Download chapter 8 – geometry

CHAPTER 8 – GEOMETRY
8.1 – Polygons
Polygons = A closed shape containing straight lines. (poly = many, gon = angles)
All polygons contain three main parts: Sides, Angles & Vertices/Corners.
Regular Polygons
Polygons that have all equal sides and all equal
angles are called Regular Polygons. Not all polygons
are regular. Below is a table of regular polygons.
8.2 – Angles in a Triangle
Triangle – A three sided polygon, all angles in a Triangle add up to 180o.
Types of Triangles
Equilateral – All sides
equal length, all angles
equal (60o).
Isosceles – Two equal
sides, Two angles at base
equal.
Scalene – All sides
different lengths, All
angles different.
60o
ao
60o
ao
60o
]
Since all angles are the same
and the sum is 180o. Each angle
is 60o
To calculate ao you minus the
top angle from 180o and divide
by 2.
Because all angles are different
you need two angles to
calculate the third.
Right-angled Triangle – If a triangle contains a right-angle in it.
*Since a right angle is 90o, the two other angles in a triangle must add up to 90 o for
the total to be 180o.
Calculating angles within a Triangle
1) The sum of all angles must equal 180o.
Write this out: 48 + 67 + 𝑥 = 180
2) Subtract known angles to find unknown angles.
𝑥 = 180 − 48 − 67
𝑥 = 65⁰
Calculating angles with more than one value of x
If there appears to have more than one value of x, you need to group up the x’s.
Example: Solve for x in the following triangle (right)
Solution
1. The sum of all angles must still equal 180o
Add up all of your angles and write it out in one
sentence.
𝑥 + 2𝑥 + 30 + 3𝑥 + 6 = 180
2. Add up both the pronumerals and the numbers.
6𝑥 + 36 = 180
3. Solve for x by subtracting then dividing
6𝑥 = 180 − 36
6𝑥 = 144
144
𝑥=
6
𝑥 = 24⁰
8.2 - Example 1
Find angle ao
1. Sum of all angles = 180o
Add up the angles.
32 + 27 + 𝑎 = 180
2. Solve for a by subtracting known angles.
𝑎 = 180 − 32 − 27
𝑎 = 121⁰
8.2 - Example 2
Find the angles a & b.
1. Add up the angles.
𝑎 + 𝑏 + 48 = 180⁰
2. In an isosceles triangle the two
base angles are equal.
𝑎 = 48⁰
Therefore all our angles are:
48 + 𝑏 + 48 = 180⁰
𝑏 = 180 − 48 − 48
𝑏 = 84⁰
8.2 - Example 3
Find angle xo
1. Add up the angles.
Since this is an isosceles triangle,
the remaining angle = xo
2. Minus the top angle from 180
180 − 21 = 159
Then divide by 2.
𝑥 + 𝑥 = 159
2𝑥 = 159
159
𝑥=
2
𝑥 = 79.5⁰
8.2 - Example 4
Find the angle xo.
1. Add up the angles
Write out in one sentence.
90 + 𝑥 + 20 + 𝑥 − 10 = 180
2. Add up the pronumerals and the numbers
2𝑥 + 100 = 180
3. Solve for x by subtracting then dividing.
2𝑥 = 180 − 100
2𝑥 = 80
80
𝑥=
2
𝑥 = 40⁰
8.3 – Exterior angles in a triangle
Interior Angles = The angles inside of a triangle
Exterior Angles = The angles outside of a triangle.
There are two main rules to understand when calculating exterior angles:
1) All angles inside of a triangle equal to 180o
2) The angles in a straight line equal to 180o
To the left, ao, bo & co are all interior angles. do is an exterior angle.
From this we can apply these formulas:
FORMULAS
1.
𝒂 + 𝒃 + 𝒄 = 𝟏𝟖𝟎⁰
2.
𝒄 + 𝒅 = 𝟏𝟖𝟎𝟎
And therefore:
3.
𝒂+𝒃=𝒅
We can then use these formulas to calculate unknown angles.
Tip – It’s always good to find every angle (marked or unmarked) so that it becomes
easier to solve.
Corresponding Angles
Angles in the same matching corner are
known as corresponding angles.
The two angles transverse to the
parallel lines below are equal.
Opposite Angles
Angles that are opposite two
intersecting lines are known as
opposite angles. The two angles
opposite each other below are equal.
8.4 – Quadrilaterals
Quadrilaterals = Four-sided polygons. All angles in a quadrilateral add up to 360o
Below is the table of properties for all the different types of quadrilaterals.
8.3 – Example 1
Find xo
1. All angles must add up to 180.
Therefore the unmarked angle is:
180 − 66 − 62 = 52
2. xo and 52o form a straight line so they must equal
180.
𝑥 + 52 = 180
𝑥 = 180 − 52
𝑥 = 128⁰
8.3 – Example 2
Solve for the pronumerals
1. All angles must add up to 180o.
Therefore the unmarked angle is:
180 − 34 − 27 = 119⁰
Since ro is opposite, it is also 119o
𝑟 = 119⁰
2. qo and 119o form a straight line
𝑞 + 119 = 180
𝑞 = 180 − 119
𝑞 = 61⁰
Since qo and po are opposite angles, they are the
same.
𝑝 = 61
8.3 – Example 3
Solve for the pronumerals
1. mo and 75o form a straight
line.
𝑚 = 180 − 75 = 105
2. ro and 75o are opposite angles
𝑟 = 750
3. Solve for qo.
𝑞 + 38 + 75 = 180
𝑞 = 180 − 75 − 38
𝑞 = 670
Since qo and po are opposite angles:
𝑝 = 670
4. Solve for no.
𝑛 + 67 + 75 = 180
𝑛 = 180 − 67 − 75
𝑛 = 38⁰
8.5 – Angles in a quadrilateral
8.5 – Example 1
Find xo.
1. All angles must =360o.
55 + 50 + 35 + 𝑥 = 360
2. Solve for x.
𝑥 = 360 − 55 − 50 − 35
𝑥 = 230⁰
Calculating angles in a quadrilateral
1. The sum of all angles in a quadrilateral = 360o
Write this out: 75 + 85 + 80 + 𝑥 = 360
2. Subtract known angles to find xo.
𝑥 = 360 − 75 − 85 − 80
𝑥 = 120⁰
Tip – Remember to take into account the unique properties of each quadrilateral
when calculating angles.
Calculating angles with more than one value of x
If there appears to have more than one value of x, you need to group up the x’s.
Example: Solve for x.
Solution
1. The sum of all the angles = 360o.
dWrite this out:
2. Group up the x’s and solve for x.
8𝑥 = 360
360
𝑥=
8
𝑥 = 45⁰
8.6 – Tessellations
Two or more shapes are said to tessellate if they fit together perfectly without any
gaps.
Tessellations use only one type of shape which can be rotated or mirrored to fit.
The best way to see whether a shape tessellates is to simply draw it out.
8.8 – Scale Drawings
Scaled drawings are drawings of the exact same shape but just on a smaller size.
Scale Factor = The ratio of the model size to the actual size.
Since ratios can also be written as fractions, the Scale Factor can also be expressed
in fractions.
For example: If the Scale factor was 1:10, then the actual size is 10 times greater
than the model size or the model is
1
10
the size of the actual.
Calculating the Scale Factor
When calculating the Scale Factor, you have to compare the same side of the
model shape to the actual shape.
*All units must be the same*
Example: Find the scale factor.
1. The Scale Factor is the ratio of:
𝒎𝒐𝒅𝒆𝒍: 𝒂𝒄𝒕𝒖𝒂𝒍
2.
8.5 – Example 2
Find xo.
1. All angles must add up to 360o.
𝑥 + 80 + 𝑥 + 50 = 360
2. Group pronumerals and numbers.
2𝑥 + 130 = 360
3. Solve for xo.
2𝑥 = 360 − 130
2𝑥 = 230
230
𝑥=
2
𝑥 = 115⁰
8.5 – Example 3
Find the pronumerals
1. All angles = 360o
42 + 𝑎 + 𝑏 + 𝑐 = 360
2. Angles ao and 42 are
opposite, therefore they must be the same.
𝑎 = 420
What we’re left with is:
42 + 42 + 𝑏 + 𝑐 = 360
𝑏 + 𝑐 + 84 = 360
𝑏 + 𝑐 = 360 − 84
𝑏 + 𝑐 = 276
3. Angles bo and co are also opposite, so they too
must be the same.
𝑏=𝑐
We can replace one with the other in our original
equation:
𝑏 + 𝑏 = 276
2𝑏 = 276
276
𝑏=
2
𝑏 = 1380
𝑐 = 138⁰
8.8 – Example 1
Find the Scale Factor of a model air plane 50cm long
when the actual size is 50m long.
1. The Scale factor = model : actual
Where the model is 50cm, actual = 50m
50𝑐𝑚 ∶ 50𝑚
2. They must be converted into the same units, in
this case into cm.
1𝑚 = 100𝑐𝑚
Therefore 50𝑚 = 50 × 100𝑐𝑚
50𝑐𝑚 ∶ 5000𝑐𝑚
3. Find the largest number that fits into both and
divide by that number.
Highest common factor = 50
Scale Factor = 1 ∶ 100
40cm
2cm
model
The model is 2cm, where the actual is 40cm
𝑆𝑐𝑎𝑙𝑒 𝐹𝑎𝑐𝑡𝑜𝑟 = 2: 40
Calculating model/actual size based on Scale Factor
actual
8.8 – Example 2
If the actual size of a ship was 150m and the scale
factor was 1:600. Find the model size
1. Equate the ratio of the scale factor to the
When calculating model or actual size based on Scale Factor you:
1. Equate the ratio of the scale factor to the model : actual
2. Change the ratio’s into fractions
3. Cross multiply and solve
Example: Find the actual size of a house whose model is 10cm to a scale factor of
1:50.
1. The model : actual is 10cm:𝑥 (still unknown), whereas the scale factor is 1:50
So 1: 50 = 10: 𝑥
2. Change the ratio’s into fractions:
1
10
=
500
𝑥
3. Cross Multiply: top-left x bottom-right = top-right x bottom-left
1 × 𝑥 = 10 × 50
𝑥 = 500𝑐𝑚
𝑥 = 5𝑚
8.9 – Enlarging or reducing
When enlarging or reducing an image, you need to use a ruler to measure.
Determining Scale Factor
1. Measure the original size
2. Measure the new size
3. The Scale Factor is the ratio between the original : new
4. If the original > new, then the image has been reduced
If the original < new, then the image has been enlarged.
8.8 – Example 3
If a road was 1400m long, what would its distance
on a map with a scale of 2:70?
1. Equate the ratio of the scale factor to the
model:actual ratio.
2: 7000 = 𝑥: 1400𝑚
2. Change the ratios into fractions.
2
𝑥
=
7000 1400
3. Cross multiply and solve.
2 × 1400 = 7000 × 𝑥
2800 = 70𝑥 𝑂𝑅 7000𝑥 = 2800
𝑥 = 2800/7000
𝑥 = 0.4m
𝑥 = 40𝑐𝑚
The distance on the map would be 40cm.
Side, Side, Side (SSS)
If the lengths of all three sides of two triangles are
identical, then the two triangles are congruent
8.10 – Congruent Triangles
Congruent Triangles = Triangles that are the same size and shape. The only
differences are that they may be of different rotation or mirrored.
o
40
Side, Angle, Side (SAS)
If the two sides and the angle in between of two
triangles are identitical = congruent
50o
Triangle A
Triangle B
Triangle A & B are congruent triangles since they have all the same properties but is
simply different in rotation.
Angle, Side, Angle (ASA)
If two angles and the side in between of two
triangles are identical then they are congruent.
Testing for Congruent Triangles
It isn’t always necessary to know the length of all sides or all angles to find out if
two triangles are congruent. The minimum number of information needed is three.
There are 3 main ways to test for congruency in triangles (details right)
1. Side, Side, Side (SSS)
2. Side, Angle, Side (SAS)
3. Angle, Side, Angle (ASA)
4. (For Right-angled Triangles) Right-Angle, Hypotenuse, Side (RHS)
8.11 – Similar Triangles
Right-Angle, Hypotenuse, Side (RHS)
If the hypotenuse and the second side of a rightangled triangle is the same = congruent.
8.11 – Example 1
Find x.
1. Pick pairs of ratios
3: 6 𝑎𝑛𝑑 7: 𝑥
Similar Triangles = Two triangles that are the same shape but not necessarily the
same size.
For two triangles to be similar they must have:
1. All corresponding angles are the same
2. All corresponding sides are of the same ratio
Triangle A
Triangle B
Triangle A and Triangle B are similar since
1. Both opposite angles are 60o
2. The ratio of 2cm : 4cm is the same ratio as 3cm:6cm.
Calculating x in similar triangles
To solve for an unknown side in two similar triangles we:
1. Choose two pairs of corresponding sides as ratios
2. Since we know the ratio of the sides must be the same for both pairs, we
Equate the two ratios to each other
3. Change the two ratios into fractions
4. Cross multiply and solve
Example: Find the value of x.
1. Choose two pairs. Make sure that
they are corresponding
(one of which includes x)
𝒙: 𝟓 𝑎𝑛𝑑 𝟏𝟐: 𝟖
2. Since we know the ratio of sides
must be the same, we can equate
them to each other.
𝑥: 5 = 12: 8
3. Change ratios into fractions
𝑥 12
=
5
8
4. Cross multiply and solve for x
𝑥 × 8 = 12 × 5
8𝑥 = 60
𝑥 = 7.5
8.11 – Example 2
Find x.
1. Pick pairs of ratios
5: 𝑥 𝑎𝑛𝑑 4: 8
*The reason why it is 4:8
Is because we are
comparing the base of the
small triangle (4) to the base of
the large triangle (8)
2. Equate ratios
5: 𝑥 = 4: 8
3. Change into fractions
5 4
=
𝑥 8
4. Cross multiply and solve for x
5×8 =4×𝑥
40 = 4𝑥 𝑂𝑅 4𝑥 = 40
40
𝑥=
4
𝑥 = 10
8.11 – Example 3
Find x.
1. Pick pairs of ratios
Our first corresponding pair is
6: 15
Our second corresponding pair is
4: 𝑥 + 4
The reason why it is x+4 is because the length of the
larger triangle is both x and 4.
2. Equate ratios
6: 15 = 4: 𝑥 + 4
3. Change into fractions
6
4
=
15 𝑥 + 4
4. Cross multiply and solve for x
6(𝑥 + 4) = 4 × 15
6(𝑥 + 4) = 60
60
𝑥+4=
6
𝑥 + 4 = 10
𝑥 = 10 − 4
𝑥=6