Download geometric reasoning

GEOMETRIC REASONING
GEOMETRIC
REASONING
GEOMETRIC
REASONING IN
SOLVING PROBLEMS
4 CREDITS(91031)
THE SKILLS YOU NEED TO KNOW:
PYTHAGORAS THEORM p 94
1. Label the two short sides of the
triangle a and b and the long side c
2. Substitute the known values into
a 2 + b 2 = c 2 and solve
TRIGONOMETRY:
FINDING AN ANGLE
p 102
Use to find an angle from two sides
1. Label each side (O, A or H)
2. Cross out one letter
3. Select SOH, CAH, or TOA
4. Draw triangle
5. Substitute lengths in with inverse
sign and brackets
6. Write equation and solve to find the
other length.
Bearings: Problems involving bearings
are very similar to trigonometry problems except angles are given as clockwise from north and written with three
numbers e.g. 030 for 30°
TRIGONOMETRY:
FINDING A LENGTH
p 98
Use to find a length from an angle and
a length
1. Label each side (O, A or H)
2. Cross out one letter
3. Select SOH, CAH, or TOA
4. Draw triangle
5. Substitute numbers in
6. Write equation and solve to find the
other length.
Bearings: Problems involving bearings
are very similar to trigonometry problems except angles are given as clockwise from north and written with three
numbers e.g. 030 for 30°
SIMILAR SHAPES
p 106
1.Substitute in numbers and solve:
Side 1 ( small ) Side 2 ( small )
=
Side 1 (large) Side 2 (large)
2.If two shapes have the same angles
then they are similar shapes and will
be in proportion to each other
PAGE 91
ANGLES OF POLYGONS
p 110
• The interior angles of a polygon add to 180(n – 2)°, where n is
the number of sides
(int ∠, sum of polygon)
• If a shape is regular (all the sides and angles are the same) to
find each interior angle divide the sum of the interior angles by
the number of sides.
• The exterior angle is the angle between any side of a shape, and
a line extended from the next side.
• The exterior angles of a polygon add to 360°
(ext ∠, sum of polygon)
a + b + c + d + e + f = 360°
ANGLES AROUND INTERSECTING LINES
p 113
1.Adjacent angles on a straight line add to 180° ( ∠ s on str. line)
a + b + c = 180°
2.Angles at a point add to 360°
( ∠ s at pt)
a + b + c + d = 360°
3.Vertically opposite angles are equal
(vert opp ∠ s)
a = c and b = d
ANGLES OF PARALLEL LINES
• Corresponding angles on parallel lines are equal
(corr ∠ s // lines)
a=b
• Alternate angles on parallel lines are equal
(alt ∠ s, // lines)
a=b
• Co-interior angles on parallel lines are supplementary
(add to 180°)
(co-int ∠ s, // lines)
a + b = 180°
PAGE 92
p 116
GEOMETRIC REASONING
ANGLES WITHIN CIRCLES
p 119
• Angles on the same arc are equal
a=b
(∠s on same arc)
• The angle at the centre is equal to twice the angle at the circumference
on the same arc
2c = d
(∠ at centre)
• The angle in a semicircle is a right angle This is a special case of the
above rule
(∠ in semicircle)
• The angle where the radius meets the tangent is 90°
(rad ⊥ tgt)
• Two tangents coming from the same point are equal (same length and
angles)
(tangs from a point)
• The angle between a chord and a tangent equals the angle in the alternate segment. a = d, c = b.
(∠ in alt seg)
• Remember any chord forms an isosceles triangle with the centre.
ABO and OBC are isosceles triangles.
• Opposite angles in a cyclic quadrilateral add to 180°
(opp ∠s cyclic quad)
a + c = 180°, b + d = 180°
• The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. a = b
(ext ∠ cyclic quad)
Note: Problems may rely on knowledge from earlier geometric reasoning sections
PAGE 93
PYTHAGORAS THEORM
SUMMARY
a +b = c
2
2
2
1. Use this when 2 sides of a right angled triangle are known and the 3rd is
required. a and b are the short sides and c is the long side.
2. Steps and example:
3. Label the two shorter sides a and
b and the longest side c
a 2 + b2 = c2
4. Substitute into a + b = c
32 + 42 = c 2
known values and solve to find the
9 + 16 = c 2 = 25
length of the other side
c=5
For a complete tutorial on this topic visit www.learncoach.co.nz
2
2
2
NCEA QUESTIONS
1.
The triangle FGH is part of a frame for a climbing
net.
OT is 90 cm long. OP is 70 cm long.
Find the length of PT, x, the distance between the
pole and a support along the ground.
3.
Ali and Rob are designing a triathlon course.
HF = 4.4m and the distance along the ground,
HG = 6.2m
Calculate the length of the side of the frame FG.
a. The swim leg is around a triangular course ABC.
2.
AB is 250 m. BC is 100 m. The angle at B is 90°.
Calculate the length of AC.
b. The cycle leg is around a triangular course DEF.
A child’s practice goal post has one pole and two
supports, as shown above, to the left. The two
supports are each 90 cm long. The pole is always
perpendicular to the ground.
The diagram above right, shows the view from the
side.
PAGE 94
EF is 2.1 km.
FD is 3.4 km.
The angle at E is 90°.
Calculate the length of DE.
GEOMETRIC REASONING
PRACTICE QUESTIONS
4.
There is a bush walk in the Waipoua State Forest
near the large kauri tree, Tane Mahuta (T).
6.
a.
The diagram shows a farm gate.
Angle WXZ is a right angle.
The width of the gate WX is 3.1 m.
The height of the gate XZ is 1.25 m.
Find the length of the brace WZ.
7.
b.
James stands at S, 8 m away from Tane Mahuta
(T).
FT, the height of the lowest branch is 12 m
above ground level.
Calculate the length of SF, the distance of James
from the lowest branch.
Lisa wanted to fly a remote control aeroplane from
A to C but there was a tree in the way so instead
she flew it from A to B to C. Calculate the distance
she flew.
8.
James walks 200 metres North from Tane
Mahuta, point T, to point N.
He then walks west until point W, ending up
600 metres from T.
Calculate how far James walks to the West,
WN.
5.
Sam the dog likes pacing around the back yard. He
goes along two fence lines then makes a diagonal
back to where he started. What is the total distance
he travels?
9.
A flag pole FP is 22 metres high.
A cable FC 36 metres long helps secure the pole.
What is the distance of point C from the base of the
tower P?
A cuboid has a base measuring 10 cm by 12 cm. The
cuboid is 14 cm high.
The cuboid manages to fit exactly inside a sphere.
What would the diameter of the sphere have to be?
PAGE 95
ANSWERS
NCEA
6. WZ 2 = WX 2 + XZ 2
WZ 2 = 3.12 + 1.252
1. HF 2 + FG 2 = HG 2
FG 2 = HG 2 − HF 2
FG 2 = 6.22 − 4.42
FG = 4.37
WZ = 3.12 + 1.252
= 3.34 m (2 dp)
(Achieved)
2. PT 2 + OP 2 = OT 2
2
2
7.
x + 70 = 90
x 2 = 902 − 702 = 3200
x = 3200 = 56.6 cm (1 dp)
2
CA = 250 + 100
Distance = 14.4 + 5 = 19.4 m (Merit)
2
2
CA = 250 + 100
8. Diag 2 = 7 2 + 92
2
= 269.3 m (1 dp)
Diag = 7 2 + 92
= 11.4 m (1 dp)
(Achieved)
b. DE 2 + EF 2 = FD 2
Distance = 11.4 + 7 + 9 = 27.4 m
DE 2 = FD 2 − EF 2
2
AB 2 = 82 + 122
AB = 42 + 32
=5 m
(Achieved)
3. a.
AB 2 + BC 2 = CA2
2
(Achieved)
AB = 82 + 122
= 14.4 m (1 dp)
2
BC = 42 + 32
2
9.
2
DE = 3.4 − 2.1
= 2.67 km (2 dp)
(Achieved)
(Merit)
The longest distance is between opposite corners
of the box A and C.
PRACTICE
4. a. SF 2 = ST 2 + FT 2
SF 2 = 82 + 122
SF = 82 + 122
(Achieved)
= 14.4 m (1 dp)
b. TW 2 = WN 2 + NT 2
WN 2 = TW 2 − NT 2
WN 2 = 6002 − 2002
2
2
WN = 600 − 200
(Achieved)
= 565.7 m (1 dp)
5.
FC 2 = FP 2 + CP 2
CP 2 = FC 2 − FP 2
CP 2 = 362 − 222
2
2
CP = 36 − 22
= 28.5 m (1 dp)
PAGE 96
(Achieved)
This can be found using two right angled triangles.
First finding length AB then finding length AC.
AB 2 = 102 + 122
AB = 102 + 122
= 15.62 cm (2 dp)
2
AC = 142 + 15.622
AC = 142 + 15.622
= 20.98 cm (2 dp)
Therefore the diameter of the sphere is 20.98 cm.
(Excellence)
GEOMETRIC REASONING
Study Tip:
Last Minute Study
If you are running out of time:
• Focus on topics that came up often in class
• Don’t bother studying topics that haven’t been covered in
class. They probably won’t be in the exam
• The LearnCoach summaries are a good place to start!
PAGE 97
TRIGONOMETRY: FINDING
A LENGTH
SUMMARY
1. Use when 1 length and one angle of a right angled triangle is know and a second
length is required.
2. Steps and example:
3. Label each side (O, A or H)
4. Cross out one side
5. Select SOH, CAH, or TOA
SOH
6. Draw triangle
7. Substitute numbers in
x = 10 × sin 30
8. Write equation and solve to find
=5 m
the other length.
• Note: Problems involving bearings are very similar to trigonometry problems
except angles are given as clockwise from north and written with three numbers
e.g. 030 for 30°
For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONS
1.
A balloon, A, is tied to the ground by the rope
labelled TA.
The wind is strong and causes the rope, TA, to
make a straight line. The balloon is 40 m above the
ground. The rope TA makes an angle of 26° with the
ground.
Calculate the length of the rope, TA.
2.
3.
a. Calculate the height of the shed.
b. Calculate the width of the shed.
A ladder has two legs AB and AC. Each leg is 1.8m
long. Angle ABD = 113°
A shed in the playground has a roof that is 3.6 m
long.
0.4 m of the roof overhangs the wall. The roof is at
an angle of 40° to the horizontal.
The walls of the shed are 3.8 m high.
Calculate the length of c.
PAGE 98
GEOMETRIC REASONING
4.
A 35 m long bridge BE crosses a river.
The width of the river is BC.
Angle EBC is 18°.
5.
An orienteering course is planned from point O. The
first leg to a point marked A is 120 m on a bearing of
030°. The second leg begins at A and ends at point
B. B is on a bearing of 120° and 110 m from A.
Calculate the distance from O to B giving reasons
for each step.
Calculate the length of EC.
PRACTICE QUESTIONS
6.
Riggers were putting up a circus tent and the first
job is to put up the main centre pole. The pole is
65 m high. 10 m from the top it has a support wire
attached. The support wire is supposed to be at an
angle of 48° to the main pole (ABC).
The house is 6 m wide.
The pitch of one side of the roof, angle EDF, is 27°.
The owners want to put a solar panel on the side EF.
What is the maximum length it could be?
9.
The top of a cliff is 40 m above sea level.
A person on a boat floating in the sea manages to
spot a person standing on the cliff when they look
at an angle of 17°
Calculate the distance away from the pole that the
support wire should be attached (AC).
7.
When watching the circus Kim is sitting at K, 25 m
away from the area under where the trapeze artists
are performing, B.
How far is the boat from the cliffs?
10.
Ali, Rob, and Sarah are doing a triathlon course.
Ali is doing the running leg of the course.
From the start point, A, she runs 800 m at a bearing
of 060°.
She then runs a distance at a bearing of 150° until
her bearing is 097° from her original position.
She looks up at them with an angle of an angle of
63°. Calculate how high up the trapeze artists are,
TB.
8.
The triangle DEF shows a cross-section of the roof
of a house.
Calculate Ali’s distance from the start point A to her
current position.
PAGE 99
ANSWERS
NCEA
1.
Using SOH
Substituting:
5. Angle OAB = 90°
corresponding angles parallel lines
Sum of angles on a straight line = 180°
40
sin 26
40
TA =
= 91.25 m (2 dp) (Merit)
sin 26°
2. a. Using SOH
Substituting:
Length OB = 1202 + 1102
= 162.8 m (1 dp)
TA
PRACTICE
6.
O
3.6
sin 40 - 0.4
O = 3.2 × sin 40 = 2.06 (2 dp)
Height = 2.06 + 3.8 = 5.86 m
b. Using CAH
Substituting:
(Merit)
Using TOA
Substituting:
AC
65
tan 48 -10
AC = 55 × tan 48 = 61.1 m (1 dp)
7.
A
Using TOA
Substituting:
(Achieved)
TB
tan 63 25
3.6
cos 40 - 0.4
A = 3.2 × cos 40 = 2.45 (2 dp)
Width = 2 × 2.45 = 4.90 m
=
TB 25
=
tan 63 49.1 m (1 dp) (Achieved)
8.
(Achieved - a or b correct)
(Merit - a and b correct, any method)
(Excellence - a and b correct,
correct working)
3.
Using SOH
Substituting:
KM
6
sin 27
KM = 6 × sin 27 = 2.72 m (2
9.
Using TOA
Substituting:
dp) (Achieved)
A
tan 17 40
Split into two right angled triangles and solve
Angle ABC = 180 – 113 = 67° (angles on a line)
Using CAH
Substituting:
A
40
=
A = 130.8 m (1 dp)
tan 17
(Achieved)
10.
cos 67 1.8
A = 1.8 × cos 67 = 0.7 (1 dp)
c = 2 × 0.70 = 1.4 m
4.
(Merit - Answer without support)
(Excellence - Answer with support)
Using SOH
Substituting:
The change in direction is 90° (Co-interior angles, parallel lines and angles at a point)
Using TOA
Substituting:
EC
sin 18
35
EC = 35 × sin 18 = 10.8 m (1
x
dp) (Achieved)
cos 37 800
800
=
x = 1001.7 m (1 dp)
cos 37
PAGE 100
(Excellence)
GEOMETRIC REASONING
Study Tip:
Evaluation
After Each Exam ask yourself:
• Where did most of the questions come from?
• Which parts ate up most of my time?
• Was I anxious during the exam? If so why?
• What could I do differently next time?
PAGE 101
TRIGONOMETRY: FINDING
AN ANGLE
SUMMARY
1. Use when 2 lengths of a right angled triangle are known and an angle is required
2. Steps and example:
3. Label each side (O, A or H)
4. Cross out one side
5. Select SOH, CAH, or TOA
SOH
6. Draw triangle
7. Substitute lengths in with inverse
sign and brackets
(4
sin-1 10)
 4
x = sin −1  
8. Write equation and solve to find
 10 
the angle.
= 23.6° (1 dp)
• Note: Problems involving bearings are very similar to trigonometry problems
except angles are given as clockwise from north and written with three numbers
e.g. 030 for 30°
For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONS
1.
The triangle FGH is part of a frame for a climbing
net.
3.
HF = 4.4 m and the distance along the ground,
HG = 6.2 m. Calculate the angle the frame makes
with the ground at FGH.
2.
A cuboid has a base measuring 5 cm by 6 cm. The
cuboid is 7 cm high.
What is the angle between the line AB shown and
the 5 cm edge of the cuboid?
PAGE 102
A child’s practice goal post has one pole and two
supports, as shown above, to the left.
The two supports are each 90 cm long.
The pole is always perpendicular to the ground.
The diagram above right, shows the view from the
side.
OT is 90 cm long.
OP is 70 cm long.
Calculate the size of angle PTO.
GEOMETRIC REASONING
4.
The diagram shows a square pyramid, with base
ABCD.
5.
Each side of the base is 220 metres long.
FG = 110 m
The height of the pyramid EF is 140 m.
Calculate the angle EGF.
An orienteering course is planned from point O. The
first leg to a point marked A is 120 m on a bearing of
030°. The second leg begins at A and ends at point
B. B is on a bearing of 120° and 110 m from A.
Calculate the bearing of the starting point O from
the finish B.
PRACTICE QUESTIONS
6.
Ben bought a new fridge but wants to use a shelf
from the old one as there are not many shelves in
the new fridge.
The problem is that the new fridge is narrower than
the old one. The old shelf, AB, is 84 cm long.
When he fits it into the new fridge it sits at an angle
with one end 4 cm above the other.
Calculate the angle the shelf AB makes with the
horizontal AC.
7.
Calculate Ali’s bearing from the start point A to the
end point of the run.
9.
A carver was given a wooden block and asked
to replicate it. The base is a horizontal regular
hexagon, with sides of 6 cm shown in the diagrams
and a height of 25 cm.
[A regular hexagon is a polygon made up from 6
equilateral triangles.]
A pool player needs to hit a ball into a corner
pocket. It sits 81 cm from the back wall and 62 cm
from the side wall. What angle does the ball need
to travel at to be sunk?
10.
Calculate the angle between each triangular face of
the block and the horizontal hexagon base.
8.
Ali and Rob are designing a triathlon course.
Ali is doing the running leg of the course.
From the start point, A, she runs 800 m at a bearing
of 060°.
She then runs 600 m at a bearing of 150°.
A plywood bike jump was created with a 1.5 m long
angled ramp and that is 0.3 m high. At what angle
does the rider ride up at?
PAGE 103
ANSWERS
7.
NCEA
1.
Using SOH
Substituting:
(4.4
sin-1 6.2)
4.4


 6.2 


G = sin −1
= 45.2°
2.
Using TOA
Substituting:
x = 62 − 32
= 5.20 m (2 dp)
OR
(Achieved)
x
tan 60
6
tan 5)
6
θ = tan −1   = 50.2° (1 dp)
5
Using SOH
Substituting:
x
sin 60
(Achieved)
Step 2:
Using TOA
Substituting:
-1
−1
7
9
4.
Using TOA
Substituting:
sin 90)
dp)
 25 
θ = tan −1 
 = 78.3° (1 dp) (Excellence)
 5.20 
(Achieved)
8.
tan-1110)
−1
 14 
 11 
5.
(1 dp) (Achieved)
Angle is a right angle
(Co-interior angles, parallel
lines and angles at a point)
Using TOA
Substituting: (600
tan-1800)
Angle OAB is a right angle (Co-interior angles, parallel lines and angles at a point)
Using TOA
Substituting:
(120
3
θ = tan −1   = 36.9° (1 dp)
4
Bearing = 60 + 36.9 = 097°
9.
tan-1110)
 120 
ABO = tan 
 = 47.5° (1 dp)
 110 
Bearing = 360 − (60 + 47.5) = 252.5°
Using TOA
Substituting:
−1
Using SOH
Substituting:
sin 84)
 1 
θ = sin   = 2.73° (2 dp)
 21 
PAGE 104
−1
 62 
 81 
Using TOA
Substituting:
dp) (Achieved)
(0.3
sin-1 1.5)
-1
−1
(62
θ = tan   = 37.4° (1
10.
(4
(Merit)
tan-1 81)
(Excellence)
PRACTICE
6.
(25
tan-1 5.2)
(140
EGF = tan   = 51.8°
6
x = 6 sin 60 = 5.20 m (2 dp)
(70
θ = sin   = 51.1° (1
3
x = 3 tan 60 = 5.20 m (2 dp)
OR
-1
3.
There are two steps to this. First
calculate the distance from the
edge of the base to the centre,
then find the angle.
Step 1: Either
(Achieved)
 0.3 
θ = sin −1
= 11.5° (1


 1.5 
dp) (Achieved)
GEOMETRIC REASONING
Study Tip:
Motivation
When you study well for an exam:
• Treat yourself for giving it your best shot
• Watch a movie with friends or get some takeaways
Small rewards motivate you to try your best in whatever you do.
PAGE 105
SIMILAR SHAPES
SUMMARY • If two polygons are similar, then:
▶▶ Corresponding angles are equal
▶▶ Corresponding sides are in proportion
1.Substitute in numbers and solve:
Side 1 ( small ) Side 2 ( small )
=
Side 1 (large) Side 2 (large)
Side 1 ( small ) Side 2 ( small )
=
Side 1 (large) Side 2 (large)
• If two shapes have the same angles then
1.5 2
they are similar shapes and will be in
=
proportion to each other
x 6
6
x = 1.5 × = 4.5 m
For a complete tutorial on this topic2visit www.learncoach.co.nz
OLD NCEA QUESTIONS
1.
A ladder has two legs AB and AC. Each leg is 1.8m
long. Angle ABD = 113°
Express b in terms of c.
PAGE 106
2.
A child’s practice goal
post has one pole and two
supports, as shown to the
right.
The two supports are each
90 cm long.
The pole is always
perpendicular
to
the
ground.
The diagram below, shows
the view from the side.
OT is 90 cm long.
OP is 70 cm long.
A support bar, QR, is added
at Q, where OQ = 30 cm.
Calculate the distance of
the Length OR.
Show your working and
explain your reasoning.
GEOMETRIC REASONING
PRACTICE QUESTIONS
3.
6.
The Egyptian pyramids originally had a cap stone
at the top which was made of a different material.
The angled side of the pyramid is 78 m long and the
base is 85 m wide. If the cap stone is 3.5 m wide,
how far down the angled side of the pyramid does
it extend?
Two crosses are shown which are similar. Each arm
of the large cross is 12 cm wide and 5 cm wide for
the small cross. If each arm on the large cross is 33
cm long, how long are the arms on the small cross?
4.
7.
At a BMX track there are two jumps next to each
other. The smaller one has a ramp that is 2.5 m long
and a 1.3 m drop off. If the larger ramp has a 6.5 m
long ramp, how high is the drop off?
Two arrows are drawn which are similar. The large
arrow has an overall length of 30 cm and a tail length
of 22 cm. If the small arrow has a tail length of
13.5 cm, what is its overall length?
8.
5.
ABCEG has similar shape DEFHI inside it.
DE = 8 cm, EF = 6 cm, and FG = 4 cm.
Find the length CD.
These two shapes are similar. Use the given lengths
to find the length BC.
PAGE 107
ANSWERS
NCEA
1. a. Side 1 ( small ) = Side 2 ( small )
5.
Side 1 (large) Side 2 (large)
b 0.475
=
1.8
c
0.86
(Merit)
b=
c
2.
OQR and OPT are similar triangles:
Side 1 ( small ) Side 2 ( small )
=
Side 1 (large) Side 2 (large)
OR OQ
=
OT OP
OR 30
=
90 70
30
y = 90 ×
= 38.6 cm (1 dp) (Merit)
70
PRACTICE
3.
4.
Side 1 ( small ) Side 2 ( small )
=
Side 1 (large) Side 2 (large)
5
x
=
12 33
5
(Merit)
33 × = 13.75 cm
x =
12
Side 1 ( small ) Side 2 ( small )
=
Side 1 (large) Side 2 (large)
13.5 l
=
22 30
13.5
l = 30 ×
= 18.41 cm (2 dp) (Merit)
22
PAGE 108
6.
7.
8.
Side 1 ( small ) Side 2 ( small )
=
Side 1 (large) Side 2 (large)
EF DE
=
EG CE
6
8
=
10 CE
10
CE = 8 × = 13.33 cm (2 dp)
6
CD = 13.3 - 8 = 5.3 cm
(Merit)
Side 1 ( small ) Side 2 ( small )
=
Side 1 (large) Side 2 (large)
3.5 x
=
85 78
3.5
x = 78 ×
= 3.21 m (2 dp) (Merit)
85
Side 1 ( small ) Side 2 ( small )
=
Side 1 (large) Side 2 (large)
2.5 1.3
=
6.5 h
6.5
h = 1.3 ×
= 3.38 m
(Merit)
2.5
Side 1 ( small ) Side 2 ( small )
=
Side 1 (large) Side 2 (large)
13.5
6
=
22
AC
22
AC = 6 ×
= 9.78 cm (2 dp)
13.5
BC = 9.8 - 6 = 3.8 cm
(Merit)
GEOMETRIC REASONING
Study Tip:
Study Enhancement
If you are spending long hours studying or
working remember to:
• Drink fluids
• Eat well
• Sleep well
• Do regular exercise and move around occasionally while studying
It’s the basics that can make some of the biggest differences.
PAGE 109
ANGLES OF POLYGONS
SUMMARY • The angles inside a polygon are called interior angles.
• A Polygon with n sides has n interior angles
• The Exterior Angle is the angle between any side of a
shape, and a line extended from the next side.
• The interior angles of a polygon add to 180(n – 2)°,
where n is the number of sides
(int ∠, sum of polygon)
e.g. to the right n = 5
Sum of interior angles = 180(5 - 2) = 540°
• If a shape is regular (all the sides and angles are the
same) to find each interior angle divide the sum of
the interior angles by the number of sides.
• The exterior angles of a polygon add to 360°
(ext ∠, sum of polygon)
a + b + c + d + e + f = 360°
For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONS
1.
ABCDE is a regular pentagon.
a. Calculate the size of angle ABC giving reasons
for each step.
b. If many objects of the same shape fit together
to form a pattern, without leaving any spaces,
the shape is said to tessellate.
Explain whether or not a regular pentagon will
tessellate, giving reasons for your answer.
PAGE 110
GEOMETRIC REASONING
PRACTICE QUESTIONS
2.
6.
A fire tower is placed at a high place overlooking a
huge pine plantation.
In the regular pentagon shown, what is the value
of a and all exterior angles and b and all interior
angles.
3.
4.
Calculate both the interiors and exterior angles of a
a. regular hexagon
b. regular decagon (10 sided figure)
c. regular nine sided figure
In the STOP sign below, what is the value of any
interior angle (assume it is a regular figure). Give
reasons.
a. Given angle b = 140°, calculate angle a, giving
reasons.
b. Given angle a = 160°, calculate angle b, giving
reasons.
c. Find angle c.
d. In a differently sized fire tower, x = 152° and y =
5.
In the soccer ball calculate angles a and b, assuming
all panels are regular.
74°, find angle c.
7.
Fencing mesh is commonly made up of either 4 or
6 sided figures. Although in reality this is not the
case, assume the figures are regular and calculate
angles a, b, c, d, and e giving reasons.
f.
Give geometric reasons why both these
geometric shapes tessellate.
PAGE 111
ANSWERS
b. Sum of interior angle of a pentagon is:
NCEA
1. a. Sum
of interior angles of a pentagon
= 3 ×180° = 540°
540
= 108°
Each interior angle is =
5
ABC = 108°
(Achieved - Attempted to find
interior angle)
(Merit - correct angle)
b. For a shape to tessellate the interior angles
must be factors of 360°.
108° is not a factor of 360°, so a pentagon will
not tessellate.
This is based on the principal of angles at a
point adding to 360°.
(Excellence)
c.
d.
PRACTICE
2.
Sum of the angles = 360° (ext ∠, sum of polygon)
360
= 72°
a(and all other exterior angles) =
5
Sum of the interior angles of a pentagon
= 180(5 − 2) = 540°
Regular pentagon has 5 equal interior angles
540
b (and all other interior angles) =
= 108°
5
(Merit)
3. a. Hexagon Interior angle = 180(6 − 2) = 720 = 120°
6
6
Hexagon exterior angle =
b. Decagon interior angle
=
180(10 − 2) 1440
=
= 144°
10
10
360
Decagon exterior angle =
c.
360
= 60°
6
10
= 36°
Nonagon interior angle
=
180(9 − 2) 1260
=
= 140°
9
9
360
= 40°
9
(Merit - 2 out of 3 correct)
4. Octagon interior angle = 180(8 − 2) = 1080 = 135° 8
8
int ∠, sum of polygon
(Merit)
Nonagon exterior angle =
5. a. a = 180(5 − 2) = 540 = 108°
5
5
b. b = 360 -108 = 252° int ∠, sum of
polygon
∠s at pt
(Merit)
6. a. Sum of interior angle of a pentagon is:
= 180(5 − 2) = 540°
90 + 90 + 110 + 140 + a = 540
a = 540 − 140 − 110 − 90 − 90
a = 110°
PAGE 112
(Merit)
= 180(5 − 2) = 540°
90 + 90 + 110 + 160 + b = 540
b = 540 − 160 − 110 − 90 − 90
b = 90°
(Merit)
Sum of interior angle of a pentagon is:
= 180(5 − 2) = 540°
108 + 108 + 120 + 120 + c = 540
c = 540 − 120 − 120 − 108 − 108
c = 84°
(Merit)
c above = c below as they are vertically opposite.
Sum of interior angle of a pentagon is:
= 180(5 − 2) = 540°
x + x + y + y + c = 540°
c = 540 − (2 × 152) − (2 × 74)
= 88°
(Merit)
7. a.
180(6 − 2) 720
=
= 120°
6
6
int ∠, sum of polygon
(Merit)
b. b = 360 - a = 360 - 120 = 240° ∠s at pt
(Merit)
c. c = 360 - 2a = 360 - 240 = 120° ∠s at pt
(Merit)
180(4 − 2) 360
=
= 90°
d. d =
4
4
(int ∠, sum of polygon)
(Merit)
e. e = 360 - d = 360 - 90 = 270° ∠s at pt
(Merit)
f. For a shape to tessellate the interior angles
must be factors of 360°.
For the hexagon the interior angles are 120°
which is a factor of 360°.
For the diamond the interior angles are 90°
which are also a factor of 360°.
This is based on the principal of angles at a
point adding to 360°.
(Excellence)
a=
GEOMETRIC REASONING
ANGLES AROUND
INTERSECTING LINES
SUMMARY Example:
Rules:
• Adjacent angles on a straight
line add to 180°
( ∠ s on str. line)
a + b + c = 180°
x = 180 – 112 = 68°
( ∠ s on str. line)
• Angles at a point add to 360°
( ∠ s at pt)
a + b + c + d = 360°
x = 360 – 98 – 155
= 107°
( ∠ s at pt)
d
• Vertically opposite angles
are equal
(vert opp ∠ s)
a = c and b = d
x = 139°
(vert opp ∠ s)
For a complete tutorial on this topic visit www.learncoach.co.nz
PRACTICE QUESTIONS
Find the size of the angles shown and give a reason (the diagrams are not to scale):
1.
2.
3.
4.
5.
6.
7.
8.
PAGE 113
ANSWERS
PRACTICE
1.
x = 180 – 95 – 45 = 40°
∠ s on str. line
2.
x = 360 – 139 – 132 = 89°
∠ s at pt
3.
x = 180 – 90 – 68 = 22°
∠ s on str. line
4.
a = 180 – 30 = 150°
b = 30°
c= 150°
∠ s on str. line
vert opp ∠ s
∠ s on str. line OR
vert opp ∠ s
(Achieved)
5.
a = 180 – 162 = 18°
b = 180 – 108 = 72°
∠ s on str. line
∠ s on str. line
6.
x = 360 – 90 – 59 = 211°
∠ s at pt
7.
x = 180 – 80 – 55 = 45°
∠ s on str. line and
vert opp ∠ s
(Achieved)
8.
x = (180 – 90) / 2 = 45°
∠ s on str. line
(Achieved)
(Achieved)
(Achieved)
(Achieved)
(Achieved)
(Achieved)
Study Tip:
Pre-Exam
Before an Important Exam:
Don’t: spend a lot of time talking with classmates who haven’t
studied.
Do: Avoid negative vibes and focus on your own preparation
and goals.
PAGE 114
GEOMETRIC REASONING
Study Tip:
Cramming
Cramming helps some people.
But don’t lose sleep. Getting your normal sleep will ensure you
are your best physically and mentally for the exam.
PAGE 115
ANGLES OF PARALLEL
LINES
SUMMARY Example:
Rules:
• Corresponding angles on parallel
lines are equal
(corr ∠ s // lines)
a=b
x = 63°
(corr ∠ s // lines)
• Alternate angles on parallel lines
are equal
(alt ∠ s, // lines)
a=b
x = 56°
(alt ∠ s, // lines)
• Co-interior angles on parallel lines
are supplementary (add to 180°)
(co-int ∠ s, // lines)
a + b = 180°
x = 180 – 40 = 140°
(co-int ∠ s, // lines)
For a complete tutorial on this topic visit www.learncoach.co.nz
OLD NCEA QUESTIONS
1.
ABCD is an isosceles trapezium. Angle CBA = 78°.
AD = BC.
3.
Metal railings are fitted to the edge of a deck.
XZ is parallel to AD.
BY is parallel to CZ.
One section of railing is shown in the diagram
below:
Calculate the size of angle EDA giving reasons for
each step of your answer.
2.
The diagram shows part of a climbing frame.
a. Find the size of angle YBC.
LM = LN.
KL is parallel to NM.
LM is parallel to KN.
Angle LNK = 54°.
Calculate the size of angle LMN, explaining the
reason for each step of your answer
PAGE 116
Give geometric reasons for each step in your
solution.
b. XYB is an isosceles triangle.
Use geometric reasoning for each step to show
that XB and YC cannot be parallel.
GEOMETRIC REASONING
4.
a. If x is 110°, find the size of angle PRQ.
PR and QR are the same length.
Angle RQS is x.
RT is parallel to PS.
Give geometric reasons.
b. Prove that angle PRT and angle RQS are equal
for all values of x.
Explain your geometric reasoning clearly and
logically.
PRACTICE QUESTIONS
Find the size of the angles shown and give a reason (the diagrams are not to scale):
5.
6.
7.
8.
9.
10.
11.
12.
a. In the diagram find angles x and y
b. Find any sets of parallel lines and give reasons
13.
Toby has bought some expanding trellis. All
pieces of wood pointing in the same direction
are parallel. In the diagram below the trellis
is fully expanded. What is angle a? Give a
reason
14.
Explain why c = a + b
PAGE 117
ANSWERS
NCEA
1. ∠ BAD = 78°
PRACTICE
Isosceles trapezium
Alt ∠ s, // lines
(Achieved - Both angles correct)
(Merit - Angles and explanations correct)
∠ EDA = 78° 2. ∠ NLM = 54°
Alt
∠ s, // lines (or rotation
about mid point of LN)
∠ LMN = 63° Base ∠ , isos Δ
(Achieved - Both angles correct)
(Merit - Angles and explanations correct)
3. a.
∠ XYB = 62°
∠ YBC= 62°
∠ s, // lines
alt ∠ s, // lines
corr
(Achieved - Both angles correct)
(Merit - Angles and explanations correct)
b. XYB is an isosceles triangle, so two of its interior
angles must be equal. If XB and YC are parallel
then:
∠ XBA= 64°
corr ∠ s, // lines
∠ YBX = 180 - ∠ YBC - ∠ XBA
=180 - 62 - 64
=54°
∠ sum Δ
∠ BXY = 180 - ∠ YBX - ∠ XYB
=180 - 54 - 62
∠ sum Δ
=64°
If this is the case then no two interior angles of
XYB are equal, which is a contradiction. XYB is
actually an isosceles triangle, therefore, XB and
YC cannot be parallel.
(Achieved - two angles correct)
(Merit - Angles and explanations correct)
5.
x = 65° corr
6.
x = 75° alt
b.
∠QRT = 180 − x
co-int ∠ s, // lines
∠PRQ = 180 − 2 × ∠RQP Base ∠ , isos Δ
= 180 − 2 × (180 − x) ∠ s on str. line
= 180 − 360 + 2 x
= −180 + 2 x
Therefore:
∠PRT = ∠PQR + ∠QRT
= −180 + 2 x + 180 − x
=x
= ∠RQS (Achieved - 2 angles correct)
(Merit - Angles and explanations
correct)
PAGE 118
∠ s, // lines
(Achieved)
60° vert opp ∠ s
(Achieved)
b. b = 180 – 60 = 120° alt ∠s, // lines OR adj
∠s on str. Line
(Achieved)
8.
x = 180 – 102 = 78° corr ∠ s, // lines and
adj ∠ s on str. line
(Achieved)
9.
x = 180 – 37 = 143° alt.
∠ s, // lines
(Achieved)
10. a. a = 180 – 39 = 141° co-int ∠ s, // lines
(Achieved)
b. b = 60° alt ∠ s, // lines
(Achieved)
11. a. a = 180 – 42 – 59 = 79° adj ∠ s on str line
(Achieved)
b. b = 59° alt. ∠ s, // lines
(Achieved)
c. c = 59 + a = 59 + 79 = 138° alt. ∠ s, // lines OR
co-int ∠ s, // lines
(Achieved)
12. a. x = 180 – 98 = 82° adj, ∠ s on str. line
∠ s on str. line
∠ s on str. line
int. ∠ s of quadrilateral
∠ s on str. line
y = 99° (Merit)
b. Lines AB and CD are parallel alt. ∠ s, // lines
Lines FG and HI are parallel
corr ∠ s // lines
(Merit)
∠ LJK = 87°
∠ JKM = 93°
∠ LMK =88°
∠ RQP = 180 - x = 70° ∠ s on str. line
∠ PRQ = 180 - 2 x 70 Base ∠ , isos Δ
= 40°
(Achieved - Both angles correct)
(Merit - Angles and explanations correct)
(Achieved)
7. a. a =
4. a.
∠ s, // lines
13.
a = 180 – 68 = 112° alt.
14. ∠ XYU
∠ s, // lines
(Merit)
= a and ∠ ZYU = b
alt. ∠ s, // lines
∠ XYZ = ∠ XYU + ∠ ZYU = a + b vert opp ∠ s
∠ XYZ = c = a + b
(Merit)
GEOMETRIC REASONING
ANGLES WITHIN CIRCLES
SUMMARY
The parts of a circle are shown below:
Rules:
Example:
• Angles on the same arc are equal
a=b
(∠s on same arc)
x = 35°
(∠s on same arc)
• The angle at the centre is equal to twice
the angle at the circumference on the
same arc 2c = d
(∠ at centre)
x = 43 x 2 = 86°
(∠ at centre)
• The angle in a semicircle is a right angle
This is a special case of the above rule
(∠ in semicircle)
x = 90 – 72
= 18°
(∠ in semicircle)
• The angle where the radius meets the
tangent is 90°
(rad ⊥ tgt)
x = 180 – 56 – 90
= 34°
(rad ⊥ tgt)
• Two tangents coming from the same
point are equal (same length and angles)
(tangs from a point)
• The angle between a chord and a tangent equals the angle in the alternate
segment. a = d, c = b.
(∠ in alt seg)
x = (180 – 52) / 2
= 64°
(tangs from a point
and base ∠ isos Δ)
• Remember any chord forms an isosceles triangle with the centre.
ABO and OBC are isosceles triangles.
x = 180 – 2 x 34
= 112°
( ∠ sum isos Δ)
• Opposite angles in a cyclic
quadrilateral add to 180°
(opp ∠s cyclic quad)
a + c = 180°, b + d = 180°
• The exterior angle of a cyclic quadrilateral is equal to the interior opposite
angle. a = b
(ext ∠ cyclic quad)
x = 32°
y = 73°
(∠ in alt seg)
x = 72°
(opp ∠s cyclic quad)
x = 96°
(ext ∠ cyclic quad)
For a complete tutorial on this topic visit www.learncoach.co.nz
PAGE 119
OLD NCEA QUESTIONS
1.
DAC is a tangent. O is the centre of the circle.
Calculate the size of angle DAB, explaining the
reason for each step of your answer.
2.
A, B, and C are points on the circumference of the
circle. O is the centre of the circle.
AB is parallel to OC. Angle CAO = x°.
5.
EFGD is a cyclic quadrilateral with angle EDG = 82°.
O is the centre of the circle.
Find the size of angle HFJ.
Explain your geometric reasoning clearly and
logically.
6. a.
Calculate the size of angle ACB in terms of x.
3.
For the diagram above:
a. Find the size of angle reflex COA, x.
Explain your reasoning.
b. Find the size of angle DCO, y.
Give geometric reasons for each step in your
solution.
4.
b.
The points A, Q, Z, N lie on the circumference
of a circle centre O.
AQ is parallel to NZ.
Find the size of angle NZQ, in terms of x.
Explain your geometric reasoning clearly and
logically.
For the diagram below, prove that angle C + angle
A = 180°.
What angle properties does a cyclic
parallelogram have?
Explain your answer with geometric reasoning.
Use the blank diagram above (where O is the
centre of the circle) if you wish.
Give geometric reasons for each step in your
solution.
PAGE 120
GEOMETRIC REASONING
PRACTICE QUESTIONS
You must give a geometric reason for each step leading to your answer in every question.
11.
7.
In the diagram, WX and YZ are parallel.
WZ and XZ are equal length.
Angle WXZ = 37°.
Calculate the size of angle YXZ.
In the diagram above, AB is a tangent to the circle
and TD is a diameter of the circle.
∠ABC = 42°.
Calculate the size of ∠CED.
12.
8.
In the diagram above, points A and T lie on a circle
with centre C.
TCF is a straight line. BF = BT.
DE is a tangent that touches the circle at T.
∠TFB = 54°. Find the size of ∠ADT.
9.
O is the centre of the circle.
Angle BAC = 39°.
Lines CB and BA are the same length. Line OB is
perpendicular to line AC. Find the size of angle OCA.
In the diagram P, Q, R and S lie on a circle, centre, O.
TR is a tangent to the circle at R.
Angle PQR = 29°.
Angle SRT = 22°.
Find the size of angle SPO.
13.
ABE and ACF are two triangles.
Angle BCD = 25°.
Angle DEF = 38°.
Find the size of angle FDE
14.
10.
O is the centre of the circle. TQ and RQ are tangents
to the circle. Angle RST is 42°.
Lines SR and ST are the same length.
Find the size of angle QRS.
The corners of ABCD lie on a circle.
The angle AOC is 132° and the angle FAB is 101°.
Find the angle CED.
PAGE 121
ANSWERS
4.
NCEA
Let c = angle C and a = angle A, then:
∠ DOB (reflex) = 2c
1.
∠ CAO = 90°
∠ AOC = 58°
∠ AOB = 122°
∠ OAB = 29°
∠ DAB = 61°
base ∠ , isos Δ and
2.
base ∠ , isos Δ
∠ sum Δ
(Merit)
5.
∠ EFG = 180 - 82 = 98°
∠ HFJ= ∠ EFG = 98°
opp ∠s cyclic quad
vert opp
∠s
(Merit)
6. a.
∠ sum Δ
∠ s at pt
∠ at centre
alt ∠s, // lines
∠ s at pt
2a + 2c = 360°
2(a + c) = 360°
360
a+c =
= 180°
2
∠ sum Δ
rad ⊥ tangent
(Achieved - Two angles correct)
(Merit - Two angles and explanations correct)
(Excellence - Full answer with coherent steps and
explanations)
∠ ACO = x
Obtuse ∠ AOC = 180 - 2x
Reflex ∠ AOC = 180 + 2x
∠ ABC = 90 + x
∠ CAB = x
∠ ACB = 90 - 2x
∠ at centre
∠ BOD (reflex) = 2a
rad ⊥ tangent
∠ sum Δ
∠s on a line
∠ AQZ = 180 - ∠ ZNA
opp ∠s cyclic quad
∠ NZQ = 180 - ∠ AQZ
Co-int ∠s, // lines
= 180 - x
b.
= 180 - (180 - x)
= x°
(Merit)
(Achieved - Two angles correct)
(Merit - Two angles and explanations correct)
(Excellence - Full answer with coherent steps and
explanations)
3. a.
∠ AOC = 2 x 72 = 144° ∠ at centre
∠ s at pt
x = Reflex ∠ COA = 360 - 144
Obtuse
= 216°
b.
(Merit)
Draw line segment OD. ODA is an isosceles
triangle:
∠ AOC = 38°
base ∠ , isos Δ
∠ ODC = 72 - 38 = 34°
y = ∠ DCO = 34° base ∠ , isos Δ
PAGE 122
(Excellence)
Let ANZQ be a cyclic parallelogram. From (a)
above we know that:
∠ ZNA = ∠ NZQ and ∠ NAQ = ∠ AQZ
AN and QZ are also parallel so we also know:
∠ ZNA = ∠ NAQ and ∠ AQZ= ∠ NZQ
Therefore all interior angles are equal.
Since the interior angles of a quadrilateral add
to (4 - 2) x 180 = 360°, each interior angle is 90°.
Therefore a cyclic parallelogram must be a
rectangle.
(Excellence)
GEOMETRIC REASONING
PRACTICE
7.
∠ BTD = 90°
∠ BDT = 90 – 42 = 48°
∠ CET = 48°
∠ DET = 90°
∠ CED = ∠ DET - ∠ CET
rad ⊥ tangent
12.
∠ sum Δ
∠ on same arc
∠ in a semi-circle
(Excellence) h
= 90 – 48 = 42°
8.
= 83°
x=
∠ RSO = 42 / 2 = 21°
∠ SRO = 21°
∠ QRS = 90 + 21 = 111°
∠ DBC = 180 – 25 – x
∠ sum Δ
∠ DBA = 180 –(155 – x)
∠s on a line
opp
Solve: 38° + x + 25° + x = 180°
2x = 117°
x = 58.5°
14.
symmetry
base ∠ s, isos Δ
rad ⊥ tangent
(Excellence)h
11.
∠ XZY = 37°
∠ XZW = (180 – 37) / 2 = 71.5°
∠ WZY = 71.5 + 37 = 108.5°
∠ WXY = 180 - 108.5 = 71.5°
∠ YXZ = 71.5 - 37 = 34.5°
∠s on a line
∠ DFA + ∠ DBA = 180°
(Excellence) h
10.
∠ DFA = 180 –(142 – x)
= 25° + x
base ∠ , isos Δ
∠ s, // lines
base ∠ s, isos Δ
sum adj ∠
opp ∠s cyclic
alt
∠s
∠ sum Δ
= 155 – x
base ∠ , isos Δ
vert opp
∠ DFE = 180 – 38 – x
= 38° + x
∠ at centre
= 51 - 39 = 12°
∠ FDE = ∠ BDC
= 142 – x
(Excellence) h
∠ BOC = 39 x 2 = 78°
∠ OCB = (180 – 78) / 2 = 51°
∠ BCA = 39°
∠ OCA = ∠ OCB- ∠ BCA
(Excellence) h
13.
∠ ABT = 180 - 2 x 54 = 72°
∠ sum isos Δ
∠ at centre
∠ ACT = 72 x 2 = 144°
∠ ATF = (180 – 144) / 2 = 18° base ∠ , isos Δ
∠ ADT = 90 - 18 = 72°
Rad ⊥ tan
9.
∠ at centre
∠ POR = 2 x 29 = 58°
∠ PSR = 180 – 29 = 151°
opp ∠s cyclic quad
∠ SRO = 90 – 22 = 68°
rad ⊥ tangent
∠ SPO = 360 – 151 - 68 - 58 ∠ sum quad
∠ ABC = 132 / 2 = 66°
∠ CDE = 66°
∠ BAD = 180 –101 = 79°
∠ DCE = 79°
∠ CED = 180 – 66 – 79
=35°
∠ s cyclic quad
(Excellence)
∠ at centre
ext
∠ s cyclic quad
∠s on a line
ext
∠ s cyclic quad
∠ sum Δ
(Excellence) h
quad
(Excellence) h
PAGE 123