Geometry » Pythagoras' theorem

Pythagoras' theorem is also known as Pythagorean theorem.

You can do Pythagoras' theorem:

1. With a scheme/table
2. With the shortest/fastest way
3. With an equation.
Directly to the examples

More theory

4. Pythagoras in space
5. Reverse Pythagoras' theorem (to check whether the angle is 90°)

All three methods use a2 + b2 = c2.

The theorem says that in a right-angled triangle the area of the two squares adjacent to the shorter sides added together is equal to the area of the area of the square adjacent to the longest side. In the figure below this means that the area of the two pink squares together is equal to the area of the blue square. The longest side is called the hypotenuse.
image that shows Pythagoras with coloured squares
On the left, you can see that the pink area is the area of the whole square minus the four equal triangles. The area of the blue area is the same, also the area of the whole square minus the four equal triangles. On the right you can see those figures slid on top of each other in such a way that the yellow triangles are on top of each other. You now know that a2 + b2 = c2 must apply in the yellow triangle.

You can also say this:
In a right-angled triangle, the sum of the squares of the two shorter sides is equal to the square of the hypotenuse.

Misleading name

Contrary to what you might think, it was not Pythagoras who was the first to came up with this theorem. This method was already known with the Sumerians and Babylonians (current Iraq) and also the Indians long before Pythagoras lived. The Greek and perhaps Pythagoras are probably the one who brought the theorem to the western world.

1. With a scheme/table

Method that is used in most modern mathematics books.
Often used to teach it to pupils who get it for the first time.
Below you can see the scheme that is used for this method.

sidesquare
shorter side   
shorter side    +  
hypotenuse 

1. In the left column you fill in the sides that are known/given.
Put a question mark at the side you need to calculate.
REMEMBER: The hypotenuse should always be at the bottom!
2. Calculate the squares of the known sides (#2).
3.Calculate the square of the unknown side (by adding or subtracting).
4.Calculate the unknown side by using the square root.
See examples.


2. With the shortest/fastest way

Use Pythagoras' theorem in the following way:

a = square root(c^2 – b^2)
b = square root(c^2 – a^2)
c = square root(a^2 + b^2)
See examples.


3. With an equation

This method is used in some maths books and is probably the method your parents learned. You really use a2 + b2 = c2 in that form.
You fill in the known sides directly in the formula and then start solving the equation.
See examples.



Examples

 
Example 1
Calculate the length of the side with the '?'.
So you have to calculate the hypotenuse.
Right-angled triangle with shorter sides 16 and 30

Scheme/table Shortest/fastest Equation

sidesquare
16 256 
30900+  
?1156

1. Fill in 16 and 30
2. Calculate 256 and 900
3. 256 + 900 = 1156
4. ? = square root(1156) = 34

c = square root(a^2 + b^2)

thus

c = square root(16^2 + 30^2) = 34    

a2 + b2 = c2

thus

162 + 302 = c2
256 + 900 = c2
1156 = c2
c = square root(1156) = 34
Example 2
Calculate the length of the side with the '?'.
You have to calculate a shorter side.

right-angled triangle with shorter side and hypotenuse 7

Scheme/table Shortest/fastest Equation

sidesquare
RQ = 3  9 
PQ = ?40+  
PR = 749

1. Fill in 3 and 7
2. Calculate 9 and 49
3. 49 – 9 = 40
4. PQ = square root(40)

a = square root(c^2 – b^2)

thus

PQ = square root(7^2 – 3^2) = square root(40)

PQ2 + QR2 = PR2

thus

PQ2 + 32 = 72
PQ2 + 9 = 49
PQ2 = 40
PQ = square root(40)

Example 3
Given are the points A(–10, 3) and B(5, 23).
Calculate the distance between these two points.

You have to calculate the hypotenuse.

First you need to calculate the horizontal and vertical distance between A and B.
A and B are horizontally 5 – –10 = 15 and vertically 23 – 3 = 20 units apart.

In a sketch it will look like this:
sketch with line AB and arrows with the horizontal and vertical distance

Scheme/table Shortest/fastest Equation

sidesquare
15 225 
20400+  
?625

1. Fill in 15 and 20
2. Calculate 225 and 400
3. 225 + 400 = 625
4. ? = square root(625) = 25

c = square root(a^2 + b^2)

thus

c = square root(15^2 + 20^2) = 25

a2 + b2 = c2

thus

152 + 202 = c2
225 + 400 = c2
625 = c2
c = square root(625) = 25


4. Pythagoras in space

Within solids you often do not have any right-angled triangles. However, it is always possible to make your own, by using a cross-section. At the theory about long diagonals is an example of how you can calculate a long diagonal in a cuboid. It is also possible to use Pythagoras' theorem in other solids like pyramids and cones.

Example 1

In cube ABCD.EFGH with edges measuring 4, point K is in the middle of BC.
Calculate EK.
cube ABCD.EFGH with line EK drawn
To calculate EK you will have to use right-angled triangles.
In the cube you see two possible right-angled triangles you can use.

Using the green cross-section:
We can calculate EK with right-angled triangle EBK.
BK = 2, but BE is unknown.
BE is calculable using the front face.

Scheme/tableShortest/fastest
sidesquare
AB = 4  16 
AE = 416+  
BE = ?32

 BE = square root(32)

 
sidesquare
BK = 2  4 
BE = square root(32)32+  
PR = ?36

 BE = square root(36) = 6

 
BE = square root(4^2 + 4^2) = square root(32)
Now we can calculate EK.

EK = square root((square root(32))^2 + 2^2) = square root(36) = 6

Using the red cross-section:
We can calculate EK with right-angled triangle EKM.
KM = 4, but KM is still unknown.
EM is calculable using the top face.

Scheme/tableShortest/fastest
sidesquare
EF = 4  16 
FM = 24+  
EM = ?20

 EM = square root(20)

 
sidesquare
KM = 4  4 
EM = square root(20)20+  
EK = ?36

 BE = square root(36) = 6

 
EM = square root(4^2 + 2^2 = square root(20)
Now we can calculate EK.

EK = square root((square root(20))^2 + 4^2 = square root(36) = 6


Example 2

Pyramid ABCD.T has a square base with sides measuring 6 cm and slanted sides measuring 8 cm.
Calculate height ST.
Pyramid ABCD.T with height ST drawn. S is the middle of the base.

To calculate ST you can use cross-section ACT.
Sketch cross-section ACT with height ST.
You can see that SCT will be the needed right-angled triangle.
AC and SC are already written in the sketch below, however you need to calculate them first.
Doorsnede ACT met lijn ST getekend.

Calculate, using the base, the length of AC first.
With AC you can calculate the length of SC.

sidesquare
AB = 6  36 
BC = 636+  
AC = ?72

 AC = square root(72) cm

   

SC is half the length of AC.
SC = 12AC = 12square root(72) cm


Now you can calculate ST in triangle SCT.

sidesquare
SC = 12square root(72)  18 
ST = ?  46+  
AC = 8  64

 ST = square root(46) ≈ 6,78 cm


5. Reverse Pythagoras' theorem

With the reverse of Pythagoras' theorem you can calculate/find out whether an angle in a triangle is greater than, less than or equal to 90°. With this method you always look at the angle opposite of the hypotenuse/longest side. In the newest maths books, this name might not be used. Although the method is explained.

It works like this:
Because a2 + b2 = c2 only works in a right-angled triangle you can use Pythagoras' theorem to check whether or not a triangle, of which you know all three sides, is right-angled or not. If not, you can also tell whether the triangle is acute-angled or obtuse-angled.

The rules:
a2 + b2 = c2. The triangle is right-angled.
a2 + b2 < c2. The triangle is obtuse-angled (c is too long and makes the angle bigger).
a2 + b2 > c2. The triangle is acute-angled (c is too short and makes the angle smaller).

Examples
Triangle ABC with AB=14, BC=12 and AC=5 and triangle PQR with DE=square root(656), EF=20 and DF=16
52 + 122 = 142
25 + 144 = 196
169 = 196
169 < 196
a2 + b2 < c2
So triangle ABC is obtuse-angled.
angle signA > 90°
162 + 202 = (square root(656))2
256 + 400 = 656
656 = 656
a2 + b2 = c2
So triangle DEF is right-angled.
angle signA = 90°

Triangle GHI with GH=square root(95), HI=13 and GI=square root(75)
(square root(75))2 + (square root(95))2 = 132
75 + 95 = 169
170 = 169
170 > 169
a2 + b2 > c2
So triangle GHI is acute-angled.
angle signA < 90°


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