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МатематикаГДЗ по геометрии 8 класс Атанасян Задание 1087
Задание 1087
\[\boxed{\mathbf{1087.ОК\ ГДЗ - домашка\ на}\ 5}\]
| \[N\] | \[R\] | \[r\] | \[a_{4}\] | \[P\] | \[S\] |
|---|---|---|---|---|---|
| \[1\] | \[3\sqrt{2}\] | \[3\] | \[6\] | \[24\] | \[36\] |
| \[2\] | \[2\sqrt{2}\] | \[2\] | \[4\] | \[16\] | \[16\] |
| \[3\] | \[4\] | \[2\sqrt{2}\] | \[4\sqrt{2}\] | \[16\sqrt{2}\] | \[32\] |
| \[4\] | \[\frac{7\sqrt{2}}{2}\] | \[3,5\] | \[7\] | \[28\] | \[49\] |
| \[5\] | \[2\sqrt{2}\] | \[2\] | \[4\] | \[16\] | \[16\] |
\[1)\ R = \frac{\sqrt{a^{2} + a^{2}}}{2} = \frac{a}{\sqrt{2}} = \frac{6}{\sqrt{2}} =\]
\[= 3\sqrt{2};\]
\[r = \frac{a}{2} = \frac{6}{2} = 3;\]
\[P = a \bullet 4 = 6 \bullet 4 = 24;\]
\[S = a^{2} = 6^{2} = 36.\]
\[2)\ a = r \bullet 2 = 4;\]
\[R = \frac{\sqrt{a^{2} + a^{2}}}{2} = \frac{a}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2};\]
\[P = a \bullet 4 = 4 \bullet 4 = 16;\]
\[S = a^{2} = 4^{2} = 16.\]
\[3)\ a = R \bullet \sqrt{2} = 4\sqrt{2};\]
\[r = \frac{a}{2} = \frac{4\sqrt{2}}{2} = 2\sqrt{2};\]
\[P = a \bullet 4 = 4\sqrt{2} \bullet 4 = 16\sqrt{2};\]
\[S = a^{2} = \left( 4\sqrt{2} \right)^{2} = 32.\]
\[4)\ a = \frac{P}{4} = 7;\]
\[R = \frac{\sqrt{a^{2} + a^{2}}}{2} = \frac{a}{\sqrt{2}} = \frac{7}{\sqrt{2}} =\]
\[= \frac{7\sqrt{2}}{2};\]
\[r = \frac{a}{2} = \frac{7}{2} = 3,5;\]
\[S = a^{2} = 7^{2} = 49.\]
\[5)\ a = \sqrt{S} = 4;\]
\[R = \frac{\sqrt{a^{2} + a^{2}}}{2} = \frac{a}{\sqrt{2}} = \frac{4}{\sqrt{2}} =\]
\[= 2\sqrt{2};\]
\[r = \frac{a}{2} = \frac{4}{2} = 2;\]
\(P = a \bullet 4 = 4 \bullet 4 = 16.\)
