Question

Write a statement of Pythagoras theorem and show that 6, 8 and 10 are Pythagoras triplet.

Answer 1

Hint: Here we will first define the statement of Pythagoras theorem and then to show that the given numbers 6, 8 and 10 are Pythagoras triples we will show that these numbers follow the Pythagoras theorem.

Complete step by step answer:
Let us start with the statement of Pythagoras theorem but stating it we should know that the Pythagoras theorem is only applicable to a right angled triangle.

So, according to Pythagoras theorem:

In a right angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of other two sides that is the sum of the squares of the lengths of the base and perpendicular of the triangle.

So, if the length of the hypotenuse of a right triangle be h, length of its base be b and the length of its perpendicular be p, then according to the Pythagoras theorem we have:

${{h}^{2}}={{p}^{2}}+{{b}^{2}}$

Now, we have to show that the numbers 6, 8 and 10 are Pythagoras triples. So, for this first we will find the squares of each of these numbers:

So, ${{6}^{2}}=36$

${{8}^{2}}=64$

${{10}^{2}}=100$

Now we can see that the sum of squares of 6 and 8 is equal to the square of 10 that is:

$\begin{align}

  & 36+64=100 \\

 & {{6}^{2}}+{{8}^{2}}={{10}^{2}} \\

\end{align}$

So, if we have a right triangle with the lengths of its base and perpendicular to be 6 units and 8 units, then the length of its hypotenuse will be 10 units.

Hence, it shows that the numbers 6, 8 and 10 form a Pythagoras triplet.

Note: Students should note that since the hypotenuse is the longest side of a right triangle so, while showing three numbers for being Pythagoras triplets we show that the sum of the squares of the two smaller numbers is equal to the square of the third of the largest.