1. Oblicz pole deltoidu o przekątnych:a) b) 2. Oblicz:7x x.... Question from @Boss46

heh Pole deltoidu (latawca) wyznaczamy ze wzoru:
$P=\frac{d_{1}*d_{2}}{2}$
gdzie $d_{1}\ i\ d_{2}$ to przekątne.
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$a)\ d_{1}=\sqrt{27};\ d_{2}=\sqrt{3}\\
\\
P=\frac{\sqrt{27}*\sqrt{3}}{2}\\
\\
P=\frac{\sqrt{27*3}}{2}\\
\\
P=\frac{\sqrt{81}}{2}\\
\\
P=\frac{\sqrt{9^{2}}}{2}\\
\\
P=\frac{9}{2}\\
\\
P=4\frac{1}{2}\ [j^{2}]$
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$b)\ d_{1}=\sqrt[3]{24};\ d_{2}=\sqrt[3]{9}\\
\\
P=\frac{\sqrt[3]{24}*\sqrt[3]{9}}{2}\\
\\
P=\frac{\sqrt[3]{24*9}}{2}\\
\\
P=\frac{\sqrt[3]{216}}{2}\\
\\
P=\frac{\sqrt[3]{6^{3}}}{2}\\
\\
P=\frac{6}{2}\\
\\
P=3\ [j^{2}]$
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$7*\sqrt{216}*\sqrt{36}=7*\sqrt{36*6}*\sqrt{6^{2}}=7*\sqrt{6^{2}*6}*6=\\
\\
=7*6\sqrt{6}*6=252\sqrt{6}$

0 votes Thanks 2

animaldk $Pole\ deltoidu\ o\ przekatnych\ e\ i\ f:\\\\P=\frac{ef}{2}$

$1.\\a)\ e=\sqrt{27};\ f=\sqrt{3}\\\\P=\frac{\sqrt{27}\cdot\sqrt3}{2}=\frac{\sqrt{27\cdot3}}{2}=\frac{\sqrt{81}}{2}=\frac{9}{2}=4,5\\\\b)\ e=\sqrt[3]{24};\ f=\sqrt[3]{9}\\\\P=\frac{\sqrt[3]{24}\cdot\sqrt[3]9}{2}=\frac{\sqrt[3]{24\cdot9}}{2}=\frac{\sqrt[3]{216}}{2}=\frac{6}{2}=3\\\\\\2.\\7\cdot\sqrt{216}\cdot\sqrt{36}=7\cdot\sqrt{36\cdot6}\cdot6=42\cdot\sqrt{36}\cdot\sqrt6=42\cdot6\sqrt6=252\sqrt6$

©DRK

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