Triangle Puzzle
Introduction. The following activities can be explored at various levels.
Starting with creative use of triangles to make interesting shapes moving on
to consider area and perimeter before considering uses of Pythagoras (KS3),
surds with trigonometric ratios (KS4) and trigonometric formulae (AS level).
The triangles below are all right angled. The dimensions of the perpendicular
sides are 1x1, 1x2 and 1x3 units. For the following activities it is best to copy
the shapes onto card and then cut them up in order to manipulate them.
The triangles can be arranged to form a number of interesting shapes, e.g.
rectangle, triangle, Christmas tree, etc..
You may like to consider the area of each triangle and note that when all the
triangles are used the different shapes all have the same area (as long as
there are no overlapping triangles).
You could investigate the smallest and largest perimeters. In doing this you
may need to measure or calculate the length of the sloping edge in each
triangle.
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For the triangle solution given below what are the dimensions of each side?
You have probably spotted by now that we are looking at the classic 3,4,5
triangle whose area is 6 square units and whose perimeter is 12 units. From
Pythagoras’ rule for right angles triangles we get
32  4 2  5 2
i.e.
9 + 16 = 25 .
The numbers 3, 4, 5 and 6 (the area of the triangle) also combine to give
33  43  53  63
i.e
27 + 64 + 125 = 216 .
(Not many people know this! Or that 216 is known as Plato’s number. 216 is
the smallest cube that can be written as the sum of three cubes. You will
probably need a spreadsheet to find other numbers with this property. )
You could consider what this tells us about the radius of the inscribed circle
(also known as the incircle) for this triangle. Can you also show that the
radius is one unit by using algebra. (Hint – you could make an equation using
the total perimeter of 12 units broken up into six sections described using the
radius r.)
There is a nice extension of this to other Pythagorian triples. What is the
radius of the incircle in a 5,12,13 triangle? Continue the investigation for 7,
24, 25, then 9, 40, 41, etc..
Going back to the original triangles each hypoteneuse can be calculated in
surd form using Pythagoras to give 2 ,
smallest triangle is 45° so we can write
1
tan 45  ,
1
sin 45 
1
,
2
5 , and
and
10 . The angle in the
cos 45 
1
.
2
If the smallest angle in the 1x2 triangle is A then similar results can be
obtained for tan, sin and cos of angle A. Similarly for angle B in the 1x3
triangle.
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Rearranging the triangles is reasonably convincing to show that
45  A  B  90 .
(Confirmed by halving the sum of the interior angles in the 3,4,5 triangle.)
Converting this to radians and using the surd trig ratios above leads nicely to:
1
1
1 
tan
 tan 1  tan 1 
.
1
2
3 2

1
Similar results can be produced for inverse sin and cos.
The smallest angle in the 3,4,5 triangle is 2A. The double angle formulae for
sin, cos and tan can be checked out for both 2A and 2B. For example:
sin 2 A  2 sin A cos A
3
1
3
 2

5
10
10
2 tan A
when applied to
1  tan 2 A
1
2 
3
 3

2 .
the smallest angle in the 3,4,5 triangle (i.e. 2A) leads to 4
1
1  
 3
Also the ‘double angle’ formulae, e.g.
tan 2 A 
Again, similar results follow for sin and cos.
These are nice demonstrations of trig formulae which are not often
demonstrated in this way but also provides useful manipulation and practice
in working with fractions and surds.
Mick Blaylock
22 May 2007
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