What is Pythagoras Theorem?
Pythagoras was a Greek philosopher born in 571 BCE in Samos, Greece.
The Pythagoras theorem definition shows the relation among the three sides of a right triangle.
The square of the hypotenuse is equal to the sum of the square of the other two sides.
Right triangles follow this rule and they are called Pythagoras theorem triangle.
Pythagoras Theorem Statement
In a rightangled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
In the given \(\Delta ABC \), we see
\[\begin{align}
\text{BC}^2 &= \text{AB}^2 +\text{AC}^2
\end{align}\]
where,
 \( \text{AB}\) is the base,
 \( \text{AC}\) is the altitude or the height, and
 \( \text{BC}\) is the hypotenuse.
Hence, the Pythagoras Theorem states that:
\[\begin{align} 
Let's explore this with the help of a simulation.
Click Go to see the Geometric proof of the Pythagoras Theorem:
Pythagoras Theorem Derivation
The Pythagoras theorem definition can be derived and proved in different ways. Let us see a few methods here.
By Algebraic method
Consider four right triangles \( \Delta ABC\) where b is the base, a is the height and c is the hypotenuse.
Arrange these four congruent right triangles in the given square, whose side is (\( \text {a + b}\)).
Start the simulation below to observe how these congruent triangles are placed and how the proof of the Pythagoras theorem is derived using the algebraic method.
By Similarity of Triangles
\(\Delta ABD \) and \(\Delta ACB\):
 \(\angle A = \angle A\) (common)
 \(\angle ADB = \angle ABC\) (both are right angles)
Thus, \(\Delta ABD \) and \(\Delta ACB\) are equiangular, which means that they are similar by AA similarity criterion.
Similarly, we can prove
\[\Delta BCD\sim \Delta ACB\]
(Note carefully the order of the vertices)
Since
\[\Delta ABD \sim \Delta ACB\]
we have:
\[\begin{align}
\frac{{AD}}{{AB}} &= \frac{{AB}}{{AC}} \hfill \\\\
\Rightarrow AD \times AC &= A{B^2}....(1) \hfill \\
\end{align} \]
Similarly,
\[\Delta BCD \sim \Delta ACB\]
gives:
\[\begin{align}\frac{{CD}}{{BC}} &= \frac{{BC}}{{AC}} \\\\
\Rightarrow CD \times AC &= B{C^2}....(2)\end{align}\]
Adding (1) and (2), we have:
\[\begin{array}{*{20}{l}}
{\left( {AD \times AC} \right) + \left( {CD \times AC} \right) = A{B^2} + B{C^2}} \\\\
{ \Rightarrow (AD + CD) \times AC = A{B^2} + B{C^2}} \\\\
{ \Rightarrow \boxed{A{C^2} = A{B^2} + B{C^2}}\left( {\because\! AD \!+\! CD \!=\! AC} \right)}
\end{array}\]
Formula  Pythagoras Theorem
The Pythagoras theorem formula states that in a right triangle \(\text {ABC}\), the square of the hypotenuse is equal to the sum of the square of the other two legs.
\[ \text{AB}^2 + \text{AC}^2 =\text{BC}^2\] 
where,
 \( \text{AB}\) is the base
 \( \text{AC}\) is the altitude or the height and
 \( \text{BC}\) is the hypotenuse.
You can refer to the Solved Examples section here for some interesting reallife pythagoras' theorem questions.
Pythagorean Theorem Calculator
Use this Pythagoras theorem calculator to see the relationship between the altitude, height and hypotenuse of the right triangle.
Key in any two values and see the third value being populated in the right triangle with this measure.
 Hypotenuse is the side opposite to the right angle and it is the longest side of the right triangle.
 Pythagorean triplet is a set of three whole numbers \(\text{a, b and c}\) that satisfy Pythagoras’ theorem.
 In a right triangle, the Pythagoras theorem formula states that \(
\text{Hypotenuse}\! =\! \!\sqrt{\text{Base}^2\!\!+\! \text{Height}^2}\!\!\)  The sides of an isosceles right triangle will be in the ratio \( 1 : 1: \sqrt{2}\)
Solved Examples
Example 1 
The side of a square kite sheet is \(8 \: \text{cm}\). Tim cut the kite along the diagonal.
What is the measure of the diagonal?
Solution:
This looks like a right triangle with the base and height as \(8 \: \text{cm}\).
The diagonal of the kite is the hypotenuse.
By Pythagorean Theorem, we know that
\[\begin{align}
\text{Hypotenuse}^2 & = \text{Base}^2 + \text{Height}^2 \\
&= 8^2 +8^2 \\
&= 64 + 64 = 128 \\
\text{Hypotenuse} & = \sqrt 128 \\
&= 8\sqrt 2 \\
&≈ 11.31
\end{align} \]
\(\therefore\) Diagonal of the square \(≈ 11.31\: \text{cm}\) 
Example 2 
Tim and Ria are bouldering experts.
Ria climbed \(8 \: \text{feet}\) on a vertical bouldering rock and Tim was standing on the floor \( 6 \: \text{feet}\) away from the base.
How far away are Tim and Ria from each other?
Solution:
We can visualize this scenario as a right triangle.
Ria had climbed \(8\: \text{feet} \) which is the height of our right triangle.
Tim is \( 6 \:\text{feet}\) away from the rock, which is the base of our triangle.
The hypotenuse is the distance between them.
Therefore,
\[\begin{align}
\text{Hypotenuse} & = \sqrt {\text{Base}^2 + \text{Height}^2} \\
&= \sqrt {6^2 +8^2} \\
&= \sqrt{36+64} \\
&=\sqrt{100} \\
&=10 \: \text{feet}
\end{align}\]
\(\therefore\) Tim and Ria are \(10\: \text{feet} \) away 
Example 3 
\(N\) is a point on a straight line segment \(AB\), and \(PN \bot AB\).
Show that \( A{P^2}  B{P^2} = A{N^2}  B{N^2}\)
Solution:
In \(\Delta(APN)\) using Pythagoras theorem, we have:
\[A{P^2} = A{N^2} + P{N^2}....(1)\]
In \(\Delta(BPN)\) using Pythagoras theorem, we have:
\[B{P^2} = B{N^2} + P{N^2}....(2)\]
Subtracting (2) from (1), we have:
\[A{P^2}  B{P^2} = A{N^2}  B{N^2}\]
\[ \therefore A{P^2}  B{P^2} = A{N^2}  B{N^2}\] 
Example 4 
Julie wanted to wash her building window which is \(12 \: \text{feet}\) off the ground.
She has a ladder which is \(13 \: \text{feet}\) long.
How far should she place the base of the ladder away from the building?
Solution:
We can visualize this scenario as a right triangle.
The window on the wall is \(12 \text{ feet} \) from the ground. This is the height of our right triangle.
The ladder which is of height \(13 \: \text{feet}\), is placed leaning against the wall.
The hypotenuse is \(13 \: \text{feet}\). We need to find the base of our right triangle.
We know that,
\[\begin{align}
\text{Hypotenuse}^2 & = \text{Base}^2 + \text{Height}^2 \\
\text{Base} &=\sqrt{ \text{Hypotenuse}^2  \text{Height}^2} \\
&= \sqrt {13^2  12^2} \\
&= \sqrt{169 144} \\
&=\sqrt{25} \\
&=5 \: \text{feet}
\end{align}\]
\(\therefore\) Ladder should be placed \(5 \:\text{feet} \) away from the base. 
Example 5 
Can a triangle with sides measuring \(8 \:\text{cm}\) \(15 \:\text{cm}\) and \(17 \:\text{cm}\) form a Pythagoras theorem triangle?
Solution:
\[\begin{align}
8^2 &= 8 \times 8 \\
&=64 \\
15^2 &= 15 \times 15\\
&=225 \\
17^2 &= 17 \times 17 \\
&= 289 \\
289 &=225 +64 & \\
\Rightarrow 17^2 &= 15^2 + 8^2
\end{align}\]
Since they follow the Pythagoras theorem, they form a right triangle.
\(\therefore 8 \:\text{cm,} \:\: 15 \:\text{cm}\:\: \text{and}\: 17 \:\text{cm} \) form a Pythagoras theorem triangle 
Example 6 
\( ABC\) is an equilateral triangle, and \(AD\) is the altitude through \(A\).
Show that \(A{D^2} = 3B{D^2}\).
Solution:
Since the triangle is equilateral, the altitude \(AD\) is also the median, and so
\[BD = CD = \frac{1}{2}BC\]
Consider \(\Delta ABD\), by the Pythagoras theorem, we have:
\[\begin{align}
A{B^2} &= A{D^2} + B{D^2} \hfill \\
\Rightarrow\! B{C^2}\! &=\! A{D^2}\! +\! B{D^2} \! \left ( {\because \!AB\! =\! BC\! =\! AC\!} \right) \hfill\\
\Rightarrow \!{\left(\! {2BD} \right)^2} \!&= \!A{D^2}\! +\! B{D^2} \!\left(\! {\because\! BD\! =\! \!\frac{1}{2}BC}\! \right) \hfill \\
\Rightarrow\! 4B{D^2}\! &=\! A{D^2}\! +\! B{D^2} \hfill \\
\Rightarrow A{D^2} &= 3B{D^2} \hfill \\
\end{align} \]
\(\therefore A{D^2} = 3B{D^2} \) 
Practice Questions
Here are a few Pythagoras' Theorem questions for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.
Common Mistakes
We often make mistakes while applying the Pythagoras theorem to triangles.
Therefore, keep these points in mind to avoid making any mistakes.
 Pythagoras theorem does not work for all triangles. It has to be a right triangle.
 The height, base and hypotenuse need to be labelled correctly. Remember that hypotenuse is the longest side of the right triangle.
 If the figure is flipped or rotated, we may end up adding instead of subtracting.
 We often tend to multiply and divide by \(2\) instead of squaring
 Prove the converse of the Pythagoras theorem, i.e., if in a triangle, the sum of the squares of two sides is equal to the square of the third, show that this triangle is rightangled.

BL and CM are medians of \(\Delta ABC\) which is rightangled at A. Prove that \[4(B{L^2} + C{M^2}) = 5B{C^2}\]

\(O\) is an arbitrary point inside a rectangle \(ABCD\). Show that
\[O{A^2} + O{C^2} = O{B^2} + O{D^2}\]
Maths Olympiad Sample Papers
IMO (International Maths Olympiad) is a competitive exam in Mathematics conducted annually for school students. It encourages children to develop their math solving skills from a competition perspective.
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 IMO Sample Paper Class 1
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Frequently Asked Questions(FAQs)
1. How do you calculate hypotenuse using Pythagoras theorem?
The square of the hypotenuse of a right triangle is equal to the sum of the square of the other two legs.
When any two values are known, we can apply the theorem and calculate the other.
Pythagoras Theorem Calculator will help you calculate this.
2. What is Pythagoras theorem example?
Find the hypotenuse of a right triangle whose altitude is \(9 \:\text{cm} \) and the base is \(12 \:\text{cm} \)
We know that
\[\begin{align}
\text{Hypotenuse}^2 & = \text{Base}^2 + \text{Height}^2 \\
& = 9^2 + 12^2 \\
&=81 + 144 \\
&=225\\
\therefore Hypotenuse &= \sqrt{225} \\
&=15 \:\text{cm}
\end{align} \]
3. What is Pythagoras theorem used for?
The Pythagoras theorem helps in computing the distance between points on the plane.
It also helps in calculating the perimeter, the surface area, the volume of geometrical shapes, and so on.
In real life, Pythagoras theorem is used in architecture and construction industries.
It is also used in survey and many realtime applications.
4. How does Pythagoras theorem work?
The Pythagoras theorem works only for a right triangle. The square of the hypotenuse is equal to the sum of the other two arms of the triangle.
In \(\Delta ABC \) right angled at A,
\[\begin{align}
\text{AB}^2 + \text{AC}^2 &=\text{BC}^2
\end{align} \]