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 \displaystyle \angle DBF ?

Possible Answers:

\displaystyle \angle FBE

\displaystyle \angle B

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

\displaystyle \angle CBF

\displaystyle \angle FBD

Correct answer:

\displaystyle \angle B

Explanation:

\displaystyle \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 \displaystyle B and each of the other two letters being points on a different side of the angle.

Example Question #1 : 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:

\displaystyle x=180-120

\displaystyle 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:

\displaystyle y=180-(90+60)

\displaystyle y=180-150

\displaystyle y=30

Example Question #2 : Analyzing Figures

Circle

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

 

Possible Answers:

x=10, y=30

x=39.5, y=11

x=13, y=10

x=11, y= 39.5

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.

\displaystyle 2x=22

\displaystyle x=11

Then:

\displaystyle 4y+2x=180

\displaystyle 4y+2(11)=180

\displaystyle 4y+22=180

\displaystyle 4y=158

\displaystyle y=39.5

Example Question #3 : Analyzing Figures

 

 

 

Rhombus

Solve for x and y using the rules of quadrilateral

Possible Answers:

x=9, y=6

x=6, y=9

x=6, y=10

x=2, y=4

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.

\displaystyle 2x+4=16

\displaystyle 2x=12

\displaystyle x=6

 

\displaystyle x+y=15

\displaystyle 6+y=15

\displaystyle y=9

Example Question #3 : Analyzing Figures

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

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

Possible Answers:

\displaystyle 6

\displaystyle 5\frac{1}{2}

\displaystyle 3 \frac{2}{3}

\displaystyle 3

\displaystyle 4\frac{1}{2}

Correct answer:

\displaystyle 3

Explanation:

If we let \displaystyle t = AX, then \displaystyle 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,

\displaystyle AX \cdot BX = CX \cdot DX

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

\displaystyle t \cdot 2t = 2 \cdot 9

\displaystyle 2t^{2} =18

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

\displaystyle t^{2} = 9

Taking the positive square root of both sides:

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

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

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