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SAT Math : Circles

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Example Questions

Example Question #1 : Circles

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point . $\overline{CX}$ is twice as long as $\overline{AX}$ , BX = 20 , and DX = 10 .

Give the length of $\overline{AX}$ .

Possible Answers:

Insufficient information is given to find the length of $\overline{AX}$ .

$10 \sqrt{3}$

$10 \sqrt{2}$

$20\sqrt{2}$

Correct answer:

Insufficient information is given to find the length of $\overline{AX}$ .

Explanation:

Let stand for the length of $\overline{AX}$ ; then the length of $\overline{CX}$ is twice this, or . The figure referenced is below:

Chords

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Substituting the appropriate quantities, then solving for :

$t \cdot 20 = 2t \cdot 10$

20 t = 20t

This statement is identically true. Therefore, without further information, we cannot determine the value of - the length of $\overline{AX}$ .

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Example Question #1 : Circles

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point . $\overline{BX}$ is 12 units longer than $\overline{AX}$ , CX = 8 , and DX = 10 .

Give the length of $\overline{AX}$ (nearest tenth, if applicable)

Possible Answers:

15.0

4.8

3.0

16.8

7.2

Correct answer:

4.8

Explanation:

Let stand for the length of $\overline{AX}$ ; then the length of $\overline{BX}$ is t+12 . The figure referenced is below:

Chords

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Substituting the appropriate quantities, then solving for :

$t \cdot (t+12) = 8 \cdot 10$

$t \cdot t+t \cdot 12 = 8 0$

$t ^{2}+12 t = 8 0$

This quadratic equation can be solved by completing the square; since the coefficient of is 12, the square can be completed by adding

$\left (\frac{12}{2} \right ) ^{2} = 6 ^{2} = 36$

to both sides:

$t ^{2}+12 t + 36 = 8 0+ 36$

Restate the trinomial as the square of a binomial:

$(t+6 )^{2 } = 116$

Take the square root of both sides:

$t+6 = \pm \sqrt{116}$

$t+6 \approx - 10.8$ or $t+6 \approx 10.8$

Either

$t+6 \approx - 10.8$ ,

in which case

$t+6 - 6 \approx - 10.8 - 6$

$t \approx -16.8$ ,

$t+6 \approx 10.8$

in which case

$t+6 - 6 \approx 10.8 - 6$

$t \approx 4.8$ ,

Since is a length, we throw out the negative value; it follows that $t \approx 4.8$ , the correct length of $\overline{AX}$ .

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Example Question #1 : Chords

A diameter $\overline{AB}$ of a circle is perpendicular to a chord $\overline{CD}$ at a point .

CD= 20, AX= 8

What is the diameter of the circle?

Possible Answers:

$20\frac{1}{2}$

$33\frac{1}{3}$

Insufficient information is given to answer the question.

Correct answer:

$20\frac{1}{2}$

Explanation:

In a circle, a diameter perpendicular to a chord bisects the chord. This makes the midpoint of $\overline{CD}$ ; consequently, $CX =DX = \frac{1}{2} CD = \frac{1}{2} \cdot 20 = 10$ .

The figure referenced is below:

Chords

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Setting AX= 8, CX =DX = 10 , and solving for :

$8 \cdot BX = 10 \cdot 10$

$8 \cdot BX = 100$

$8 \cdot BX \div 8 = 100 \div 8$

BX = 12.5

AB = AX + BX = 8 + 12.5 = 20.5 ,

the correct length.

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Example Question #1 : Circles

Two chords of a circle, $\overline{AB}$ and $\overline{CD}$ , intersect at a point .

CX = AX + 6

BX = 12

DX = 10

Give the length of $\overline{AX}$ .

Possible Answers:

$3 \sqrt{14}$

$2 \sqrt{30}$

Insufficient information is given to answer the question.

Correct answer:

Explanation:

Let t = AX , in which case CX = t+ 6 ; the figure referenced is below (not drawn to scale).

Chords

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$AX \cdot BX = CX \cdot DX$

Setting AX= t, CX= t+ 6, BX = 12, DX = 10 , and solving for :

$AX \cdot BX = CX \cdot DX$

$t \cdot 12 = (t+6) \cdot 10$

12t = 10t + 60

12t- 10t = 10t + 60 - 10t

2t = 60

$\frac{2t}{2} = \frac{60}{2}$

t = 30 ,

which is the length of $\overline{AX}$ .

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Example Question #2 : Circles

A diameter $\overline{AB}$ of a circle is perpendicular to a chord $\overline{CD}$ at point . AX = 12 and BX = 20 . Give the length of $\overline{CD}$ (nearest tenth, if applicable).

Possible Answers:

31.0

16.0

15.5

insufficient information is given to determine the length of $\overline{CD}$ .

32.0

Correct answer:

31.0

Explanation:

A diameter of a circle perpendicular to a chord bisects the chord. Therefore, the point of intersection is the midpoint of $\overline{CD}$ , and

$CX = DX = \frac{1}{2} CD$ .

Let stand for the common length of $\overline{CX}$ and $\overline{DX}$ ,

The figure referenced is below.

Chords

If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,

$CX \cdot DX = AX \cdot BX$

Set AX = 12 and BX = 20 , and CX = DX = t ; substitute and solve for :

$t \cdot t = 12 \cdot 20$

$t^{2 } = 240$

$t = \sqrt{240} \approx 15.5$

This is the length of $\overline{CX}$ ; the length of $\overline{CD}$ is twice this, so

$CD = 2 \cdot t \approx 2 \cdot 15.5 = 31.0$

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Example Question #1 : How To Find The Length Of A Chord

Secant

Figure is not drawn to scale

In the provided diagram, the ratio of the length of $\overarc{BD}$ to that of $\overarc{AC}$ is 7 to 2. Evaluate the measure of $\angle BND$ .

Possible Answers:

$25^{\circ }$

Cannot be determined

$45^{\circ }$

$40^{\circ }$

$20^{\circ }$

Correct answer:

Cannot be determined

Explanation:

The measure of the angle formed by the two secants to the circle from a point outside the circle is equal to half the difference of the two arcs they intercept; that is,

$m \angle BND = \frac{1}{2} ( \overarc{BD} - \overarc{AC})$

The ratio of the degree measure of $\overarc{BD}$ to that of $\overarc{AC}$ is that of their lengths, which is 7 to 2. Therefore,

$\frac{m \overarc{BD} }{m \overarc{AC} } = \frac{7}{2}$

Letting $t = m \overarc{AC}$ :

$\frac{m \overarc{BD} }{t } = \frac{7}{2}$

$\frac{m \overarc{BD} }{t } \cdot t = \frac{7}{2} \cdot t$

$m \overarc{BD} = \frac{7}{2} t$

Therefore, in terms of :

$m \angle BND = \frac{1}{2} \left ( \frac{7}{2} t - t \right ) = \frac{1}{2} \left ( \frac{5}{2} t \right ) =\frac{5}{4} t$

Without further information, however, we cannot determine the value of or that of $m \angle BND$ . Therefore, the given information is insufficient.

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Example Question #1 : How To Find The Angle Of Clock Hands

It is 4 o’clock. What is the measure of the angle formed between the hour hand and the minute hand?

Possible Answers:

$60^\circ$

$180^\circ$

$90^\circ$

$30^\circ$

$120^\circ$

Correct answer:

$120^\circ$

Explanation:

At four o’clock the minute hand is on the 12 and the hour hand is on the 4. The angle formed is 4/12 of the total number of degrees in a circle, 360.

4/12 * 360 = 120 degrees

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Example Question #1 : How To Find The Angle Of Clock Hands

If a clock reads 8:15 PM, what angle do the hands make?

Possible Answers:

150^o

90^o

138.5^o

157.5^o

240^o

Correct answer:

157.5^o

Explanation:

A clock is a circle, and a circle always contains 360 degrees. Since there are 60 minutes on a clock, each minute mark is 6 degrees.

$\frac{360^o\ \text{total}}{60\ \text{minutes total}}=6\ \text{degrees per minute}$

The minute hand on the clock will point at 15 minutes, allowing us to calculate it's position on the circle.

$(15\ min)(6)=90^o$

Since there are 12 hours on the clock, each hour mark is 30 degrees.

$\frac{360^o\ \text{total}}{12\ \text{hours total}}=30\ \text{degrees per hour}$

We can calculate where the hour hand will be at 8:00.

$(8\ hr)(30)=240^o$

However, the hour hand will actually be between the 8 and the 9, since we are looking at 8:15 rather than an absolute hour mark. 15 minutes is equal to one-fourth of an hour. Use the same equation to find the additional position of the hour hand.

$240^o+(\frac{1}{4}\ hr)(30)$

240^o+7.5^o

247.5^o

We are looking for the angle between the two hands of the clock. The will be equal to the difference between the two angle measures.

$\angle=247.5^o-90^o=157.5^o$

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Example Question #1 : How To Find The Angle Of Clock Hands

What is the measure of the smaller angle formed by the hands of an analog watch if the hour hand is on the 10 and the minute hand is on the 2?

Possible Answers:

90°

45°

30°

120°

56°

Correct answer:

120°

Explanation:

A analog clock is divided up into 12 sectors, based on the numbers 1–12. One sector represents 30 degrees (360/12 = 30). If the hour hand is directly on the 10, and the minute hand is on the 2, that means there are 4 sectors of 30 degrees between then, thus they are 120 degrees apart (30 * 4 = 120).

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Example Question #1 : How To Find The Angle Of Clock Hands

At 5:00 , what angle is between the hour and minute hand on a clock?

Possible Answers:

$180^\circ$

$160^\circ$

$5^\circ$

$120^\circ$

$150^\circ$

Correct answer:

$150^\circ$

Explanation:

At 5:00 , the hour hand is on the and the minute hand is at the . There are spaces on a clock, and these hands are separated by spaces.

Thus, the angle between them is $\frac{5}{12}$ the degrees of the entire clcok, which is $360^\circ$ .

Therefore, we multiply these to get our answer.

$\frac{5}{12}\times \frac{360}{1}$

We can cancel out as we multiply to get:

$\frac{5}{1}\times \frac{30}{1}=150^\circ$

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SAT Math : Circles

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #1 : Circles

Example Question #1 : Circles

Example Question #1 : Chords

Example Question #1 : Circles

Example Question #2 : Circles

Example Question #1 : How To Find The Length Of A Chord

Example Question #1 : How To Find The Angle Of Clock Hands

Example Question #1 : How To Find The Angle Of Clock Hands

Example Question #1 : How To Find The Angle Of Clock Hands

Example Question #1 : How To Find The Angle Of Clock Hands

All SAT Math Resources

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