TAKS TutorialTAKS Tutorial

Geometry Objectives 6 – 8 Geometry Objectives 6 – 8

Part 2Part 2

The Geometry tested on the The Geometry tested on the Exit Level TAKS test covers Exit Level TAKS test covers

High School Geometry.High School Geometry.Topics to be covered in today’s tutorial areTopics to be covered in today’s tutorial are Special Right TrianglesSpecial Right Triangles Segment-Angle RelationshipsSegment-Angle Relationships ArclengthArclength Parallel & Perpendicular LinesParallel & Perpendicular Lines Dimension Changes Dimension Changes Similarity & Scale FactorSimilarity & Scale Factor Views of 3D figuresViews of 3D figures

Special Right Triangles have Special Right Triangles have relationships that are relationships that are

important enough to test.important enough to test.The formula chart now has those The formula chart now has those

relationships listed. Please be sure to relationships listed. Please be sure to consult the chart when you come consult the chart when you come upon one of those test questions.upon one of those test questions.

60o

30o

3x

x x2

45o

45o 2x

x

x

On the chart, the angles and the On the chart, the angles and the lengths of the sides opposite them are lengths of the sides opposite them are listed in order of size.listed in order of size.

The side across from the 30The side across from the 30oo angle (the angle (the smallest angle) is the shortest side, the leg x.smallest angle) is the shortest side, the leg x.

The side across from the 60The side across from the 60oo angle (the mid- angle (the mid-sized angle) is the mid-sized side, the leg sized angle) is the mid-sized side, the leg ..

The side across from the 90The side across from the 90oo angle (the angle (the largest angle) is the longest side, the largest angle) is the longest side, the hypotenuse, 2x.hypotenuse, 2x.

3x

60o

30o

3x

x x2

On the chart, the angles and the On the chart, the angles and the lengths of the sides opposite them lengths of the sides opposite them are listed in order of size.are listed in order of size.

The sides across from each of the 45The sides across from each of the 45oo angles (the smallest angles) are the angles (the smallest angles) are the shortest sides, the legs x.shortest sides, the legs x.

Notice, the angles are the same size Notice, the angles are the same size (congruent) and so are the sides across (congruent) and so are the sides across from them.from them.

The side across from the 90The side across from the 90oo angle (the angle (the largest angle) is the longest side, the largest angle) is the longest side, the hypotenuse, .hypotenuse, .

2x

45o

45o 2x

x

x

11 ft

11 ft11 ft

Notice that the hose is perpendicular to one side of the garden.

Now you have a right triangle to work with.

You have two different ways that you can find the length of the hose.

Option 1: 30Option 1: 30oo-60-60oo-90-90oo RelationshipRelationship

11 ft

11 ft11 ft

The side of the garden forms the hypotenuse of the right triangle.

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.

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

5.5 ft

Option 1: 30Option 1: 30oo-60-60oo-90-90oo RelationshipRelationship

11 ft

11 ft11 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

Option 2: Pythagorean Option 2: Pythagorean TheoremTheorem

11 ft

11 ft11 ft

The hose is an altitude for the equilateral triangle. It cuts the perpendicular side in half, making the short side of the right triangle 11/2 ft or 5.5 ft

5.5 ft

If you don’t remember the Pythagorean Theorem, it is on the formula chart:

a

c

b

timeCalculator

5.511

)5.5()11(

)11()5.5(

22

222

222

a

a

a

This next problem cannot be This next problem cannot be done using the Pythagorean done using the Pythagorean

Theorem because only one of Theorem because only one of the three triangle sides is the three triangle sides is

known.known.You have no choice but to use the You have no choice but to use the

relationships among the sides of a relationships among the sides of a 3030oo-60-60oo-90-90oo triangle. triangle.

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You should know that the measures of the depression and of elevation are the same.

Angle of depression—Mr.

Ryan looking down from the

horizon

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.

30o

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Since the height of the plane is across from the 30o angle, 2400 ft is the shortest leg of the right triangle.

30o

Recall from the math chart, the shortest leg is x.

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!

?

20062006

Because the figure is a cube, all of its sides have the same length.

s

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

45o

45o

20062006

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”.

s

s

We now know both sides of the shaded rectangle.

45o

45o

2s

2

)2)((2sA

ssA

lwA

Since the radii of circles must be the same length (or you wouldn’t have a circle!), both of these triangles are isosceles.

Because the radii are perpendicular, we have 45o-45o-90o triangles.

45o 45o

To find the length of a leg in a 45o-45o-90o triangle, we divide the hypotenuse by the square root of 2.

The missing lengths that make up segment LT are also radii.

45o 45o

22

22

2

2

5355.32

5

3.5355

3.53552 3.5355

20062006If you bring your own, you may use map pencils or highlighters on the test.There are lots of angles here. Use the same color to denote which angles are congruent. (No map pencils? Use a symbol.)

You know the 3 angles of a triangle add up to be 180o.

m3 + m8 + m9 = 180o

This fact is not an answer choice. So now look at your symbols/colors. Start replacing these angles with ones they are equal to until you find the right match.

m6 is in each answer choice, so start by replacing the 8 with 6.

m3 + m6 + m9 = 180o

m1 + m6 + m11 = 180o

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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

Solve each equation:

2x + 3 = 3x + 1 subtract 2x and subtract 1 from each side

2 = x

x = 2

Solve each equation:

y + 2 = 4y - 7 subtract y and add 7 to each side

9 = 3y divide by 3 so 3 = y

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 and 3x + 1 = 3(2) + 1 = 7 so the diagonal MQ is 14

x = 2 y = 3

20062006

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 Let’s concentrate on the information they gave us about information they gave us about

the door handle.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.

inches34.54C

)5.5)(14.3(2r2C

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.

125.8

1

360

45

20062006

.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

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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.

The diameter of the plate is 12 inches. The circumference formula is on the math chart if you don’t remember it.

inches 37.68 C

3.14(12) dC

Remember, we only want the part of the plate where the peas are located.

The entire circle goes around 360o. Subtract the other two sections of the plate to find the central angle of the peas section: 360 – 203 – 105 = 52o.

144444.360

52

We just want .14444(37.68) = 5.442 inches of the plate edge.

In Algebra I, you studied In Algebra I, you studied parallel and perpendicular parallel and perpendicular

lines. In Geometry, you built on lines. In Geometry, you built on that knowledge. Algebra II that knowledge. Algebra II expected you to remember expected you to remember

that material.that material.Now, TAKS is going to test that Now, TAKS is going to test that

knowledge.knowledge.

Parallel lines are going in Parallel lines are going in the same direction. Since the same direction. Since

slope is the indicator of the slope is the indicator of the line’s direction, the line’s direction, the slopes slopes

of parallel lines must be the of parallel lines must be the samesame..Both of these parallel

lines have a slope of -2

Perpendicular lines must go in opposite Perpendicular lines must go in opposite directions so their slopes must have directions so their slopes must have opposite signs. opposite signs. However, they must be positioned so that However, they must be positioned so that the angles formed at their intersection are the angles formed at their intersection are 9090oo. To ensure that measurement, the . To ensure that measurement, the slopes must also be reciprocals.-----slopes must also be reciprocals.-----Opposite Reciprocal slopes!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 A vertical line and a horizontal line are perpendicular to each line are perpendicular to each

other.other.The horizontal line has a slope of zero The horizontal line has a slope of zero

while the vertical line has undefined while the vertical line has undefined slope.slope.

Parallel lines must have the same slope.

Look for a line with a slope of -4/5.

X

X

X

Positive slope

Positive slope

Down 4, right 5, and I hit the line.This is it!

Down 4, right 5, and I pass the line.

X

XX 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 y-intercept is changed!

Do check this answer! You have one equation in slope-intercept form already. You know the slope and y-intercept of the other line. Graph both on the calculator.

See why the slopes must be opposite reciprocals to be

perpendicular? These angles are definitely not right angles!

same x-int

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This slope is negative. We need a positive slope.

All of these lines go through (0, 4).

X X

XPerpendicular lines have opposite reciprocals for slopes. We need positive 2/1.

Yes

Similar figures are figures with Similar figures are figures with the same shape, although they the same shape, although they

may be different sizes.may be different sizes.

To have the same shape, all of the To have the same shape, all of the corresponding angles in both figures must corresponding angles in both figures must have the same degree measure.have the same degree measure.

Even though the figures might be different Even though the figures might be different sizes, the ratio of the sides must be the sizes, the ratio of the sides must be the same for all corresponding parts.same for all corresponding parts.

Congruent figures are indeed similar. The Congruent figures are indeed similar. The ratio of their corresponding parts is 1:1.ratio of their corresponding parts is 1:1.

20032003 Problem part 1 continued on next slide

on previous slide

We need to compare each answer We need to compare each answer choice with the original triangle. choice with the original triangle.

We’ll start with F.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 It is easiest to count vertically and horizontally, so let’s count what would horizontally, so let’s count what would

be the slope of the diagonal sides.be the slope of the diagonal sides.

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.

Looks like this is the similar triangle—all the sides have the same ratio. Triangle F is half of the original triangle.

We really should check all of the We really should check all of the triangles. Do not jump to conclusions triangles. Do not jump to conclusions

too fast.too fast.

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.

Triangle G goes down 3, right 2 while the original goes down 4, right 2. These corresponding sides are not in the ratio 1:1.

Now, to look at Triangle H.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!One last answer choice to check!

Triangle J doesn’t even have the same shape as the original triangle.

The horizontal sides are in the ratio 2:4 which is 1:2

3 right, 1 down does not give us a side that is half the size of the original triangle.

Definitely, F is the correct answer choice.

20042004

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.

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For choice F:

3

1

2

1

4

1

not true. is12

4

6

3

8

2

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For choice G:

3

2

3

1

2

1

not true. is12

8

6

2

8

4

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For choice H:

2

1

6

1

4

1

not true. is12

6

6

1

8

2

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That leaves J: But check it first!

2

1

2

1

2

1

true.IS12

6

6

3

8

4

20062006 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.

A = ½bh

10 = ½(b)(4)

10 = 2b

5 = b

20062006 Now that we know that the base of the smaller triangle is 5, we can compare bases and find the scale factor.

A = ½bh

10 = ½(b)(4)

10 = 2b

5 = b

75.15

75.8

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

20062006 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.

As you can see, if only the height doubles, then the Volume would double.

If you look at the formula:

V = Bh = r2h for the original

If the height doubles V = Bh = r2(2h) = 2r2h means that the volume is doubled.

To understand the next problem, To understand the next problem, let’s have a little lesson: Start with a let’s have a little lesson: Start with a

cube…cube…Next, double the length of each side…Next, double the length of each side…

Now, let’s triple the lengths…

s 2s 3s

Looking at Looking at perimeterperimeter……Looking at the lengths of the sides:

s 2s 3sRatio: 2 to 1 2 times as large

Ratio: 3 to 1 3 times as large

Looking at the perimeter of the base:

P = 4s P = 4(2s) = 2(4s) P = 4(3s) = 3(4s)

4 sides that are s units long

4 sides that are 2s units long

4 sides that are 3s units long

2 times longer than the original

3 times longer than the original

original original

Measurements that are Measurements that are 1-1-DDimensional: length, side, width, imensional: length, side, width,

height, altitude, perimeter, radius, height, altitude, perimeter, radius, diameter, circumference, slant diameter, circumference, slant

height all have the same increase or height all have the same increase or decrease when there is a dimension decrease when there is a dimension

change.change.In other words, if the length of the sides of In other words, if the length of the sides of

a cube doubles, the perimeter of the base a cube doubles, the perimeter of the base doubles. If the radius of a circle triples, doubles. If the radius of a circle triples, the circumference triples. If the perimeter the circumference triples. If the perimeter of a rectangle quadruples, then both the of a rectangle quadruples, then both the length and the width quadrupled.length and the width quadrupled.

Looking at Looking at AreaArea……Looking at the lengths of the sides:

s 2s 3sRatio: 2 to 1 2 times as large

Ratio: 3 to 1 3 times as large

Looking at the surface area:

A = 6s2 A = 6(2s)2 = 6(4s2) = 4(6s2)

A = 6(3s)2 = 6(9s2) = 9(6s2)

6 sides that are s2 in area

6 sides that are 2s by 2s in area

6 sides that are 3s by 3s in area

4 times more surface area than the original

9 times more surface area than the original

Ratio: 4 to 1 4 times as large

Ratio: 9 to 1 9 times as large

Measurements that are Measurements that are 2-D2-Dimensional: any imensional: any kind of area, all have the kind of area, all have the squaresquare of the of the increase or decrease when there is a increase or decrease when there is a dimension change.dimension change.In other words, if the length of the sides of a cube In other words, if the length of the sides of a cube

doubles, the area of the base quadruples. If the doubles, the area of the base quadruples. If the radius of a circle triples, the area of the circle radius of a circle triples, the area of the circle becomes 9 times greater. If the perimeter of a becomes 9 times greater. If the perimeter of a rectangle quadruples, then its area increases by rectangle quadruples, then its area increases by a multiple of 16.a multiple of 16.

On the other hand, if the area of a square On the other hand, if the area of a square increases by 4, the length of the side doubled increases by 4, the length of the side doubled (square root of 4). If the total area of a pyramid (square root of 4). If the total area of a pyramid increased by a factor of 9, the perimeter of its increased by a factor of 9, the perimeter of its base increased by a factor of 3 (square root of base increased by a factor of 3 (square root of 9).9).

Looking at Looking at VolumeVolume……Looking at the lengths of the sides:

s 2s 3sRatio: 2 to 1 2 times as large

Ratio: 3 to 1 3 times as large

Looking at volume:

V = s3 V = (2s)3 = 8s3

A = (3s)3 = 27s3

1 cube 8 cubes that are s3 in volume

27 cubes that are s3 in volume

8 times more volume than the original 27 times more volume

than the original

Ratio: 8 to 1 8 times as large

Ratio: 27 to 1 27 times as large

Measurements that are Measurements that are 3-D3-Dimensional: imensional: any kind of volume, all have the any kind of volume, all have the cubecube of of the increase or decrease when there is a the increase or decrease when there is a dimension change.dimension change.In other words, if the length of the sides of a cube In other words, if the length of the sides of a cube

doubles, its volume is increase 8 times. If the doubles, its volume is increase 8 times. If the radius of a sphere triples, the volume of the radius of a sphere triples, the volume of the sphere becomes 27 times greater. If the sphere becomes 27 times greater. If the perimeter of the base of a rectangular solid perimeter of the base of a rectangular solid quadruples, then its volume increases by a quadruples, then its volume increases by a multiple of 64.multiple of 64.

On the other hand, if the volume of a cube On the other hand, if the volume of a cube increases by 8, the length of the side doubled increases by 8, the length of the side doubled (cube root of 8). If the volume of a pyramid (cube root of 8). If the volume of a pyramid increased by 27, the perimeter of its base increased by 27, the perimeter of its base increased by a factor of 3 (cube root of 27).increased by a factor of 3 (cube root of 27).

Let’s look again at the cubes we drew…Let’s look again at the cubes we drew…

s 2s 3s

Increase (2)1: 2 times as large

Increase (3)1 : 3 times as large

1-D: same increase

3-D: increase cubed

The dimension is the exponent for the increase.

2-D: increase squared Increase (2)2

4 times as largeIncrease (3)2 9 times as large

Increase (2)3 8 times as large

Increase (3)3 27 times as large

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.

We need to square root the increase of the 2-D to find the increase of the 1-D.

24

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-Let’s now look at figures with 2-D and 3-D in mind.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.

X

X

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

20062006Either, 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.

X

20062006Either, 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.

X

X

20062006Either, 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 D will not work since the long edges of the triangular base will be too long for the sides to which it must attach.

X

XX

Practice ProblemsPractice Problems

All sides are same length.

Divide perimeter by 3.

37/3 = 12.33333 cm is the length of each side.

12.3333

12.3333 12.3333

60o

30o

Altitude divides the perpendicular side in half 12.333333/2 =6.16666666

6.16666809.10316666.6

10.6809

A = bh/2 = (12.3333)(10.6809)/2

A = 65.86537

Longer leg is shorter leg times square root of 3

5

071.725

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too short

too short

too short

Needs to be longer than 5280 ft + 10,560

ft which means longer than 15,840 ft

20042004

20042004

sum of all 5 angles = 3(180) = 540o

90o

90o90o

130o

540 – 90 – 90 – 90 – 130 = 140o

1 4 0

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360o

Regular hexagon—6 congruent sides means six congruent intercepted arcs means six congruent central angles.

Y = 360/6 = 60o6 0

Original V = lwh = 24 dm3

Changed V = (½l)(½w)(½h)

= 1/8lwh

1/8(24 dm3) = 3 dm3

8 ft

5 ft

3 ft