9.04 Triples and 3D problems (Extended)
Pythagorean triple
The numbers $3$3, $4$4 and $5$5 have a special property. If we square the first two numbers and add them they will be equal to the square of the largest number.
| $3^2+4^2$32+42 | $=$= | $5^2$52 |
| $9+16$9+16 | $=$= | $25$25 |
| $25$25 | $=$= | $25$25 |
$a^2+b^2=c^2$a2+b2=c2
That is if we substitute $a$a with $3$3, $b$b with $4$4, and $c$c with $5$5 both sides will be equal. Any group of three whole numbers that satisfy the equation are called a Pythagorean triple or a Pythagorean triad. We can check any three numbers by substituting the two smaller numbers for $a$a and $b$b and the largest number for $c$c.
Below are three Pythagorean triples:
| $\left(3,4,5\right)$(3,4,5) | $\left(5,12,13\right)$(5,12,13) | $\left(8,15,17\right)$(8,15,17) |
If you know one of the triples you can make another one by multiplying each number by a constant. For example if we multiply the triple $\left(3,4,5\right)$(3,4,5) by $2$2 we will have $\left(6,8,10\right)$(6,8,10).
$6^2+8^2=10^2$62+82=102
is also true so it will be a Pythagorean triple. The triples introduced above are now shown below with two of their multiples in each column:
| $\left(3,4,5\right)$(3,4,5) | $\left(5,12,13\right)$(5,12,13) | $\left(8,15,17\right)$(8,15,17) |
| $\left(6,8,10\right)$(6,8,10) | $\left(10,24,26\right)$(10,24,26) | $\left(16,30,34\right)$(16,30,34) |
| $\left(30,40,50\right)$(30,40,50) | $\left(50,120,130\right)$(50,120,130) | $\left(80,150,170\right)$(80,150,170) |
The three numbers of a triple are often given from smallest to largest, however sometimes it may have the first two numbers swapped, and sometimes it is in no particular order. As long as you know which number is the largest you can check if three numbers are a Pythagoran triple.
$\left(3,4,5\right)$(3,4,5), $\left(4,3,5\right)$(4,3,5), $\left(5,3,4\right)$(5,3,4)
The biggest number is $5$5 in each case, this will be $c$c.
Worked example
Example 1
Is $\left(5,7,12\right)$(5,7,12) a Pythagorean triple?
Think: A Pythagorean triple must satisfy $a^2+b^2=c^2$a2+b2=c2.
Do: Substitute $5$5 and $7$7 into $a^2+b^2$a2+b2 and see if it is equal to substituting $12$12 into $c^2$c2:
| $a^2+b^2$a2+b2 | $=$= | $5^2+7^2$52+72 |
Substsituting for left-hand side |
| $$ | $=$= | $25+49$25+49 |
|
| $$ | $=$= | $74$74 |
Evaluating |
| $c^2$c2 | $=$= | $12^2$122 |
Substsituting for right-hand side |
| $$ | $=$= | $144$144 |
Evaluating |
| $74$74 | $\ne$≠ | $144$144 |
Comparing the two sides |
| $a^2+b^2$a2+b2 | $\ne$≠ | $c^2$c2 |
|
The numbers $\left(5,7,12\right)$(5,7,12) are not a Pythagorean triple.
Reflect: We couldn't assume $a^2+b^2=c^2$a2+b2=c2 was true. We had to test each side separately and then see if they were equal.
A Pythagorean triple is any three whole numbers that satisfy
$a^2+b^2=c^2$a2+b2=c2
where $c$c is the largest number.
Right-angled triangles
In a right-angled triangle the largest angle in the triangle is $90^\circ$90°. The side across from the right angle will be the largest side. We call this side the hypotenuse.
All three sides of a right-angled triangle are related by the equation shown below:
The two smaller sides will be called $a$a and $b$b, and the hypotenuse (the longest side) will be $c$c. Earlier we looked at Pythagorean triples which satisfy the same equation. Any triangle with sides that are a Pythagorean triple will be a right-angled triangle.
Worked example
Example 2
Is the following triangle a right-angled triangle?
Think: If its three side lengths satisfy $a^2+b^2=c^2$a2+b2=c2 then the triangle will be a right-angled triangle.
Do:
| $a^2+b^2$a2+b2 | $=$= | $15^2+20^2$152+202 |
Calculating left-hand side |
| $$ | $=$= | $225+400$225+400 |
|
| $$ | $=$= | $625$625 |
Evaluating |
| $c^2$c2 | $=$= | $25^2$252 |
Calculating right-hand side |
| $$ | $=$= | $625$625 |
Evaluating |
| $a^2+b^2$a2+b2 | $=$= | $c^2$c2 |
|
Reflect: We can also skip some of the working out if we realise that $\left(15,20,25\right)$(15,20,25) is a multiple of $\left(3,4,5\right)$(3,4,5), because each number has been multiplied by $5$5. This means it will also be a Pythagorean triple and the triangle will be a right-angled triangle.
Pythagoras' theorem relates the three sides of a right-angled triangle, $a$a and $b$b are the two smaller sides, and the longest side, called the hypotenuse, is $c$c.
We can also test to see if a triangle is right-angled by checking to see if its three sides satisfy $a^2+b^2=c^2$a2+b2=c2.
Practice questions
Question 1
Using your knowledge of common Pythagorean triples or otherwise, is $\left(12,5,13\right)$(12,5,13) a Pythagorean triple?
Yes
ANo
B
Question 2
Which side of the triangle in the diagram is the hypotenuse?
$X$X
A$Y$Y
B$Z$Z
C
Question 3
Use Pythagoras' theorem to determine whether this is a right-angled triangle.
Let $a$a and $b$b represent the two shorter side lengths. First find the value of $a^2+b^2$a2+b2.
Let $c$c represent the length of the longest side. Find the value of $c^2$c2.
Is the triangle a right-angled triangle?
Yes
ANo
B
Pythagoras' theorem gives us a relationship between the three sides of a right-angled triangle, allowing us to use any two sides to find the third. This theorem can be used for any right-angled triangle, even those in a three dimensional context.
Right-angled triangles in 3D space
We can identify right-angled triangles in 3D space the same way that we do in 2D space, by looking for a right angle.
In 3D space, right angles occur between lines that are perpendicular in the same 2D plane. In other words, if two lines meet at a right-angle in some 2D slice of 3D space, the triangle formed with these two lines is a right-angled triangle.
Exploration
Consider the cube with its vertices labelled like so:
If we take the 2D slice through the vertices $A$A, $B$B, $C$C and $D$D we can see that any triangle determined by three of these vertices will be right-angled, since this slice gives us a square.
In fact, any of the 2D slices that corresponds to a face of the cube will give us multiple possible right-angled triangles.
What if we take the 2D slice through the vertices $F$F, $H$H, $B$B and $D$D?
Taking this slice gives us a rectangle, which means that we still form a right-angled triangle from any three of these vertices.
If we return to 3D space, we can see how these right-angled triangles look in 3D. Taking perspective into account when using a 3D view, these triangles do indeed have right angles.
We can perform a similar exercise with other 3D solids like cones and prisms to find other right-angled triangles in 3D space.
Finding the lengths of 3D diagonals
In most straight-edged solids, the diagonals formed by joining two non-adjacent vertices will often be the hypotenuse of some right-angled triangle in that solid.
In order to find the length of such a diagonal, we can simply use Pythagoras' theorem to calculate the hypotenuse, provided we know the lengths of the other two sides.
In the cases where we don't know both the other side lengths, we may need to use Pythagoras' theorem on a different right-angled triangle in the solid, as seen in the example below.
Worked example
example 3
A cube has a side length of $3$3. What is the length of the diagonal between vertices $G$G and $C$C?
Think: To find the length of the diagonal $GC$GC, we can use Pythagoras' theorem in the right-angled triangle $\triangle GCA$△GCA. We know that $GA$GA has a length of $3$3, but we don't know the length of $AC$AC. To find the length of $AC$AC, we can use Pythagoras' theorem in the right-angled triangle $\triangle ADC$△ADC. Since we know the lengths of both $AD$AD and $DC$DC, we can start here.
Do: Using Pythagoras' theorem in $\triangle ADC$△ADC tells us that:
$AC$AC $=$= $\sqrt{3^2+3^2}$√32+32 $=$= $\sqrt{18}$√18
We can then use this value for Pythagoras' theorem in $\triangle GCA$△GCA. This tells us that:
$GC$GC $=$= $\sqrt{3^2+\left(\sqrt{18}\right)^2}$√32+(√18)2 $=$= $\sqrt{27}$√27
Rounding to three decimal places, the diagonal $GC$GC has a length of $5.196$5.196.
Reflect: To find the length of the diagonal, we found a right-angled triangle in 3D space which had the diagonal as its hypotenuse. For any missing values needed for our calculations, we simply found another right-angled triangle.
By repeating this until we have enough starting information, we used Pythagoras' theorem multiple times to find the desired lengths.
When multiple applications of Pythagoras' theorem are required, we should never round our answer before the final step. This is because any rounding error in an early step will carry over into later steps, introducing unnecessary error into the final answer.
It is for this reason that the example above used the exact result from the first step when calculating the final answer.
Practice questions
Question 4
A square prism has sides of length $5$5cm, $5$5cm and $9$9cm as shown.
If the diagonal $HF$HF has a length of $z$z cm, calculate the exact length of $z$z, leaving your answer in surd form.
Now, we want to find $y$y, the length of the diagonal $DF$DF.
Calculate $y$y to two decimal places.
Question 5
Consider the triangular prism below.
Find the exact length of $CE$CE. Leave your answer in surd form.
Find the length of $CX$CX correct to two decimal places.
Find the length of $BE$BE correct to two decimal places.
Question 6
A soft drink can has a height of $13$13cm and a radius of $5$5cm. Find $L$L, the length of the longest straw that can fit into the can (so that the straw is not bent and fits entirely inside the can).
Give your answer rounded down to the nearest cm, to ensure it fits inside the can.
