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SAT II Math II : Analyzing Figures

Study concepts, example questions & explanations for SAT II Math II

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Example Questions

Example Question #1 : Analyzing Figures

Thingy_5

Refer to the above diagram. Which of the following is not a valid name for $\angle DBF$ ?

Possible Answers:

$\angle FBE$

$\angle CBF$

$\angle FBD$

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

$\angle B$

Correct answer:

$\angle B$

Explanation:

$\angle B$ 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.

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Example Question #2 : Analyzing Figures

Triangle

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

Possible Answers:

x=30, y=60

x=60, y=30

x=30, y=30

x=45, y=45

Correct answer:

x=60, y=30

Explanation:

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:

x=180-120

x=60

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

y=180-(90+60)

y=180-150

y=30

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Example Question #3 : Analyzing Figures

Circle

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

Possible Answers:

x=11, y= 39.5

x=13, y=10

x=39.5, y=11

x=10, y=30

Correct answer:

x=11, y= 39.5

Explanation:

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.

2x=22

x=11

Then:

4y+2x=180

4y+2(11)=180

4y+22=180

4y=158

y=39.5

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Example Question #4 : Analyzing Figures

Rhombus

Solve for x and y using the rules of quadrilateral

Possible Answers:

x=9, y=6

x=6, y=10

x=2, y=4

x=6, y=9

Correct answer:

x=6, y=9

Explanation:

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.

2x+4=16

2x=12

x=6

x+y=15

6+y=15

y=9

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Example Question #4 : Analyzing Figures

Chords $\overline{AB}$ and $\overline{CD}$ intersect at point . $\overline{BX}$ is twice as long as $\overline{AX}$ ; CX = 2 and DX = 9 .

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

Possible Answers:

$5\frac{1}{2}$

$4\frac{1}{2}$

$3 \frac{2}{3}$

Correct answer:

Explanation:

If we let t = AX , then BX = 2t .

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

Chords

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,

$AX \cdot BX = CX \cdot DX$

Setting AX = t, BX = 2t, CX = 2, DX = 8 , and solving for :

$t \cdot 2t = 2 \cdot 9$

$2t^{2} =18$

$2t^{2} \div 2 =18 \div 2$

$t^{2} = 9$

Taking the positive square root of both sides:

$t = \sqrt{9} = 3$ ,

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

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SAT II Math II : Analyzing Figures

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #1 : Analyzing Figures

Example Question #2 : Analyzing Figures

Example Question #3 : Analyzing Figures

Example Question #4 : Analyzing Figures

Example Question #4 : Analyzing Figures

All SAT II Math II Resources

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