Question

We calculate the value of the hypotenuse of the right triangle whose legs measure 3 and 4cm respectively

Help please it's due today​

Answer 1

`\quad \huge \quad \quad \boxed{ \tt \:Answer }`

`\qquad \tt \rightarrow \: hypotenuse = 5 \:\:cm`

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`\large \tt Solution \: :`

`\textsf{Pythagoras Theorem }`

`\qquad \tt \rightarrow \: h {}^(2) = p {}^(2) + b {}^(2)`

[ h = hypotenuse, p = perpendicular and b = base ]

`\qquad \tt \rightarrow \: {h}^(2) = {3}^(2) + {4}^(2)`

`\qquad \tt \rightarrow \: {h}^(2) = 9 + 16`

`\qquad \tt \rightarrow \: h {}^(2) = 25`

`\qquad \tt \rightarrow \: h = √(25)`

`\qquad \tt \rightarrow \: h = 5 \: \: cm`

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

Answer 2

`{\huge \underline{{ \fbox \color{red}{A}}{\fbox \color{green}{n}}{\fbox \color{purple}{s}}{\fbox \color{brown}{w}}{\fbox \color{yellow}{e}}{\fbox \color{gray}{r } }}}`

Answer :- The result is 5cm

`\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\bf \red {c {}^(2) = a {}^(2) + b {}^(2)} }`

`\: \: \: \: \: \: \boxed{ \bf \green{c {}^(2) = (3cm) {}^(2) + (4cm) {}^(2)}}`

`\: \: \: \: \: \: \: \: \boxed{ \bf \gray{c {}^(2) = 9cm {}^(2) + 16cm {}^(2) }}`

`\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \blue {c {}^(2) = 25cm {}^(2) }}`

`\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \green{\\ot{\sqrt {c {}^{ \\ot{2} }}} = \sqrt{25cm {}^(2) }}}`

`\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \red{c = 5cm}}`