Question

Write a Pythagoras triplet such that one of its number is:
(i) 6
(ii) 14
(iii) 16
(iv) 18

Answer 1

Hint: Pythagoras theorem in a right-angle triangle is given as ${{\left( \text{Hypotenuse} \right)}^{\text{2}}}\text{=}{{\left( \text{Base} \right)}^{\text{2}}}\text{+}{{\left( \text{Perpendicular} \right)}^{\text{2}}}$ .So, any three numbers a, b, c will be termed as Pythagorean triplet if any of the relations ${{a}^{2}}={{b}^{2}}+{{c}^{2}}$ or ${{b}^{2}}={{a}^{2}}+{{c}^{2}}$ or ${{c}^{2}}={{a}^{2}}+{{b}^{2}}$ is true.

Complete step-by-step solution -
As we know the Pythagoras theorem is defined for right angle triangles only. And it is given as ${{\left( \text{Hypotenuse} \right)}^{\text{2}}}\text{=}{{\left( \text{Base} \right)}^{\text{2}}}\text{+}{{\left( \text{Perpendicular} \right)}^{\text{2}}}$
$\Rightarrow {{R}^{2}}={{P}^{2}}+{{Q}^{2}}..........\left( i \right)$
Where, R is representing the hypotenuses side of the right angle and P and Q are base and perpendicular.
Now, coming to the question, as we need to determine the Pythagoras triplets of which one number is given in the subparts of the problem.
(i) 6
We need to predict the other two numbers by hit and trial method. Let us suppose P = 6 and Q = 8 to the equation (i)
We get
${{6}^{2}}+{{8}^{2}}=36+64=100={{\left( 10 \right)}^{2}}$
Hence, ${{6}^{2}}+{{8}^{2}}$ can be written as ${{10}^{2}}$ .It means the Pythagoras triplet can be given as (6, 8, 10) as it is following the expression given in equation(i) as
${{6}^{2}}+{{8}^{2}}={{10}^{2}}$
So, triplet is given as (6, 8, 10)
(ii) 14
Let us suppose the two values P and Q as
$P=10$ and $Q=4\sqrt{6}$
Now, we can put P and Q to the equation and hence, we get
$\begin{align}
  & {{P}^{2}}+{{Q}^{2}}={{\left( 10 \right)}^{2}}+{{\left( 4\sqrt{6} \right)}^{2}} \\
 & {{P}^{2}}+{{Q}^{2}}=100+96=196 \\
 & {{P}^{2}}+{{Q}^{2}}={{\left( 14 \right)}^{2}} \\
\end{align}$
So, we can put R = 14 to the equation(i). Hence the triplet related to given number is given as $\left( 4\sqrt{6},10,14 \right)$
(iii) 16
Now, here we need to predict two other numbers which follow the Pythagoras rule as well.
So, suppose
P = 16 and Q = 12
Putting values of P and Q to the equation (i), we get
\[\begin{align}
  & {{\left( P \right)}^{2}}+{{\left( Q \right)}^{2}}={{\left( 16 \right)}^{2}}+{{\left( 12 \right)}^{2}} \\
 & =256+144=400
\end{align}\]
\[\Rightarrow {{P}^{2}}+{{Q}^{2}}={{\left( 20 \right)}^{2}}\]
Hence, we can put R = 20 to the equation (i). So we get that the triplet (16, 12, 20) will also follow the Pythagoras theorem. Hence the triplet related to the given number is (16, 12, 20)
(iv) 18
Let us suppose P = 10 and $Q=4\sqrt{14}$
So, putting values of P and Q to the equation(i), we get
\[\begin{align}
  & {{P}^{2}}+{{Q}^{2}}={{\left( 10 \right)}^{2}}+{{\left( 4\sqrt{14} \right)}^{2}} \\
 & =100+16\times 14=100+224 \\
 & {{P}^{2}}+{{Q}^{2}}=324={{18}^{2}}
\end{align}\]
Hence, we get value R from the equation (i) as 18. So, the triplet $\left( 10,4\sqrt{14},18 \right)$ will be a triplet, which follows the Pythagoras theorem. So $\left( 10,4\sqrt{14},18 \right)$ will be a Pythagoras triplet whose one number is 18.

Note: One may suppose any value of P, Q and R, other than given in the solution. As there can be an infinite combination of a number with other numbers, which can follow the relation of Pythagoras theorem. So, answers may be different than given in the solution. One may go wrong if he/she does not place the positions of the terms of the Pythagoras equation correctly. One may apply the formula as
${{\left( Hypotenuse \right)}^{2}}+{{\left( Base \right)}^{2}}={{\left( Perpendicular \right)}^{2}}$ which is wrong. So, be clear with the corresponding positions of the terms given in the Pythagoras equation.