Mensuration of Plane and Solid Shapes for SS 1

Mensuration

Performance Objectives

By the end of the lesson, students should be able to:

1. find length of an arc of a circle using formula;
2. find perimeter of sectors and segments;
3. find area of sector and segment of a circle;
4. explain the relationship between the area of sector of a circle and the surface area of a cone;
5. calculate the surface area and volume of solid shapes;
6. calculate the surface area and volume of frustum of a cone and pyramid;
7. find surface area and volume of compound shapes.

Introduction to Mensuration

Mensuration is the branch of mathematics that deals with the measurement of geometric figures and their parameters such as length, area, and volume. It involves calculating the size, dimensions, and other properties of both two-dimensional (plane) and three-dimensional (solid) shapes.

Understanding mensuration is important for various real-life applications, including construction, engineering, design, and everyday tasks like calculating the amount of paint needed for a wall or the capacity of a water tank.

Mensuration of Plane Shapes

Plane shapes are two-dimensional figures that lie entirely on a flat surface. In this section, the focus is on parts of a circle: arcs, sectors, and segments.

Length of an Arc of a Circle

An arc is a portion of the circumference of a circle. The length of an arc depends on the radius of the circle and the angle subtended by the arc at the centre of the circle.

The formula for the length of an arc is derived from the fact that the ratio of the arc length to the total circumference is equal to the ratio of the angle subtended by the arc to the total angle in a circle ($360^\circ$ or $2\pi$ radians).

The formula for the length of an arc ($L$) is:

$$ L = \frac{\theta}{360^\circ} \times 2\pi r $$

where:

1. $\theta$ is the angle subtended by the arc at the centre in degrees.
2. $r$ is the radius of the circle.
3. $\pi$ (pi) is a mathematical constant approximately equal to $3.142$ or $\frac{22}{7}$.

If the angle is given in radians, the formula is:

$$ L = r\theta $$

where $\theta$ is the angle in radians.

Worked Example 1.1

A circle has a radius of $7 \text{ cm}$. Find the length of an arc that subtends an angle of $60^\circ$ at the centre. (Use $\pi = \frac{22}{7}$)

Solution:

1. Given: $r = 7 \text{ cm}$, $\theta = 60^\circ$.
2. Formula for arc length: $L = \frac{\theta}{360^\circ} \times 2\pi r$.
3. Substitute the values:

$$ L = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7 $$
$$ L = \frac{1}{6} \times 2 \times 22 $$
$$ L = \frac{44}{6} $$
$$ L = \frac{22}{3} $$
$$ L = 7.33 \text{ cm (to 2 decimal places)} $$

The length of the arc is $7.33 \text{ cm}$.

Worked Example 1.2

The radius of a circle is $10 \text{ cm}$. Calculate the length of an arc that subtends an angle of $1.5$ radians at the centre.

Solution:

1. Given: $r = 10 \text{ cm}$, $\theta = 1.5 \text{ radians}$.
2. Formula for arc length (in radians): $L = r\theta$.
3. Substitute the values:

$$ L = 10 \times 1.5 $$
$$ L = 15 \text{ cm} $$

The length of the arc is $15 \text{ cm}$.

Perimeter of a Sector of a Circle

A sector of a circle is a region bounded by two radii and the arc connecting their endpoints. It resembles a slice of a pie.

The perimeter of a sector is the sum of the lengths of the two radii and the arc length.

The formula for the perimeter of a sector ($P_{sector}$) is:

$$ P_{sector} = r + r + L $$
$$ P_{sector} = 2r + \frac{\theta}{360^\circ} \times 2\pi r $$

where:

1. $r$ is the radius of the circle.
2. $L$ is the length of the arc.
3. $\theta$ is the angle subtended by the arc at the centre in degrees.

Worked Example 1.3

Find the perimeter of a sector of a circle with a radius of $14 \text{ cm}$ and a central angle of $90^\circ$. (Use $\pi = \frac{22}{7}$)

Solution:

1. Given: $r = 14 \text{ cm}$, $\theta = 90^\circ$.
2. First, calculate the arc length ($L$):

$$ L = \frac{\theta}{360^\circ} \times 2\pi r $$
$$ L = \frac{90^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 14 $$
$$ L = \frac{1}{4} \times 2 \times 22 \times 2 $$
$$ L = \frac{1}{4} \times 88 $$
$$ L = 22 \text{ cm} $$

3. Now, calculate the perimeter of the sector:

$$ P_{sector} = 2r + L $$
$$ P_{sector} = 2(14) + 22 $$
$$ P_{sector} = 28 + 22 $$
$$ P_{sector} = 50 \text{ cm} $$

The perimeter of the sector is $50 \text{ cm}$.

Perimeter of a Segment of a Circle

A segment of a circle is a region bounded by an arc and the chord connecting the endpoints of the arc. It is the area cut off from the rest of the circle by a chord.

The perimeter of a segment is the sum of the length of the arc and the length of the chord.

The formula for the perimeter of a segment ($P_{segment}$) is:

$$ P_{segment} = L + \text{chord length} $$

The chord length can be found using trigonometry. If the central angle is $\theta$ and the radius is $r$, the chord length ($c$) is given by:

$$ c = 2r \sin\left(\frac{\theta}{2}\right) $$

So, the perimeter of a segment is:

$$ P_{segment} = \frac{\theta}{360^\circ} \times 2\pi r + 2r \sin\left(\frac{\theta}{2}\right) $$

where:

1. $r$ is the radius of the circle.
2. $L$ is the length of the arc.
3. $\theta$ is the angle subtended by the arc at the centre in degrees.

Worked Example 1.4

A segment of a circle has a radius of $10 \text{ cm}$ and a central angle of $120^\circ$. Find its perimeter. (Use $\pi = 3.142$)

Solution:

1. Given: $r = 10 \text{ cm}$, $\theta = 120^\circ$.
2. Calculate the arc length ($L$):

$$ L = \frac{\theta}{360^\circ} \times 2\pi r $$
$$ L = \frac{120^\circ}{360^\circ} \times 2 \times 3.142 \times 10 $$
$$ L = \frac{1}{3} \times 20 \times 3.142 $$
$$ L = \frac{62.84}{3} $$
$$ L \approx 20.947 \text{ cm} $$

3. Calculate the chord length ($c$):

$$ c = 2r \sin\left(\frac{\theta}{2}\right) $$
$$ c = 2(10) \sin\left(\frac{120^\circ}{2}\right) $$
$$ c = 20 \sin(60^\circ) $$
$$ c = 20 \times 0.8660 $$
$$ c \approx 17.32 \text{ cm} $$

4. Calculate the perimeter of the segment:

$$ P_{segment} = L + c $$
$$ P_{segment} = 20.947 + 17.32 $$
$$ P_{segment} \approx 38.27 \text{ cm (to 2 decimal places)} $$

The perimeter of the segment is approximately $38.27 \text{ cm}$.

Area of a Sector of a Circle

The area of a sector is the region enclosed by the two radii and the arc. Similar to arc length, the area of a sector is a fraction of the total area of the circle.

The formula for the area of a sector ($A_{sector}$) is:

$$ A_{sector} = \frac{\theta}{360^\circ} \times \pi r^2 $$

where:

1. $\theta$ is the angle subtended by the arc at the centre in degrees.
2. $r$ is the radius of the circle.

If the angle is given in radians, the formula is:

$$ A_{sector} = \frac{1}{2} r^2\theta $$

where $\theta$ is the angle in radians.

Worked Example 1.5

Calculate the area of a sector of a circle with a radius of $6 \text{ cm}$ and a central angle of $100^\circ$. (Use $\pi = 3.142$)

Solution:

1. Given: $r = 6 \text{ cm}$, $\theta = 100^\circ$.
2. Formula for area of a sector: $A_{sector} = \frac{\theta}{360^\circ} \times \pi r^2$.
3. Substitute the values:

$$ A_{sector} = \frac{100^\circ}{360^\circ} \times 3.142 \times (6)^2 $$
$$ A_{sector} = \frac{10}{36} \times 3.142 \times 36 $$
$$ A_{sector} = 10 \times 3.142 $$
$$ A_{sector} = 31.42 \text{ cm}^2 $$

The area of the sector is $31.42 \text{ cm}^2$.

Worked Example 1.6

A sector of a circle has a radius of $8 \text{ cm}$ and a central angle of $0.8$ radians. Find its area.

Solution:

1. Given: $r = 8 \text{ cm}$, $\theta = 0.8 \text{ radians}$.
2. Formula for area of a sector (in radians): $A_{sector} = \frac{1}{2} r^2\theta$.
3. Substitute the values:

$$ A_{sector} = \frac{1}{2} \times (8)^2 \times 0.8 $$
$$ A_{sector} = \frac{1}{2} \times 64 \times 0.8 $$
$$ A_{sector} = 32 \times 0.8 $$
$$ A_{sector} = 25.6 \text{ cm}^2 $$

The area of the sector is $25.6 \text{ cm}^2$.

Area of a Segment of a Circle

The area of a segment is the area of the region bounded by an arc and its corresponding chord. It can be found by subtracting the area of the triangle formed by the two radii and the chord from the area of the sector.

The formula for the area of a segment ($A_{segment}$) is:

$$ A_{segment} = A_{sector} – A_{triangle} $$

The area of the triangle formed by two radii ($r$) and the chord, with a central angle $\theta$, is given by:

$$ A_{triangle} = \frac{1}{2} r^2 \sin\theta $$

So, the area of a segment is:

$$ A_{segment} = \frac{\theta}{360^\circ} \times \pi r^2 – \frac{1}{2} r^2 \sin\theta $$

where:

1. $\theta$ is the angle subtended by the arc at the centre in degrees.
2. $r$ is the radius of the circle.

Worked Example 1.7

Find the area of a segment of a circle with a radius of $12 \text{ cm}$ and a central angle of $60^\circ$. (Use $\pi = 3.142$)

Solution:

1. Given: $r = 12 \text{ cm}$, $\theta = 60^\circ$.
2. Calculate the area of the sector ($A_{sector}$):

$$ A_{sector} = \frac{\theta}{360^\circ} \times \pi r^2 $$
$$ A_{sector} = \frac{60^\circ}{360^\circ} \times 3.142 \times (12)^2 $$
$$ A_{sector} = \frac{1}{6} \times 3.142 \times 144 $$
$$ A_{sector} = 3.142 \times 24 $$
$$ A_{sector} = 75.392 \text{ cm}^2 $$

3. Calculate the area of the triangle ($A_{triangle}$):

$$ A_{triangle} = \frac{1}{2} r^2 \sin\theta $$
$$ A_{triangle} = \frac{1}{2} \times (12)^2 \times \sin(60^\circ) $$
$$ A_{triangle} = \frac{1}{2} \times 144 \times 0.8660 $$
$$ A_{triangle} = 72 \times 0.8660 $$
$$ A_{triangle} = 62.352 \text{ cm}^2 $$

4. Calculate the area of the segment:

$$ A_{segment} = A_{sector} – A_{triangle} $$
$$ A_{segment} = 75.392 – 62.352 $$
$$ A_{segment} = 13.04 \text{ cm}^2 $$

The area of the segment is $13.04 \text{ cm}^2$.

Relationship Between the Area of a Sector of a Circle and the Surface Area of a Cone

A cone is a three-dimensional solid shape with a circular base and a single vertex. The curved surface of a cone can be formed by rolling up a sector of a circle.

When a sector of a circle is cut out and its two straight edges (radii) are joined together, it forms the curved surface of a cone. In this transformation:

1. The radius of the sector ($R$) becomes the slant height ($l$) of the cone.
2. The arc length of the sector becomes the circumference of the base of the cone ($2\pi r_{cone}$), where $r_{cone}$ is the radius of the cone’s base.

Therefore, the area of the sector is equal to the curved surface area of the cone.

Let the radius of the sector be $R$ and the central angle be $\theta$. The area of the sector is $A_{sector} = \frac{\theta}{360^\circ} \times \pi R^2$.

Let the slant height of the cone be $l$ and the radius of its base be $r_{cone}$. The curved surface area of the cone is $A_{curved\_cone} = \pi r_{cone} l$.

Since $R = l$ and the arc length ($L = \frac{\theta}{360^\circ} \times 2\pi R$) equals the base circumference ($2\pi r_{cone}$), we have:

$$ \frac{\theta}{360^\circ} \times 2\pi R = 2\pi r_{cone} $$
$$ \frac{\theta}{360^\circ} R = r_{cone} $$

Now, substitute $r_{cone}$ into the curved surface area formula:

$$ A_{curved\_cone} = \pi \left(\frac{\theta}{360^\circ} R\right) l $$

Since $l = R$,

$$ A_{curved\_cone} = \pi \left(\frac{\theta}{360^\circ} R\right) R $$
$$ A_{curved\_cone} = \frac{\theta}{360^\circ} \pi R^2 $$

This shows that the curved surface area of the cone is indeed equal to the area of the sector from which it was formed.

Mensuration of Solid Shapes

Solid shapes are three-dimensional figures that occupy space. For solid shapes, we calculate their surface area (the total area of all their faces) and their volume (the amount of space they occupy).

Cube

A cube is a three-dimensional solid object bounded by six square faces, facets or sides, with three meeting at each vertex. All its edges are of equal length.

Let the length of an edge of the cube be $s$.

1. Surface Area of a Cube: A cube has 6 identical square faces. The area of one square face is $s \times s = s^2$.

$$ \text{Surface Area} = 6s^2 $$

2. Volume of a Cube: The volume of a cube is the product of its length, width, and height. Since all are equal to $s$:

$$ \text{Volume} = s \times s \times s = s^3 $$

Worked Example 2.1

A cube has an edge length of $5 \text{ cm}$. Calculate its surface area and volume.

Solution:

1. Given: $s = 5 \text{ cm}$.
2. Surface Area:

$$ \text{Surface Area} = 6s^2 = 6 \times (5)^2 = 6 \times 25 = 150 \text{ cm}^2 $$

3. Volume:

$$ \text{Volume} = s^3 = (5)^3 = 125 \text{ cm}^3 $$

The surface area of the cube is $150 \text{ cm}^2$ and its volume is $125 \text{ cm}^3$.

Cuboid (Rectangular Prism)

A cuboid is a three-dimensional solid object bounded by six rectangular faces. It has length, width, and height, which are generally different.

Let the length be $l$, the width be $w$, and the height be $h$.

1. Surface Area of a Cuboid: A cuboid has 3 pairs of identical rectangular faces.
   • Area of top and bottom faces: $2 \times (l \times w)$
   • Area of front and back faces: $2 \times (l \times h)$
   • Area of side faces: $2 \times (w \times h)$

   $$ \text{Surface Area} = 2(lw + lh + wh) $$

2. Volume of a Cuboid: The volume of a cuboid is the product of its length, width, and height.

$$ \text{Volume} = l \times w \times h $$

Worked Example 2.2

A cuboid has a length of $8 \text{ cm}$, a width of $3 \text{ cm}$, and a height of $4 \text{ cm}$. Calculate its surface area and volume.

Solution:

1. Given: $l = 8 \text{ cm}$, $w = 3 \text{ cm}$, $h = 4 \text{ cm}$.
2. Surface Area:

$$ \text{Surface Area} = 2(lw + lh + wh) $$
$$ \text{Surface Area} = 2((8 \times 3) + (8 \times 4) + (3 \times 4)) $$
$$ \text{Surface Area} = 2(24 + 32 + 12) $$
$$ \text{Surface Area} = 2(68) $$
$$ \text{Surface Area} = 136 \text{ cm}^2 $$

3. Volume:

$$ \text{Volume} = l \times w \times h $$
$$ \text{Volume} = 8 \times 3 \times 4 $$
$$ \text{Volume} = 96 \text{ cm}^3 $$

The surface area of the cuboid is $136 \text{ cm}^2$ and its volume is $96 \text{ cm}^3$.

Cylinder

A cylinder is a three-dimensional solid with two parallel circular bases and a curved surface connecting them.

Let the radius of the base be $r$ and the height be $h$.

1. Surface Area of a Cylinder:
   • Area of two circular bases: $2 \times (\pi r^2)$
   • Area of the curved surface: The curved surface, when unrolled, forms a rectangle with length equal to the circumference of the base ($2\pi r$) and width equal to the height ($h$). So, its area is $2\pi r h$.

   $$ \text{Surface Area} = 2\pi r^2 + 2\pi rh $$
   $$ \text{Surface Area} = 2\pi r(r + h) $$

2. Volume of a Cylinder: The volume of a cylinder is the area of its base multiplied by its height.

$$ \text{Volume} = \pi r^2 h $$

Worked Example 2.3

A cylinder has a radius of $7 \text{ cm}$ and a height of $10 \text{ cm}$. Calculate its surface area and volume. (Use $\pi = \frac{22}{7}$)

Solution:

1. Given: $r = 7 \text{ cm}$, $h = 10 \text{ cm}$.
2. Surface Area:

$$ \text{Surface Area} = 2\pi r(r + h) $$
$$ \text{Surface Area} = 2 \times \frac{22}{7} \times 7 \times (7 + 10) $$
$$ \text{Surface Area} = 2 \times 22 \times 17 $$
$$ \text{Surface Area} = 44 \times 17 $$
$$ \text{Surface Area} = 748 \text{ cm}^2 $$

3. Volume:

$$ \text{Volume} = \pi r^2 h $$
$$ \text{Volume} = \frac{22}{7} \times (7)^2 \times 10 $$
$$ \text{Volume} = \frac{22}{7} \times 49 \times 10 $$
$$ \text{Volume} = 22 \times 7 \times 10 $$
$$ \text{Volume} = 1540 \text{ cm}^3 $$

The surface area of the cylinder is $748 \text{ cm}^2$ and its volume is $1540 \text{ cm}^3$.

Cone

A cone is a three-dimensional geometric shape that tapers smoothly from a flat base (usually circular) to a point called the apex or vertex.

Let the radius of the base be $r$, the height be $h$, and the slant height be $l$. The slant height can be found using the Pythagorean theorem: $l = \sqrt{r^2 + h^2}$.

1. Surface Area of a Cone:
   • Area of the circular base: $\pi r^2$
   • Area of the curved surface: $\pi r l$

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

2. Volume of a Cone: The volume of a cone is one-third of the volume of a cylinder with the same base radius and height.

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

Worked Example 2.4

A cone has a base radius of $3 \text{ cm}$ and a perpendicular height of $4 \text{ cm}$. Calculate its surface area and volume. (Use $\pi = 3.142$)

Solution:

1. Given: $r = 3 \text{ cm}$, $h = 4 \text{ cm}$.
2. First, calculate the slant height ($l$):

$$ l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ cm} $$

3. Surface Area:

$$ \text{Surface Area} = \pi r(r + l) $$
$$ \text{Surface Area} = 3.142 \times 3 \times (3 + 5) $$
$$ \text{Surface Area} = 3.142 \times 3 \times 8 $$
$$ \text{Surface Area} = 3.142 \times 24 $$
$$ \text{Surface Area} = 75.392 \text{ cm}^2 $$

4. Volume:

$$ \text{Volume} = \frac{1}{3} \pi r^2 h $$
$$ \text{Volume} = \frac{1}{3} \times 3.142 \times (3)^2 \times 4 $$
$$ \text{Volume} = \frac{1}{3} \times 3.142 \times 9 \times 4 $$
$$ \text{Volume} = 3.142 \times 3 \times 4 $$
$$ \text{Volume} = 3.142 \times 12 $$
$$ \text{Volume} = 37.704 \text{ cm}^3 $$

The surface area of the cone is $75.39 \text{ cm}^2$ (to 2 d.p.) and its volume is $37.70 \text{ cm}^3$ (to 2 d.p.).

Prism

A prism is a polyhedron comprising an n-sided polygonal base, a second base which is a translated copy of the first, and n other faces (necessarily all parallelograms) joining corresponding sides of the two bases. The shape of the base determines the name of the prism (e.g., triangular prism, rectangular prism, pentagonal prism).

Let $A_{base}$ be the area of the base, $P_{base}$ be the perimeter of the base, and $h$ be the height of the prism.

1. Surface Area of a Prism: The total surface area is the sum of the areas of the two bases and the lateral surface area (area of the rectangular/parallelogram faces).

$$ \text{Surface Area} = 2 \times A_{base} + P_{base} \times h $$

2. Volume of a Prism: The volume of any prism is the area of its base multiplied by its height.

$$ \text{Volume} = A_{base} \times h $$

Worked Example 2.5 (Triangular Prism)

A triangular prism has a base that is a right-angled triangle with sides $3 \text{ cm}$, $4 \text{ cm}$, and $5 \text{ cm}$. The height of the prism is $10 \text{ cm}$. Calculate its surface area and volume.

Solution:

1. Given: Base sides $3 \text{ cm}, 4 \text{ cm}, 5 \text{ cm}$. Height of prism $h = 10 \text{ cm}$.
2. The base is a right-angled triangle, so the perpendicular sides are $3 \text{ cm}$ and $4 \text{ cm}$.
3. Area of the base ($A_{base}$):

$$ A_{base} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6 \text{ cm}^2 $$

4. Perimeter of the base ($P_{base}$):

$$ P_{base} = 3 + 4 + 5 = 12 \text{ cm} $$

5. Surface Area:

$$ \text{Surface Area} = 2 \times A_{base} + P_{base} \times h $$
$$ \text{Surface Area} = 2 \times 6 + 12 \times 10 $$
$$ \text{Surface Area} = 12 + 120 $$
$$ \text{Surface Area} = 132 \text{ cm}^2 $$

6. Volume:

$$ \text{Volume} = A_{base} \times h $$
$$ \text{Volume} = 6 \times 10 $$
$$ \text{Volume} = 60 \text{ cm}^3 $$

The surface area of the triangular prism is $132 \text{ cm}^2$ and its volume is $60 \text{ cm}^3$.

Pyramid

A pyramid is a polyhedron formed by connecting a polygonal base and a point, called the apex. Each base edge and apex form a triangle, called a lateral face. The shape of the base determines the name of the pyramid (e.g., square pyramid, triangular pyramid).

Let $A_{base}$ be the area of the base, $P_{base}$ be the perimeter of the base, $h$ be the perpendicular height, and $l_s$ be the slant height of the lateral faces (for regular pyramids).

1. Surface Area of a Pyramid: The total surface area is the sum of the area of the base and the areas of all the triangular lateral faces. For a regular pyramid, all lateral faces are identical triangles.

$$ \text{Surface Area} = A_{base} + \text{Area of lateral faces} $$
$$ \text{Surface Area} = A_{base} + \frac{1}{2} P_{base} l_s $$

2. Volume of a Pyramid: The volume of any pyramid is one-third of the area of its base multiplied by its perpendicular height.

$$ \text{Volume} = \frac{1}{3} A_{base} h $$

Worked Example 2.6 (Square Pyramid)

A square pyramid has a base side length of $6 \text{ cm}$ and a perpendicular height of $4 \text{ cm}$. Calculate its surface area and volume.

Solution:

1. Given: Base side $s = 6 \text{ cm}$, perpendicular height $h = 4 \text{ cm}$.
2. Area of the base ($A_{base}$):

$$ A_{base} = s^2 = 6^2 = 36 \text{ cm}^2 $$

3. Perimeter of the base ($P_{base}$):

$$ P_{base} = 4s = 4 \times 6 = 24 \text{ cm} $$

4. To find the slant height ($l_s$), consider a right-angled triangle formed by the perpendicular height ($h$), half the base side ($\frac{s}{2}$), and the slant height ($l_s$).

$$ l_s = \sqrt{h^2 + \left(\frac{s}{2}\right)^2} = \sqrt{4^2 + \left(\frac{6}{2}\right)^2} = \sqrt{16 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ cm} $$

5. Surface Area:

$$ \text{Surface Area} = A_{base} + \frac{1}{2} P_{base} l_s $$
$$ \text{Surface Area} = 36 + \frac{1}{2} \times 24 \times 5 $$
$$ \text{Surface Area} = 36 + 12 \times 5 $$
$$ \text{Surface Area} = 36 + 60 $$
$$ \text{Surface Area} = 96 \text{ cm}^2 $$

6. Volume:

$$ \text{Volume} = \frac{1}{3} A_{base} h $$
$$ \text{Volume} = \frac{1}{3} \times 36 \times 4 $$
$$ \text{Volume} = 12 \times 4 $$
$$ \text{Volume} = 48 \text{ cm}^3 $$

The surface area of the square pyramid is $96 \text{ cm}^2$ and its volume is $48 \text{ cm}^3$.

Mensuration of Frustum of a Cone and Pyramid

A frustum is a portion of a solid (like a cone or pyramid) that remains when the top part is cut off by a plane parallel to the base. It has two parallel bases, one larger and one smaller.

Frustum of a Cone

A frustum of a cone is formed when a cone is cut by a plane parallel to its base, and the smaller cone at the top is removed. It has two circular bases of different radii.

Let $R$ be the radius of the larger base, $r$ be the radius of the smaller base, $H$ be the height of the original cone, $h$ be the height of the frustum, and $L$ be the slant height of the frustum.

The slant height of the frustum can be found using the formula: $L = \sqrt{h^2 + (R-r)^2}$.

1. Surface Area of a Frustum of a Cone: The total surface area is the sum of the areas of the two circular bases and the curved surface area of the frustum.

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

2. Volume of a Frustum of a Cone: The volume of the frustum is the volume of the original large cone minus the volume of the small cone that was removed.

$$ \text{Volume} = \frac{1}{3} \pi h (R^2 + Rr + r^2) $$

Worked Example 3.1

A frustum of a cone has a larger base radius of $10 \text{ cm}$, a smaller base radius of $4 \text{ cm}$, and a height of $8 \text{ cm}$. Calculate its surface area and volume. (Use $\pi = 3.142$)

Solution:

1. Given: $R = 10 \text{ cm}$, $r = 4 \text{ cm}$, $h = 8 \text{ cm}$.
2. First, calculate the slant height ($L$) of the frustum:

$$ L = \sqrt{h^2 + (R-r)^2} = \sqrt{8^2 + (10-4)^2} = \sqrt{64 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ cm} $$

3. Surface Area:

$$ \text{Surface Area} = \pi R^2 + \pi r^2 + \pi L(R + r) $$
$$ \text{Surface Area} = 3.142(10^2) + 3.142(4^2) + 3.142(10)(10 + 4) $$
$$ \text{Surface Area} = 3.142(100) + 3.142(16) + 3.142(10)(14) $$
$$ \text{Surface Area} = 314.2 + 50.272 + 439.88 $$
$$ \text{Surface Area} = 804.352 \text{ cm}^2 $$

4. Volume:

$$ \text{Volume} = \frac{1}{3} \pi h (R^2 + Rr + r^2) $$
$$ \text{Volume} = \frac{1}{3} \times 3.142 \times 8 \times (10^2 + (10 \times 4) + 4^2) $$
$$ \text{Volume} = \frac{1}{3} \times 3.142 \times 8 \times (100 + 40 + 16) $$
$$ \text{Volume} = \frac{1}{3} \times 3.142 \times 8 \times 156 $$
$$ \text{Volume} = 3.142 \times 8 \times 52 $$
$$ \text{Volume} = 1306.928 \text{ cm}^3 $$

The surface area of the frustum of the cone is $804.35 \text{ cm}^2$ (to 2 d.p.) and its volume is $1306.93 \text{ cm}^3$ (to 2 d.p.).

Frustum of a Pyramid

A frustum of a pyramid is formed when a pyramid is cut by a plane parallel to its base, and the smaller pyramid at the top is removed. It has two parallel polygonal bases of different sizes.

Let $A_1$ be the area of the larger base, $A_2$ be the area of the smaller base, $h$ be the height of the frustum, $P_1$ be the perimeter of the larger base, $P_2$ be the perimeter of the smaller base, and $l_s$ be the slant height of the frustum’s lateral faces.

1. Surface Area of a Frustum of a Pyramid: The total surface area is the sum of the areas of the two bases and the areas of the trapezoidal lateral faces.

$$ \text{Surface Area} = A_1 + A_2 + \text{Area of lateral faces} $$

For a regular frustum, the lateral faces are identical trapezoids. The area of one trapezoidal face is $\frac{1}{2} (s_1 + s_2) l_s$, where $s_1$ and $s_2$ are the lengths of the parallel sides of the trapezoid (which are the base edges of the larger and smaller bases, respectively).

2. Volume of a Frustum of a Pyramid: The volume of the frustum is the volume of the original large pyramid minus the volume of the small pyramid that was removed.

$$ \text{Volume} = \frac{1}{3} h (A_1 + \sqrt{A_1 A_2} + A_2) $$

Worked Example 3.2 (Frustum of a Square Pyramid)

A frustum of a square pyramid has a larger base with side length $10 \text{ cm}$ and a smaller base with side length $4 \text{ cm}$. The height of the frustum is $6 \text{ cm}$. Calculate its volume.

Solution:

1. Given: Larger base side $s_1 = 10 \text{ cm}$, smaller base side $s_2 = 4 \text{ cm}$, height $h = 6 \text{ cm}$.
2. Area of the larger base ($A_1$):

$$ A_1 = s_1^2 = 10^2 = 100 \text{ cm}^2 $$

3. Area of the smaller base ($A_2$):

$$ A_2 = s_2^2 = 4^2 = 16 \text{ cm}^2 $$

4. Volume of the frustum:

$$ \text{Volume} = \frac{1}{3} h (A_1 + \sqrt{A_1 A_2} + A_2) $$
$$ \text{Volume} = \frac{1}{3} \times 6 \times (100 + \sqrt{100 \times 16} + 16) $$
$$ \text{Volume} = 2 \times (100 + \sqrt{1600} + 16) $$
$$ \text{Volume} = 2 \times (100 + 40 + 16) $$
$$ \text{Volume} = 2 \times 156 $$
$$ \text{Volume} = 312 \text{ cm}^3 $$

The volume of the frustum of the square pyramid is $312 \text{ cm}^3$.

Surface Area and Volume of Compound Shapes

Compound shapes (also known as composite shapes) are shapes made up of two or more basic geometric shapes. To find the surface area or volume of a compound shape, it is usually broken down into its simpler component shapes.

Calculating Surface Area of Compound Shapes

To find the surface area of a compound shape, identify all the exposed surfaces of the individual shapes. Remember that surfaces where the shapes join are internal and should not be included in the total surface area.

The general steps are:

1. Identify the basic shapes that make up the compound shape.
2. Calculate the surface area of each component shape.
3. Subtract any areas that are overlapping or internal (where the shapes are joined).
4. Add the remaining surface areas together.

Calculating Volume of Compound Shapes

To find the volume of a compound shape, simply add the volumes of all the individual component shapes. Volume is additive, so there are no overlapping volumes to subtract.

The general steps are:

1. Identify the basic shapes that make up the compound shape.
2. Calculate the volume of each component shape.
3. Add the volumes together.

Worked Example 4.1 (Volume of a Compound Shape)

A solid is made up of a cylinder with a hemisphere on top. The cylinder has a radius of $3 \text{ cm}$ and a height of $5 \text{ cm}$. Calculate the total volume of the solid. (Use $\pi = 3.142$)

Note: The volume of a hemisphere is $\frac{2}{3}\pi r^3$.

Solution:

1. Identify the component shapes: a cylinder and a hemisphere.
2. Given: Cylinder radius $r = 3 \text{ cm}$, cylinder height $h = 5 \text{ cm}$. Hemisphere radius $r = 3 \text{ cm}$.
3. Volume of the cylinder:

$$ V_{cylinder} = \pi r^2 h $$
$$ V_{cylinder} = 3.142 \times (3)^2 \times 5 $$
$$ V_{cylinder} = 3.142 \times 9 \times 5 $$
$$ V_{cylinder} = 3.142 \times 45 $$
$$ V_{cylinder} = 141.39 \text{ cm}^3 $$

4. Volume of the hemisphere:

$$ V_{hemisphere} = \frac{2}{3} \pi r^3 $$
$$ V_{hemisphere} = \frac{2}{3} \times 3.142 \times (3)^3 $$
$$ V_{hemisphere} = \frac{2}{3} \times 3.142 \times 27 $$
$$ V_{hemisphere} = 2 \times 3.142 \times 9 $$
$$ V_{hemisphere} = 56.556 \text{ cm}^3 $$

5. Total Volume:

$$ V_{total} = V_{cylinder} + V_{hemisphere} $$
$$ V_{total} = 141.39 + 56.556 $$
$$ V_{total} = 197.946 \text{ cm}^3 $$

The total volume of the solid is $197.95 \text{ cm}^3$ (to 2 d.p.).

Worked Example 4.2 (Surface Area of a Compound Shape)

A solid is formed by placing a cone on top of a cylinder. The cylinder has a radius of $5 \text{ cm}$ and a height of $12 \text{ cm}$. The cone has the same base radius of $5 \text{ cm}$ and a height of $12 \text{ cm}$. Calculate the total surface area of the solid. (Use $\pi = 3.142$)

Solution:

1. Identify the exposed surfaces:
   • Base of the cylinder (circular area).
   • Curved surface of the cylinder.
   • Curved surface of the cone.
   • The circular area where the cone and cylinder meet is internal and not part of the total surface area.
2. Given: Cylinder radius $r = 5 \text{ cm}$, cylinder height $h_{cyl} = 12 \text{ cm}$. Cone radius $r = 5 \text{ cm}$, cone height $h_{cone} = 12 \text{ cm}$.
3. Calculate the slant height ($l$) of the cone:

$$ l = \sqrt{r^2 + h_{cone}^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm} $$

4. Area of the base of the cylinder:

$$ A_{base} = \pi r^2 = 3.142 \times (5)^2 = 3.142 \times 25 = 78.55 \text{ cm}^2 $$

5. Curved surface area of the cylinder:

$$ A_{curved\_cyl} = 2\pi r h_{cyl} = 2 \times 3.142 \times 5 \times 12 = 3.142 \times 120 = 377.04 \text{ cm}^2 $$

6. Curved surface area of the cone:

$$ A_{curved\_cone} = \pi r l = 3.142 \times 5 \times 13 = 3.142 \times 65 = 204.23 \text{ cm}^2 $$

7. Total Surface Area:

$$ A_{total} = A_{base} + A_{curved\_cyl} + A_{curved\_cone} $$
$$ A_{total} = 78.55 + 377.04 + 204.23 $$
$$ A_{total} = 659.82 \text{ cm}^2 $$

The total surface area of the solid is $659.82 \text{ cm}^2$.