### All SAT Math Resources

## Example Questions

### Example Question #1 : Circles

Two chords of a circle, and , intersect at a point . is twice as long as , , and .

Give the length of .

**Possible Answers:**

Insufficient information is given to find the length of .

**Correct answer:**

Insufficient information is given to find the length of .

Let stand for the length of ; then the length of is twice this, or . The figure referenced is below:

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,

Substituting the appropriate quantities, then solving for :

This statement is identically true. Therefore, without further information, we cannot determine the value of - the length of .

### Example Question #1 : Chords

Two chords of a circle, and , intersect at a point . is 12 units longer than , , and .

Give the length of (nearest tenth, if applicable)

**Possible Answers:**

**Correct answer:**

Let stand for the length of ; then the length of is . The figure referenced is below:

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,

Substituting the appropriate quantities, then solving for :

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

to both sides:

Restate the trinomial as the square of a binomial:

Take the square root of both sides:

or

Either

,

in which case

,

or

in which case

,

Since is a length, we throw out the negative value; it follows that , the correct length of .

### Example Question #2 : Chords

A diameter of a circle is perpendicular to a chord at a point .

What is the diameter of the circle?

**Possible Answers:**

Insufficient information is given to answer the question.

**Correct answer:**

In a circle, a diameter perpendicular to a chord bisects the chord. This makes the midpoint of ; consequently, .

The figure referenced is below:

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,

Setting , and solving for :

,

the correct length.

### Example Question #3 : Chords

Two chords of a circle, and , intersect at a point .

Give the length of .

**Possible Answers:**

Insufficient information is given to answer the question.

**Correct answer:**

Let , in which case ; the figure referenced is below (not drawn to scale).

Setting , and solving for :

,

which is the length of .

### Example Question #4 : Chords

A diameter of a circle is perpendicular to a chord at point . and . Give the length of (nearest tenth, if applicable).

**Possible Answers:**

insufficient information is given to determine the length of .

**Correct answer:**

A diameter of a circle perpendicular to a chord bisects the chord. Therefore, the point of intersection is the midpoint of , and

.

Let stand for the common length of and ,

The figure referenced is below.

Set and , and ; substitute and solve for :

This is the length of ; the length of is twice this, so

### Example Question #5 : Chords

Figure is not drawn to scale

In the provided diagram, the ratio of the length of to that of is 7 to 2. Evaluate the measure of .

**Possible Answers:**

Cannot be determined

**Correct answer:**

Cannot be determined

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,

The ratio of the degree measure of to that of is that of their lengths, which is 7 to 2. Therefore,

Letting :

Therefore, in terms of :

Without further information, however, we cannot determine the value of or that of . Therefore, the given information is insufficient.

Certified Tutor

### All SAT Math Resources

### Incompatible Browser

Please upgrade or download one of the following browsers to use Instant Tutoring: