Geometry of Circles and Composite Solids for the ESAT

Core Formulae for Circles and Cylinders

To succeed in the ESAT, you must memorise the following fundamental formulae for circles and right circular cylinders. These will not be provided in the formula sheet. The radius is denoted as $r$, the diameter as $d$ ($d = 2r$), and the vertical height as $h$.

1. Circumference of a circle: $C = 2\pi r = \pi d$
2. Area of a circle: $A = \pi r^2$
3. Volume of a right circular cylinder: $V = \pi r^2 h$

Formulae Provided in the Examination

For more complex solids, such as spheres, pyramids, and cones, the following formulae will be provided in the question or an examiner guide if required for a calculation:

• Sphere of radius $r$: Volume $= \frac{4}{3}\pi r^3$; Surface Area $= 4\pi r^2$.
• Cone of base radius $r$, perpendicular height $h$, and slant height $l$: Volume $= \frac{1}{3}\pi r^2 h$; Curved Surface Area $= \pi r l$.
• Pyramid of base area $A$ and perpendicular height $h$: Volume $= \frac{1}{3}Ah$.

Calculating Perimeters of Simple and Compound Shapes

The perimeter is the total distance around the outside of a 2-dimensional shape. For compound shapes involving circles, you must identify arcs (fractions of the circumference) and add them to the straight-line segments.

Worked Example: Perimeter of a Window

A window is formed by a rectangle $ABCD$ with a semicircle $BEC$ drawn on the side $BC$. Side $BC = 2x$ cm and side $AB = 4x$ cm. The total perimeter of the window $ABECD$ is given as $24(10 + \pi)$. Find the length of side $CD$.

Step 1: Define the components of the perimeter. The perimeter $P$ consists of the three straight sides of the rectangle ($AB$, $CD$, and $AD$) plus the arc length of the semicircle $BEC$.

Step 2: Calculate the arc length. The diameter of the semicircle is $BC = 2x$, so the radius is $x$. The arc length of a full circle is $2\pi r$, so a semicircle is $\pi r$. Thus, arc $BEC = \pi x$.

Step 3: Sum the components. $P = AB + CD + AD + \text{arc } BEC$. Substituting the variables: $P = 4x + 4x + 2x + \pi x = 10x + \pi x = x(10 + \pi)$.

Step 4: Solve for $x$. We are told $x(10 + \pi) = 24(10 + \pi)$, which implies $x = 24$.

Step 5: Find $CD$. $CD = 4x = 4 \times 24 = 96$ cm.

Calculating Areas of Circles and Compound Shapes

Areas of composite shapes are found by adding the areas of individual shapes or subtracting the area of a shape cut out from another.

Worked Example: Concentric Circles

A circle of radius 4 cm is cut from a circle of radius 8 cm. What fraction of the area of the larger circle remains?

Method 1 (Calculation): The area of the larger circle is $\pi (8)^2 = 64\pi$. The area of the smaller circle is $\pi (4)^2 = 16\pi$. The remaining area is $64\pi - 16\pi = 48\pi$. The fraction remaining is $\frac{48\pi}{64\pi} = \frac{3}{4}$.

Method 2 (Scale Factors): The ratio of the radii is $4:8$ or $1:2$. Since area is proportional to the square of the linear scale factor, the area ratio is $1^2:2^2$ or $1:4$. If the small circle is $\frac{1}{4}$ of the area, then the remaining part is $1 - \frac{1}{4} = \frac{3}{4}$.

Surface Area and Volume of Composite Solids

When calculating the properties of composite solids, you must be careful not to include internal faces in the total surface area.

Worked Example: Ice Cream Cone Model

A model consists of a solid cone (base radius 10 cm, height 25 cm) attached to a solid hemisphere of radius 10 cm. Find the surface area and volume of the model.

Step 1: Find the slant height $l$ of the cone. Using Pythagoras' theorem: $l^2 = r^2 + h^2 = 10^2 + 25^2 = 100 + 625 = 725$. Therefore, $l = \sqrt{725} = 5\sqrt{29}$ cm.

Step 2: Calculate Surface Area. The total surface area $S$ is the curved surface of the cone plus the surface of the hemisphere. Note the circular base is internal and excluded. $S = \pi r l + 2\pi r^2$. Substituting: $S = \pi (10)(5\sqrt{29}) + 2\pi (10)^2 = 50\sqrt{29}\pi + 200\pi = 50(\sqrt{29} + 4)\pi$ cm².

Step 3: Calculate Volume. The total volume $V$ is the volume of the cone plus the volume of the hemisphere. $V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$. Substituting: $V = \frac{1}{3}\pi (100)(25) + \frac{2}{3}\pi (1000) = \frac{2500\pi}{3} + \frac{2000\pi}{3} = \frac{4500\pi}{3} = 1500\pi$ cm³.