How to find the circumference of a circle - SSAT Upper Level Quantitative

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If you have a circular yard and need to put up a fence around the outside, you would use the formula $\pi $r^{2}$$ to figure out the amount of fence you need.

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To figure out the amount of the fence around a circular yard, you need to find the circumference of the yard. The equation for the circumference of a cirle is $2\pi r$ and not $\pi $r^{2}$$ .

A circle on the coordinate plane has equation

.

Which of the following gives the circumference of the circle?

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The equation of a circle on the coordinate plane is

,

where is the radius. Therefore,

$r ^{2} = 75$

and

$r = $\sqrt{ 75}$ = $\sqrt{25}$ \times $\sqrt{ 3}$ = 5 $\sqrt{ 3}$$ .

The circumference of a circle is $2 \pi$ times is radius, which here would be

$C = 2 \pi \times 5 $\sqrt{3}$ = 10\pi $\sqrt{3}$$ .

A circle on the coordinate plane has equation

Which of the following gives the circumference of the circle?

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The equation of a circle on the coordinate plane is

where is the radius. Therefore,

and

$r = $\sqrt{49}$ = 7$ .

The circumference of a circle is $2 \pi$ times is radius, which here would be

$C= 2 \pi \cdot 7 = 14 \pi$

Refer to the above diagram. Give the length of arc $\widehat{AB}$ .

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The figure is a sector of a circle with radius 8; the sector has degree measure $180 ^{\circ } - 45 ^{\circ }= 135 ^{\circ }$ . The length of the arc is

$A = $\frac{ $135^{\circ }$$}{360 ^{\circ }} \cdot 2\pi r$

$A = $\frac{ $135^{\circ }$$}{360 ^{\circ }} \cdot 2\pi \cdot 8$

$A = $\frac{3}{8}$ \cdot 16\pi$

$A = 6\pi$

A $60 ^{\circ }$ central angle of a circle has a chord with length 9. Give the circumference of the circle.

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The figure below shows $\angle AOB$ , which matches this description, along with its chord $\overline{AB}$ :

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved equilateral, so . This is the radius, so the circumference is

$C = 2 \pi r = 2 \pi \cdot 9 = 18 \pi$

A $90 ^{\circ }$ central angle of a circle has a chord with length . Give the circumference of the circle.

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The figure below shows $\angle AOB$ , which matches this description, along with its chord $\overline{AB}$ :

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved a 45-45-90 triangle with hypotenuse 40. By the 45-45-90 Theorem, its legs, both radii, have length that can be determined by dividing this by $\sqrt{2}$ , so

$A O = $\frac{AB }{\sqrt{2}$} = $\frac{40}{\sqrt{2}$} = $\frac{40\cdot\sqrt{2}$}{$\sqrt{2}$\cdot $\sqrt{2}$}= $\frac{40 \sqrt{2}$}{2}=20 $\sqrt{2}$$

This is the radius, so the circumference is

$C = 2 \pi r = 2 \pi \cdot 20 $\sqrt{2}$= 40 \pi $\sqrt{2}$$

A $120 ^{\circ }$ central angle of a circle has a chord with length 24. Give the circumference of the circle.

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The figure below shows $\angle AOB$ , which matches this description, along with its chord $\overline{AB}$ and triangle bisector $\overline{OM}$ .

We will concentrate on $\Delta AOM$ , which is a 30-60-90 triangle.

Chord $\overline{AB}$ has length 24, so $\overline{AM}$ has length half this, or 12.

By the 30-60-90 Theorem,

$OM = AM \div $\sqrt{3}$= 12 \div $\sqrt{3}$= $\frac{12}{ \sqrt{3}$}$

and

$AO = 2 \cdot OM = 2 \cdot $\frac{12}{ \sqrt{3}$} = $\frac{24}{ \sqrt{3}$} = $\frac{24 \cdot \sqrt{3}$}{ $\sqrt{3}$ \cdot $\sqrt{3}$ } =\frac{24 \cdot \sqrt{3}$}{ 3 } = 8$\sqrt{3}$$

This is the radius, so the circumference is

$C = 2 \pi r = 2 \pi \cdot 8$\sqrt{3}$= 16 \pi $\sqrt{3}$$

Give the circumference of a circle that circumscribes a 30-60-90 triangle whose longer leg has length $7$\frac{1}{2}$$ .

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If a right triangle is inscribed inside a circle, then the arc intercepted by the right angle is a semicircle, making the hypotenuse of triangle a diameter.

The length of the shorter leg of a 30-60-90 triangle is that of the longer leg divided by $\sqrt{3}$ , so the shorter leg will have length

$7$\frac{1}{2}$ \div $\sqrt{3}$ = $\frac{15}{2}$ \times $\frac{1}{\sqrt{3}$} = $\frac{15}{2}$ \times $\frac{\sqrt{3}$}{3} =\frac{5\sqrt{3}$}{2}$

The hypotenuse will have length twice that of its short leg, so the hypotenuse of this triangle will have twice this length, or

$2 \times $\frac{5\sqrt{3}$}{2} = 5$\sqrt{3}$$

This is the diameter, so multiply this by $\pi$ to get the circumference:

$5$\sqrt{3}$ \cdot \pi = 5 \pi $\sqrt{3}$$

Give the circumference of a circle that circumscribes a triangle whose shorter leg has length $6$\frac{1}{2}$$ .

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If a right triangle is inscribed inside a circle, then the arc intercepted by the right angle is a semicircle, making the hypotenuse of triangle a diameter.

The length of a hypotenuse of a 30-60-90 triangle is twice that of its short leg, so the hypotenuse of this triangle will be twice $6$\frac{1}{2}$$ , or

$6$\frac{1}{2}$ \times 2 = 13$ .

This is the diameter, also, so the circumference is $\pi$ times this, or $C =13 \pi$ .

Give the circumference of a circle that circumscribes a right triangle with legs of length 18 and 24.

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If a right triangle is inscribed inside a circle, then the arc intercepted by the right angle is a semicircle, making the hypotenuse of triangle a diameter.

The length of the hypotenuse of this triangle can be calculated using the Pythagorean Theorem:

$d = \sqrt ${18^{2}$$+24^{2}$} = $\sqrt{324+576}$ = $\sqrt{900}$ = 30$

This is the diameter, also, so the circumference is $C = \pi d = 30 \pi$ .

Give the circumference of a circle that circumscribes an equilateral triangle with perimeter .

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An equilateral triangle of perimeter 84 has sidelength one-third of this, or 28.

Construct this triangle and its circumscribed circle, as well as a perpendicular bisector to one side and a radius to one of that side's endpoints:

Each side of the triangle has measure 28, so . Also, the triangle formed by the segments, by symmetry, is a 30-60-90 triangle. Therefore, by the 30-60-90 Theorem,

$BO = $\frac{AB}{\sqrt{3}$} = $\frac{14}{\sqrt{3}$} = $\frac{14\cdot \sqrt{3}$}{$\sqrt{3}$\cdot $\sqrt{3}$} = $\frac{14 \sqrt{3}$}{3}$

and $AO = 2 \cdot BO = 2 \cdot $\frac{14 \sqrt{3}$}{3}= $\frac{28 \sqrt{3}$}{3}$ .

This is the radius, so the circumference is $2 \pi$ times this, or

$$\frac{28 \sqrt{3}$}{3} \times 2 \pi = $\frac{56 \pi \sqrt{3}$}{3}$

Give the ratio of the circumference of a circle that circumscribes an equilateral triangle to that of a circle that is inscribed inside the same triangle.

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If a (perpendicular) radius of the inscribed circle is constructed to the triangle, and a radius of the circumscribed circle is constructed to a neighboring vertex, a right triangle is formed. By symmetry, it can be shown that this is a 30-60-90 triangle, and, subsequently,

$BO = 2 \cdot AO$

If we let , the circumference of the inscribed circle is $C =2 \pi r$ .

Then , and the circumference of the circumscribed circle is $2 \pi \cdot 2r = 4 \pi r = 2C$ .

The ratio of the circumferences is therefore 2 to 1.

Give the circumference of a circle that is inscribed in an equilateral triangle with perimeter 60.

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An equilateral triangle of perimeter 60 has sidelength one-third of this, or 20.

Construct this triangle and its inscribed circle, as well as a radius to one side - which, by symmetry, is a perpendicular bisector - and a segment to one of that side's endpoints:

Each side of the triangle has measure 20, so . Also, the triangle formed by the segments, by symmetry, is a 30-60-90 triangle. Therefore,

$BO = $\frac{AB}{\sqrt{3}$} = $\frac{10}{\sqrt{3}$} = $\frac{10\cdot \sqrt{3}$}{$\sqrt{3}$\cdot $\sqrt{3}$} = $\frac{10 \sqrt{3}$}{3}$

which is the radius of the circle. The cricumference of the circle is $2 \pi$ times this, or

$$\frac{10 \sqrt{3}$}{3} \times 2 \pi = $\frac{20 \pi \sqrt{3}$}{3}$

Find the circumference of a circle with a diameter of 14.

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Write the formula to find the circumference of a circle.

$C=2\pi r=\pi D$

Substitute the diameter into the formula.

$c=\pi(14)= 14\pi$

Find the circumference of the circle with an area of $9\pi$ .

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Write the formula for the area of the circle. The radius is needed to find the circumference of the circle.

$A=\pi $r^2$$

Substitute the area.

$9\pi=\pi $r^2$$

The radius of the circle is 3. Substitute this in the circumference formula.

$C=2\pi r= 2\pi(3)=6\pi$

The correct answer is: $6\pi$

The track at Madison High School is a perfect circle of radius 600 feet, and is shown in the above figure. Boris wants to run around the track for one mile. If Boris starts at point A and runs clockwise, which of the following is closest to the point at which he will stop running?

(Assume the five points are evenly spaced)

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A circle of radius 600 feet will have a circumference of

$C = 2 \pi r \approx 2 \cdot 3.14 \cdot 600 = 3,768$ feet.

Boris will run one mile, or 5,280 feet, which will be about

$$\frac{5,280}{3,768}$ \approx 1.4$ tiimes the circumference of the track, or, equivalently, once around the track, plus another two-fifths of the circle. This means he will end up closest to point C.

The track at Cesar Chavez High School is a perfect circle of radius 500 feet, and is shown in the above figure. Rafael starts at point C, runs around the track clockwise twice, and continues to run clockwise until he makes it to point D. Which of the following comes closest to the number of miles Rafael has run?

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The circumference of a circle with radius 500 feet is

$C = 2 \pi r \approx 2 \cdot 3.14 \cdot 500 = 3,140$ feet.

Rafael runs this distance twice, then he runs from Point C to D, which is about one-fifth of this distance. Therefore, Rafael's run will be about

$3,140 \cdot 2 $\frac{1}{5}$ = 6,908$ feet.

Divide by 5,280 to convert to miles:

$6,908 \div 5,280 \approx 1.3$ miles.

Of the given choices, $1 $\frac{1}{2 }$$ miles is the closest to the correct length.

The track at Douglas MacArthur High School is shown above; it is the composite of a square and a semicircle.

Jennifer wants to run one mile. If she begins at Point A and begins running clockwise, where will she be when she is finished?

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First, it is necessary to know the length of the semicircle connecting Points B and D, which has diameter 400 feet; this length is

$$\frac{1}{2}$ C = $\frac{1}{2}$ \cdot \pi d \approx $\frac{1}{2}$ \cdot 3.14 \cdot 400 = 628$ feet.

The distance around the track is about

feet.

Divide this into 5,280 feet (one mile):

$5,280 \div 1,828 = 2\textup{ R }1,624$

This means that Jennifer will run two times around the track to return to Point A. She will have 1,624 feet to go, so she will do the following:

She will run 400 feet from A to B, leaving feet;

She will run 628 feet from B to D, leaving feet;

She will run 400 feet from D to E, leaving feet;

She will then run the remaining 196 feet from Point E toward Point A.

The correct response is that she will be between Point E and Point A.

The track at Eisenhower High School is shown above; it is comprised of a square and a semicircle.

Mike begins at Point A, runs five times around the track clockwise, and continues further until he reaches Point C. Which of the following comes closest to the distance Mike runs?

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First, it is necessary to know the length of the semicircle connecting Points B and D, which has diameter 500 feet; this length is

$$\frac{1}{2}$ C = $\frac{1}{2}$ \cdot \pi d \approx $\frac{1}{2}$ \cdot 3.14 \cdot 500 = 785$ feet.

The distance around the track is about

feet.

Mike runs around the track five times, which is a distance of about

$2,285 \times 5 = 11,425$ feet.

He then proceeds to Point B, which is an additional 500 feet, and to Point C, which is half the semicircle, or

$$\frac{1}{2}$ \cdot 785 \approx 393$ feet.

Therefore, Mike runs about

feet.

Divide by 5,280 to convert to miles:

$12,318 \div 5,280 \approx 2.3$ miles.

Of the choices, the closest is $2$\frac{1}{2}$$ miles.

Find the circumference of a circle with a diameter of 12.

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There are two possible formulas for finding the circumference of a circle. They are as follows:

$C=2\Pi \cdot r$

And:

$C=\Pi \cdot d$

Where C is circumference, r is radius, and d is diameter.

For this problem, we are given the diameter is 12, meaning we can plug our numbers into the second equation. Since pi is an irrational constant, it is okay to leave the answer in terms of pi.

$C=\Pi \cdot 12= 12\Pi$