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TAKS Tutorial
Geometry Objectives 6 – 8
Part 2
The Geometry tested on the Exit
Level TAKS test covers High School
Geometry.
Topics to be covered in today’s tutorial are
 Special Right Triangles
 Segment-Angle Relationships
 Arclength
 Parallel & Perpendicular Lines
 Dimension Changes
 Similarity & Scale Factor
 Views of 3D figures
Special Right Triangles have
relationships that are important
enough to test.
The formula chart now has those
relationships listed. Please be sure to
consult the chart when you come upon
one of those test questions.
x
60o
2x
45o
30o
x 3
x 2
x
45o
x
On the chart, the angles and the lengths of
the sides opposite them are listed in order
of size.
60
x
o
2x
30o
x 3



The side across from the 30o angle (the
smallest angle) is the shortest side, the leg x.
The side across from the 60o angle (the midsized angle) is the mid-sized side, the leg x 3
.
The side across from the 90o angle (the largest
angle) is the longest side, the hypotenuse, 2x.
On the chart, the angles and the
lengths of the sides opposite them are
listed in order of size. 45 x 2
o
x
45o
x



The sides across from each of the 45o angles
(the smallest angles) are the shortest sides,
the legs x.
Notice, the angles are the same size
(congruent) and so are the sides across from
them.
The side across from the 90o angle (the x 2
largest angle) is the longest side, the
hypotenuse,
.
Notice that the hose is
perpendicular to one
side of the garden.
11 ft
11 ft
11 ft
You have two different ways that
you can find the length of the
Now you have a right
hose.
triangle to work with.
Option 1: 30o-60o-90o Relationship
The side of the garden
forms the hypotenuse of
the right triangle.
11 ft
11 ft
5.5 ft
In an equilateral
triangle, each angle
measures 60o. The
hose divides its angle
in half, making it 30o.
And, of course, the
right angle is 90o.
11 ft
According to the formula on the chart,
the hypotenuse measures 2x, double
the length of the shortest side, x.
We need to divide 11 by 2, and find
x = 5.5 ft
Option 1: 30o-60o-90o Relationship
11 ft
11 ft
Again, according to the formula on the
chart, the longer leg, which is the
hose, measures x times the square
root of 3. We need to multiply:
5.5 ft
11 ft
Option 2: Pythagorean Theorem
a  (5.5)  (11)
2
a  (11)  (5.5)
2
2
2
2
2
a  11  5.5
2
2
11 ft
11 ft
a
b
5.5 ft
The hose is an altitude
the
Calculatorfortime
equilateral triangle. It cuts the
11 ft
perpendicular side in half, making the
c
short side of the right triangle 11/2 ft or
5.5 ft
If you don’t remember the Pythagorean Theorem, it is on the
formula chart:
This next problem cannot be done
using the Pythagorean Theorem
because only one of the three
triangle sides is known.
You have no choice but to use the
relationships among the sides of a 30o60o-90o triangle.
2003
Angle of
depression—Mr.
Ryan looking
down from the
horizon
You should know that the
measures of the depression
and of elevation are the same.
30o
Angle of elevation—
someone looking up at
Mr. Ryan from the
same location that Mr.
Ryan is looking at sees
Mr. Ryan at the same
angle measurement.
2003
Since the height of the plane is
across from the 30o angle,
2400 ft is the shortest leg of the
right triangle.
Recall from the math chart,
the shortest leg is x.
?
30o
We are looking for the horizontal distance, which
is the length of the longer leg. That leg
measures 2400 times the square root of 3. Get
out the calculator!
2006
Because the
figure is a cube,
all of its sides
have the same
length.
45o
s
45o
s
We now know one side of the shaded
rectangle. We cannot find its area until we
know the length of the other side. A = lw
To find the remaining side of the rectangle, we
need to look at the right triangle shown, which
happens to be isosceles and 45o-45o-90o
2006
Using the special
triangle
relationship,
when the leg of a
45-45-90 triangle
is x, the
hypotenuse is x
times the square
root of 2. Our “x”
just happens to
be “s”.
45o
s 2
s
45o
s
We now know both sides
of the shaded rectangle.
A  lw
A  ( s )( s 2 )
As
2
2
Since the radii of circles
must be the same
length (or you wouldn’t
have a circle!), both of
these triangles are
isosceles.
45o
45o
Because the radii are
perpendicular, we have
45o-45o-90o triangles.
To find the length of a
leg in a 45o-45o-90o
triangle, we divide the
hypotenuse by the
square root of 2. 2 2  2
2
3.5355
2
45o
45o
2
2
The missing lengths that
make up segment LT are
also radii.
3.5355
3.5355
5
2
 3.5355
2006
If you bring
your
m3 +own,
m8 you
+ m9 = 180o
may use map
m3
+ m6
pencils
or + m9 = 180o
This
fact is not
highlighters
an
onanswer
the test.
choice.
There So
arenow
lots of
You
know the 3 angles of a triangle
o
look
at
your
m1
+
m6
+
m11
=
180
angles here.
o.
add
up
to
be
180
symbols/colors.
Use the same
Start
replacing
color
to denote
these
angles
m6 is in each
which
angles
answer choice,
with
they
areones
congruent.
so start by
are
equal
to
until
(No map
replacing the 8
you
find
the
right
pencils? Use a
with 6.
match.
symbol.)
2006x = 2
y=3
You are expected to remember that the diagonals of a
parallelogram bisect each other.
That means:
2x + 3 = 3x + 1
and y + 2 = 4y – 7
Solveeach
each equation:
equation:
Solve
y ++23==4y
and
add
7 to each
sideeach side
2x
3x-+71 subtract
subtracty 2x
and
subtract
1 from
29==x3y
divide by 3
so 3 = y
x=2
y=3
Now substitute those values into the expression on the
diagonals. Make sure they are equal and then add them
together to find the length of the diagonal.
That means: 2x + 3 = 2(2) + 3 = 7
so the diagonal MQ is 14
and 3x + 1 = 3(2) + 1 = 7
2006
An arc is a fractional part of the
circumference of a circle.
First, we need to find the circumference of the circle
made if the door handle turned all the way around.
Let’s concentrate on the
information they gave us about the
door handle.
The radius of the circle is 5.5
inches. The circumference
formula is on the math chart
if you don’t remember it.
C  2  r  2(3.14)(5.5)
C  34.54 inches
45
1
  .125
360 8
Remember that the handle is NOT
going all the way around so we just
want part of the circumference.
The handle is going around 45o
of the 360o of the complete turn.
2006
.125(34.54 inches) = 4.3175 inches
All the way around the circle is 34.54
inches, but the door handle only moves 45o
of the circle which is 4.3175 inches
2004
C  d  3.14(12)
C  37.68 inches
An
Remember,
arc52
is a .fractional
we only want
part of
thethe
part of
144444
360 where
circumference
the plate
ofthe
a circle.
peas are located.
First,
we need
to goes
find the
circumference
of thethe
circle
The entire
circle
around
360o. Subtract
other two
made if the door handle turned all the way around.
sections of the plate to find the central angle of the peas
o.
section:
360of–the
203plate
– 105
The
diameter
is =
1252
inches.
The
circumference
formula is on =
the
mathinches
chart ifofyou
We just want .14444(37.68)
5.442
the don’t
plate edge.
remember it.
In Algebra I, you studied parallel
and perpendicular lines. In
Geometry, you built on that
knowledge. Algebra II expected
you to remember that material.
Now, TAKS is going to test that knowledge.
Parallel lines are going in the
same direction. Since slope is
the indicator of the line’s
direction, the slopes of parallel
lines must be the same.
Both of these parallel
lines have a slope of -2
Perpendicular lines must go in opposite directions
so their slopes must have opposite signs.
However, they must be positioned so that the
angles formed at their intersection are 90o. To
ensure that measurement, the slopes must also
be reciprocals.-----Opposite Reciprocal slopes!
Here, the slopes are -2 and
½. Note, they have opposite
signs and 2 and ½ are
reciprocals.
A vertical line and a horizontal line
are perpendicular to each other.
The horizontal line has a slope of zero while
the vertical line has undefined slope.
Down 4, right 5, and I
Parallel
pass the line.
X
X
Positive slope
Positive slope
Down 4, right 5, and I hit
the line.This is it!
X
lines
must
have the
same
slope.
Look for
a line
with a
slope of
-4/5.
X
X
X
See why the slopes must be
opposite reciprocals to be
perpendicular? These angles are
definitely not right angles!
Do check this answer! You
have one equation in slopeintercept form already. You
know the slope and yintercept of the other line.
Graph both on the calculator.
same
x-int
Parallel lines have the
same slope. These slopes
have opposite signs.
Even though the slopes have
opposite signs, the fractions
are NOT reciprocals.
They distinctly said the yintercept is changed!
2003
All of these lines
go through (0, 4).
Yes
X
X
X
This slope is
negative. We need
a positive slope.
Perpendicular
lines have
opposite
reciprocals for
slopes. We need
positive 2/1.
Similar figures are figures with the
same shape, although they may be
different sizes.



To have the same shape, all of the
corresponding angles in both figures must
have the same degree measure.
Even though the figures might be different
sizes, the ratio of the sides must be the
same for all corresponding parts.
Congruent figures are indeed similar. The
ratio of their corresponding parts is 1:1.
2003
Problem part 1 continued on next slide
on previous slide
We need to compare each answer choice
with the original triangle. We’ll start with
F.
Let’s look at the horizontal side
first. The original triangle has that
side 4 units long.
The horizontal side of F is 2 units
long. The ratio of the scale factor is
2:4 or 1:2. That means, all the sides
of triangle F must be half the size of
the original triangle.
It is easiest to count vertically and
horizontally, so let’s count what would be the
slope
oflikethe
sides.
Looks
this isdiagonal
the similar triangle—all
the sides have the same ratio. Triangle F
is half of the original triangle.
Down 4, right 2 for this side. The
corresponding side on triangle F
should be up 2 left 1 (if a rotation)
or up 2 right 1 (if a reflection).
That works! Either way
Now, to check out the remaining
side on the original triangle.
We really should check all of the
Triangle G goes down 3, right 2
triangles. Do not jump while
to conclusions
too 4,
the original goes down
2. These corresponding
fast. right
sides are not in the ratio 1:1.
Both the original triangle and triangle G have a
horizontal side that is 4 units long. That means triangle
G needs to be exactly the same size as the original. It
definitely is not.
Now, to look at Triangle H.
The horizontal side of triangle H is 6
units long. The ratio of the sides then is
6:4 or 3:2. Triangle H needs to be 1 ½
times as large as the original.
5 up, 3 right gives the ratio
5:3, NOT 3:2. These two
triangles are not similar,
either.
One last answer choice to check!
Triangle J doesn’t even
have the same shape as
the original triangle.
3 right, 1 down does not give
us a side that is half the size of
the original triangle.
The horizontal sides are in the
ratio 2:4 which is 1:2
Definitely, F is the correct
answer choice.
2004
The dimensions are given in terms
of length by width by height.
The given rectangular solid is
8 units by 6 units by 12 units.
You need to look at corresponding ratios;
they must all reduce to the same ratio.
Remember: in similar figures, corresponding
sides must be proportional.
2004
For choice F:
2 3
4
 
is not true.
8 6 12
1 1 1
 
4 2 3
2004
For choice G:
4 2
8
 
is not true.
8 6 12
1 1 2
 
2 3 3
2004
For choice H:
2 1
6
 
is not true.
8 6 12
1 1 1
 
4 6 2
2004
That leaves J: But check it first!
4 3
6
 
IS true.
8 6 12
1 1 1
 
2 2 2
2006
A = ½bh
10 = ½(b)(4)
10 = 2b
5=b
You were told that
the triangles ARE
similar. That means
that the sides and
altitude are in
proportion.
We are going to need to
use some algebra to find
the base of ΔHKM. We
need the base so that we
can compare them and
find the ratio which is the
scale factor used to get
ΔRTV.
2006
A = ½bh
10 = ½(b)(4)
10 = 2b
5=b
Now that we know
that the base of the
smaller triangle is
5, we can compare
bases and find the
scale factor.
8.75
 1.75
5
We know now that the
altitude of the larger
triangle is 1.75 times
greater than the altitude of
the smaller triangle:
4(1.75) = 7 units
2006
Let’s use the Area
of a triangle formula
again—this time to
find the area of the
larger triangle RTV.
A = ½bh
A = ½(8.75)(7)
A = 30.625
We know now that the
altitude of the larger
triangle is 1.75 times
greater than the altitude of
the smaller triangle:
4(1.75) = 7 units
Start with a
visual—draw a
picture of what
you see.
If you look at the formula:
V = Bh = r2h for the original
If the height doubles
V = Bh = r2(2h) = 2r2h
means that the volume is doubled.
As you can see, if only
the height doubles, then
the Volume would double.
To understand the next problem, let’s
have a little lesson: Start with a cube…
Next, double the length of each side…
Now, let’s triple the lengths…
s
2s
3s
Looking at perimeter…
Looking at the lengths of the sides:
s
2s
3s
Ratio: 2 to 1
2 times as large
Ratio: 3 to 1
3 times as large
Looking at the perimeter of the base:
4 sides that are
s units long
P = 4s
4 sides that are
2s units long
P = 4(2s) = 2(4s)
2 times longer
than the original
original
4 sides that are
3s units long original
P = 4(3s) = 3(4s)
3 times longer
than the original
Measurements that are 1-Dimensional:
length, side, width, height, altitude,
perimeter, radius, diameter,
circumference, slant height all have the
same increase or decrease when there is
a dimension change.
In other words, if the length of the sides of
a cube doubles, the perimeter of the base
doubles. If the radius of a circle triples,
the circumference triples. If the perimeter
of a rectangle quadruples, then both the
length and the width quadrupled.
Looking at Area…
Looking at the lengths of the sides:
s
2s
Ratio: 2 to 1
2 times as large
3s
Ratio: 3 to 1
3 times as large
Looking at the surface area:
6 sides that
are s2 in area
A=
6s2
6 sides that are
2s by 2s in area
6 sides that are
3s by 3s in area
A = 6(2s)2
= 6(4s2)
= 4(6s2)
A = 6(3s)2
= 6(9s2)
= 9(6s2)
4 times more surface
area than the original
Ratio: 4 to 1
4 times as large
9 times more surface
area than the original
Ratio: 9 to 1
9 times as large
Measurements that are 2-Dimensional: any kind
of area, all have the square of the increase or
decrease when there is a dimension change.
In other words, if the length of the sides of a
cube doubles, the area of the base quadruples.
If the radius of a circle triples, the area of the
circle becomes 9 times greater. If the
perimeter of a rectangle quadruples, then its
area increases by a multiple of 16.
On the other hand, if the area of a square
increases by 4, the length of the side doubled
(square root of 4). If the total area of a
pyramid increased by a factor of 9, the
perimeter of its base increased by a factor of 3
(square root of 9).
Looking at Volume…
Looking at the lengths of the sides:
s
2s
3s
Ratio: 2 to 1
2 times as large
Ratio: 3 to 1
3 times as large
8 cubes that are
s3 in volume
27 cubes that
are s3 in volume
Looking at volume:
1 cube
V=
s3
V = (2s)3
= 8s3
8 times more volume
than the original
Ratio: 8 to 1
8 times as large
A = (3s)3
= 27s3
27 times more volume
than the original
Ratio: 27 to 1
27 times as large
Measurements that are 3-Dimensional: any
kind of volume, all have the cube of the
increase or decrease when there is a
dimension change.
In other words, if the length of the sides of a cube
doubles, its volume is increase 8 times. If the radius of
a sphere triples, the volume of the sphere becomes 27
times greater. If the perimeter of the base of a
rectangular solid quadruples, then its volume increases
by a multiple of 64.
On the other hand, if the volume of a cube increases by 8,
the length of the side doubled (cube root of 8). If the
volume of a pyramid increased by 27, the perimeter of
its base increased by a factor of 3 (cube root of 27).
Let’s look again at the cubes we drew…
s
1-D: same increase
2-D: increase squared
3-D: increase cubed
2s
Increase (2)1:
2 times as large
Increase (2)2
4 times as large
Increase (2)3
8 times as large
3s
Increase (3)1 :
3 times as large
Increase (3)2
9 times as large
Increase (3)3
27 times as large
The dimension is the exponent for the increase.
The catch: All the sides must increase by the same amount for this to happen!
Area is 2-D
It’s increase is
a factor of 4
Length is 1-D.
4 2
We need to
square root the
increase of the
2-D to find the
increase of the
1-D.
Diameter is 1-D. It
increases 1.5 times
Volume is 3-D. Since a
diameter change is all that is
needed to change the
volume, we need to cube the
increase…
(1.5)3 = 3.375
The volume
increases
3.375 times its
original size.
Let’s now look at figures with 2-D and 3-D in
mind.
Looking at the
Front view, you
should see 4
cubes, 2 cubes,
and 1 cube right
in the front row.
That would eliminate
F and J since they
each have 3 cubes
in the middle of the
front row.
The difference
between
choices G and
H is the middle
of the right
view. There
should be NO
cubes in the
middle far right.
X
X
X
2006
X
Either, imagine taking
a scissors and cutting
the prism at its edges
and laying it flat OR
imagine folding the
nets into the prism.
Answer choice A
will not work
since the short
edge of the
triangular base
will be too short
for the side to
which it must
attach.
2006
X
X
Either, imagine taking
a scissors and cutting
the prism at its edges
and laying it flat OR
imagine folding the
nets into the prism.
Answer choice B
will not work
since the short
edge of the
triangular base
will be too short
for the side to
which it must
attach.
2006
Either, imagine taking
a scissors and cutting
the prism at its edges
and laying it flat OR
imagine folding the
nets into the prism.
X
X
X
Answer choice D
will not work
since the long
edges of the
triangular base
will be too long
for the sides to
which it must
attach.
Practice Problems
All sides are same length.
Divide perimeter by 3.
37/3 = 12.33333 cm is the
length of each side.
30o
12.3333
12.3333
10.6809
6.1666
12.3333
60o
Altitude divides the
perpendicular side in half
12.333333/2 =6.16666666
Longer leg is shorter leg
times square root of 3
6.16666 3  10.6809
A = bh/2 = (12.3333)(10.6809)/2
A = 65.86537
5
5 2  7.071
2003
2003
too short
too short
too short
Needs to be longer
than 5280 ft + 10,560
ft which means
longer than 15,840 ft
2004
2004
1
4
130o
0
90o
sum of all 5
angles =
3(180) =
540o
90o
540 – 90 – 90 – 90 – 130
= 140o
90o
2003
2003
360o
6 0
Y = 360/6 =
60o
Regular hexagon—6 congruent
sides means six congruent
intercepted arcs means six
congruent central angles.
Original
V = lwh = 24 dm3
Changed V = (½l)(½w)(½h)
= 1/8lwh
1/8(24 dm3) = 3 dm3
8 ft
5 ft
3 ft