DarkFire30
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We calculate the value of the hypotenuse of the right triangle whose legs measure 3 and 4cm respectively
Help please it's due today​

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Murdrock

[tex]{\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 } }}}[/tex]

Answer :- The result is 5cm

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{\bf \red {c {}^{2} = a {}^{2} + b {}^{2}} }[/tex]

[tex] \: \: \: \: \: \: \boxed{ \bf \green{c {}^{2} = (3cm) {}^{2} + (4cm) {}^{2}}}[/tex]

[tex] \: \: \: \: \: \: \: \: \boxed{ \bf \gray{c {}^{2} = 9cm {}^{2} + 16cm {}^{2} }}[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \blue {c {}^{2} = 25cm {}^{2} }}[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \green{\not{\sqrt {c {}^{ \not{2} }}} = \sqrt{25cm {}^{2} }}}[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \red{c = 5cm}}[/tex]

AǫᴜᴀWɪᴢ

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \: hypotenuse = 5 \:\:cm [/tex]

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[tex] \large \tt Solution \: : [/tex]

[tex] \textsf{Pythagoras Theorem } [/tex]

[tex]\qquad \tt \rightarrow \: h {}^{2} = p {}^{2} + b {}^{2} [/tex]

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

[tex]\qquad \tt \rightarrow \: {h}^{2} = {3}^{2} + {4}^{2} [/tex]

[tex]\qquad \tt \rightarrow \: {h}^{2} = 9 + 16[/tex]

[tex]\qquad \tt \rightarrow \: h {}^{2} = 25[/tex]

[tex]\qquad \tt \rightarrow \: h = \sqrt{25} [/tex]

[tex]\qquad \tt \rightarrow \: h = 5 \: \: cm[/tex]

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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