Geometry and Formulae for the TMUA

In the TMUA, you are expected to have a fluent command of the geometric properties of 2-dimensional and 3-dimensional shapes. While some advanced formulae are provided in the exam, the fundamental properties of circles and cylinders must be memorised and applied with precision.

Circle Formulae: Circumference and Area

The most fundamental geometric relationships involve the circle. You must know the following formulae by heart:

1. Circumference of a circle: $C = 2 ext{\pi}r = ext{\pi}d$. This relationship defines the ratio of the circumference to the diameter as the constant $ext{\pi}$.
2. Area of a circle: $A = ext{\pi}r^2$.

When working with circular geometry, pay close attention to whether a question provides the radius ($r$) or the diameter ($d$). Using the diameter in the area formula without first halving it is a frequent source of error.

Worked Example: Perimeter of a Semi-Circle

Find the perimeter of a semi-circle with a radius of $5$ units. Give your answer in terms of $ext{\pi}$.

Step 1: Calculate the length of the curved arc. The circumference of a full circle is $2 ext{\pi}r$, so the arc of a semi-circle is $ext{\pi}r$. Here, $Arc = 5 ext{\pi}$.

Step 2: Calculate the length of the straight edge, which is the diameter of the circle. $d = 2r = 10$.

Step 3: Sum the components. $Perimeter = 5 ext{\pi} + 10 = 5( ext{\pi} + 2)$ units.

Right Circular Cylinders

A right circular cylinder is a 3-dimensional solid with two congruent circular bases that are parallel and connected by a curved surface. You must know the formula for its volume:

Volume of a right circular cylinder: $V = ext{\pi}r^2h$.

This formula is derived by multiplying the area of the circular base ($ext{\pi}r^2$) by the perpendicular height ($h$). If you are asked to calculate the total surface area of a cylinder, remember to sum the areas of the two circular ends and the curved surface area:

$Total�Surface�Area = 2 ext{\pi}r^2 + 2 ext{\pi}rh$.

Three-Dimensional Solids: Spheres, Pyramids, and Cones

For more complex solids such as spheres, pyramids, and cones, the TMUA specification notes that formulae will be provided if needed. However, being familiar with their structure is necessary for identifying how to use those formulae in composite problems.

1. Spheres: Defined by a single radius $r$. Metrics include volume and surface area.
2. Pyramids: Defined by a base area and a perpendicular vertical height.
3. Cones: A special type of pyramid with a circular base. Metrics involve the radius $r$, the vertical height $h$, and sometimes the slant height $l$.

Perimeters and Areas of Composite Shapes

Composite shapes are formed by combining or subtracting standard 2-dimensional figures. To calculate their perimeters or areas, you should decompose the shape into its constituent parts.

Worked Example: Composite Area

A shape is formed by removing a semi-circle of diameter $4$ cm from one side of a square with side length $10$ cm. Calculate the area of the remaining shape.

Step 1: Find the area of the square. $Area_{square} = 10 � ext{ �} � 10 = 100$ $cm^2$.

Step 2: Find the area of the semi-circle. The radius is $2$ cm (half the diameter). Area_{semi} = rac{1}{2} ext{\pi} (2^2) = 2 ext{\pi} $cm^2$.

Step 3: Subtract the semi-circle from the square. $Total�Area = 100 - 2 ext{\pi}$ $cm^2$.

Surface Area and Volume of Composite Solids

Similar to 2D shapes, composite solids require breaking the object down into standard components such as cylinders, cones, and spheres.

Worked Example: Volume of a Capsule

A capsule is formed by joining two hemispheres of radius $r$ to the ends of a cylinder with radius $r$ and height $H$. Find the volume of the capsule.

Step 1: The two hemispheres together form a single sphere of radius $r$. The volume of a sphere is rac{4}{3} ext{\pi}r^3.

Step 2: The central cylinder has a volume of $ext{\pi}r^2H$.

Step 3: Combine the volumes. V_{total} = ext{\pi}r^2H + rac{4}{3} ext{\pi}r^3 = ext{\pi}r^2(H + rac{4}{3}r).