Arc Lengths, Sector Areas, and Geometric Similarity

Circle Sectors and Their Properties

A sector of a circle with centre O is a specific region of the circle enclosed by two radii and an arc. When two radii are drawn from the centre to the edge, they divide the circle into two distinct parts: the major sector and the minor sector.

The Sector Angle

The central angle formed between the two radii that bound the sector is known as the sector angle.

Calculating Sector Area

The area of a sector is a fraction of the total area of the circle. If the sector angle is given as $x^\circ$, the area can be calculated using the formula:

$\text{Area} = \frac{x}{360} \times \pi r^2$

Worked Example: Sector Area

A sector of a circle, C, has a central angle of $70^\circ$ and a radius of $12$ cm. To find the area:

1. Use the formula: $\text{Area} = \frac{70}{360} \times \pi \times 12^2$
2. Simplify the calculation: $\text{Area} = \frac{7}{36} \times 144\pi$
3. Final result: $28\pi$ cm²

Calculating Arc Length and Perimeter

The length of the arc bounding the sector is also a fraction of the total circumference. If the sector angle is $x^\circ$, the arc length is:

$\text{Arc length} = \frac{x}{360} \times 2\pi r$

Worked Example: Sector Perimeter

Consider a sector of a circle, C, with an angle of $80^\circ$ and a radius of $15$ cm. What is the total perimeter of this sector?

1. Calculate the arc length: $\frac{80}{360} \times 2 \times \pi \times 15 = \frac{20\pi}{3}$ cm
2. The perimeter consists of the arc plus the two radii: $P = 2r + \text{arc length}$
3. Calculate: $P = (2 \times 15) + \frac{20\pi}{3} = 30 + \frac{20\pi}{3} = 10(3 + \frac{2\pi}{3})$ cm

Worked Example: Finding the Angle from Arc Length

A sector with a radius of $18$ cm has an arc length of $15\pi$ cm. We can find the unknown angle $x$ as follows:

1. Set up the equation: $\frac{x}{360} \times 2 \times \pi \times 18 = 15\pi$
2. Simplify the left side: $\frac{x\pi}{10} = 15\pi$
3. Solve for $x$: $x = 150$. The sector angle is $150^\circ$.

Congruence and Similarity

In geometry, we distinguish between figures that are exactly the same and those that are simply scaled versions of each other.

1. Congruent figures: These are identical in both shape and size. All corresponding angles are equal, and all corresponding lengths are equal.
2. Similar figures: These share the same shape but differ in size. One figure is an enlargement of the other. Corresponding angles remain identical, while corresponding lengths exist in a constant ratio.

All equilateral triangles, spheres, and cubes are inherently similar to others of their kind.

Investigating Congruence

Consider two right-angled triangles, ABC and PQR, both having a hypotenuse of $10$ cm. Is triangle ABC always congruent to triangle PQR?

According to Pythagoras' theorem, the sum of the squares of the other two sides must equal the square of the hypotenuse: $b^2 + c^2 = 10^2 = q^2 + r^2$. If $b=q$ and $c=r$, the triangles are congruent. However, we could have $b^2=64$ and $c^2=36$ for one triangle, while the other has $q^2=r^2=50$. In this case, the triangles are not congruent. Thus, the statement is only sometimes true.

Similarity and Scale Factors

When a shape is enlarged by a scale factor of $n$, the relationships for area and volume change as follows:

• Length: Multiplied by $n$
• Area: Multiplied by $n^2$
• Volume: Multiplied by $n^3$

Worked Example: Similar Triangles

In triangle ADE, point B lies on AD and point C lies on AE, with BC parallel to DE. Given $AB = 5$ cm, $BD = 10$ cm, and $BC = 7$ cm.

First, we show similarity. Both triangles share angle A. Angles ABC and ADE are equal (corresponding angles), and angles ACB and AED are also equal (corresponding angles). Since all angles are the same, triangle ABC is similar to triangle ADE.

To find the length of DE, we find the ratio of corresponding sides. Side $AD = AB + BD = 5 + 10 = 15$ cm. The ratio $AB:AD$ is $5:15$, which simplifies to $1:3$. Therefore, $DE = 3 \times BC = 3 \times 7 = 21$ cm.

Worked Example: Area and Volume Scaling

An individual chocolate bar has ingredients weighing $25$ g and a wrapper with an area of $60$ cm². A 'family bar' is made where the length, width, and height are all twice that of the original bar.

1. Area of wrapper: Since lengths are multiplied by $2$, the area is multiplied by $2^2$. The new area is $60 \times 4 = 240$ cm².
2. Mass of ingredients: Mass is proportional to volume. Since lengths are multiplied by $2$, the volume is multiplied by $2^3$. The new mass is $25 \times 8 = 200$ g.