2
Pythagoras’ Theorem
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Pythagoras’ Theorem
The converse of Pythagoras’ Theorem
Problem solving using Pythagoras’ Theorem
Three-dimensional problems
More difficult problems (Extension)
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Contents:
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34
PYTHAGORAS’ THEOREM (Chapter 2)
Right angles (90o angles) are used in the construction of buildings and in the division of areas
of land into rectangular regions.
The ancient Egyptians used a rope with 12
equally spaced knots to form a triangle with sides
in the ratio 3 : 4 : 5: This triangle has a right
angle between the sides of length 3 and 4 units,
and is, in fact, the simplest right angled triangle
with sides of integer length.
corner
take hold of knots at arrows
make rope taut
line of one side
of building
OPENING PROBLEM
Karrie is playing golf in the US Open but hits a wayward tee shot on the
opening hole. Her caddy paces out some distances and finds that Karrie has
hit the ball 250 m, but is 70 m from the line of sight from the tee to the hole.
A marker which is 150 m from the pin is further up the fairway as shown:
0
Tee
Caddy
150 m marker
Hole
70 m
250 m
Ball
Consider the following questions:
1 How far is the caddy away from the tee?
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2 From where the caddy stands on the fairway, what distance is left to the 150 m marker
if he knows the hole is 430 m long?
3 How far does Karrie need to hit her ball with her second shot to reach the hole?
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35
PYTHAGORAS’ THEOREM (Chapter 2)
INTRODUCTION
A right angled triangle is a triangle which
has a right angle as one of its angles.
se
tenu
o
hyp
The side opposite the right angle is called
the hypotenuse and is the longest side of the
triangle.
legs
The other two sides are called the legs of the
triangle.
Around 500 BC, the Greek mathematician
Pythagoras formulated a rule which connects
the lengths of the sides of all right angled triangles. It is thought that he discovered the
rule while studying tessellations of tiles on
bathroom floors. Such patterns, like the one
illustrated, were common on the walls and
floors of bathrooms in ancient Greece.
A
PYTHAGORAS’ THEOREM
c
In a right angled triangle, with
hypotenuse c and legs a and b,
a
c2 = a2 + b2 .
b
In geometric form, the Theorem of Pythagoras is:
GEOMETRY
PACKAGE
In any right angled triangle, the area of
the square on the hypotenuse is equal to
the sum of the areas of the squares on the
other two sides.
c2
c
Can you see how
Pythagoras may have
discovered the rule by
looking at the tile
pattern above?
a
c
a
b
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36
PYTHAGORAS’ THEOREM (Chapter 2)
Over 400 different proofs of the Pythagorean Theorem exist. Here is one of them:
Proof:
a
On a square we draw 4 identical (congruent) right
angled triangles, as illustrated. A smaller square in
the centre is formed.
b
b
c
Suppose the legs are of length a and b and the
hypotenuse has length c:
c
Since total area of large square
= 4 ВЈ area of one triangle + area of smaller square,
c
a
c
(a + b)2 = 4( 12 ab) + c2
2
2
a
b
2
) a + 2ab + b = 2ab + c
) a2 + b2 = c2
a
b
When using Pythagoras’ Theorem we often see surds, which are square root
p
numbers like 7:
Note:
Example 1
Self Tutor
Find the length of the hypotenuse in:
x cm
2 cm
3 cm
The hypotenuse is opposite the right angle and has length x cm.
If x2 =pk, then
x = В§ k` , p
but
we reject ВЎ k`
as lengths must
be positive!
x2 = 32 + 22
x2 = 9 + 4
x2 = 13
p
fas x > 0g
i.e., x = 13
p
) the hypotenuse is 13 cm.
)
)
)
EXERCISE 2A
1 Find the length of the hypotenuse in the following triangles, leaving your answer in surd
(square root) form if appropriate:
a
b
c
4 cm
x km
x cm
x cm
7 cm
8 km
5 cm
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13 km
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PYTHAGORAS’ THEOREM (Chapter 2)
Example 2
37
Self Tutor
The hypotenuse has length 6 cm.
) x2 + 52 = 62 fPythagorasg
) x2 + 25 = 36
) x2 = 11
p
) x = 11 fas x > 0g
p
) third side is 11 cm long.
Find the length of the third
side of:
6 cm
x cm
5 cm
2 Find the length of the third side of the following right angled triangles.
Where appropriate leave your answer in surd (square root) form.
a
b
c
x km
11 cm
6 cm
x cm
1.9 km
2.8 km
x cm
9.5 cm
Example 3
Self Tutor
The hypotenuse has length x cm.
p
) x2 = 22 + ( 10)2 fPythagorasg
) x2 = 4 + 10
) x2 = 14
p
) x = В§ 14
p
But x > 0, ) x = 14.
Find x in the following:
~`1`0 cm
2 cm
x cm
3 Find x in the following:
a
b
c
3 cm
~`7 cm
~`2 cm
x cm
x cm
x cm
~`1`0 cm
~`5 cm
Example 4
2
x +
Solve for x:
Qw_ cm
)
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x cm
1
4
Self Tutor
2
=1
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fPythagorasg
=1
x2 =
)
1 cm
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3
4
q
x = В§ 34
q
x = 34
fas x > 0g
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38
PYTHAGORAS’ THEOREM (Chapter 2)
4 Solve for x:
a
b
c
1 cm
m
x cm
Qw_ cm
Qw_ cm
Г–
`3
2
x cm
1m
xm
Ew_ cm
Example 5
Self Tutor
(2x)2
) 4x2
) 3x2
) x2
= x2 + 62 fPythagorasg
= x2 + 36
= 36
= 12
p
) x = В§ 12
p
But x > 0, ) x = 12.
Find the value of x:
2x m
xm
6m
5 Find the value of x:
a
b
9 cm
c
26 cm
2x cm
x cm
2x cm
2x m
3x m
~`2`0 m
3x cm
Example 6
Self Tutor
5 cm
A
Find the value of any unknowns:
x cm
y cm
D
6 cm
B
1 cm
C
In triangle ABC, the hypotenuse is x cm.
fPythagorasg
) x2 = 52 + 12
2
) x = 26
p
) x = В§ 26
p
) x = 26
fx > 0g
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In ВўACD, the hypotenuse is 6 cm.
p
fPythagorasg
) y 2 + ( 26)2 = 62
2
) y + 26 = 36
) y 2 = 10
p
) y = В§ 10
p
fy > 0g
) y = 10
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39
PYTHAGORAS’ THEOREM (Chapter 2)
6 Find the value of any unknowns:
a
b
c
1 cm
2 cm
x cm
3 cm
y cm
7 cm
3 cm
4 cm
x cm
y cm
y cm
2 cm
2 cm
x cm
7 Find x:
a
b
3 cm
(x-2)ВЎcm
4 cm
5 cm
13 cm
x cm
8 Find the length of AC in:
A
9m
5m
B
D
C
9 Find the distance AB, in the following figures.
(Hint: It is necessary to draw an additional line or two on the figure in each case.)
a
b
c
C
D
1m
M
3 cm
4 cm
B
4m
N
5m
7m
6m
3m
B
A
B
A
A
B THE CONVERSE OF PYTHAGORAS’ THEOREM
If we have a triangle whose three sides have known lengths, we can use the converse of
Pythagoras’ Theorem to test whether (or not) it is right angled.
THE CONVERSE OF PYTHAGORAS’ THEOREM
2
2
2
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If a triangle has sides of length a, b and c units and a + b = c ,
then the triangle is right angled.
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GEOMETRY
PACKAGE
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40
PYTHAGORAS’ THEOREM (Chapter 2)
Example 7
Self Tutor
Is the triangle with sides 6 cm, 8 cm and 5 cm right angled?
52 + 62
= 25 + 36
= 61
The two shorter sides have lengths 5 cm and 6 cm, and
But 82 = 64
) 52 + 62 6= 82
and hence the triangle is not right angled.
EXERCISE 2B
1 The following figures are not drawn accurately. Which of the triangles are right angled?
a
b
c
7 cm
9 cm
9 cm
12 cm
5 cm
5 cm
4 cm
8 cm
15 cm
d
e
f
3 cm
~`2`7 m
~`7 cm
~`4`8 m
8m
15 m
17 m
~`1`2 cm
~`7`5 m
2 If any of the following triangles (not drawn accurately) is right angled, find the right
angle:
A
C
a
b
c
A
2 cm
8m
1 cm
B
~`2`4 km
B
C
~`5 cm
5 km
~`2`0`8 m
B
C
12 m
7 km
A
PYTHAGOREAN TRIPLES
The simplest right angled triangle with sides of integer
length is the 3-4-5 triangle.
The numbers 3, 4, and 5 satisfy the rule 32 + 42 = 52 .
5
3
4
The set of integers fa, b, cg is a Pythagorean triple if it obeys the rule
a2 + b2 = c2 :
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f5, 12, 13g, f7, 24, 25g, f8, 15, 17g.
Other examples are:
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PYTHAGORAS’ THEOREM (Chapter 2)
Example 8
41
Self Tutor
Show that f5, 12, 13g
is a Pythagorean triple.
We find the square of the largest number first.
132 = 169
and 5 + 122 = 25 + 144 = 169
2
52 + 122 = 132
)
i.e., f5, 12, 13g is a Pythagorean triple.
3 Determine if the following are Pythagorean triples:
a f8, 15, 17g
b f6, 8, 10g
d f14, 48, 50g
e f1, 2, 3g
c
f
Example 9
f5, 6, 7g
f20, 48, 52g
Self Tutor
Find k if f9, k, 15g is a Pythagorean triple.
Let 92 + k2 = 152
) 81 + k2 = 225
) k2 = 144
p
) k = В§ 144
) k = 12
fPythagorasg
fk > 0g
4 Find k if the following are Pythagorean triples:
a f8, 15, kg
b fk, 24, 26g
d f15, 20, kg
e fk, 45, 51g
c
f
f14, k, 50g
f11, k, 61g
5 For what values of n does fn, n + 1, n + 2g form a Pythagorean triple?
6 Show that fn, n + 1, n + 3g cannot form a Pythagorean triple.
INVESTIGATION
PYTHAGOREAN TRIPLES SPREADSHEET
Well known Pythagorean triples are f3, 4, 5g, f5, 12, 13g,
f7, 24, 25g and f8, 15, 17g.
Formulae can be used to generate Pythagorean triples.
SPREADSHEET
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An example is 2n + 1, 2n2 + 2n, 2n2 + 2n + 1 where n is a positive integer.
A spreadsheet will quickly generate sets of Pythagorean triples using such formulae.
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PYTHAGORAS’ THEOREM (Chapter 2)
What to do:
1 Open a new spreadsheet
and enter the following:
fill down
2 Highlight the formulae in B2, C2 and D2 and fill
down to Row 3. You should now have generated
two sets of triples:
3 Highlight the formulae in Row 3 and fill down
to Row 11 to generate 10 sets of triples.
4 Check that each set of numbers is actually a triple by adding two more columns to
your spreadsheet.
In E1 enter the heading
�aˆ2+bˆ2’ and in F1 enter the
heading �cˆ2’.
In E2 enter the formula
=B2Л†2+C2Л†2 and in F2 enter
the formula =D2Л†2.
5 Highlight the formulae in E2 and F2 and fill down to Row 11. Is each set of numbers
a Pythagorean triple? [Hint: Does a2 + b2 = c2 ?]
6 Your task is to prove that the formulae f2n + 1, 2n2 + 2n, 2n2 + 2n + 1g will
produce sets of Pythagorean triples for positive integer values of n.
We let a = 2n + 1, b = 2n2 + 2n and c = 2n2 + 2n + 1.
Simplify c2 ВЎb2 = (2n2 +2n+1)2 ВЎ(2n2 +2n)2 using the difference of two squares
factorisation, and hence show that it equals (2n + 1)2 = a2 .
C
PROBLEM SOLVING USING
PYTHAGORAS’ THEOREM
Right angled triangles occur frequently in problem solving and often the presence of right
angled triangles indicates that Pythagoras’ Theorem is likely to be used.
SPECIAL GEOMETRICAL FIGURES
All of these figures contain right angled triangles where Pythagoras’ Theorem applies:
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rectangle
Construct a diagonal to form a right angled
triangle.
d
25
In a rectangle, a right angle exists between
adjacent sides.
l
5
na
o
iag
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PYTHAGORAS’ THEOREM (Chapter 2)
43
In a square and a rhombus, the
diagonals bisect each other at
right angles.
rhombus
square
In an isosceles triangle and an
equilateral triangle, the altitude
bisects the base at right angles.
isosceles triangle
equilateral triangle
Things to remember
ВІ
ВІ
ВІ
ВІ
ВІ
ВІ
Draw a neat, clear diagram of the situation.
Mark on known lengths and right angles.
Use a symbol, such as x, to represent the unknown length.
Write down Pythagoras’ Theorem for the given information.
Solve the equation.
Write your answer in sentence form (where necessary).
Example 10
Self Tutor
A rectangular gate is 3 m wide and has a 3:5 m diagonal. How high is the gate?
Let x m be the height of the gate.
fPythagorasg
Now (3:5)2 = x2 + 32
) 12:25 = x2 + 9
) 3:25 = x2
p
) x = 3:25
fas x > 0g
) x + 1:80
Thus the gate is approximately 1:80 m high.
3m
xm
3.5 m
EXERCISE 2C.1
1 A rectangle has sides of length 8 cm and 3 cm. Find the length of its diagonals.
2 The longer side of a rectangle is three times the length of the shorter side. If the length
of the diagonal is 10 cm, find the dimensions of the rectangle.
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3 A rectangle with diagonals of length 20 cm has sides in the ratio 2 : 1. Find the
a perimeter
b area of the rectangle.
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44
PYTHAGORAS’ THEOREM (Chapter 2)
Example 11
Self Tutor
A rhombus has diagonals of length 6 cm and 8 cm.
Find the length of its sides.
The diagonals of a rhombus bisect at right angles.
Let a side be x cm.
fPythagorasg
) x2 = 32 + 42
2
) x = 9 + 16
) x2 = 25
p
) x = В§ 25
) x=5
fx > 0g
x cm
3 cm
4 cm
i.e., the sides are 5 cm in length.
4 A rhombus has sides of length 6 cm. One of its diagonals is 10 cm long. Find the length
of the other diagonal.
5 A square has diagonals of length 10 cm. Find the length of its sides.
6 A rhombus has diagonals of length 8 cm and 10 cm. Find its perimeter.
Example 12
Self Tutor
A man travels due east by bicycle at 16 kmph. His son travels due south on his
bicycle at 20 kmph. How far apart are they after 4 hours, if they both leave point
A at the same time?
64 km
A
80 km
After 4 hours the man has travelled 4 ВЈ 16 = 64 km
and his son has travelled 4 ВЈ 20 = 80 km.
Thus x2 = 642 + 802 fPythagorasg
i.e., x2 = 4096 + 6400
) x2 = 10 496
p
) x = 10 496 fas x > 0g
) x + 102
) they are 102 km apart after 4 hours.
man
x km
N
W
son
E
S
7 A yacht sails 5 km due west and then 8 km due south.
How far is it from its starting point?
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8 Town A is 50 km south of town B and town C is 120
km east of town B. Is it quicker to travel directly from
A to C by car at 90 kmph or from A to C via B in a
train travelling at 120 kmph?
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PYTHAGORAS’ THEOREM (Chapter 2)
45
9 Two runners set off from town A at the same time. If
one runs due east to town B and the other runs due
south to town C at twice the speed, they arrive at B and
C respectively two hours later. If B and C are 50 km
apart, find the speed at which each runner travelled.
Example 13
Self Tutor
An equilateral triangle has sides of length 6 cm. Find its area.
The altitude bisects the base at right angles.
) a2 + 32 = 62
) a2 + 9 = 36
) a2 = 27
p
) a = В§ 27
p
) a = 27
6 cm
a cm
fPythagorasg
fa > 0g
1
2
base ВЈ height
p
= 12 ВЈ 6 ВЈ 27
p
= 3 27 cm2
+ 15:6 cm2
f1 d.p.g
Now, area =
3 cm
So, area is 15:6 cm2 :
10 Find any unknowns in the following:
a
b
c
45В°
x cm
2 cm
h cm
7 cm
x cm
60В°
1 cm
y cm
30В°
x cm
yВ°
12 cm
11 An equilateral triangle has sides of length 12 cm. Find the length of one of its altitudes.
12 An isosceles triangle has equal sides of length 8 cm and a base of length 6 cm. Find the
area of the triangle.
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13 When an extension ladder rests against a wall it
reaches 4 m up the wall. The ladder is extended
a further 0:8 m without moving the foot of the ladder and it now rests against the wall 1 m further up.
How long is the extended ladder?
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1m
4m
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46
PYTHAGORAS’ THEOREM (Chapter 2)
p
14 An equilateral triangle has area 16 3 cm2 . Find the length of its sides.
15 Revisit the Opening Problem on page 34 and answer the questions posed.
TRUE BEARINGS
When using true bearings we measure the direction of travel by comparing it with the true
north direction. Measurements are always taken in the clockwise direction.
Imagine you are standing at point A, facing north. You turn
clockwise through an angle until you face B. The bearing
of B from A is the angle through which you have turned.
That is, the bearing of B from A is the measure of the angle
between AB and the �north’ line through A.
north
72В°
B
north
A
In the diagram at right, the bearing of B from A is 72o
from true north. We write this as 72o T or 072o .
B
If we want to find the true bearing of A from B, we place
ourselves at point B and face north and then measure the
clockwise angle through which we have to turn so that we
face A. The true bearing of A from B is 252o .
252В°
A
Example 14
Self Tutor
A helicopter travels from base station S on a true bearing of
074o for 112 km to outpost A. It then travels 134 km on a true
bearing of 164o to outpost B. How far is outpost B from base
station S?
Let SB be x km.
From the diagram alongside, in triangle SAB
]SAB = 90o .
x2 = 1122 + 1342
) x2 = 12 544 + 17 956
) x2 = 30 500
p
) x = 30 500
) x + 175
In bearings
problems, notice the
use of the properties
of parallel lines for
finding angles.
N
fPythagorasg
N
112 km
74В°
A
164В°
74В°
16В°
S
134 km
fas x > 0g
x km
i.e., outpost B is 175 km from base station S.
B
EXERCISE 2C.2
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1 Two bushwalkers set off from base camp at the same time. If one walks on a true bearing
of 049o at an average speed of 5 kmph and the other walks on a true bearing of 319o at
an average speed of 4 kmph, find their distance apart after 3 hours.
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PYTHAGORAS’ THEOREM (Chapter 2)
47
2 James is about to tackle an orienteering course. He has been given these instructions:
ВІ the course is triangular and starts and finishes at S
ВІ the first checkpoint A is in a direction 056o from S
ВІ the second checkpoint B is in a direction 146o from A
ВІ the distance from A to B is twice the distance from S to A
ВІ the distance from B to S is 2:6 km.
Find the length of the orienteering course.
3 A fighter plane and a helicopter set off from airbase A at the same time. If the helicopter
travels on a bearing of 152o and the fighter plane travels on a bearing of 242o at three
times the speed, they arrive at bases B and C respectively 2 hours later. If B and C are
1200 km apart, find the average speed of the helicopter.
D
THREE-DIMENSIONAL PROBLEMS
Pythagoras’ Theorem is often used when finding lengths in three-dimensional solids.
Example 15
Self Tutor
A 50 m rope is attached inside an empty cylindrical
wheat silo of diameter 12 m as shown. How high
is the wheat silo?
12 m
50 m
12 m
Let the height be h m.
)
)
hm
50 m
h2 + 122 = 502
h2 + 144 = 2500
) h2 = 2356
p
) h = 2356
) h + 48:5
i.e., the wheat silo is
fPythagorasg
fas h > 0g
fto 1 dec. placeg
48:5 m high.
EXERCISE 2D
1 A cone has a slant height of 17 cm and a base radius of 8 cm. How high is the cone?
2 Find the length of the longest nail that could be put entirely within a cylindrical can of
radius 3 cm and height 8 cm.
3 A 20 cm nail just fits inside a cylindrical can. Three identical spherical balls need to fit
entirely within the can. What is the maximum radius of each ball?
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In three-dimensional problem solving questions we often need the theorem of Pythagoras
twice. We look for right angled triangles which have two sides of known length.
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48
PYTHAGORAS’ THEOREM (Chapter 2)
Example 16
Self Tutor
A room is 6 m by 4 m at floor level and the floor to ceiling height is 3 m. Find the
distance from a floor corner point to the opposite corner point on the ceiling.
The required distance is AD. We join BD.
fPythagorasg
In ВўBCD, x2 = 42 + 62
2
2
2
In ВўABD, y = x + 3
fPythagorasg
ym
2
2
2
2
) y = 4 +6 +3
) y2 = 61
p
D
) y = В§ 61
p
) the required distance is 61 + 7:81 m.
A
3m
B
4m
xm
6m
C
But y > 0
4 A cube has sides of length 3 cm. Find
the length of a diagonal of the cube.
diagonal
5 A room is 7 m by 5 m and has a height of 3 m. Find the distance from a corner point
on the floor to the opposite corner of the ceiling.
6 A rectangular box is 2 cm by 3 cm by 2 cm (internally). Find the length of the longest
toothpick that can be placed within the box.
7 Determine the length of the longest piece of timber which could be stored in a rectangular
shed 6 m by 5 m by 2 m high.
Example 17
Self Tutor
A pyramid of height 40 m has a square base with edges 50 m.
Determine the length of the slant edges.
Let a slant edge have length s m.
Let half a diagonal have length x m.
x2 + x2 = 502
) 2x2 = 2500
) x2 = 1250
Using
xm
40 m
sm
xm
50 m
fPythagorasg
s2 = x2 + 402 fPythagorasg
) s2 = 1250 + 1600
2
40 m ) s = 2850
p
fas s > 0g
) s = 2850
) s + 53:4
fto 1 dec. placeg
Using
xm
sm
50 m
xm
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i.e., each slant edge is 53:4 m long.
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PYTHAGORAS’ THEOREM (Chapter 2)
49
E
8 ABCDE is a square-based pyramid. E, the apex of the pyramid
is vertically above M, the point of intersection of AC and BD.
If an Egyptian Pharoah wished to build a square-based pyramid
with all edges 100 m, how high (to the nearest metre) would
the pyramid reach above the desert sands?
C
D
M
A
B
9 A symmetrical square-based pyramid has height 10 cm and
slant edges of 15 cm. Find the dimensions of its square base.
B
10 A cube has sides of length 2 m. B is at the centre of one face,
and A is an opposite vertex. Find the direct distance from A
to B.
A
E
MORE DIFFICULT PROBLEMS (EXTENSION)
PRINTABLE
EXERCISE
Click on the icon to obtain a printable exercise on more difficult
problems requiring a solution using Pythagoras’ Theorem.
REVIEW SET 2A
1 Find the lengths of the unknown sides in the following triangles:
2 cm
4 cm
a
b
c
5 cm
x cm
x cm
9 cm
7 cm
2x cm
x cm
A
2 Is the following triangle right angled?
Give evidence.
1
4
C
Г–
``17
B
3 Show that f5, 11, 13g is not a Pythagorean triple.
4 A rectangle has diagonal 15 cm and one side 8 cm. Find the perimeter of the rectangle.
5 An isosceles triangle has equal sides of length 12 cm and a base of length 8 cm. Find
the area of the triangle.
6 A boat leaves X and travels due east for 10 km. It then sails 10 km south to Y. Find
the distance and bearing of X from Y.
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7 What is the length of the longest toothpick which can be placed inside a rectangular
box that is 3 cm ВЈ 5 cm ВЈ 8 cm?
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50
PYTHAGORAS’ THEOREM (Chapter 2)
8 Two rally car drivers set off from town C
at the same time. A travels in a direction
63oT at 120 kmph and B travels in a
direction 333oT at 135 kmph. How far
apart are they after one hour?
REVIEW SET 2B
1 Find the value of x in the following:
a
b
x cm
xm
`Г–7 cm
5 cm
2x
c
`Г–`42
5m
5x
6m
B
2 Show that the following triangle is right angled
and state which vertex is the right angle:
5
2
C
A
~`2`9
3 A rectangular gate is twice as wide as it is high. If a diagonal strut is 3:2 m long, find
the height of the gate to the nearest millimetre.
4 If a softball diamond has sides of length 30 m, determine the distance a fielder must
throw the ball from second base to reach home base.
5 Town B is 27 km in a direction 134o T from town A, and town C is 21 km in a direction
224o T from town B. Find the distance between A and C.
6 If a 15 m ladder reaches twice as far up a vertical wall as the base is out from the wall,
determine the distance up the wall to the top of the ladder.
7 Can an 11 m long piece of timber be placed in a rectangular shed of dimensions 8 m by
7 m by 3 m? Give evidence.
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8 Straight roads from towns A and B intersect at right angles at X. A and B are 52
km apart. Mika leaves A at the same time
as Toshi leaves B. Mika cycles at 24 km/h
and Toshi jogs at 10 km/h, towards X, and
they arrive at X at the same time. For how
long were they travelling?
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