Question

Can a right triangle with sides $6\mathrm{cm}$,$10\mathrm{cm}$ and $8\mathrm{cm}$ be formed? Give reason.

Answer 1

Pythagoras Theorem:

A right triangle with sides $6\mathrm{cm}$,$10\mathrm{cm}$ and $8\mathrm{cm}$

Let,

$$\begin{gathered} \mathrm{a}=10\mathrm{cm} \\ \mathrm{b}=8\mathrm{cm} \\ \mathrm{c}=6\mathrm{cm} \end{gathered}$$

To form a right-angled triangle we know the condition that the square on one side must be equal to the sum of the square of the other two sides of the right-angle triangle.

$\begin{array}{rcl}{\mathrm{hyp}}^2& =& {\mathrm{adj}}^2+{\mathrm{opp}}^2\\ {\mathrm{a}}^2& =& {\mathrm{b}}^2+{\mathrm{c}}^2\\ {10}^2& =& 8^2+6^2\\ 100& =& 64+36\\ 100& =& 100\end{array}$

$\begin{array}{rcl}\because \mathrm{LHS}& =& \mathrm{RHS}\end{array}$

Since the square on one side is equal to the sum of the square of the other two sides.

Hence, the right-angle triangle is formed by the given sides $6\mathrm{cm}$, $10\mathrm{cm}$ and $8\mathrm{cm}$.