Chapter 9.6, Problem 19PSC

a

To determine

To prove: The Pythagoras triple, through Pythagoras rule.

Explanation of Solution

Given information :

Given a right angle with sides n, $\frac{{\text{n}}^2+1}{2}$ and $\frac{{\text{n}}^2-1}{2}$ , where is any odd number.

Formula used :

Pythagoras Rule

Proof :

Pythagoras rule says

  ${\left(\text{Hypotenuse}\right)}^2={\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2$

If I take, n = 3,

  $\frac{{\text{n}}^2+1}{2}$ = $\begin{array}{l}\frac{3^2+1}{2}=\frac{10}{2}=5\text{ and }\\ \end{array}$

  $\frac{{\text{n}}^2-1}{2}=\frac{3^2-1}{2}=\frac{8}{2}=4$

(3, 4, 5) is one of the Pythagoras triple which satisfies rule of Pythagoras.

It can be verified as:

  $\begin{array}{l}{\left(\text{Hypotenuse}\right)}^2={\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2\\ \Rightarrow {\left(\text{5}\right)}^2={\left(3\right)}^2+{\left(\text{4}\right)}^2\end{array}$

b

To determine

To prove: The Pythagoras triple, through Euclid’s rule

Explanation of Solution

Given information :

Given a right angle with sides, $\frac{\text{m}+\text{n}}{2},\frac{\text{m}-\text{n}}{2}\text{and}\sqrt{\text{mn}}$ where m, n are either both are even or both are odd numbers

Formula used :

Euclid’s rule

Proof :

Euclid says

  ${\text{c}}^2={\text{a}}^2+{\text{b}}^2$

If I take, m = 5 and n = 3,

  $\begin{array}{l}\frac{\text{m}+\text{n}}{2}=\frac{5+3}{2}=4\\ \frac{\text{m}-\text{n}}{2}\text{=}\frac{5-3}{2}\text{=1}\\ \text{and}\sqrt{\text{mn}}=\sqrt{\text{5}\text{.3}}=\sqrt{\text{15}}\end{array}$

(1, $\sqrt{15}$ , 4) is one of the Pythagoras triple which satisfies rule of Euclid

It can be verified as:

  $\begin{array}{l}{c}^2=a^2+b^2\\ \Rightarrow {\left(4\right)}^2={\left(1\right)}^2+{\left(\sqrt{15}\right)}^2\end{array}$

c

To determine

To prove: The Pythagoras triple, through rule of Plato

Explanation of Solution

Given information :

Given a right angle with sides, $\frac{{\text{m}}^2}{4}+1,\frac{{\text{m}}^2}{4}-1\text{ and m}$ where m is any even number

Formula used :

Plato’s rule

Proof :

If m be any even number, then

  $\frac{{\text{m}}^2}{4}+1,\frac{{\text{m}}^2}{4}-1\text{ and m}$

If we take, m = 6, then

  $\begin{array}{l}\frac{{\text{m}}^2}{4}+1=\frac{6^2}{4}+1=\frac{36}{4}+1=10\\ \frac{{\text{m}}^2}{4}-1\text{ =}\frac{6^2}{4}-1=\frac{36}{4}-\text{1=8 and}\\ \text{ m}=6\end{array}$

It can be verified as:

  $\begin{array}{l}{c}^2=a^2+b^2\\ \Rightarrow {\left(10\right)}^2={\left(6\right)}^2+{\left(8\right)}^2\end{array}$

d

To determine

To prove: The Pythagoras triple, through Rule of Maseres

Explanation of Solution

Given information :

Given a right angle with sides $m^2+n^2$ , $m^2-n^2$ and 2mn where m, n are integers

Formula used :

Rule of Maseres

Proof :

If I take, m = 5 and n = 3, we have

  $m^2+n^2=5^2+3^2=25+9=34$

  $m^2-n^2=5^2-3^2=25-9=16$ and

2mn = 2.5.3=30

For verification,

In right angle, we have

  $\begin{array}{l}{\left(\text{Hypotenuse}\right)}^2={\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2\\ \Rightarrow {\left(34\right)}^2={\left(16\right)}^2+{\left(9\right)}^2\end{array}$