Contents

Circle

This lesson covers the definition of a circle, the elements of a circle, the approximation of π\pi, the formulas for circumference and area, and a few contextual examples.

3.7 Explain central angles, inscribed angles, arc length, and the area of a sector, along with their relationships.

4.7 Solve problems related to central angles, inscribed angles, arc length, and the area of a sector.

After studying this material, students are expected to be able to:

  1. Derive the formula for the circumference of a circle from contextual situations.
  2. Derive the formula for the area of a circle from contextual situations.
  3. Explain the relationship between a central angle and arc length.
  4. Explain the relationship between a central angle and sector area.
  5. Solve problems involving circumference and area of circles.

Definition of a circle

A circle is a simple closed curve whose points are all the same distance from one fixed point. That fixed point is called the center of the circle. The distance from the center to a point on the circle is called the radius.

The region enclosed by the circle is called the circular region.

Elements of a circle

Important elements of a circle include:

  1. Center, the point equidistant from all points on the circle, for example point OO.
  2. Radius, a line segment from the center to a point on the circle, for example OA\overline{OA}, OB\overline{OB}, and OC\overline{OC}.
  3. Diameter, a line segment connecting two points on the circle and passing through the center, for example AB\overline{AB}.
  4. Arc, a curved part of the circle, for example arc AB^\widehat{AB}.
  5. Chord, a line segment connecting two points on the circle, for example AC\overline{AC}.
  6. Sector, the region bounded by two radii and an arc, for example sector BOCBOC.
  7. Segment, the region bounded by a chord and its arc, for example segment ADCADC.
  8. Apothem, the shortest segment from the center to a chord, for example OE\overline{OE}.

Circumference of a circle

The circumference of a circle is the length of the path around the edge of the circle.

To approximate π\pi, try the following activity:

  1. Find several circular objects such as coins, wheels, or cans.
  2. Measure the circumference and the diameter of each object.
  3. Compute circumference / diameter.
  4. Record the results in a table.
NoObjectCircumferenceDiameterCircumference / Diameter
1First object
2Second object
3Third object

With careful measurement, the ratio will approach a constant number, namely π\pi.

Common approximations are:

  • π3.14\pi \approx 3.14
  • π227\pi \approx \frac{22}{7}

From the activity above:

π=KdK=πd \pi = \frac{K}{d} \quad \Longleftrightarrow \quad K = \pi d

Since d=2rd = 2r, then:

K=πd=2πr K = \pi d = 2\pi r

So the circumference formula is:

K=πdorK=2πr K = \pi d \quad \text{or} \quad K = 2\pi r

where:

  • KK = circumference
  • dd = diameter
  • rr = radius

Usually, π=227\pi = \frac{22}{7} is used when the radius or diameter is a multiple of 7, while π=3.14\pi = 3.14 is used in other cases.

  1. Find the circumference of a circle with radius 2121 cm.
  2. A motorcycle wheel has radius 2828 cm.
    • Find the circumference.
    • Find the distance traveled after 100100 rotations.
  3. A circular garden has diameter 105105 m. If palm trees are planted every 66 meters along the boundary, how many trees are needed?
  4. Find the perimeter of the shaded region in the following two figures.

Circumference example 1

Circumference example 2

K=2πr=2×227×21=132 K = 2\pi r = 2 \times \frac{22}{7} \times 21 = 132

So the circumference is 132132 cm.

K=2πr=2×227×28=176 K = 2\pi r = 2 \times \frac{22}{7} \times 28 = 176

So the circumference is 176176 cm.

Distance traveled after 100100 rotations:

176×100=17,600 cm=176 m 176 \times 100 = 17{,}600 \text{ cm} = 176 \text{ m}
  1. Garden boundary:
K=πd=227×105=330 \begin{aligned} K &= \pi d \\ &= \frac{22}{7} \times 105 \\ &= 330 \end{aligned}

Number of trees:

n=3306=55 \begin{aligned} n &= \frac{330}{6} \\ &= 55 \end{aligned}

So 5555 palm trees are needed.

  1. First figure:
K=12K+14K+14K+2r=K+2r=2πr+2r=2×3.14×10+2×10=82.8 \begin{aligned} K &= \frac{1}{2}K_\odot + \frac{1}{4}K_\odot + \frac{1}{4}K_\odot + 2r \\ &= K_\odot + 2r \\ &= 2\pi r + 2r \\ &= 2 \times 3.14 \times 10 + 2 \times 10 \\ &= 82.8 \end{aligned}

So the perimeter of the shaded region is 82.882.8 cm.

Second figure:

K=12Klarge+Ksmall=12Klarge+12Klarge=Klarge=227×21=66 \begin{aligned} K &= \frac{1}{2}K_{\text{large}} + K_{\text{small}} \\ &= \frac{1}{2}K_{\text{large}} + \frac{1}{2}K_{\text{large}} \\ &= K_{\text{large}} \\ &= \frac{22}{7} \times 21 \\ &= 66 \end{aligned}

So the perimeter of the shaded region is 6666 cm.

Area of a circle

The area of a circle is the area enclosed by its circumference.

One way to derive the formula is:

  1. Draw a circle.
  2. Divide it into two equal parts.
  3. Split it further into several sectors.
  4. Rearrange the sectors so that the shape approaches a rectangle.

Deriving the area formula

As the number of sectors increases, the new shape approaches a rectangle with:

  • length =12= \frac{1}{2} circumference
  • width =r= r

Thus:

L=p×l=12×circumference×r=12×2πr×r=πr2 \begin{aligned} L &= p \times l \\ &= \frac{1}{2} \times \text{circumference} \times r \\ &= \frac{1}{2} \times 2\pi r \times r \\ &= \pi r^2 \end{aligned}

In terms of diameter:

L=πr2=π(d2)2=14πd2 \begin{aligned} L &= \pi r^2 \\ &= \pi \left(\frac{d}{2}\right)^2 \\ &= \frac{1}{4}\pi d^2 \end{aligned}

So the area formula is:

L=πr2orL=14πd2 L = \pi r^2 \quad \text{or} \quad L = \frac{1}{4}\pi d^2

where:

  • LL = area
  • rr = radius
  • dd = diameter
  1. Find the area of a circle if:
    • the radius is 1010 cm
    • the diameter is 77 cm
  2. Find the radius if the area is 154 cm2154\text{ cm}^2.
  3. Find the area of the shaded region in the following three figures.

Area example 1

Area example 2

Area example 3

  1. If r=10r = 10 cm:
L=πr2=3.14×102=314 \begin{aligned} L &= \pi r^2 \\ &= 3.14 \times 10^2 \\ &= 314 \end{aligned}

So the area is 314 cm2314\text{ cm}^2.

If d=7d = 7 cm:

L=14πd2=14×227×72=38.5 \begin{aligned} L &= \frac{1}{4}\pi d^2 \\ &= \frac{1}{4} \times \frac{22}{7} \times 7^2 \\ &= 38.5 \end{aligned}

So the area is 38.5 cm238.5\text{ cm}^2.

154=227r2r2=49r=7 \begin{aligned} 154 &= \frac{22}{7}r^2 \\ r^2 &= 49 \\ r &= 7 \end{aligned}

So the radius is 77 cm.

  1. First figure:

Area of the large square:

14×14=196 14 \times 14 = 196

Shaded area:

L=7×7+2×227×(72)2=49+77=126 \begin{aligned} L &= 7 \times 7 + 2 \times \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \\ &= 49 + 77 \\ &= 126 \end{aligned}

Unshaded area:

196126=70 196 - 126 = 70

So the unshaded area is 70 cm270\text{ cm}^2.

Second figure:

L=7×714×227×72=4938.5=10.5 \begin{aligned} L &= 7 \times 7 - \frac{1}{4} \times \frac{22}{7} \times 7^2 \\ &= 49 - 38.5 \\ &= 10.5 \end{aligned}

So the shaded area is 10.5 cm210.5\text{ cm}^2.

Third figure:

Lsegment=14×227×14212×14×14=15498=56 \begin{aligned} L_{\text{segment}} &= \frac{1}{4} \times \frac{22}{7} \times 14^2 - \frac{1}{2} \times 14 \times 14 \\ &= 154 - 98 \\ &= 56 \end{aligned}

Since there are two equal segments:

Ltotal=2×56=112 L_{\text{total}} = 2 \times 56 = 112

So the total shaded area is 112 cm2112\text{ cm}^2.

The following image is a circle optical illusion previously created using LaTeX\LaTeX.

Circle optical illusion

This simulation shows the parts of a circle. Turn the elements on or off to inspect them.

Move the radius slider to see how the circumference and area change.

The old interactive worksheet in this article depended on a third-party embed that is no longer stable with the current theme. It has been removed to keep the page rendering intact.

If needed, the evaluation can be replaced later with:

  1. a stable Hugo shortcode,
  2. an external worksheet link,
  3. or a local quiz component.

This completes the circle lesson and its main elements. Suggestions and corrections can be added through the available feedback channel.

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