetting the \\(\frac{1}{3}\\): Pyramids and cones need the extra factor of one-third.
• Mixing radius and diameter: If you are given the diameter, divide by 2 to get the radius.
• Incorrect units: Volume must be written in cubic units, such as \\(cm^3\\) or \\(m^3\\).

10. Problem-Solving Strategy

1. Identify the solid: prism, cylinder, pyramid, or cone.
2. Find the base area \\(B\\).
3. Use the correct formula.
4. Substitute carefully, including units.
5. Check whether the shape needs the factor \\(\frac{1}{3}\\).

11. Quick Formula List

• Prism: \\(V = Bh\\)
• Rectangular prism: \\(V = lwh\\)
• Cylinder: \\(V = \pi r^2 h\\)
• Pyramid: \\(V = \frac{1}{3}Bh\\)
• Cone: \\(V = \frac{1}{3}\pi r^2 h\\)

12. Summary

Volume tells us how much space a solid occupies. For prisms and cylinders, volume equals base area times height. For pyramids and cones, volume is one-third of that amount.

If two solids have the same base area and height, then a pyramid has one-third the volume of the prism, and a cone has one-third the volume of the cylinder. When solving problems, always identify the shape, use the correct base area, and make sure the height is perpendicular.

Properties of Spheres

In spatial geometry, a sphere is a perfectly round 3-dimensional shape. Every point on the surface of a sphere is the same distance from its center. That fixed distance is called the radius.

Spheres appear in many real-life situations, such as balls, bubbles, planets, and storage tanks. To solve geometry problems involving spheres, we usually need to calculate their surface area or volume.

This lesson will teach you the main properties of spheres and hemispheres, the formulas you need, and how to apply them in worked examples.

1. Key parts of a sphere

• Center: the middle point of the sphere
• Radius \\(r\\): the distance from the center to any point on the surface
• Diameter \\(d\\): the distance across the sphere through the center

The diameter is always twice the radius:

$$d = 2r$$

So if you know one, you can find the other:

$$r = \frac{d}{2}$$

2. Surface area of a sphere

The surface area tells us how much material is needed to cover the outside of a sphere.

The formula is:

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

This means the surface area depends on the square of the radius. If the radius gets larger, the surface area increases quickly.

3. Volume of a sphere

The volume tells us how much space is inside the sphere.

The formula is:

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

This formula uses the cube of the radius, so small changes in radius can cause big changes in volume.

4. Hemisphere

A hemisphere is half of a sphere. It is formed when a sphere is cut into two equal parts through its center.

Because it is half of a sphere, its volume is half the volume of a sphere:

$$V_{\text{hemisphere}} = \frac{1}{2}\left(\frac{4}{3}\pi r^3\right) = \frac{2}{3}\pi r^3$$

For surface area, you must be careful. A hemisphere has two possible surface area questions:

• Curved surface area only: just the rounded outside
• Total surface area: the curved part plus the flat circular base

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

$$CSA = 2\pi r^2$$

The area of the circular base is:

$$\pi r^2$$

So the total surface area of a hemisphere is:

$$TSA = 2\pi r^2 + \pi r^2 = 3\pi r^2$$

5. Important formula list

• Diameter of a sphere: \\(d = 2r\\)
• Radius of a sphere: \\(r = \frac{d}{2}\\)
• Surface area of a sphere: \\(SA = 4\pi r^2\\)
• Volume of a sphere: \\(V = \frac{4}{3}\pi r^3\\)
• Volume of a hemisphere: \\(V = \frac{2}{3}\pi r^3\\)
• Curved surface area of a hemisphere: \\(CSA = 2\pi r^2\\)
• Total surface area of a hemisphere: \\(TSA = 3\pi r^2\\)

6. Units and answers

Always pay attention to units:

• Lengths are measured in units such as cm, m, or mm.
• Surface area is measured in square units, such as \\(\text{cm}^2\\) or \\(\text{m}^2\\).
• Volume is measured in cubic units, such as \\(\text{cm}^3\\) or \\(\text{m}^3\\).

If a question asks for an exact answer, leave your answer in terms of \\(\pi\\). If it asks for a decimal answer, use a calculator and round as instructed.

7. Worked Example 1: Finding the surface area of a sphere

Question: Find the surface area of a sphere with radius \\(5\text{ cm}\\).

Step 1: Write the formula.

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

Step 2: Substitute \\(r = 5\\).

$$SA = 4\pi(5)^2$$

$$SA = 4\pi(25)$$

$$SA = 100\pi$$

Step 3: Give the answer.

The exact surface area is:

$$100\pi \text{ cm}^2$$

As a decimal:

$$100\pi \approx 314.16 \text{ cm}^2$$

Worked Example 2: Finding the volume from the diameter

Question: A sphere has diameter \\(12\text{ m}\\). Find its volume.

Step 1: Find the radius.

$$r = \frac{d}{2} = \frac{12}{2} = 6\text{ m}$$

Step 2: Use the volume formula.

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

$$V = \frac{4}{3}\pi(6)^3$$

$$V = \frac{4}{3}\pi(216)$$

$$V = 288\pi$$

Step 3: State the answer.

The exact volume is:

$$288\pi \text{ m}^3$$

As a decimal:

$$288\pi \approx 904.78 \text{ m}^3$$

Worked Example 3: Hemisphere total surface area

Question: Find the total surface area of a hemisphere with radius \\(4\text{ cm}\\).

Step 1: Choose the correct formula.

Because this is total surface area, we include the curved surface and the flat base:

$$TSA = 3\pi r^2$$

Step 2: Substitute \\(r = 4\\).

$$TSA = 3\pi(4)^2$$

$$TSA = 3\pi(16)$$

$$TSA = 48\pi$$

Step 3: Give the answer.

The total surface area is:

$$48\pi \text{ cm}^2$$

As a decimal:

$$48\pi \approx 150.80 \text{ cm}^2$$

Worked Example 4: Applied problem with a hemisphere

Question: A bowl is shaped like a hemisphere with radius \\(7\text{ cm}\\). How much water can it hold?

This is asking for the volume of the hemisphere.

Step 1: Use the hemisphere volume formula.

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

Step 2: Substitute \\(r = 7\\).

$$V = \frac{2}{3}\pi(7)^3$$

$$V = \frac{2}{3}\pi(343)$$

$$V = \frac{686}{3}\pi$$

Step 3: Write the answer.

The exact volume is:

$$\frac{686}{3}\pi \text{ cm}^3$$

As a decimal:

$$\frac{686}{3}\pi \approx 718.38 \text{ cm}^3$$

So the bowl can hold about \\(718.38\text{ cm}^3\\) of water.

8. Common mistakes to avoid

• Using the diameter instead of the radius: Most formulas use \\(r\\), not \\(d\\). Always check first.
• Mixing up area and volume formulas: Surface area uses \\(r^2\\), but volume uses \\(r^3\\).
• Forgetting the base of a hemisphere: Total surface area is not just the curved part.
• Wrong units: Use square units for area and cubic units for volume.
• Rounding too early: Keep values exact until the final step when possible.

9. How scaling affects a sphere

Sometimes a question changes the size of a sphere. It is important to know how this affects surface area and volume.

If the radius is multiplied by a scale factor \\(k\\):

• The surface area is multiplied by \\(k^2\\).
• The volume is multiplied by \\(k^3\\).

For example, if the radius doubles:

• Surface area becomes \\(2^2 = 4\\) times as large.
• Volume becomes \\(2^3 = 8\\) times as large.

This is very useful in applied geometry problems.

10. Quick check for understanding

1. What is the relationship between diameter and radius?
2. What formula gives the surface area of a sphere?
3. What is the difference between curved surface area and total surface area for a hemisphere?
4. If the radius of a sphere triples, by what factor does the volume increase?

Answers:

1. \\(d = 2r\\)
2. \\(SA = 4\pi r^2\\)
3. Curved surface area includes only the rounded part; total surface area includes the rounded part and the circular base.
4. The volume increases by \\(3^3 = 27\\).

11. Summary

A sphere is a 3-dimensional shape where every point on the surface is the same distance from the center. The most important measurements are the radius and diameter, with \\(d = 2r\\).

To solve problems with spheres, use:

• $$SA = 4\pi r^2$$
• $$V = \frac{4}{3}\pi r^3$$

For hemispheres, remember that volume is half the sphere’s volume, and surface area depends on whether the flat circular base is included. Careful formula choice, correct units, and checking whether you are given radius or diameter will help you avoid mistakes.

Mensuration of Composite Solids

In mensuration, we measure lengths, areas, and volumes of geometric figures. When a solid is made by combining two or more simple solids, or by removing one solid from another, it is called a composite solid.

Examples of composite solids include a cylinder with a hemisphere on top, a cuboid with a cone attached, or a block with a cylindrical hole drilled through it. In these shapes, we often need to find the volume and surface area.

This lesson explains how to break complex solids into simpler parts, how to handle hidden or removed surfaces correctly, and how to calculate volume and surface area step by step.

1. What is a composite solid?

A composite solid is a 3-dimensional object formed by:

• joining simple solids, or
• cutting out one solid from another.

The simple solids you usually work with in Grade 11 are:

• cube
• cuboid (rectangular prism)
• cylinder
• cone
• sphere
• hemisphere
• prism
• pyramid

To solve mensuration problems involving composite solids, the main idea is:

1. Identify the simple solids involved.
2. Decide whether they are added or removed.
3. Use the correct formulas for each part.
4. Be careful with common surfaces that are hidden inside the solid.

2. Important formulas

Here are the formulas you will use most often.

Cuboid

Volume:

$$V = lwh$$

Total surface area:

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

Cube

Volume:

$$V = a^3$$

Total surface area:

$$TSA = 6a^2$$

Cylinder

Volume:

$$V = \pi r^2 h$$

Curved surface area:

$$CSA = 2\pi rh$$

Total surface area:

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

Cone

Volume:

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

Curved surface area:

$$CSA = \pi rl$$

Total surface area:

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

where the slant height is

$$l = \sqrt{r^2 + h^2}$$

Sphere

Volume:

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

Surface area:

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

Hemisphere

Volume:

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

Curved surface area:

$$CSA = 2\pi r^2$$

Total surface area:

$$TSA = 3\pi r^2$$

3. Volume of a composite solid

The volume of a composite solid is usually the easiest part. You simply:

• add the volumes of parts that are joined, and
• subtract the volumes of parts that are removed.

So in general:

$$\text{Volume of composite solid} = \text{sum of added volumes} - \text{sum of removed volumes}$$

4. Surface area of a composite solid

Surface area needs more care than volume. Not every face of the smaller solids appears on the outside of the final shape.

When two solids are attached, the touching surfaces become internal surfaces. These hidden surfaces are not included in the external surface area.

When a hole or cavity is made, the newly exposed inside surface may need to be included, depending on the question. If the hole is open to the outside, its inner wall is part of the surface area. But the part removed from contact inside the object is not counted twice.

Useful surface area rule:

$$\text{External surface area} = \text{sum of surface areas of parts} - \text{hidden joined areas}$$

If a common face is fully hidden after joining, it usually gets removed twice, because it was counted once in each solid.

5. Step-by-step method

For any composite solid, follow this method:

1. Draw or imagine the solid clearly.
2. Split it into familiar shapes.
3. Mark all dimensions carefully.
4. Find the volume or surface area of each part.
5. Add or subtract as needed.
6. Check whether any surfaces are hidden, internal, or removed.
7. Write the final answer with correct units.

6. Worked Example 1: Cylinder with a hemisphere on top

A solid consists of a cylinder of radius \(3\text{ cm}\) and height \(10\text{ cm}\), with a hemisphere of the same radius attached on top. Find:

• the total volume
• the external surface area

Step 1: Identify the parts

• One cylinder
• One hemisphere

Step 2: Volume

Volume of cylinder:

$$V_1 = \pi r^2 h = \pi(3)^2(10) = 90\pi$$

Volume of hemisphere:

$$V_2 = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi(3)^3 = \frac{2}{3}\pi(27) = 18\pi$$

Total volume:

$$V = 90\pi + 18\pi = 108\pi \text{ cm}^3$$

So,

$$V = 108\pi \text{ cm}^3$$

Step 3: Surface area

The circular top of the cylinder is covered by the hemisphere, so that circle is hidden and should not be counted.

External surface area consists of:

• curved surface of cylinder
• curved surface of hemisphere
• base of cylinder

Curved surface area of cylinder:

$$2\pi rh = 2\pi(3)(10) = 60\pi$$

Curved surface area of hemisphere:

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

Base of cylinder:

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

Total external surface area:

$$SA = 60\pi + 18\pi + 9\pi = 87\pi \text{ cm}^2$$

Answer:

• Volume = \(108\pi \text{ cm}^3\)
• External surface area = \(87\pi \text{ cm}^2\)

7. Worked Example 2: Cuboid with a cylindrical hole through it

A rectangular block has length \(12\text{ cm}\), width \(8\text{ cm}\), and height \(6\text{ cm}\). A cylindrical hole of radius \(2\text{ cm}\) is drilled straight through the height of the block. Find:

• the remaining volume
• the total surface area of the new solid

Step 1: Volume of the original cuboid

$$V_{\text{cuboid}} = lwh = 12 \times 8 \times 6 = 576\text{ cm}^3$$

Step 2: Volume of the cylindrical hole

The cylinder has radius \(2\text{ cm}\) and height \(6\text{ cm}\).

$$V_{\text{hole}} = \pi r^2 h = \pi(2)^2(6) = 24\pi$$

Step 3: Remaining volume

$$V = 576 - 24\pi \text{ cm}^3$$

Step 4: Surface area

First find the original surface area of the cuboid:

$$TSA = 2(lw + lh + wh)$$ $$= 2(12\times 8 + 12\times 6 + 8\times 6)$$ $$= 2(96 + 72 + 48)$$ $$= 2(216) = 432\text{ cm}^2$$

Now consider the effect of the hole:

• Two circular areas are removed from the top and bottom faces.
• The curved inner wall of the cylindrical hole is added.

Area removed from top and bottom:

$$2\pi r^2 = 2\pi(2)^2 = 8\pi$$

Curved surface area added inside hole:

$$2\pi rh = 2\pi(2)(6) = 24\pi$$

New total surface area:

$$SA = 432 - 8\pi + 24\pi = 432 + 16\pi \text{ cm}^2$$

Answer:

• Remaining volume = \(576 - 24\pi \text{ cm}^3\)
• Total surface area = \(432 + 16\pi \text{ cm}^2\)

8. Worked Example 3: A cone on top of a cylinder

A toy is made by attaching a cone to the top of a cylinder. Both have radius \(4\text{ cm}\). The cylinder has height \(9\text{ cm}\), and the cone has height \(3\text{ cm}\). Find:

• the total volume
• the external surface area

Step 1: Volume of cylinder

$$V_1 = \pi r^2 h = \pi(4)^2(9) = 144\pi$$

Step 2: Volume of cone

$$V_2 = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(4)^2(3) = 16\pi$$

Total volume:

$$V = 144\pi + 16\pi = 160\pi \text{ cm}^3$$

Step 3: Surface area

The common circular face between the cone and cylinder is hidden, so it is not included.

We need:

• curved surface area of cylinder
• curved surface area of cone
• base of cylinder

First find the slant height of the cone:

$$l = \sqrt{r^2 + h^2} = \sqrt{4^2 + 3^2} = \sqrt{16+9} = 5$$

Curved surface area of cylinder:

$$2\pi rh = 2\pi(4)(9) = 72\pi$$

Curved surface area of cone:

$$\pi rl = \pi(4)(5) = 20\pi$$

Base of cylinder:

$$\pi r^2 = \pi(4)^2 = 16\pi$$

Total external surface area:

$$SA = 72\pi + 20\pi + 16\pi = 108\pi \text{ cm}^2$$

Answer:

• Volume = \(160\pi \text{ cm}^3\)
• External surface area = \(108\pi \text{ cm}^2\)

9. Worked Example 4: A hemisphere removed from a cube

A solid cube has side length \(10\text{ cm}\). A hemispherical cavity of radius \(3\text{ cm}\) is scooped out from the top face. Find:

• the volume of the remaining solid
• the total surface area of the remaining solid

Step 1: Volume of cube

$$V_{\text{cube}} = 10^3 = 1000\text{ cm}^3$$

Step 2: Volume of hemisphere removed

$$V_{\text{hemi}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi(3)^3 = 18\pi$$

Remaining volume:

$$V = 1000 - 18\pi \text{ cm}^3$$

Step 3: Surface area

Original surface area of cube:

$$SA_{\text{cube}} = 6a^2 = 6(10^2) = 600\text{ cm}^2$$

When the hemispherical cavity is made:

• the circular opening on the top face is removed from the flat surface,
• the curved inside surface of the hemisphere is added.

Circular area removed from top face:

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

Curved surface area of hemisphere added:

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

Total surface area of remaining solid:

$$SA = 600 - 9\pi + 18\pi = 600 + 9\pi \text{ cm}^2$$

Answer:

• Remaining volume = \(1000 - 18\pi \text{ cm}^3\)
• Total surface area = \(600 + 9\pi \text{ cm}^2\)

10. Common mistakes to avoid

• Counting hidden faces: If two solids are joined, the touching faces are inside the shape and should not be counted in the external surface area.
• Forgetting inside surfaces of holes or cavities: The inner curved surface may be part of the final surface area.
• Using total surface area instead of curved surface area: For joined solids, only the exposed parts count.
• Mixing dimensions: Make sure each radius, height, length, or width belongs to the correct solid.
• Wrong units: Volume uses cubic units such as \(\text{cm}^3\), while surface area uses square units such as \(\text{cm}^2\).

11. Quick strategy for exam questions

1. Write the names of the simple solids present.
2. Decide whether the question asks for volume, surface area, or both.
3. For volume: add or subtract volumes.
4. For surface area: include only surfaces you can actually touch from outside, unless an inner cavity is open and exposed.
5. Check whether any circles, rectangles, or curved surfaces are hidden due to joining.
6. Leave answers in terms of \(\pi\) unless a decimal value is requested.

12. Brief summary

A composite solid is formed by joining or removing simple 3D solids. To find its volume, split the object into known shapes and add or subtract their volumes. To find its surface area, count only the exposed surfaces and exclude hidden joined faces, while including any exposed inner surfaces of holes or cavities.

With practice, the key skill is learning to visualize the solid carefully. Once you identify what is added, what is removed, and what is hidden, the calculations become much easier.