Perimeter and Area of Circles, Sectors and Segments

Try this

In “Puzzle 1”, drag the slider to $100%$ . What shape does the shaded region become? Do you need $π$ for its area?
While you drag, watch the “Shaded area” readout. Does it ever change?
Before the move, the two pieces have areas $π$ and $4 - π$ , both involving $π$ . Where does the $π$ go when you add them?
Switch to “Puzzle 2” and “Puzzle 3”. Which piece moves each time? Is it a translation or a rotation?

Translations, rotations and reflections do not change area. So before computing a shaded area, ask: can I move the scattered pieces together into a shape I know? This is called cut and paste.

Three basic formulas

A circle of radius $r$ has circumference $C = 2 π r$ and area $S = π r^{2}$ . A sector with central angle $n^{\circ}$ is $\frac{n}{360}$ of the whole circle, so

l = \frac{nπ r}{180} S_{sector} = \frac{nπ r ^{2}}{360} = \frac{1}{2} l r

A segment is the region between a chord and its arc. It has no formula of its own: you build it from a sector and a triangle.

Drag

B

to change the central angle. Once the angle passes

18 0^{\circ}

, how are the segment and the triangle related?

Predict first

Keep the radius fixed and change the central angle from $6 0^{\circ}$ to $12 0^{\circ}$ . What happens to the area of the segment?

Exactly

2

timesLess than

2

timesMuch more than

2

times

Choose an answer first; then you can see the result.

Why you are allowed to move pieces

Example: a segment plus a triangle makes a sector

Step 1 / 5

Look at the shading
The half disc has diameter $A B$ and radius $O A = O B = 2$ ; $M$ is the midpoint of the semicircular arc. The shading has two pieces: the segment on chord $A M$ , and triangle $O M B$ .
Brute force works too
Spot the congruence
Turn it over
Compare the answers

Practice

Exercise 1Basic

A sector has radius $6$ and central angle $6 0^{\circ}$ . Its arc length is $k π$ . Find $k$ .

Exercise 2Basic

A sector has arc length $4 π$ and radius $6$ . Its area is $k π$ . Find $k$ .

Exercise 3Advanced

A large circle has center $O$ and radius $4$ , and $A B$ is a horizontal diameter. Draw two small circles with diameters $O A$ and $O B$ . The shaded region has two parts: the part of the upper half of the large circle that lies outside both small circles, and the lower halves of the two small circles. The shaded area is $k π$ . Find $k$ .

Exercise 4Challenge

In a circle of radius $2$ , a chord subtends a central angle of $9 0^{\circ}$ . The chord splits the circle into a large segment and a small one. The area of the large segment minus the area of the small one is $k π + 4$ . Find $k$ .

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Always taking “sector minus triangle” for a segment

In a circle of radius

2

, the segment on a

27 0^{\circ}

arc has area

\frac{270 π \cdot 2 ^{2}}{360} - \frac{1}{2} \cdot 2 \cdot 2 = 3 π - 2

What went wrong? Think first, then open

Diagnosis When the arc is more than half the circle, the segment contains the center, so it is the sector plus a triangle. Subtracting the triangle throws away a piece that belongs to the segment.

Arc less than half the circle: segment

=

sector

-

triangle. Arc more than half: segment

=

sector

+

triangle. Here that gives

3 π + 2

. You can check it by taking the small segment away from the whole circle:

4 π - (π - 2) = 3 π + 2

“First ask whether the center is inside the segment or outside it”

Exhibit

Mixing up the arc-length and area formulas

A sector with central angle

6 0^{\circ}

and radius

6

has area

\frac{60 π \times 6}{180} = 2 π

What went wrong? Think first, then open

Diagnosis $\frac{nπ r}{180}$ gives the arc length, which is a length. An area must have $r^{2}$ in it.

Sector area

= \frac{nπ r ^{2}}{360} = \frac{60 π \times 36}{360} = 6 π

. Or find the arc length

l = 2 π

first, then use

S = \frac{1}{2} l r = \frac{1}{2} \times 2 π \times 6 = 6 π

“Arc length is a fraction of the circumference; sector area is the same fraction of the circle's area”