..
[tex]D=sisi \times \sqrt{2} [/tex]
[tex]~~~~=12 \: cm \times \sqrt{2} [/tex]
[tex]~~~~=\boxed{12 \sqrt{2} \: cm }[/tex]
[tex]\begin{array}{lr}\texttt{Rate 1.0 Jika Kalian Iri dengan}\\\\ \texttt{Yang Mulia Maharaja Danial Alf'at}\end{array}[/tex] ☝️
[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at}}\boxed{\colorbox{ccddff}{12/03/23}}[/tex]
Penjelasan dengan langkah-langkah:
d = √(s^2 + s^2)
d = √((12×12) + (12×12))
d = √144 + 144
d = √288
d = 12√2
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Teorema Pythagoras
[Diagonal Bujur Sangkar]
..
[tex]D=sisi \times \sqrt{2} [/tex]
[tex]~~~~=12 \: cm \times \sqrt{2} [/tex]
[tex]~~~~=\boxed{12 \sqrt{2} \: cm }[/tex]
..
[tex]\begin{array}{lr}\texttt{Rate 1.0 Jika Kalian Iri dengan}\\\\ \texttt{Yang Mulia Maharaja Danial Alf'at}\end{array}[/tex] ☝️
[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at}}\boxed{\colorbox{ccddff}{12/03/23}}[/tex]
Penjelasan dengan langkah-langkah:
d = √(s^2 + s^2)
d = √((12×12) + (12×12))
d = √144 + 144
d = √288
d = 12√2