Comprehensive Study Guide: Area and Properties of Circles, Sectors, and Segments

Definition of Area: The area of a circle is defined based on its radius ( $r$ ). The standard formula is provided as:
$A = \pi r^2$
Definition of Circumference: The distance around the circle, or the perimeter, is defined by the formula:
$C = 2\pi r$
Radius from Area: To find the radius of a circle when the area is known, the equation is rearranged:
$r = \sqrt{\frac{A}{\pi}}$
Area from Circumference: To find the area given a circumference ( $C$ ), first determine the radius ( $r = \frac{C}{2\pi}$ ) and then apply the area formula.

Definition of a Sector: A sector is a portion of a circle enclosed by two radii and an arc. It is conceptually a "slice" of the circle.
Formula for Sector Area: The area of a sector depends on the measure of the central angle ( $m$ ) in degrees:
$A_{\text{sector}} = \frac{m}{360} \pi r^2$
Calculated Examples:
For a circle with radius $r = 9$ and a central angle of $40^\circ$ : - $A = \frac{40}{360} \pi (9)^2 = \frac{1}{9} \pi (81) = 9\pi$
For a circle with radius $r = 6$ and a central angle of $60^\circ$ : - $A = \frac{60}{360} \pi (6)^2 = \frac{1}{6} \pi (36) = 6\pi$
For a circle with radius $r = 10$ and a central angle of $45^\circ$ : - $A = \frac{45}{360} \pi (10)^2 = \frac{1}{8} \pi (100) = 12.5\pi$
For a circle with radius $r = 16$ and a central angle of $270^\circ$ : - $A = \frac{270}{360} \pi (16)^2 = \frac{3}{4} \pi (256) = 192\pi$
Finding Arc Measure from Area: If the sector area is $24\pi$ and the total circle area is $60\pi$ , the ratio of the areas determines the degree measure:
$\frac{24\pi}{60\pi} = \frac{m}{360} \rightarrow \frac{2}{5} = \frac{m}{360} \rightarrow m = 144^\circ$

Definition of a Segment: A segment is a region of a circle bounded by a chord and an arc. It is calculated by subtracting the area of the triangle formed by the two radii and the chord from the area of the sector.
General Formula:
$A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}}$
Case Study: 60-Degree Central Angle:
Radius $r = 6$ , Central Angle $60^\circ$ . Since the triangle is equilateral ( $s = 6$ ): - $A_{\text{sector}} = 6\pi$ - $A_{\text{triangle}} = \frac{s^2 \sqrt{3}}{4} = \frac{36 \sqrt{3}}{4} = 9\sqrt{3}$ - $A_{\text{segment}} = 6\pi - 9\sqrt{3}$
Case Study: 90-Degree Central Angle:
Given a radius of $r = 6\sqrt{2}$ and a sector angle of $90^\circ$ : - $A_{\text{sector}} = \frac{90}{360} \pi (6\sqrt{2})^2 = \frac{1}{4} \pi (72) = 18\pi$ - $A_{\text{triangle}} = \frac{1}{2} bh = \frac{1}{2} (6\sqrt{2})(6\sqrt{2}) = \frac{1}{2} (72) = 36$ - $A_{\text{segment}} = 18\pi - 36$
Case Study: Equilateral Triangle Inscribed:
Given an equilateral triangle with a "radius" (distance from center to vertex) of 6: - Segment Area Calculation: $27\sqrt{3} - 9\pi$

Definition: An annulus is the region between two concentric circles with different radii.
General Formula Derivation: If the inner circle has radius $r$ and the outer circle has radius $R$ :
$A = \pi R^2 - \pi r^2 = \pi(R^2 - r^2)$
Example Calculation: For a "washer" where the inner radius is $3$ and the outer radius is $5$ :
$A = \pi(5^2 - 3^2) = \pi(25 - 9) = 16\pi$

Lawn Sprinkler Coverage:
Two sprinklers each spray a circular region with a radius of $3\,m$ on a rectangular lawn ( $10\,m \times 12\,m$ ).
Total area watered: $2 \times (\pi \times 3^2) = 18\pi \,m^2$
Total lawn area: $10 \times 12 = 120 \,m^2$
Area not watered (shaded): $120 - 18\pi \,m^2$
Concentric Target Intervals:
A target has a bulls-eye with a diameter of $5\,cm$ ( $r = 2.5\,cm$ ). Each subsequent band has a width of $2.5\,cm$ .
Area of central bulls-eye: $\pi (2.5)^2 = 6.25\pi \,cm^2$
Area of first band: $\pi (5.0)^2 - \pi (2.5)^2 = 25\pi - 6.25\pi = 18.75\pi \,cm^2$
Wankel Rotary Engine Geometry:
Based on an equilateral triangle where each arc is centered at the opposite vertex.
Calculates both the perimeter (sum of three circular arcs) and the area of the resulting rotor shape.
Lunes of Hippocrates:
These are the shaded crescent regions formed by arcs drawn from the midpoints of the sides of a triangle reaching to the vertices.
Finding the area of these lunes often involves the principle that the sum of the areas of the lunes on the two legs of a right triangle is equal to the area of the triangle itself.
Inscribed Circle in a Rhombus:
Given a rhombus with diagonals of $30$ and $40$ .
Area of rhombus: $\frac{1}{2} (30)(40) = 600$ .
The radius of the inscribed circle must be found to determine the area of the shaded region surrounding it.

Question 1: Find the area and circumference of a circle with radius 1, 8, and 15.
Response: For $r=1$ , $A=\pi$ , $C=2\pi$ . For $r=8$ , $A=64\pi$ , $C=16\pi$ . For $r=15$ , $A=225\pi$ , $C=30\pi$ .
Question 2: Find the radius of a circle whose area is $16\pi$ and $169\pi$ .
Response: For $A=16\pi$ , $r=4$ . For $A=169\pi$ , $r=13$ .
Question 3: Find the circumference of a circle whose area is $100\,cm^2$ .
Response: $100 = \pi r^2 \rightarrow r = \frac{10}{\sqrt{\pi}}$ . Then $C = 2\pi (\frac{10}{\sqrt{\pi}}) = 20\sqrt{\pi} \,cm$ .
Question 4: Find the area of a circle whose circumference is $18\pi\,dm$ .
Response: $18\pi = 2\pi r \rightarrow r = 9$ . Area $A = \pi (9)^2 = 81\pi \,dm^2$ .
Question 5: What observation is possible when comparing shaded regions for $x=6$ and $x=10$ in certain polygon-circle configurations?
Response: The ratio of the area of the shaded region to the total area remains constant if the geometric proportions are maintained regardless of the value of $x$ .