Free SAT Math Geometry and Trigonometry

SAT Math · Geometry and Trigonometry · free lesson

Circles — Arc, Sector, and Circumference

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The topic in one minute

Press Play: the idea and a worked example, explained aloud (about 1:05).

Arcs, sectors and circumference

A sector is a slice of a circle. Its arc is the piece of the edge.

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Read what is said

Circles show up on the SAT in questions about arcs, sectors and circumference. An arc is a piece of the edge, and a sector is a slice. Here the radius is six and the central angle is sixty degrees.
Two formulas give the size of a circle. The SAT usually leaves pi in the answer, so you work out only the number in front of pi.
A central angle takes a fraction of the full turn of three hundred sixty degrees. Here we have a radius of nine and a central angle of eighty degrees.
The arc is that fraction of the circumference. Multiply the fraction by two r, and the number in front of pi works out to four.
The sector is the same fraction of the area. Both answers are two ninths of the whole circle.
Read whether you are given the radius or the diameter. A diameter of fourteen means a radius of seven. Squaring the diameter would give one hundred ninety-six, four times too much.
Remember that an arc and a sector are each the same fraction of the whole circle, the central angle over three hundred sixty. Always use the radius, not the diameter. Now practice it with new numbers.

The idea

Around the circle and inside it

Every point of a circle is the same distance from the center. That distance is the radius $r$ . A diameter goes across the circle through the center, so it is two radii long. Two formulas give the size of a circle:

the circumference, the distance around, is $C = 2\pi r$ ;
the area, the space inside, is $A = \pi r^2$ .

The SAT usually leaves $\pi$ in the answer. So work out only the number in front of $\pi$ . For a circle of radius 5:

\begin{aligned} C &= 2r \\ &= 2 \times 5 \\ &= 10 \\ A &= r^{2} \\ &= 5^{2} \\ &= 25 \end{aligned}

The circumference is $10$ $\pi$ and the area is $25$ $\pi$ .

Warning.

Read whether the question gives the radius or the diameter. The formulas use the radius. A diameter of 10 means a radius of 5.

A part of the circle

An arc is a piece of the circumference. A sector is the slice between two radii, shaped like a piece of pie. The angle between the two radii, at the center, is the central angle. A full turn is 360 degrees, so a central angle of 60 degrees takes $\frac{60}{360}$ of the circle: that fraction of the circumference and the same fraction of the area.

Rule.

Arc length = $\frac{\text{angle}}{360}$ of the circumference. Sector area = $\frac{\text{angle}}{360}$ of the area of the circle.

In a circle of radius 6, take a central angle of 60 degrees. The circumference is $12\pi$ and the area is $36\pi$ , so the numbers in front of $\pi$ are:

\begin{aligned} f &= \frac{\theta}{360} \\ &= \frac{60}{360} \\ &= \frac{1}{6} \\ s &= \frac{\theta}{360} \times 2r \\ &= \frac{60}{360} \times 2 \times 6 \\ &= 2 \\ A &= \frac{\theta}{360}r^{2} \\ &= \frac{60}{360} \times 6^{2} \\ &= 6 \end{aligned}

The arc is $2$ $\pi$ long and the sector has area $6$ $\pi$ .

Watch the arc grow with its angle. At 180 degrees it is half of the circle, and at 360 degrees it is all of it.

Now choose the radius and the angle yourself. The title gives the arc length and the sector area.

Angles in a circle

An arc is also measured in degrees, and its measure equals its central angle. An inscribed angle has its corner on the circle instead of at the center. It is half of the central angle that cuts off the same arc. Below, arc $AB$ and the central angle $AOB$ are both 100 degrees, and the inscribed angle $ACB$ is 50 degrees.

Two more facts come up with radii. A triangle made by two radii and a chord is isosceles, because the two radii are equal, so its two angles on the chord are equal. And a tangent, a line that touches the circle at one point, is perpendicular to the radius drawn to that point.

Radians

A radian is another unit for angles. A half turn, 180 degrees, is $\pi$ radians. To change degrees to radians multiply by $\frac{\pi}{180}$ , and to change back multiply by $\frac{180}{\pi}$ . For 150 degrees the number in front of $\pi$ is:

\begin{aligned} k &= \frac{x}{180} \\ &= \frac{150}{180} \\ &= \frac{5}{6} \end{aligned}

So 150 degrees is $\frac{5}{6}$ $\pi$ radians. With the angle $\theta$ in radians, the arc formula becomes very short: arc length $= r\theta$ . In a circle of radius 12 that angle cuts off an arc of:

\begin{aligned} s &= rk \\ &= 12 \times \frac{5}{6} \\ &= 10 \\ s &= \frac{2rx}{360} \\ &= \frac{2 \times 12 \times 150}{360} \\ &= 10 \end{aligned}

The arc is $10$ $\pi$ long, and the degree method gives the same number.

Note.

On the SAT these questions are of medium difficulty. The usual ones: the area is $k\pi$ and you give $k$ ; how many times as large one circle's area is as another's (square the ratio of the radii); an arc from its central angle; a base angle or the chord of the triangle made by two radii. The answer is nearly always a fraction of the whole circle.

Worked examples

Each example is one of the kinds of question the test asks about circles. The numbers in the working are found by the engine. Where an answer has $\pi$ in it, the working finds the number in front of $\pi$ .

Example.

1. Area from a diameter. A circle has a diameter of 14. Its area is $k\pi$ . What is $k$ ?

The area formula $A = \pi r^2$ needs the radius, and the question gives the diameter. Halve it first, then square.

\begin{aligned} r &= \frac{d}{2} \\ &= \frac{14}{2} \\ &= 7 \\ k &= r^{2} \\ &= 7^{2} \\ &= 49 \end{aligned}

The area is $49$ $\pi$ , so $k$ is $49$ . Squaring the diameter instead would give 196, four times too much.

Example.

2. An arc and a sector. In a circle of radius 9 with center $O$ , the central angle $AOB$ measures 80 degrees. Find the length of arc $AB$ and the area of sector $AOB$ .

The angle is $\frac{80}{360}$ of a full turn, so the arc is that fraction of the circumference $2\pi \cdot 9$ and the sector is that fraction of the area $\pi \cdot 9^2$ .

\begin{aligned} f &= \frac{\theta}{360} \\ &= \frac{80}{360} \\ &= \frac{2}{9} \\ s &= \frac{\theta}{360} \times 2r \\ &= \frac{80}{360} \times 2 \times 9 \\ &= 4 \\ A &= \frac{\theta}{360}r^{2} \\ &= \frac{80}{360} \times 9^{2} \\ &= 18 \end{aligned}

The arc is $4$ $\pi$ long and the sector has area $18$ $\pi$ . Both are $\frac{2}{9}$ of the whole.

Example.

3. Comparing two circles. Circle $P$ has a radius of $3n$ and circle $Q$ has a radius of $n$ , where $n$ is a positive number. The area of circle $P$ is how many times the area of circle $Q$ ?

Area uses the square of the radius. Square each radius and compare; the $\pi$ is the same in both and cancels.

\begin{aligned} a &= 3^{2}n^{2} \\ &= 9n^{2} \\ k &= 3^{2} \\ &= 9 \end{aligned}

Circle $P$ has area ${9n^{2}}\pi$ and circle $Q$ has area $n^2\pi$ , so the first is $9$ times the second. The radius is 3 times as long, but the area is not 3 times as large: the ratio of the radii is squared.

Example.

4. The triangle made by two radii. Points $A$ and $B$ lie on a circle with center $O$ and radius 9. Angle $AOB$ measures 104 degrees, and the perimeter of triangle $AOB$ is 32.2. Find angle $OAB$ and the length of the chord $AB$ .

$OA$ and $OB$ are both radii, so the triangle is isosceles and the angles at $A$ and $B$ are equal. The three angles add to 180 degrees. The perimeter is two radii and the chord.

\begin{aligned} x &= \frac{180 - \theta}{2} \\ &= \frac{180 - 104}{2} \\ &= 38 \\ c &= P - 2r \\ &= 32.2 - 2 \times 9 \\ &= 14.2 \end{aligned}

Angle $OAB$ is $38$ degrees and the chord is $14.2$ long. Neither part needed $\pi$ .

Example.

5. Two diameters and a ratio. A circle has a circumference of 60. $AC$ and $BD$ are diameters. The length of arc $AB$ is to the length of arc $BC$ as 2 is to 3. What is the length of arc $AB$ ?

$AC$ is a diameter, so going from $A$ through $B$ to $C$ is half of the circle. The two arcs share that half, not the whole circumference. Split the half into 2 + 3 equal parts.

\begin{aligned} h &= \frac{C}{2} \\ &= \frac{60}{2} \\ &= 30 \\ 2p + 3p &= h \\ 2p + 3p &= 30 \\ p &= 6 \\ s &= 2p \\ &= 2 \times 6 \\ &= 12 \\ t &= 3p \\ &= 3 \times 6 \\ &= 18 \end{aligned}

Arc $AB$ is $12$ long and arc $BC$ is $18$ long. Together they make $30$ , half of the circumference.

Example.

6. Radians and an inscribed angle. In a circle of radius 15 with center $O$ , the central angle $AOB$ measures $\frac{2\pi}{3}$ radians. Point $C$ is on the circle, on the far side from arc $AB$ . Find the length of arc $AB$ , the measure of angle $AOB$ in degrees, and the measure of angle $ACB$ in degrees.

With the angle in radians, arc length is $r\theta$ : multiply the radius by the angle. To change radians to degrees, replace $\pi$ by 180. The inscribed angle $ACB$ cuts off the same arc as the central angle, so it is half of it.

\begin{aligned} s &= rk \\ &= 15 \times \frac{2}{3} \\ &= 10 \\ x &= k \times 180 \\ &= \frac{2}{3} \times 180 \\ &= 120 \\ y &= \frac{x}{2} \\ &= \frac{120}{2} \\ &= 60 \end{aligned}

The arc is $10$ $\pi$ long, angle $AOB$ is $120$ degrees and angle $ACB$ is $60$ degrees.

Practice

Answer SAT-style questions on circles — arc, sector, and circumference. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise circles — arc, sector, and circumference with new numbers