TAKS Tutorial

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TAKS Tutorial

• Geometry Objectives 6 8
• Part 1

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Welcome to part 1 of the Geometry Objectives
tutorial.

• Today, we will cover the topics
• Using Algebra to solve problems involving basic
  geometric definitions
• Pythagorean Theorem
• Patterns
• Transformations
• Coordinate Geometry

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Part 2 of the tutorial over these objectives will
be on Tues, April 15 for 10th grade and Wed,
April 16 for Exit Level

• Remember, 8th grade geometry is covered on the
  10th grade TAKS and High School Geometry is
  covered on the Exit Level.

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Topics covered next time for the Exit Level will
be

• Special Right Triangles
• Segment-Angle Relationships
• Circle segments Arclength
• Parallel Perpendicular Lines
• Dimensional Changes
• 2D 3D Figures Relationships
• Please be sure to attend this very important
  session on April 16!

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The topic of Measurement has already been
covered.

• You may find the power point lessons and their
  corresponding worksheets on Mrs. Nelsons
  website.

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This problem was on the 2006 test.
With this figure they provide a table of values.
This problem is so much easier if you look in the
table for patterns!
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Lets study the table.

• The value of x, water depth, increases by 1 each
  time.
• The surface of the water is the circle at the top
  of the cone
• Its difficult to see the pattern to the
  y-values, Area of water surface

Lets get some common denominatorseasier to see
patterns!
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Lets study the table.

• Now that all of the denominators are the same,
  look at the numerators.
• They each have ?.
• And the x-value squared gives the coefficient of
  ?.

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Now, lets check out the answer choices.

• Which choice shows x2 multiplied to pi over 16?

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This problem was on the 2003 test.
Notice, this time you didnt get a table of
values---so make one of your own!
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• What is the independent variable?
• What is the dependent variable?

Stage
Circles
Stage
1
1
2
3
of circles
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3
?
n
Now, look for that pattern. Or better yet, look
at the options!
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Use your calculator and the options until you
find a match for this graph.

• Enter each option into y on the calculator, go
  to the table, and find the match.

FYI 2n is entered as 2x
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This problem was also on the 2003 test. It covers
a basic Geometry definition with an Algebra twist.
Complementary angles are 2 angles whose measures
add up to be 90o.
Now, eliminate the two options that do not use
90o.
So, you have to ask yourself, what do I know
about complementary angles?
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m?A m?B 90
Replacing m?A with x, I get x m?B 90
Subtracting x from both sides of the equation, I
get m?B 90 x
Since complementary angles add up to be 90o, I
can write
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April 2006
What do you notice about the relationship of the
hose to the side of the garden?
The hose is perpendicular to one side, forming 2
congruent right triangles.
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As soon as you recognize that you have a right
triangle, you should be thinking of how you can
apply the Pythagorean Theorem.

• Check the formula chart. The Pythagorean Theorem
  formula is there.

a
b
a2 b2 c2
c
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We were told that each side of the equilateral
triangle is 11 ft.

• That means that
• c 11 ft and
• a ½ of 11 5.5 ft

a2 b2 c2
(5.5)2 b2 (11)2 b2 (11)2
(5.5)2 b2 121 30.25 b2
90.75
a
b
5.5
c
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Find the square root of 90.75!
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Here are the answer choices
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Transformations are always tested.

• You could be asked about
• Translations
• Reflection
• Rotation
• Dilation
• So, you have to be prepared for any one of them.
  Remember, dilations can also include any kind of
  similar polygon problem or one that talks about
  scale factor.

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This one was on the 2006 test.
Note The 10th grade TAKS usually only have one
kind of transformation per problem and reflects
over the x-axis or the y-axis.
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Use your calculator and graph y -x so you know
what it looks like. Find some points from the
table and draw the line on your paper.
Note You are only looking for the image of point
M, so use that point only. You do not need to use
the entire parallelogram.
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You need to move from M perpendicular to the
line. The line has slope -1, so you move with
slope 1.
From (6, 3), you need to translate (slide) down 4
units.
From M to the line was 4.5 diagonals. Now, move
4.5 diagonals past the line.
You now are at the point (6, -1)
That will now bring you to (6, 3). Note that
answer is there just in case you stop too soon.
You are not finished with the problem yet!
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Look for the point (6, -1) in the answer choices.
These are the coordinates of M
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This problem was on the 2003 test. See how they
have you work several problems all in one? Arent
they sneaky?
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Lets check out the first option Rotate the
figure 90o around M.

• Here is the visual. The rectangle is filled in.
• Next, it is rotated 90o around M.
• No vertex is at the origin (0, 0) so eliminate
  choice A.

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Lets check out the second answer choice
Reflect the figure across the line x 1.

• Here is the visual. The rectangle is filled in
  and the line x 1 is marked.
• Next, the figure is reflected across x 1.
• No vertex is at the origin (0, 0) so eliminate
  choice B.

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Lets check out answer choice C Reflect the
figure across the line y 2.5.

• Here is the visual. The rectangle is filled in
  and the line y 2.5 is marked.
• Next, the figure is reflected across y 2.5.
• No vertex is at the origin (0, 0) so eliminate
  choice C.

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That leaves answer choice D Dont
assume!Translate the figure left 6 and then down
5.

• Here is the visual. The rectangle is filled in.
• Next, the figure is translated left 6.
• Then, it is translated down 5.
• Point N is now at the origin (0, 0) so choice D
  is the answer.

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This problem was on the 2006 test.
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• Study the graph.
• Find the point (0, 0).
• Look for point S.
• Draw a ray from (0, 0) through S.
• The point you are looking for MUST be on this ray
• This point MUST, since the scale factor is 2,
  move the distance between the origin and S past S.

Down 3 and right 2
Down 3 and right 2 a second time
That brings us to (4, -6)
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And the answer is
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There are a few problems on you paper for you to
practice what we learned today.
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