SAT Hub

Circles

** Circumference (c):** The distance around a circle (the perimeter of the circle)

Chord:

Pi (П):

Because we don't know pi's exact value, we use two approximations: \(3.14\leftarrow OR\rightarrow \frac{22}{7}\)

The formula for finding the circumference of a circle is: *circumference = 2 • П • radius*

To find the AREA of a circle, use the following formula: *area = П • radius2*

*Solved Example 1*

Points X and Y are two different points on a circle. Point M is located so that line segment XM and line segment YM have equal length. Which of the following could be true?

I. M is the center of the circle

II. M is on arc XY

III. M is outside of the circle

** A.** I only

** Solution: **Option I considers the possibility that M could be the center of the circle if lines XM and YM are equal. We know this must be true because M being the center point of the circle would make lines XM and YM radii of the circle, which would mean that they were equal.

Option III presents us with the possibility that M lies somewhere on the outside of the circle. So long as M lies at a distance halfway between X and Y, this scenario would still work. So option III is also correct.

This means that all of our options (I, II, and III) are possible. **The correct answer is D.**

A full revolution, or turn, of a spinning wheel is equivalent to the wheel going around once. A point on the edge of the wheel travels one circumference in one revolution.

For example, if a wheel spins at 3 revolutions per second, a point on the edge travels a distance equal to 3 circumferences per second. If the wheel has a diameter of 4 feet, then the point travels at a rate of 3 x 4π = 12π feet per second.

** Solved Example 2**Sophia is riding her bicycle across a basketball court. She knows her bicycle wheel has a radius of 1 ft and that the basketball court is about 94 ft long. Approximately how many times does her bicycle wheel rotate over the course of her ride across the basketball court?

A.

** Solution: **Each time Sophia’s bicycle wheel makes a full rotation, she travels a linear distance of one circumference across the court. The circumference of her wheel is C = dП = 2rП = 2(1)П = 2П

Now divide the length of the entire basketball court by the diameter of the wheel in order to find the number of revolutions/turns/circumferences Sophia’s wheel must complete to get her to the other side.

\(\frac{94\: ft}{2\pi \: ft/rev}\approx 30\: revolutions\)

**The correct answer is B**

Often, the SAT or ACT will ask you to solve for a portion of the distance on a circle, instead of the entire circumference. This portion is termed an ARC.

Arc AXB is the arc from A to B, passing through the point X. To find its length, first find the circumference of the circle.

C = 2πr = 24π

Since the arc is defined by the central angle of 60*°*, and the entire circle is 360*°*, then the arc is \(\frac{60}{360}=\frac{1}{6}\) of the circle

Therefore, the measure of arc AXB= \(\frac{1}{6}\left ( 24\pi \right )=4\pi\)

In the figure above, square RSTU is inscribed in the circle. What is the degree measure of arc ST? 45

A.

** Solution:** If you draw a line connecting points R and T, you will have a perfect semi-circle, or 180

**Now, the arc we are looking for spans exactly half of that semi-circle This means that the arc degree measure of ST is 90*°*. **The correct answer is C.**

If you know the length of the radius and the central (or inscribed) angle, you can find the perimeter of the sector.

In the figure above, the smaller circles each have a radius of 3. They are tangent to the larger circle at Points A and C and are tangent to each other at point B, which is the center of the larger circle. What is the perimeter of the shaded region? 6π

A.

** Solution:** Let's start with the two circles in the middle. We know that each circle has a radius of 3 and that our shaded perimeter spans exactly half of each circle. So the circumference for each small circle is: C = πr = 3π

And there are two small circles, so we must **double** this number: 2 x 3π = 6π

So the interior perimeter is 6π.

Now, let’s find the outer **perimeter, which is the circumference for half the larger circle.** If the radius of each of the small circles is 3, then that means the diameter of each small circle is 6.

This means that the full circumference of the larger semicircle is: C = πr = 6π

Our outer perimeter equals 6π, and our inner perimeter equals 6π. To get the full perimeter, we must add them together.

6π + 6π = 12π

**The correct answer is D, 12π.**

Point O is the center of both the circles in the figure above. If the circumference of the larger circle is 36 and the radius of the small circle is half the radius of the larger circle, what is the length of the darkened arc? 10

A.

** Solution:** We are trying to find the length of an arc circumference, which means that we need two pieces of information - the

Well, we already have the degree measure, so we’re halfway there, but now we need the radius (or diameter) of the smaller circle. We are told that it is half the radius of the larger circle, so we must find the radius of the larger circle first.

All that we are told about the larger circle is that it has a circumference of 36. Luckily, **we can find its radius from its circumference.**

C = 2πr

36 = 2πr

r = \(\frac{18}{\pi }\)

Because we know that **the smaller circle has a radius that is half the length of the radius of the larger circle**, we know that the radius of the smaller circle is:

\(r=\frac{1}{2}\) x \(\frac{18}{\pi }=\frac{9}{\pi }\)

\(c_{arc}=2\pi r\frac{(arc\: degree)}{360^{\circ}}\)

\(c_{arc}=2\pi \frac{9}{\pi }\frac{(80^{\circ})}{360^{\circ}}=4\)

**The correct answer is D.**

Often, the SAT will ask you to solve for the area of a sector of a circle, instead of the area of the entire circle. You can find the area of a sector by determining the fraction of the entire area that the sector occupies. **To determine this fraction, look at the central angle that defines the sector.**

** Example: **What is the area of sector ACB (the striped region) below?

First, find the area of the entire circle: A = πr2 = π32 = 9π

Then, use the central angle to determine what fraction of the entire circle is represented by the sector. Since the sector is defined by the central angle of 60*°*, and the entire circle is 360*°*, the sector occupies 60*°*/360*°*, or one-sixth, of the area of the circle. Therefore, the area of sector ACB = (9π) = 1.5π

*Solved Example 6*

Dave claims to have only eaten 120*°*, of a circular apple pie that has a radius of 7.8 cm. When looking at the leftover pie from the top, it is found that 23.14π cm2 of the flat plate underneath the pie is exposed in the area that Dave ate. Did Dave truly eat 120*°*, of the pie?Yes, he ate exactly 120o, of the pie.

A.

\(A_{pie}=\pi\) x \((7.8\: cm)^{2}\)

\(A_{pie}=60.84\pi \: cm^{2}\)

Now, we must divide the area of pie that Dave ate (which is given directly in the question) by the area of the whole pie to determine the proportion of pie he really ate.

\(\frac{23.14\pi }{60.84\pi }=0.38\)

0.38 of a pie will translate to = 0.38 x 360

Therefore, Dave actually ate more than 120

*Solved Example 7*

The circle has an area of 25π and is divided into 8 congruent regions. What is the perimeter of one of these regions? 10 - 25 π

A.

Solution:

Area = \(\pi r^{2}=25\pi\)

r = 5

Now, we must find the arc measurement of each wedge. To do so, let us find the full circumference measurement and divide by the number of wedges (in this case, 8).

The full circumference is C = 2πr = 10π which, divided by 8, is: \(Arc_{Wedge}=\frac{10\pi }{8}=\frac{5\pi }{4}\)

Now, let us add that arc measurement to twice the radius value of the circle in order to get the full perimeter of one of the wedges.

\(5+5+\frac{5\pi }{4}\)

\(10+\frac{5\pi }{4}\)

So, the correct Answer is C.

The SAT may ask you to apply your knowledge of circle geometry to situations specific to clocks. Don’t let these questions throw you off: at this point, your knowledge of circles includes everything you need to solve them, even if they seem somewhat unfamiliar.

** Solved Example 8**What is the measure, in degrees, of the acute angle formed by the hands of a twelve-hour clock that reads exactly 3:10?

A. 30°

B. 35°

C. 40°

D. 45°

Solution:

Not so fast. One has to account for the 10 minutes that have passed. 10 minutes is 1/6 of an hour, so the hour hand has also moved 1/6 of the distance between the 4 and the 4, which adds 5 (1/6 of 30). The total measure of the angle, therefore, is 35, and

*Example 9*

On a standard analog clock, what is the angle between the hands when the clock reads 11:20? Give the smaller of the two angles. 60°

A.

Solution:

At 11:20, the hour hand has gone one-third of the way between the 11 and the 12. Thus, there are two-thirds of 30° between the hour hand and the 12. (2/3) 30° = 20°. There are 60° between 12 and 2, where the minute hands is, which means there’s a total of 20° + 60° = 80° between the hands.

The standard equation for a circle is in the form **\((x-h)^{2}+(y-k)^{2}=r^{2}\)** where the circle with radius **r** has its center at **(h, k)**.

In which quadrant(s) will the circle with equation \((x+8)^{2}+(y-15)^{2}=49\) be contained?I only

A.

** Solution: **Begin by finding the center of the circle. The center of the circle is at (-8, 15), which is in Quadrant II. The radius of this circle is 7.

Now, we need to consider the four points seven units in the positive/negative x-direction and positive/negative y-direction. These will be on the line of the circle and mark the farthest points in those directions. If all of those points are located in Quadrant II, we know that the entire circle is located in that quadrant.

Let’s consider those points now:

All those points are in Quadrant II, and **B is the correct answer.**

*Solved Example 11*

Find the centre and radius of the circle \(x^{2}+y^{2}+8x+7=0\) Centre is (-4,0) and radius is 9

A.

Solution

\(x^{2}+8x+\) _ _ _ \(+y^{2}=-7\)

We need to fill in the ___ to complete the square. The middle term is 8 and therefore the last variable should be 16. Adding 16 on both sides:

\(x^{2}+8x+16+y^{2}=-7+16\)

\((x+4)^{2}+y^{2}=9\)

Comparing this equation to the equation of a circle: \((x-h)^{2}+(y-k)^{2}=r^{2}\)

We get (h,k) -> (-4,0) and r = 3.

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