### All SAT II Math II Resources

## Example Questions

### Example Question #64 : Geometry

Refer to the above diagram. Which of the following is *not* a valid name for ?

**Possible Answers:**

All of the other choices give valid names for the angle.

**Correct answer:**

is the correct choice. A single letter - the vertex - can be used for an angle if and only if that angle is the only one with that vertex. This is not the case here. The three-letter names in the other choices all follow the convention of the middle letter being vertex and each of the other two letters being points on a different side of the angle.

### Example Question #65 : Geometry

Use the rules of triangles to solve for x and y.

**Possible Answers:**

x=30, y=60

x=45, y=45

x=30, y=30

x=60, y=30

**Correct answer:**

x=60, y=30

Using the rules of triangles and lines we know that the degree of a straight line is 180. Knowing this we can find x by creating and solving the following equation:

Now using the fact that the interior angles of a triangle add to 180 we can create the following equation and solve for y:

### Example Question #66 : Geometry

Use the facts of circles to solve for x and y.

**Possible Answers:**

x=11, y= 39.5

x=39.5, y=11

x=13, y=10

x=10, y=30

**Correct answer:**

x=11, y= 39.5

In this question we use the rule that oppisite angles are congruent and a line is 180 degrees. Knowing these two facts we can first solve for x then solve for y.

Then:

### Example Question #71 : 2 Dimensional Geometry

**Solve for x and y using the rules of quadrilateral**

**Possible Answers:**

x=6, y=10

x=2, y=4

x=6, y=9

x=9, y=6

**Correct answer:**

x=6, y=9

By using the rules of quadrilaterals we know that oppisite sides are congruent on a rhombus. Therefore, we set up an equation to solve for x. Then we will use that number and substitute it in for x and solve for y.

### Example Question #72 : 2 Dimensional Geometry

Chords and intersect at point . is twice as long as ; and .

Give the length of .

**Possible Answers:**

**Correct answer:**

If we let , then .

The figure referenced is below (not drawn to scale):

If two chords intersect inside the circle, then the cut each other so that for each chord, the product of the lengths of the two parts is the same; in other words,

Setting , and solving for :

Taking the positive square root of both sides:

,

the correct length of .

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